1.7 Practice Exam Problems

Practice – Mass Percent

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What is the mass percent of cobalt in Vitamin B12 (C₆₃H₈₈CoN₁₄O₁₄P)?

Solution

• The mass of a Vitamin B12 molecule is 1355.38 amu.
• The mass of a cobalt atom is 58.93 amu.

\[\begin{align*} \%~\mathrm{Co} &= \dfrac{58.93~\mathrm{amu}}{1355.38~\mathrm{amu}} \times 100\% = 4.348\% \end{align*}\]

Practice – Chemical Formulas

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You need the molecular formula of an unknown sample. Your lab partner is able to find that the sample’s molecular mass is 142.04 g mol^–1. You are able to find that the sample is composed of 32.37% sodium, 22.57% sulfur, and 45.06% oxygen. Give the structural and empirical formula and name this compound.

Solution

Find the mass of each element.

\[\begin{align*} m_{\mathrm{Na}} &= \left ( \dfrac{32.37\%}{100} \right ) \left ( \dfrac{142.04~\mathrm{g}}{} \right ) = 45.98~\mathrm{g} \\[2ex] m_{\mathrm{S}} &= \left ( \dfrac{22.57\%}{100} \right ) \left ( \dfrac{142.04~\mathrm{g}}{} \right ) = 32.06~\mathrm{g} \\[2ex] m_{\mathrm{O}} &= \left ( \dfrac{45.06\%}{100} \right ) \left ( \dfrac{142.04~\mathrm{g}}{} \right ) = 64.00~\mathrm{g} \end{align*}\]

Find the mol of atoms needed to equal the calculated masses by using the molar masses of each element.

\[\begin{align*} n_\mathrm{Na} &= \dfrac{45.98~\mathrm{g}}{22.99~\mathrm{g~mol^{-1}}} = 2~\mathrm{mol} \\[2ex] n_\mathrm{S} &= \dfrac{32.06~\mathrm{g}}{32.06~\mathrm{g~mol^{-1}}} = 1~\mathrm{mol} \\[2ex] n_\mathrm{O} &= \dfrac{64.00~\mathrm{g}}{16.00~\mathrm{g~mol^{-1}}} = 4~\mathrm{mol} \end{align*}\]

Create a structural formula.

\[\begin{align*} \mathrm{Na}_{2}\mathrm{S}\mathrm{O}_{4} \end{align*}\]

Practice – Chemical Formulas

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Give the empirical structure and molecular structure of a compound composed of 84.1% carbon, 15.9% hydrogen and a molecular weight of 114.23 g mol^–1.

Solution

Find the mass of each element.

\[\begin{align*} m_{\mathrm{C}} &= \left ( \dfrac{84.1\%}{100} \right ) \left ( \dfrac{114.23~\mathrm{g}}{} \right ) = 96.07~\mathrm{g} \\[2ex] m_{\mathrm{H}} &= \left ( \dfrac{15.9\%}{100} \right ) \left ( \dfrac{114.23~\mathrm{g}}{} \right ) = 18.16~\mathrm{g} \end{align*}\]

Find the mol of atoms needed to equal the calculated masses by using the molar masses of each element.

\[\begin{align*} n_\mathrm{C} &= \dfrac{96.07~\mathrm{g}}{12.01~\mathrm{g~mol^{-1}}} = 8~\mathrm{mol} \\[2ex] n_\mathrm{H} &= \dfrac{18.16~\mathrm{g}}{1.01~\mathrm{g~mol^{-1}}} = 18~\mathrm{mol} \end{align*}\]

Create a molecular formula.

\[\begin{align*} \mathrm{C}_{8}\mathrm{H}_{18} \end{align*}\]

Create an empirical formula by reducing the subscripts in the molecular formula.

\[\begin{align*} \mathrm{C}_{8}\mathrm{H}_{18} \rightarrow \mathrm{C}_{8/2}\mathrm{H}_{18/2} \rightarrow \mathrm{C_4H_9} \end{align*}\]

Practice – Dilution

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Willy Wonka’s bubble machine has very precise requirements when it comes to the chemicals that are put into it. His bubble machine requires soap, water, and glycerine (C₃H₈O₃). Willy finds 200.0 mL of a premade glycerine solution in a volumetric flask, but it says that there are 60.00 g of glycerine. His bubble machine requires there to be 236.43 mL of 1.123 M glycerine added.

1. Is the premade solution the correct concentration?
2. What volume of stock solution (in mL) does he need to dilute to 236.43 mL to get the final concentration of 1.123 M?
3. How much water (in mL) is needed?
4. How much glycerine (in g) should be present in the required glycerine solution?

Solution

Part A

Find the molar concentration of the premade solution. The molar mass of glycerine is 92.09 g mol^–1.

\[\begin{align*} c &= \dfrac{\mathrm{mol~solute}}{\mathrm{L~solution}} \\[2ex] &= \dfrac{\left (60.00~\mathrm{g}\right )\left ( \dfrac{\mathrm{mol}}{92.09~\mathrm{g}} \right ) } {\left ( 200.0~\mathrm{mL} \right ) \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right )} \\[2ex] &= 3.25768~M \\[2ex] &= 3.258~M \end{align*}\]

The concentration of the premade solution is too high.

Part B

Use the dilution equation to find the volume of stock solution needed. Be sure to use the unrounded result from Part A.

\[\begin{align*} M_1V_1 &= M_2V_2 \\[2ex] V_1 &= \dfrac{M_2V_2}{M_1} \\[2ex] &= \dfrac{\left ( 1.123~\mathrm{mol~L^{-1}} \right ) \left ( 236.43~\mathrm{mL} \right )}{3.25768~M} \\[2ex] &= 81.50~\mathrm{mL} \end{align*}\]

Part C

The final volume of the correct solution is said to be 236.43 mL. Simply subtract the volume of stock solution to determine how much water is needed for the dilution.

\[\begin{align*} V_{\mathrm{water}} &= V_{\mathrm{total}} - V_{\mathrm{stock}} \\[2ex] &= 236.43~\mathrm{mL} - 81.50~\mathrm{mL} \\[2ex] &= 154.93~\mathrm{mL} \end{align*}\]

Part D

Find the mass (in g) of glycerine in a 236.43 mL 1.123 M glycerine solution.

\[\begin{align*} m_{\mathrm{glycerine}} &= MV \\[2ex] &= \left ( \dfrac{1.123~\mathrm{mol}}{\mathrm{L}} \right ) \left ( \dfrac{236.43~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}}\right ) \left ( \dfrac{92.09~\mathrm{g}}{\mathrm{mol}} \right ) \\[2ex] &= 24.45~\mathrm{g} \end{align*}\]

Practice – Nomenclature

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Give the correct name or formula for each of the following:

1	tin(II) chloride
2	Fe2O3
3	Hg2Cl2
4	potassium fluoride
5	CoCl3
6	hydrochloric acid
7	Cl2O3
8	sodium chloride
9	trisulfur dinitride
10	HF
11	CuCO3
12	CoSO4·6H2O
13	copper(II) hydroxide
14	sodium phosphate
15	magnesium hydroxide
16	lithium chromate
17	sodium hypochlorite
18	ammonium chromate

Solution

1	tin(II) chloride	SnCl2
2	Fe2O3	iron(III) oxide
3	Hg2Cl2	mercury(I) chloride
4	potassium fluoride	KF
5	CoCl3	cobalt(III) chloride
6	hydrochloric acid	HCl(aq)
7	Cl2O3	dichlorine trioxide
8	sodium chloride	NaCl
9	trisulfur dinitride	S3N2
10	HF	hydrogen fluoride
11	CuCO3	copper(II) carbonate
12	CoSO4·6H2O	cobalt(II) sulfate hexahydrate
13	copper(II) hydroxide	Cu(OH)2
14	sodium phosphate	Na3PO4
15	magnesium hydroxide	Mg(OH)2
16	lithium chromate	Li2CrO4
17	sodium hypochlorite	NaOCl
18	ammonium chromate	(NH4)2CrO4

Practice – Reaction Yield

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Steam reforming of methane (methane = CH₄) is the most important industrial process for the production of hydrogen gas, represented by the following balanced reaction:
     H₂O(g) + CH₄(g) → CO(g) + 3H₂(g)
The United States produces around 9.0×10⁶ tons of hydrogen gas (H₂) per year using this process (1 ton = 907 kg). The yield of the process is 65%. How much methane (in kg) is consumed in this process per year?

Solution

Determine the mass of H₂ produced per year in kg.

\[\begin{align*} m_{\mathrm{H_2}} &= \left ( \dfrac{9.0\times 10^{6}~\mathrm{tons}}{} \right ) \left ( \dfrac{907~\mathrm{kg}}{1~\mathrm{ton}} \right ) \\[2ex] &= 8.163\times 10^{9}~\mathrm{kg} \end{align*}\]

The problem states that the reaction is only 65% efficient (i.e. the actual yield of the reaction is 65%). We must determine the yield of the reaction as if it were 100% efficient (i.e. theoretical yield) in order to determine the amount of CH₄ that went into the process.

\[\begin{align*} 8.163\times 10^{9}~\mathrm{kg} \left ( \dfrac{100}{65} \right ) &= 1.25585\times 10^{10}~\mathrm{kg} \end{align*}\]

Determine the mass (in kg) of CH₄ required to make this much H₂ per year.

\[\begin{align*} m_{\mathrm{CH_4}} &= \left ( \dfrac{1.25585\times 10^{10}~\mathrm{kg~H_2}}{} \right ) \left ( \dfrac{10^3~\mathrm{g}}{\mathrm{kg}} \right ) \left ( \dfrac{1~\mathrm{mol~H_2}}{2.02~\mathrm{g}} \right ) \left ( \dfrac{1~\mathrm{mol~CH_4}}{3~\mathrm{mol~H_2}} \right ) \left ( \dfrac{16.04~\mathrm{g}}{1~\mathrm{mol~CH_4}} \right ) \left ( \dfrac{\mathrm{kg}}{10^3~\mathrm{g}} \right ) \\[2ex] &= 3.324\times 10^{10}~\mathrm{kg} \\[2ex] &= 3.3\times 10^{10}~\mathrm{kg} \end{align*}\]

or more compactly using kg kmol^–1

\[\begin{align*} m_{\mathrm{CH_4}} &= \left ( \dfrac{1.25585\times 10^{10}~\mathrm{kg~H_2}}{} \right ) \left ( \dfrac{1~\mathrm{kmol~H_2}}{2.02~\mathrm{kg}} \right ) \left ( \dfrac{1~\mathrm{kmol~CH_4}}{3~\mathrm{kmol~H_2}} \right ) \left ( \dfrac{16.04~\mathrm{kg}}{1~\mathrm{kmol~CH_4}} \right ) \\[2ex] &= 3.324\times 10^{10}~\mathrm{kg} \\[2ex] &= 3.3\times 10^{10}~\mathrm{kg} \end{align*}\]

Practice – Reaction Yield

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A student combines 23 mL of 0.114 M aqueous aluminum sulfate and 6.75 mL of 0.89 M aqueous potassium hydroxide. A double-displacement reaction occurs. Assuming 100% yield, what mass (in g) of solid aluminum hydroxide is formed?

Solution

Write a balanced chemical equation for this double-displacement reaction.

\[\mathrm{Al_2(SO_4)_3}(aq) + 6\mathrm{KOH}(aq) \longrightarrow 2\mathrm{Al(OH)_3}(s) + 3\mathrm{K_2SO_4}(aq)\] Determine the number of mol of each reactant.

\[\begin{align*} n_{\mathrm{Al_2(SO_4)_3}} &= \left ( \dfrac{23~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right ) \left ( \dfrac{0.114~\mathrm{mol}}{\mathrm{L}} \right ) \\[2ex] &= 0.002622~\mathrm{mol} \\[4ex] n_{\mathrm{K(OH)_3}} &= \left ( \dfrac{6.75~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right ) \left ( \dfrac{0.89~\mathrm{mol}}{\mathrm{L}} \right ) \\[2ex] &= 0.0060075~\mathrm{mol} \\[4ex] \end{align*}\]

Identify the limiting reactant by determining the theoretical yield of a product for each reactant. This process assumes the other reactants are in excess.

Yield for Al₂(SO₄)₃

\[\begin{align*} n_{\mathrm{Al(OH)_3}} &= \left ( \dfrac{0.002622~\mathrm{mol~Al_2(SO_4)_3}}{} \right )+ \left ( \dfrac{2~\mathrm{mol~Al(OH)_3}}{1~\mathrm{mol~Al_2(SO_4)_3}} \right ) \\[2ex] &= 0.40908~\mathrm{mol} \end{align*}\]

Yield for KOH

\[\begin{align*} n_{\mathrm{Al(OH)_3}} &= \left ( \dfrac{0.0060075~\mathrm{mol~KOH}}{} \right )+ \left ( \dfrac{2~\mathrm{mol~Al(OH)_3}}{6~\mathrm{mol~KOH}} \right ) \\[2ex] &= 0.0020025~\mathrm{mol} \end{align*}\]

KOH is the limiting reactant.

Determine the mass of Al(OH)₃ that can be made.

\[\begin{align*} m_{\mathrm{Al(OH)_3}} &= \left ( \dfrac{0.0020025~\mathrm{mol~Al_2(SO_4)_3}}{} \right ) \left ( \dfrac{78.01~\mathrm{g}}{\mathrm{mol}} \right ) \\[2ex] &= 0.16~\mathrm{g} \end{align*}\]

Practice – Oxidation Numbers

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Consider each of the compounds below. Which underlined element has the largest (in the positive direction) oxidation number?

1. MgO₂
2. CH₄
3. F₂
4. S^2–
5. NH₄Cl

Solution

The oxidation state for each underlined element is as follows:

1. –2
2. –4
3. 0
4. –2
5. –3

C is the correct answer.

Practice – Ionic Equations

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Which of the following statements is incorrect?

1. AgNO₃, Hg₂(NO₃)₃, and Pb(NO₃)₂ are soluble.
2. In the reaction between MgSO₄ and CaI₂, Mg²⁺ is a spectator ion.
3. The net ionic equation for the reaction of H₂CO₃ and NaOH is H⁺ + OH^– → H₂O.
4. Hydrofluoric acid is a weak acid.
5. A proton and H⁺ represent to the same thing.

Solution

Answer: C

The net ionic equation shown is for a strong acid reacting with a strong base. Here, a weak acid reacts with a strong base. The net ionic reaction should be

     H₂CO₃(aq) + 2OH^–(aq) → 2H₂O(l) + CO₃^2–(aq)

Practice – Oxidation-Reduction Reactions

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In a typical combustion reaction, what happens to oxygen?

1. It is oxidized
2. It is reduced
3. It is both oxidized and reduced
4. It is neither oxidized or reduced
5. It is impossible to say from the given information

Solution

Answer: B

Here is a simple combustion reaction involving methane.

     CH₄ + 2O₂ → CO₂ + 2H₂O

Here, oxygen in O₂ has an oxidation state 0. Oxygen in CO₂ has an oxidation state of –2. Oxygen in H₂O has an oxidation state of –2. Therefore, oxygen is reduced in a combustion reaction.

Practice – Chemical Formulas

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A sample of caffeine is found to consist of 17.77 g carbon, 1.86 g hydrogen, 10.04 g nitrogen, and 5.92 g oxygen. Given that the molar mass of caffeine is about 194 g mol^–1, what is the molecular formula of caffeine?

Solution

Find the mol of each element.

\[\begin{align*} n_{\mathrm{C}} &= \left ( \dfrac{17.77~\mathrm{g~C}}{} \right ) \left ( \dfrac{\mathrm{mol}}{12.01~\mathrm{g}} \right ) = 1.4796~\mathrm{mol} \\[2ex] n_{\mathrm{H}} &= \left ( \dfrac{1.86~\mathrm{g~H}}{} \right ) \left ( \dfrac{\mathrm{mol}}{1.01~\mathrm{g}} \right ) = 1.8416~\mathrm{mol} \\[2ex] n_{\mathrm{N}} &= \left ( \dfrac{10.04~\mathrm{g~N}}{} \right ) \left ( \dfrac{\mathrm{mol}}{14.01~\mathrm{g}} \right ) = 0.7166~\mathrm{mol} \\[2ex] n_{\mathrm{O}} &= \left ( \dfrac{5.92~\mathrm{g~N}}{} \right ) \left ( \dfrac{\mathrm{mol}}{16.00~\mathrm{g}} \right ) = 0.3700~\mathrm{mol} \\[2ex] \end{align*}\]

Find the ratio of the molar mass of caffeine to the masses of the collected elements.

\[\begin{align*} \dfrac{194~\mathrm{g~mol^{-1}}}{\left ( 17.77~\mathrm{g} + 1.86 ~\mathrm{g} + 10.04 ~\mathrm{g} + 5.92~\mathrm{g} \right )} = 5.451~\mathrm{mol^{-1}} \end{align*}\]

Multiply the moles of each element by the calculated ratio.

\[\begin{align*} \#~\mathrm{C} &= 5.451~\mathrm{mol^{-1}} \times 1.4796~\mathrm{mol} \approx 8 \\[2ex] \#~\mathrm{H} &= 5.451~\mathrm{mol^{-1}} \times 1.8416~\mathrm{mol} \approx 10 \\[2ex] \#~\mathrm{N} &= 5.451~\mathrm{mol^{-1}} \times 0.7166~\mathrm{mol} \approx 4 \\[2ex] \#~\mathrm{O} &= 5.451~\mathrm{mol^{-1}} \times 0.3700~\mathrm{mol} \approx 2\\[2ex] \end{align*}\]

Write the molecular formula.

\[\begin{align*} \mathrm{C}_8\mathrm{H}_{10}\mathrm{N}_4\mathrm{O}_2 \end{align*}\]

Practice – Acid-Base Reaction

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A sample of 40.25 mL hydroiodic acid requires a 19.20 mL 3.14 M calcium hydroxide solution to neutralize it. What is the concentration (in M) of the hydroiodic acid solution?

Solution

This neutralization reaction involves a strong acid and a strong base. Neutralization occurs when the hydronium ions supplied by the strong acid (HI) completely reacts with the hydroxide ions supplied by the strong base (Ca(OH)₂).

Write out the dissociation of aqueous calcium hydroxide

\[\begin{align*} \mathrm{Ca(OH)_2}(aq) \longrightarrow \mathrm{Ca^{2+}}(aq) + \mathrm{OH^-}(aq) \end{align*}\]

Determine the moles of hydroxide ions in the strong base solution.

\[\begin{align*} n_{\mathrm{OH^-}} &= \left ( \dfrac{3.14~\mathrm{mol~Ca(OH)_2}}{\mathrm{L}} \right ) \left ( \dfrac{19.20~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right ) \left ( \dfrac{2~\mathrm{mol~OH^-}}{\mathrm{mol~Ca(OH)_2}} \right )\\[2ex] &= 0.120576~\mathrm{mol} \end{align*}\]

For complete neutralization,

\[n_{\mathrm{H_3O^+}} = n_{\mathrm{OH^-}}\]

meaning there was 0.120576 mol of H₃O⁺ present.

Since HI is a strong acid,

\[n_{\mathrm{HI}} = n_{\mathrm{H_3O^+}}\] meaning there was initially 0.120576 mol of HI.

Determine the concentration of HI.

\[\begin{align*} c_{\mathrm{HI}} &= \left ( \dfrac{0.120576~\mathrm{mol}}{40.25~\mathrm{mL}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \\[2ex] &= 2.996~M \\[2ex] &\approx 3.00~M \end{align*}\]

Practice – VSEPR

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Which of the following molecular geometries can be nonpolar?

1. Bent
2. Trigonal pyramidal
3. Seesaw
4. T-shaped
5. Trigonal planar

Solution

Answer: E

Practice – Bond Enthalpies

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Given the bond enthapies (in kJ mol^–1) for H–H, O=O, and H–O are 436.4, 498.7, and 460.0, respectively, what is the bond enthalpy (in kJ mol^–1) for the reaction of H₂ and O₂ from H₂?

1. –15.1
2. 15.1
3. 468.5
4. –468.5
5. 0

Solution

Answer: D

Write a balanced reaction.

\[\mathrm{2H_2O} \longrightarrow \mathrm{2H_2} + \mathrm{O_2}\]

1. H₂ has one H–H bond (436.4 kJ mol^–1 each)
2. O₂ has one O=O bond (498.7 kJ mol^–1 each)
3. H₂O has two H–O bonds (460.0 kJ mol^–1 each)

Determine the enthalpy of reaction.

\[\begin{align*} \Delta H_{\mathrm{rxn}} &= \Sigma \left (\Delta H_{\mathrm{products}} \right ) - ~~\Sigma \left (\Delta H_{\mathrm{reactants}} \right )\\[2ex] &= \left [ \left ( 2 \times \Delta H (\mathrm{H-H}) \right ) + \left ( 1 \times \Delta H (\mathrm{O=O}) \right ) \right ] - \\ &\phantom{=} ~~ \left [ \left ( 4 \times \Delta H (\mathrm{H-O}) \right ) \right ] \\[2ex] &= \left [ \left ( 2 \times 436.4 \right ) + \left ( 1 \times 498.7 \right ) \right ] \\ &\phantom{=} ~~ \left [ \left ( 4 \times 460.0 \right ) \right ] \\[2ex] &= -468.5~\mathrm{kJ~mol^{-1}} \end{align*}\]

Practice – VSEPR

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Which bond angle (in deg) would be the most reasonable between the two hydrogen atoms in CH₂O?

1. 109.5
2. 116
3. 120
4. 122
5. 180

Solution

Answer: B

Practice – Gases

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Choose the false statement.

1. Molecular speed has a broader range at high temperatures than at low temperatures
2. Heavier molecules have lower root-mean-square speed than molecules with lower molecular mass.
3. Ideal gas molecules do not exert attractive or repulsive forces on one another.
4. The constant, R, used in the formula for root-mean-square is the same constant, R, used in the ideal gas equation, just with different units.
5. all statemenmts are true

Solution

Answer: E

Practice – Orbitals

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Choose the correct statement.

1. When two s orbitals overlap side to side, a pi bond is formed.
2. When two p orbitals overlap head to head, a sigma bond is formed.
3. The electron density of a pi bond lies along the axis between two atoms.
4. Only p orbital overlap can form pi bonds.
5. Only two of the three p orbitals of a given subshell in a given atom lie perpendicular to each other.

Solution

Answer: B

Practice – Intermolecular Forces

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Rank the following compounds from lowest to highest expected boiling point: HF, F₂, LiF

1. HF, F₂, LiF
2. F₂, HF, LiF
3. LiF, HF, F₂
4. HF, LiF, F₂
5. F₂, LiF, HF

Solution

Answer: B

Practice – Gases

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6.0 mol NO and 8.0 mol O₂ are combined in a sealed 13.0 L container and react according to the given reaction:
     NO + 2O₂ → 2NO₂
Assuming 100% yield, what volume (in L) of gas is in the container following the reaction?

1. 6.0
2. 8.0
3. 10.0
4. 13.0
5. 14.0

Solution

Answer: D

Practice – Lewis Structures

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How many lone pairs of electrons are on the central atom in IF₂^–?

1. 0
2. 1
3. 2
4. 3
5. 4

Solution

Answer: D

Practice – Valence Bond Theory

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How many σ bonds, π bonds, and lone electron pairs are there in the following molecule?

1. 6 σ bonds, 2 π bonds, 0 lone pairs
2. 15 σ bonds, 2 π bonds, 0 lone pairs
3. 16 σ bonds, 1 π bonds, 2 lone pairs
4. 15 σ bonds, 2 π bonds, 3 lone pairs
5. 6 σ bonds, 2 π bonds, 3 lone pairs

Solution

Answer: D

Practice – VSEPR

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Give the sum of the number of electron domains on the central atom in each of the following molecules: BH₃, NH₃, SO₂, H₂O

1. 10
2. 11
3. 12
4. 13
5. 14

Solution

Answer: E

BH₃ – 3 domains

NH₃ – 4 domains

SO₂ – 3 domains

H₂O – 4 domains

Total

\[3+4+3+4 = 14\]

Practice – Vapor Pressure

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A 42.5 mL sample of NO₂ gas was collected over water. The atmospheric pressure in the room at the time was 0.9980 atm and the water was 30.0 °C. Given that the vapor pressure of water at 20 °C, 25 °C, and 30 °C is 17.5, 23.8, and 31.8 torr, respectively, how many grams of NO₂ gas were collected?

1. 0.00165
2. 0.0165
3. 0.0750
4. 0.0760
5. 0.750

Solution

Answer: C

Convert the pressure of the water at 30 °C from torr to atm.

\[\begin{align*} P_{\mathrm{H_2O}} &= \left ( \dfrac{31.8~\mathrm{torr}}{} \right ) \left ( \dfrac{1~\mathrm{atm}}{760~\mathrm{torr}} \right ) \\[2ex] &= 0.0418~\mathrm{atm} \end{align*}\]

Find the pressure of NO₂.

\[\begin{align*} P_{\mathrm{tot}} &= P_{\mathrm{NO_2}} + P_{\mathrm{H_2O}} \rightarrow \\[2ex] P_{\mathrm{NO_2}} &= P_{\mathrm{tot}} - P_{\mathrm{H_2O}} \\[2ex] &= 0.9980~\mathrm{atm} - 0.0418~\mathrm{atm} \\[2ex] &= 0.9562~\mathrm{atm} \end{align*}\]

Find the amount (in mol) of NO₂.

\[\begin{align*} PV &= nRT \\[2ex] n_{\mathrm{NO_2}} &= \dfrac{PV}{RT} \\[2ex] &= \dfrac{(0.9562~\mathrm{atm})(0.0425~\mathrm{L})}{(0.08206~\mathrm{L~atm~mol^{-1}~K^{-1}})(303.15~\mathrm{K})}\\[2ex] &= 0.00163~\mathrm{mol} \end{align*}\]

Find the mass (in g) of NO₂.

\[\begin{align*} m_{\mathrm{NO_2}} &= \left ( \dfrac{0.00163~\mathrm{mol}}{} \right ) \left ( \dfrac{46.01~\mathrm{g}}{\mathrm{mol}} \right ) \\[2ex] &= 0.0750~\mathrm{g} \end{align*}\]

Practice – Electronegativity

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In which of the following molecules would the central atom be considered partially negative?

1. PCl₅
2. AlH₃
3. CO₂
4. all of the above
5. none of the above

Solution

Answer: E

PCl

P is partially positive because Cl is more electronegative.

AlH₃ – Central atom has a partial positive charge

Al is partially positive because H is more electronegative.

CO₂ – Central atom has a partial positive charge

C is partially positive because O is more electronegative.

Practice – Ideal Gas

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A balloon contains a mixture of 1.00 mol of each gas: N₂, O₂, H₂, and H₂O. The balloon has a volume of 3.00 L and is at 25.0 °C. Treating the gas as an ideal gas, what pressure (in atm) does the H₂ exert?

1. 1.02
2. 2.04
3. 4.08
4. 8.16
5. 32.6

Solution

Answer: D

Use the ideal gas law to solve for the pressure of H₂.

\[\begin{align*} PV &= nRT \\[2ex] P_{\mathrm{H_2}} &= \dfrac{nRT}{V} \\[2ex] &= \dfrac{(1.00~\mathrm{mol}) (0.08206~\mathrm{L~atm~mol^{-1}~K^{-1}}) (298.15~\mathrm{K})} {3.00~\mathrm{L}} \\[2ex] &= 8.16~\mathrm{atm} \end{align*}\]

Practice – VSEPR

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Which of the following molecular geometries can result in a polar molecule?

1. linear
2. square pyramidal
3. bent
4. octahedral
5. all of the above

Solution

Answer: E

Any molecular geometry can result in a polar molecule.

Practice – Valence Bond Theory

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Which of the following compounds has an sp² hybridized nitrogen?

1. NH₃
2. CN^–
3. N₂
4. NO₃^–
5. none of these

Solution

Answer: D

NH₃

N is sp³ hybridized.

CN^–

N is sp hybridized.

N₂

N is sp hybridized.

NO₃^–

N is sp² hybridized

Practice – VSEPR

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An s orbital and two p orbitals are hybridized. Choose the most correct of the following statements.

1. Exactly 1 sp² orbtial is formed
2. Three sp orbitals are formed
3. The molecular geometry around the atom with the hybridized orbitals is trigonal planar
4. Each hybridized orbital has two lobes, one much larger than the other
5. Each hybridized orbital may only form pi bonds

Solution

Answer: D

Practice – VSEPR

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A certain molecule contains 5 atoms and has a tetrahedral molecular geometry. Two of the four terminal atoms are of one type and have an electronegativity of 3.2 The last two terminal atoms are of the second type but also have an electronegativity of 3.2. Is the molecule polar?

1. Yes, because there are different terminal atoms
2. No, because all of the terminal atoms have the same electronegativity
3. Not enough information: it depends on the electronegativity of the central atom
4. Not enough information, it depends on which specific atoms are involved

Solution

Answer: B

If the terminal atoms in a tetrahedral molecule have the same electronegativity, the molecule will not be polar (i.e. it will not have a dipole).

Practice – Electron Configuration

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Which of the following pairs is not isoelectronic?

1. F^– and Mg²⁺
2. Ca and Ti²⁺
3. S^2– and Ar
4. Cr and Mn⁺
5. all of the above are isoelectronic pairs

Solution

Answer: B

Practice – Electron Configuration

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Elements A, B, and C have electronegativities of 3.5, 2.5, and 0.2, respectively. Which of the following bonds would be considered polar covalent?

1. A–A
2. A–B
3. A–C
4. B–C
5. none of the above

Solution

Answer: B

Recall that bonds can be classified by the difference in the electronegativities of the atoms involved in that bond.

1. < 0.5 → pure covalent
2. 0.5 to 2.0 → polar covalent
3. > 2.0 → ionic

Therefore,

1. A–A is pure covalent (nonpolar): 3.5 - 3.5 = 0
2. A–B is polar covalent: 3.5 - 2.5 = 1.0
3. A–C is ionic: 3.5 - 0.2 = 3.3
4. B–C is ionic: 2.5 - 0.2 = 2.3

Practice – VSEPR

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On the molecule CH₂Cl₂, which direction does the dipole point?

1. Towards the hydrogens (away from the chlorines)
2. Towards the chlorines (away from the hydrogens)
3. Towards the carbon
4. Towards the lone pair
5. There is no dipole

Solution

Answer: B

CH₂Cl₂ has a tetrahedral molecular geometry. The gray arrows denote the direction of the polar bonds and the blue arrow denotes the direction of the dipole.

Practice – Intermolecular Forces

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Which of the following molecules would you expect to have the lowest boiling point?

1. N₂
2. O₂
3. F₂
4. Cl₂
5. Br₂

Solution

Answer: A

The given molecules are homonuclear diatomics and therefore have no dipole. These molecules only exhibit dispersion forces. The smallest molecule, N₂ is expected to have the lowest boiling point. The normal boiling points for each are given below:

1. N₂: –195.8 °C
2. O₂: –183 °C
3. F₂: –188 °C
4. Cl₂: –34.6 °C
5. Br₂: 58.8 °C

Practice – Molar Mass of Ideal Gas

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60.54 g of a sample of gas takes up 67.2 L at STP. What is the gas?

1. H
2. N₂
3. Ne
4. Ar
5. not enough information

Solution

Answer: C

Recall that 1 mol of an ideal gas at STP occupies 22.4 L of space (shown below):

\[\begin{align*} PV &= nRT \rightarrow \\[2ex] V &= \dfrac{nRT}{P} \\[2ex] &= \dfrac{(1~\mathrm{mol})(0.08206~\mathrm{L~atm~mol^{-1}~K^{-1}})(273.15~\mathrm{K})}{1~\mathrm{atm}} \\[2ex] &= 22.4~\mathrm{L} \end{align*}\]

Determine how many moles of gas is present using this information.

\[\begin{align*} n_{\mathrm{gas}} &= \left ( \dfrac{67.2~\mathrm{L}}{} \right ) \left ( \dfrac{1~\mathrm{mol}}{22.4~\mathrm{L}} \right ) \\[2ex] &= 3.00~\mathrm{mol} \end{align*}\]

Use the mass of the gas to determine the molar mass and identity of the gas.

\[\begin{align*} M_{\mathrm{gas}} &= \dfrac{60.54~\mathrm{g}}{3~\mathrm{mol}} \\[2ex] &= 20.18~\mathrm{g~mol^{-1}} \end{align*}\]

The molar mass of Ne is 20.18 g mol^–1.

Practice – Charge Notation

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Which of the following notations is incorrect?

1. Li⁺Cl^–
2. H^δ+Cl^δ–
3. C⁺O^–
4. all of the above are incorrect
5. only (b) and (c) are correct

Solution

Answer: C

Carbon monoxide, CO, is not ionic and the notation should only use partial (δ) charges.

Practice – VSEPR

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Which of the following molecular geometries is incorrect?

1. AlCl₃: trigonal planar
2. SO₄^2–: tetrahedral
3. CO₂: linear
4. O₃: linear
5. NH₃: trigonal pyramidal

Solution

Answer: D

Given below are the Lewis structures for each molecule.

AlCl₃ – trigonal planar

SO₄^2– – tetrahedral

CO₂ – linear

O₃ – bent

NH₃ – trigonal pyramidal

Practice – Ideal Gas

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How much more dense (in g L^–1) is O₂ gas than He gas at room temperature (25.000 °C) and exactly 1 atm of pressure?

1. 0.164
2. 0.490
3. 0.654
4. 1.144
5. 1.308

Solution

Answer: D

Begin with the ideal gas and rearrange. Note that M is molar mass and M × n/V is density.

\[\begin{align*} PV &= nRT \\[2ex] \dfrac{n}{V} &= \dfrac{P}{RT} \\[2ex] \dfrac{nM}{V} &= \dfrac{PM}{RT} \\[2ex] d &= \dfrac{PM}{RT} \end{align*}\]

Use the equation obtained above to determine the density of each gas.

\[\begin{align*} d_{\mathrm{O_2}} &= \dfrac{PM}{RT} \\[2ex] &= \dfrac{(1~\mathrm{atm})(32.00~\mathrm{g~mol^{-1}})} {(0.08206~\mathrm{L~atm~mol^{-1}~K^{-1}})(298.15~\mathrm{K})}\\[2ex] &= 1.308~\mathrm{g~L^{-1}}\\[4ex] d_{\mathrm{He}} &= \dfrac{PM}{RT} \\[2ex] &= \dfrac{(1~\mathrm{atm})(4.003~\mathrm{g~mol^{-1}})} {(0.08206~\mathrm{L~atm~mol^{-1}~K^{-1}})(298.15~\mathrm{K})}\\[2ex] &= 0.164~\mathrm{g~L^{-1}} \end{align*}\]

Take the difference of the two densities.

\[\begin{align*} \Delta d &= d_{\mathrm{O_2}} - d_{\mathrm{He}} \\[2ex] &= \left ( 1.308 - 0.164 \right )~\mathrm{g~L^{-1}} \\[2ex] &= 1.144~\mathrm{g~L^{-1}} \end{align*}\]

Practice – Valence Bond Theory

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How many σ bonds and how many π bonds are in carbonyl sulfide, COS?

1. 0 σ bonds, 2 π bonds
2. 1 σ bonds, 3 π bonds
3. 2 σ bonds, 0 π bonds
4. 2 σ bonds, 2 π bonds
5. 4 σ bonds, 0 π bonds

Solution

Answer: D

COS has two double bonds. Each double bond consists of one σ and one π bond.

Practice – Ideal Gas

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A barometer in a sealed room reads 735.0 torr. A thermometer in the same room gives the air temperature as 18.30 °C. If the temperature in the room is raised by 10.20 °C, what would the new pressure be (in atm)?

1. 0.539
2. 1.001
3. 1.506
4. 761
5. 1140

Solution

Answer: B

Use Guy-Lussac’s Law to determine the new pressure at a different temperature and convert torr to atm (1 atm = 760 torr).

\[\begin{align*} \dfrac{P_1}{T_1} &= \dfrac{P_2}{T_2} \rightarrow \\[2ex] P_1 &= \dfrac{P_1T_2}{T_1} \\[2ex] &= \left [ \dfrac{(735.0~\mathrm{torr})(301.65~\mathrm{K})}{291.45~\mathrm{K}} \right ] \left ( \dfrac{1~\mathrm{atm}}{760~\mathrm{torr}} \right ) \\[2ex] &= 1.001~\mathrm{atm} \end{align*}\]

Practice – VSEPR

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Which of the following compounds could never be polar?

1. A compound with the same electron domain geometry and molecular geometry
2. A compound with no lone pairs of electrons
3. A compound with no polar bonds and no lone pairs
4. A compound made up of two atoms of the same type bonded to a central atom of a different type
5. A compound with an even number of lone electron pairs

Solution

Answer: C

Both (a) and (b) could be polar if composed of more than one element. Option (d) could be polar (e.g. water). Option (e) is not correct because something like nitrogen oxide (NO) contains an even number of lone electron pairs and is polar. Option (c) is the correct answer.

Practice – Ideal Gas Yield

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A 0.650 g sample of AlBr₃ is reacted with 0.190 L of Cl₂ gas at STP. What is the volume (in L) of Br₂ produced?
     2AlBr₃(s) + 3Cl₂(g) → 2AlCl₃(s) + 3Br₂(g)

1. 0.00244
2. 0.00366
3. 0.00482
4. 0.0820
5. 0.190

Solution

Answer: D

Determine the moles of each reactant. Note that when determining the moles of Cl₂(g), we use 22.4 L mol^–1 since the gas is at STP.

\[\begin{align*} n_{\mathrm{AlBr_3}} &= \left ( \dfrac{0.650~\mathrm{g}}{} \right ) \left ( \dfrac{1~\mathrm{mol}}{266.68~\mathrm{g}} \right )\\[2ex] &= 0.00244~\mathrm{mol} \\[4ex] n_{\mathrm{Cl_2}} &= \left ( \dfrac{0.190~\mathrm{L}}{} \right ) \left ( \dfrac{1~\mathrm{mol}}{22.4~\mathrm{L}} \right )\\[2ex] &= 0.00848~\mathrm{mol} \\[4ex] \end{align*}\]

Determine the limiting reactant.

\[\begin{align*} n_{\mathrm{Br_2}} &= \left ( \dfrac{0.00244~\mathrm{mol~AlBr_3}}{} \right ) \left ( \dfrac{3~\mathrm{mol~Br_2}}{3~\mathrm{mol~AlBr_3}} \right ) \\[2ex] &= 0.00366~\mathrm{mol} \\[4ex] n_{\mathrm{Br_2}} &= \left ( \dfrac{0.00848~\mathrm{mol~Cl_2}}{} \right ) \left ( \dfrac{3~\mathrm{mol~Br_2}}{3~\mathrm{mol~Cl_2}} \right ) \\[2ex] &= 0.00848~\mathrm{mol} \end{align*}\]

AlBr₃ is the limiting reactant.

Use the ideal gas law to determine the volume (in L) of Br₂(g) produced.

\[\begin{align*} PV &= nRT \rightarrow \\[2ex] V &= \dfrac{nRT}{V} \\[2ex] &= \dfrac{(0.00366~\mathrm{mol})(0.08206~\mathrm{L~atm~mol^{-1}~K^{-1}})(273.15~\mathrm{K})} {1~\mathrm{atm}} \\[2ex] &= 0.0820~\mathrm{L} \end{align*}\]

Practice – Periodic Trends

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Which of the following statements are true?

I. Atoms with high electronegativity form negative ions more often than positive ones
II. In general, electronegativity increases from left to right across a period
III. In general, electronegativity decreases from top to bottom down a column
IV. In general, noble gases have higher electronegativity than elements in other groups

1. Only I
2. I and II
3. I, II, and III
4. II, III, and IV
5. All of these are true

Solution

Answer: C

Practice – Ideal Gas

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A weather balloon is filled under a pressure of 1 atm to a volume of 270 ft³ and rises through the atmosphere until it reaches a volume of 22,449 ft³ and bursts. What pressure (in atm) was the balloon under when it burst? Assume constant temperature.

1. 1.65×10^–7
2. 0.0120
3. 0.0360
4. 83.1
5. 249

Solution

Answer: B

Use Boyle’s Law to determine the new pressure

\[\begin{align*} P_1V_1 &= P_2V_2 \rightarrow \\[2ex] P_2 &= \dfrac{P_1V_1}{V_2} \\[2ex] &= \dfrac{(1~\mathrm{atm})(270~\mathrm{ft^3})} {22449~\mathrm{ft^3}} \\[2ex] &= 0.0120~\mathrm{atm} \end{align*}\]

Practice – Ideal Gas

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Choose the true statement.

1. The pressure a gas exerts is dependent on the other gases in the mixture.
2. When 3.0 L of N₂ and 4.0 L of Ar are combined in a 8.0 L container, the volume of the mixture is 7.0 L.
3. The law of partial pressures is named after Boyle.
4. Mole fractions have units of moles.
5. Mole fraction is equivalent to partial pressure divided by total pressure.

Solution

Answer: E

Practice – Molecular Orbital Theory

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Which of the following statements is false?

1. Antibonding molecular orbitals arise from the destructive combination of atomic orbitals
2. A bond order of 0 means that no bond will form between the two atoms in questions
3. Each molecular orbital can hold 2 electrons with opposite spin
4. Hund’s rule and the Pauli exclusion principle apply to molecular orbitals as well as atomic orbitals
5. The π_{2p orbitals are higher in energy than the σ2p orbital for N2}

Solution

Answer: E

Practice – VSEPR

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Which direction would the dipole on CH₂O point?

1. There is no dipole moment
2. Towards the oxygen
3. Towards the carbon
4. Towards a specific hydrogen
5. Towards the hydrogens (away from the oxygen)

Solution

Answer: B

Practice – Bond Energy

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Which of the following statements is true?

1. Formation of ionic bonds releases a large amount of energy
2. Lattice energy is the amount of energy required to convert a mole of ionic solid into the gas phase.
3. Lattice energy increases as distance between charges decreases
4. The magnitude of charges has no impact on lattice energy
5. All of these statements are false

Solution

Answer: E

Practice – Ideal Gas Yield

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3.0 L of H₂ gas and 4.5 L of O₂ gas are combined at constant temperature and pressure. They react according to the following equation:
     H₂ + O₂ → H₂O
Assuming the reaction has a 100% yield, what volume (in L) of water vapor will form?

1. 3.0
2. 4.5
3. 6.0
4. 7.5
5. 10.5

Solution

Answer: A

Use Avogadro’s law to determine the limiting reactant by calculating how much H₂ would form from each reactant if reacted completely.

If we react all 3.0 L of H₂, 3.0 L of H₂O forms since 2 mol H₂ → 2 mol H₂O.

If we react all 4.5 L of O₂, 9.0 L of H₂O forms since 1 mol O₂ → 2 mol H₂O.

H₂ is the limiting reactant so 3.0 L H₂ forms.

Practice – Gases

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Describe each of these situations:
I. On a space shuttle in orbit, oxygen gas escapes through a small hole in the side of a tank into space.
II. A sample of chlorine gas and a sample of xenon gas are released into a container in which they mix together.

1. I: diffusion, II: effusion
2. I: diffusion, II: diffusion
3. I: effusion, II: effusion
4. I: effusion, II: diffusion
5. none of these are correct

Solution

Answer: D

Practice – Structure and Bonding

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Select the correct statement.

1. In real life, molecules switch between their different resonance structures very quickly
2. A free radical is an electron in the excited state
3. Dative bonds form between Lewis acids and Lewis bases
4. Coordinate covalent bonds form between two atoms which have an electronegativitity difference of less than 0.5
5. The Born-Haber Cycle is an application of Hess’s Law that breaks down the various thermochemical steps in the formation of solid ionic compounds.

Solution

Answer: C

Practice – Intermolecular Forces

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Which of the following compounds is not matched to its most dominant intermolecular force?

1. CH₄: van der Waals
2. HF: hydrogen bonding
3. HCl dipole-dipole
4. K₂SO₄: ion-ion
5. MgCl: London dispersion forces

Solution

Answer: E

Practice – Gas Laws

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What is the name of the law that states V₁T₁ = V₂T₂?

1. Guy-Lussac’s Law
2. Boyle’s Law
3. Charles’ Law
4. Avogadro’s Law
5. Dalton’s Law

Solution

Answer: C

Practice – Gases

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Arrange the following compounds in order of increasing root-mean-square speed: CO₂, Kr, He, F₂

1. CO₂, Kr, He, F₂
2. F₂, He, Kr, CO₂
3. He, F₂, CO₂, Kr
4. Kr, CO₂, F₂, He
5. none of the above

Solution

Answer: D

Practice – Gases

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Which of the following laws concern volume?

I. Avogadro’s Law II. Boyle’s Law III. Charles’ Law IV. Guy-Lussac’s Law

1. Only II
2. I, II, and III
3. I, II, and IV
4. II, III, and IV
5. all of them

Solution

Answer: B

Practice – Molecular Orbital Theory

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Use molecular orbital theory to determine how many σ bonds and how many π bonds are present in C₂

1. 1 σ bond, 0 π bonds
2. 0 σ bond, 1 π bonds
3. 1 σ bond, 1 π bonds
4. 2 σ bond, 0 π bonds
5. 0 σ bond, 2 π bonds

Solution

Answer: E

Practice – Ideal Gas

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A sample of gas is in a sealed container. What happens to the absolute temperature if the volume is doubled and the pressure is halved?

1. It is one fourth of what it was before
2. It is one half of what it was before
3. It is the same as it was before
4. It is twice what it was before
5. It is four times what it was before

Solution

Answer: C

Practice – Gases

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Find the gas constant, R, in units of mL torr mol^–1 K^–1.

1. 1.080×10^–7
2. 0.08206
3. 2.782
4. 2782
5. 6.236×10⁴

Solution

Answer: E

\[\begin{align*} R &= \left ( \dfrac{0.08206~\mathrm{L~atm}}{\mathrm{mol~K}} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{1~\mathrm{L}} \right ) \left ( \dfrac{760~\mathrm{torr}}{1~\mathrm{atm}} \right ) \\[2ex] &= 62365.6~\mathrm{mL~torr~mol^{-1}~K^{-1}} \\[2ex] &= 6.236\times 10^{4}~\mathrm{mL~torr~mol^{-1}~K^{-1}} \end{align*}\]

Practice – Gases

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Real gases behave most like ideal gases at

1. high T and high P
2. high T and low P
3. low T and high P
4. low T and low P
5. all T and all P

Solution

Answer: B

Practice – Gases

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Which of the following are inversely proportional?

1. P and T
2. P and n
3. P and V
4. V and T
5. V and n

Solution

Answer: C

Practice – Gases

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A particular gas has a density of 0.035 g mL^–1 at STP. Assuming pressure is held constant, what will the volume (in L) of 8.0 g of the gas be if the temperature is increased to 25.0 °C?

1. 0.035
2. 0.21
3. 0.23
4. 0.25
5. 3.5

Solution

Answer: D

Find the volume of the gas at STP.

\[\begin{align*} V_{\mathrm{gas}} &= \left ( \dfrac{8.0~\mathrm{g}}{} \right ) \left ( \dfrac{1~\mathrm{mL}}{0.035~\mathrm{g}} \right ) \left ( \dfrac{1~\mathrm{L}}{10^3~\mathrm{mL}} \right ) \\[2ex] &= 0.23~\mathrm{L} \end{align*}\]

Determine the new volume of the gas at 25 °C using Charles’ Law.

\[\begin{align*} \dfrac{V_1}{T_1} &= \dfrac{V_2}{T_2} \rightarrow \\[2ex] V_2 &= \dfrac{V_1T_2}{T_1} \\[2ex] &= \dfrac{(0.23~\mathrm{L})(298.15~\mathrm{K})}{273.15~\mathrm{K}} \\[2ex] &= 0.25~\mathrm{L} \end{align*}\]

Practice – Gases

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How much faster will Ne effuse than Ar?

1. ≈0.51 times as fast
2. ≈ 0.71 times as fast
3. They effuse about the same
4. ≈ 1.41 times as fast
5. ≈ 1.98 times as fast

Solution

Answer: D

\[\begin{align*} \dfrac{u_{\mathrm{rms}}(\mathrm{Ne})}{u_{\mathrm{rms}}(\mathrm{Ar})} = \sqrt{\dfrac{M_{\mathrm{Ar}}}{M_{\mathrm{Ne}}}} = \sqrt{\dfrac{39.95~\mathrm{g~mol^{-1}}}{20.18~\mathrm{g~mol^{-1}}}} = 1.41 \end{align*}\]

Practice – Gases

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Which of these is not equivalent to 1 atm?

1. 101.325 Pa
2. 760 mmHg
3. 760 torr
4. 1.01325 N
5. 14.7 psi

Solution

Answer: D

The Newton, N, is not a unit of pressure.

Practice – Gases

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What is the density (in g mL^–1) of iodine gas at STP?

1. 0.01037
2. 5.187
3. 10.37
4. 126.9
5. 253.8

Solution

Answer: A

\[\begin{align*} d &= \dfrac{PM}{RT} \\[2ex] &= \left ( \dfrac{(1~\mathrm{atm})(2.538~\mathrm{g~mol^{-1}})} {(0.08206~\mathrm{L~atm~mol^{-1}~K^{-1}})(298.15~\mathrm{K})} \right ) \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right ) \\[2ex] &= 0.01037~\mathrm{g~mL^{-1}} \end{align*}\]

Practice – Gases

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A sealed 250 mL jar at 25 °C contains 3.20 g O₂ gas and 1.40 g N₂ gas. What is the total pressure (in atm) inside the jar?

1. 0.150
2. 4.60
3. 4.89
4. 9.79
5. 14.7

Solution

Answer: E

Find the moles of each gas.

\[\begin{align*} n_{\mathrm{O_2}} &= \left ( \dfrac{3.20~\mathrm{g}}{} \right ) \left ( \dfrac{\mathrm{mol}}{32.00~\mathrm{g}} \right ) \\[2ex] &= 0.100~\mathrm{mol}\\[4ex] n_{\mathrm{N_2}} &= \left ( \dfrac{1.40~\mathrm{g}}{} \right ) \left ( \dfrac{\mathrm{mol}}{28.02~\mathrm{g}} \right ) \\[2ex] &= 0.0500~\mathrm{mol} \end{align*}\]

Find the pressure of each gas.

\[\begin{align*} P_{\mathrm{O_2}} &= \dfrac{nRT}{V} \\[2ex] &= \dfrac{(0.100~\mathrm{mol})(0.08206~\mathrm{L~atm~mol^{-1}~K^{-1}})(298.15~\mathrm{K})} {0.250~\mathrm{L}} \\[2ex] &= 9.79~\mathrm{atm}\\[4ex] P_{\mathrm{N_2}} &= \dfrac{nRT}{V} \\[2ex] &= \dfrac{(0.0500~\mathrm{mol})(0.08206~\mathrm{L~atm~mol^{-1}~K^{-1}})(298.15~\mathrm{K})} {0.250~\mathrm{L}} \\[2ex] &= 4.89~\mathrm{atm}\\[2ex] \end{align*}\]

Add the pressures together.

\[\begin{align*} P_{\mathrm{tot}} &= P_{\mathrm{O_2}} + P_{\mathrm{N_2}} \\[2ex] &= (9.79 + 4.89)~\mathrm{atm} \\[2ex] &= 14.68~\mathrm{atm} \end{align*}\]

Practice – Gases

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Find the mole fraction of argon in a mixture of argon and xenon in which the total pressure is 4.0 atm and xenon has a partial pressure of 1.2 atm.

1. 0.30
2. 0.70
3. 1.2
4. 2.8
5. 4.0

Solution

Answer: B

Find the pressure of argon.

\[\begin{align*} P_{\mathrm{Ar}} &= P_{\mathrm{tot}} - P_{\mathrm{Xe}} \\[2ex] &= (4.0 - 1.2)~\mathrm{atm} \\[2ex] &= 2.8~\mathrm{atm} \end{align*}\]

Find the mole fraction of argon.

\[\begin{align*} \chi_{\mathrm{Ar}} &= \dfrac{P_{\mathrm{Ar}}}{P_{\mathrm{tot}}} \\[2ex] &= \dfrac{2.8~\mathrm{atm}}{4.0~\mathrm{atm}} \\[2ex] &= 0.70 \end{align*}\]

Practice – Enthalpy and Heat Transfer

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What is the ΔE (in J) of a system that releases 12.4 J of heat and does 4.2 J of work on the surroundings?

Solution

The system loses energy so q is negative. The system performs work on the surroundings so w is negative.

\[\begin{align*} \Delta E &= q + w \\ &= -12.4~\mathrm{J} + -4.2~\mathrm{J}\\ &= -16.6~\mathrm{J} \end{align*}\]

Practice – Specific Heat and Heat Transfer

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How much heat (in kJ) is needed to raise the temperature of 200.0 g of water from 25.0 °C to 88.0 °C?

Solution

\[\begin{align*} q &= mc_l\Delta T \\ &= \left ( 200.0~\mathrm{g} \right ) \left ( 4.184~\mathrm{J~g^{-1}~^{\circ}C^{-1}} \right ) \left ( 88.0~^{\circ}\mathrm{C} - 25.0~^{\circ}\mathrm{C} \right ) \left ( \dfrac{\mathrm{kJ}}{10^3~\mathrm{J}} \right ) \\ &= 52.7~\mathrm{kJ} \end{align*}\]

Practice – Enthalpy of Formation

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ΔH° for the given reaction is –1107 kJ.
     2Ba(s) + O₂(g) → 2BaO(s)
How much heat (in kJ) is released when 5.75 g of Ba(s) reacts completely with oxygen to form BaO(s)?

Solution

\[\begin{align*} \Delta H &= \left ( \Delta H^{\circ} \right ) \left ( n_{\mathrm{Ba}} \right ) \\[2ex] &= \left ( \dfrac{-1107~\mathrm{kJ}}{2~\mathrm{mol~Ba}} \right ) \left ( \dfrac{5.75~\mathrm{g}}{} \right ) \left ( \dfrac{\mathrm{mol}}{137.33~\mathrm{g}} \right ) \\[2ex] &= -23.2~\mathrm{kJ} \end{align*}\]

The system loses 23.2 kJ worth of heat (i.e. –23.2 kJ). Therefore, a positive 23.2 kJ worth of heat is released to the surroundings.

Practice – Specific Heat Capacity

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The specific heat capacity of lead is 0.13 J g^–1 °C^–1. How much heat (in J) is required to raise the temperature of 15 g of lead from 22 °C to 37 °C?

Solution

\[\begin{align*} q &= mc_s\Delta T \\ &= \left ( 15~\mathrm{g} \right ) \left ( 0.13~\mathrm{J~g^{-1}~^{\circ}C^{-1}}\right ) \left ( 37~^{\circ}\mathrm{C} - 22~^{\circ}\mathrm{C} \right ) \\ &= 29~\mathrm{J} \end{align*}\]

Practice – Coffee Cup Calorimetry

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The ΔH for the solution process when solid sodium hydroxide dissolves in water is 44.4 kJ mol^–1. When a 13.9 g sample of NaOH dissolves in 250.0 g of water to make a solution in a coffee-cup calorimeter, the temperature increases from an initial temperature of 23.0 °C. Determine the final temperature (in °C) of the system. Assume the solution has the same specific heat as liquid water (i.e. 4.184 J mol^–1 K^–1).

Solution

Determine the ΔH for the dissolution of 13.9 g of NaOH in water.

\[\mathrm{NaOH}(s) \xrightarrow{\mathrm{H_2O}} \mathrm{Na^+}(aq) + \mathrm{OH^-}(aq)\]  

\[\begin{align*} \Delta H &= \left ( \Delta H^{\circ} \right ) \left ( n_{\mathrm{NaOH}} \right ) \\[2ex] &= \left ( \dfrac{44.4~\mathrm{kJ}}{1~\mathrm{mol~NaOH}} \right ) \left ( \dfrac{13.9~\mathrm{g}}{} \right ) \left ( \dfrac{\mathrm{mol}}{40.0~\mathrm{g}} \right ) \\[2ex] &= 15.43~\mathrm{kJ} \end{align*}\]

Convert kJ to J.

\[\begin{align*} 15.43~\mathrm{kJ} \left ( \dfrac{10^3~\mathrm{J}}{\mathrm{kJ}} \right ) = 15430~\mathrm{J} \end{align*}\]

Recall that ΔH = q under constant pressure conditions. Find the final temperature of the system.

\[\begin{align*} q &= mc_s\Delta T \rightarrow \\[2ex] T_{f} &= \dfrac{q}{mc_s} + T_i \\[2ex] &= \left [ \dfrac{15430~\mathrm{J}} {\left ( 250.0~\mathrm{g} + 13.9~\mathrm{g} \right ) \left ( 4.184~\mathrm{J~g^{-1}~^{\circ}C} \right ) } \right ] + 23.0~^{\circ}\mathrm{C} \\[2ex] &= 37.0~^{\circ}\mathrm{C} \end{align*}\]

Practice – Enthalpy

---

Determine the enthalpy of reaction (in kJ) for the following reaction
     4NO(g) → 2NO₂(g) + N₂(g)
given the reaction data:
     N₂(g) + O₂(g) → 2NO(g)      ΔH = 180.7 kJ
     2NO(g) + O₂(g) → 2NO₂(g)      ΔH = –113.1 kJ

Solution

Transform and combine the given reference reactions to obtain the reaction of interest. Transform ΔH appropriately.

Steps:

1. Reverse the first reaction to get N₂(g) on the right.
2. Add the second reaction as-is to get 2NO₂(g) on the right.

\[\begin{align*} \mathrm{2NO}(g) &\longrightarrow \mathrm{N_2}(g) + \mathrm{O_2}(g) \quad &&\Delta H = -180.7~\mathrm{kJ} \\ \mathrm{2NO}(g) + \mathrm{O_2}(g) &\longrightarrow \mathrm{2NO_2}(g) \quad &&\Delta H = -113.1~\mathrm{kJ}\\ \end{align*}\]

Adding the reactions together gives

\[\begin{align*} 4\mathrm{NO}(g) \longrightarrow \mathrm{2NO_2}(g) + \mathrm{N_2}(g) \quad \Delta H = -293.8~\mathrm{kJ}\\ \end{align*}\]

Practice – Enthalpy

---

Determine the ΔH_rxn for the following reaction
     4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(l)
using the given heats of formation (ΔH_f° in kJ mol^–1)
     H₂O(l): –286
     NO(g): 90
     NO₂(g): 34
     HNO₃(aq): –207
     NH₃(g): –46

Solution

It should be known that the ΔH_f° for any substance in its standard form is 0 kJ mol^–1. Therefore, this value does not need to be provided in order to complete this problem.

Ensure that the target reaction is balanced.

Determine the enthalpy of reaction.

\[\begin{align*} \Delta H_{\mathrm{rxn}} &= \Sigma \left (\Delta H_{\mathrm{products}} \right ) - ~~\Sigma \left (\Delta H_{\mathrm{reactants}} \right )\\[2ex] &= \left [ \left ( 4 \times \Delta H_f^{\circ}(\mathrm{NO}(g)) \right ) + \left ( 6 \times \Delta H_f^{\circ}(\mathrm{H_2O}(l)) \right ) \right ] - \\ &\phantom{=} ~~ \left [ \left ( 4 \times \Delta H_f^{\circ}(\mathrm{NH_3}(g)) \right ) + \left ( 5 \times \Delta H_f^{\circ}(\mathrm{O_2}(g)) \right ) \right ] \\[2ex] &= \left [ \left ( 4 \times 90 \right ) + \left ( 6 \times -286 \right ) \right ] - \\ &\phantom{=} ~~ \left [ \left ( 4 \times -46 \right ) + \left ( 5 \times 0 \right ) \right ]~\mathrm{kJ~mol^{-1}} \\[2ex] &= -1172~\mathrm{kJ~mol^{-1}} \end{align*}\]

Practice – Internal Energy

---

How can internal energy be increased?

1. transferring heat from the surroundings to the system
2. transferring heat from the system to the surroundings
3. doing work on the system

Solution

A and C both increase the internal energy of a system.

Practice – Enthalpy of Formation

---

Which of the following reactions has a ΔH_rxn° equal to the ΔH_f° of the product?

1. N₂(g) + 3H₂(g) → 2NH₃(g)
2. ½N₂(g) + O₂(g) → NO₂(g)
3. 6C(s) + 6H(g) → C₆H₆(l)
4. P(g) + 4H(g) + Br(g) → PH₄Br(l)
5. 12C(g) + 11H₂(g) + 11O(g) → C₆H₂₃O₁₁(g)

Solution

Answer: B

Concept: Thermochemistry

Practice – Magnetism

---

Which of the following elements is paramagnetic?

1. K
2. Mg
3. Ne
4. Zn

Solution

Answer: A

Concept: Molecular orbital theory

Practice – Electron Configuration

---

What element has a ground state electron configuration of [Kr]5s²4d¹?

1. Y
2. Sc
3. Zr
4. La

Solution

Answer: A

Concept: Electron configuration

Practice – Electron Configuration

---

What is the correct electron configuration for aluminum?

1. 1s²2s¹
2. 1s²2s²2p⁴3s²3p³
3. 1s²2s²2p⁶3s²3p¹
4. 1s²2s²2p²3s²3p²3d²4s¹
5. 1s²2s²2p²3s²3p²4s²5s¹

Solution

Answer: C

Concept: Electron configuration

Practice – Frequency and Wavelength

---

What is the frequency of light (in s^–1) that has a wavelength of 1.23×10^–6 cm?

Solution

Convert wavelength to frequency.

\[\begin{align*} c &= \lambda\nu \rightarrow \\[2ex] \nu &= \dfrac{c}{\lambda} \\[2ex] &= \left ( \dfrac{2.998\times 10^{8}~\mathrm{m~s^{-1}}} {1.23\times 10^{-6}~\mathrm{cm} \left ( \dfrac{\mathrm{m}}{100~\mathrm{cm}}\right )} \right ) \\[2ex] &= 2.44 \times 10^{16}~\mathrm{s^{-1}} \end{align*}\]

Practice – Electron Configuration

---

What is the electron configuration for the Co²⁺ ion?

1. [Ar]4s¹3d⁶
2. [Ar]3d⁷
3. [Ar]3d⁵
4. [Ar]4s²3d⁹
5. [Ne]3s²3p¹⁰

Solution

Answer: B

Concept: Electron configuration

Practice – Periodic Trends

---

Determine how electronegativity changes as you move
     i. move left to right within a period and
     ii. move top to bottom within a group
, respectively.

Solution

1. increases
   
2. decreases

Practice – Periodic Trends

---

Atomic radii generally increases as one moves

1. down a group and from left to right across a period
2. up a group and from left to right across a period
3. down a group and from left to right across a period
4. up a group and from right to left across a period
5. down a group; the period position has no effect

Solution

Answer: A

Concept: Periodic trends

Practice – Periodic Trends

---

Which of the following atoms has the largest radius?

1. Sr
2. Ca
3. K
4. Rb
5. Y

Solution

Answer: D

Concept: Periodic trends

Practice – Electron Configuration

---

Which species is isoelectronic with
     i. argon
     ii. neon
, respectively?

1. Cl^–; F^–
2. Cl^–; Cl⁺
3. F⁺; F^–
4. Ne^–; Kr⁺
5. Ne^–; Ar⁺

Solution

Answer: A

Concept: Electron configuration

Practice – Specific Heat Capacity

---

A 65.0 g piece of iron at 525 °C is put into 635 g of water at 15.0 °C. What is the final temperature (in °C) of the iron and water? The specific heat of iron is 0.451 J g^–1 °C^–1.

Solution

This question assumes that the reader has committed to memory the specific heat of water (4.184 J mol^–1 °C^–1).

\[\begin{align*} -q_{\mathrm{iron}} &= q_{\mathrm{water}} \\[2ex] -mc_{s,\mathrm{iron}}\Delta T &= mc_{l,\mathrm{water}}\Delta T \\[2ex] -\left (65.0~\mathrm{g}\right ) \left ( 0.451~\mathrm{J~g^{-1}~^{\circ}C^{-1}} \right ) \left ( \Delta T \right ) &= \left ( 635~\mathrm{g} \right ) \left ( 4.184~\mathrm{J~g^{-1}~^{\circ}C^{-1}} \right ) \left ( \Delta T \right ) \\[2ex] -29.315~\mathrm{J~^{\circ}C^{-1}} \left ( T_f - T_{i,\mathrm{iron}} \right ) &= 2656.84~\mathrm{J~^{\circ}C^{-1}} \left ( T_f - T_{i,\mathrm{water}} \right ) \\[2ex] \left ( -29.315~\mathrm{J~^{\circ}C^{-1}} \times T_f \right ) - \left ( -29.315~\mathrm{J~^{\circ}C^{-1}} \times 525~^{\circ}\mathrm{C} \right ) &= \left ( 2656.84~\mathrm{J~^{\circ}C^{-1}} \times T_f \right ) - \left ( 2656.84~\mathrm{J~^{\circ}C^{-1}} \times 15.0~^{\circ}\mathrm{C} \right ) \\[2ex] \left ( -29.315~\mathrm{J~^{\circ}C^{-1}} \times T_f \right ) + \left ( 15390.375~\mathrm{J} \right ) &= \left ( 2656.84~\mathrm{J~^{\circ}C^{-1}} \times T_f \right ) - \left ( 39852.6~\mathrm{J} \right ) \\[2ex] \left ( -29.315~\mathrm{J~^{\circ}C^{-1}} \times T_f \right ) - \left ( 2656.84~\mathrm{J~^{\circ}C^{-1}} \times T_f \right ) &= -55242.975~\mathrm{J}\\[2ex] -2686.155~\mathrm{J~^{\circ}\mathrm{C}^{-1}} \times T_f &= -55252.975~\mathrm{J}\\[2ex] T_f &= 20.6~^{\circ}\mathrm{C} \end{align*}\]

Practice – Enthalpy of Formation

---

Consider the following reaction.
     2Mg(s) + O₂(g) → 2MgO(s)    ΔH = –1204 kJ mol^–1

1. Is the reaction endothermic or exothermic?
2. What is the heat transferred when 2.4 g of Mg(s) reacts at a constant pressure?

Solution

Part A

The reaction is exothermic. ΔH is a negative value.

Part B

Determine the change in heat of the system.

\[\begin{align*} \Delta H_{\mathrm{rxn}} &= \left ( \Delta H \right ) \left ( n_{\mathrm{Mg}} \right ) \\[2ex] &= \left ( \dfrac{-1204~\mathrm{kJ}}{\mathrm{2~mol}} \right ) \left ( \dfrac{2.4~\mathrm{g}}{} \right ) \left ( \dfrac{\mathrm{mol}}{24.31~\mathrm{g}} \right ) \\[2ex] &= -59.43~\mathrm{kJ} \end{align*}\]

The heat released is a positive quantity. Therefore, 59.43 kJ was released.

Practice – Wavelength and Energy

---

Calculate the wavelength (in nm) for orange light if the energy of one photon is 3.18×10^–19 J.

Solution

\[\begin{align*} E &= \dfrac{hc}{\lambda} \rightarrow \\[2ex] \lambda &= \dfrac{hc}{E} \\[2ex] &= \left ( \dfrac{\left (6.626\times 10^{-34}~\mathrm{J~s} \right ) \left (2.998\times 10^{8}~\mathrm{m~s^{-1}}\right ) } {3.18\times 10^{-19}~\mathrm{J}} \right ) \left ( \dfrac{10^9\mathrm{nm}}{\mathrm{m}} \right )\\[2ex] &= 625~\mathrm{nm} \end{align*}\]

Practice – Wavelength and Energy

---

An electron in hydrogen drops from n=7 to n=3. What is the wavelength of the photon (in µm)? What is the energy (in J) of the photon emitted.

Solution

Find the wavelength using the Rydberg equation.

\[\begin{align*} \dfrac{1}{\lambda} &= R_{\infty} \left ( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right ) \\[2ex] \lambda &= \left [ R_{\infty} \left ( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right ) \right ]^{-1} \\[2ex] &= \left [ 1.097\times 10^{7} \mathrm{m}^{-1} \left ( \dfrac{1}{3^2} - \dfrac{1}{7^2} \right ) \right ]^{-1} \\[2ex] &= 1.005\times 10^{-6}~\mathrm{m} \end{align*}\]

Convert the wavelength to µm.

\[\begin{align*} \left ( \dfrac{1.005\times 10^{-6}~\mathrm{m}}{} \right ) \left ( \dfrac{10^6~\mu\mathrm{m}}{\mathrm{m}} \right ) = 1.00~\mu\mathrm{m} \end{align*}\]

Find the energy of the photon.

\[\begin{align*} E &= \dfrac{hc}{\lambda} \\[2ex] &= \left ( \dfrac{\left (6.626\times 10^{-34}~\mathrm{J~s} \right ) \left (2.998\times 10^{8}~\mathrm{m~s^{-1}}\right ) } {1.005\times 10^{-6}~\mathrm{m}} \right )\\[2ex] &= 1.98\times 10^{-19}~\mathrm{J} \end{align*}\]

Practice – Quantum Numbers

---

An electron in a certain atom is in the n=2 quantum level. What are the possible values of l and m_l for this electron?

Solution

n=2, l=0, m_l=–1,0,+1

Practice – Electron Configuration

---

How many unpaired electrons do the following have?

1. O
2. Co
3. Ba
4. Pb

Solution

1. 2
2. 3
3. 0
4. 2

Practice – Heat Transfer

---

A 98.7 °C rod with a mass of 3.253 g was placed into 36 mL of water at 22.4 °C inside of an insulated cup calorimeter. Once the water equilibrated, the final temperature was 38.75 °C. What is the specific heat (in J g^–1 °C^–1) of the rod assuming a complete heat transfer to only the water?

Solution

Assume that the water has a density of 1.0 g mL^–1. Therefore, there is 36 g of water present. Also, this problem assumes that the reader has committed the specific heat of water (4.184 J g^–1 °C^–1) to memory.

\[\begin{align*} -q_{\mathrm{rod}} &= q_{\mathrm{water}} \\[2ex] -mc_{s,\mathrm{rod}}\Delta T &= mc_{l,\mathrm{water}}\Delta T \\[2ex] -\left (3.253~\mathrm{g}\right ) c_{s,\mathrm{rod}} \left ( 38.75~^{\circ}\mathrm{C} - 98.7~^{\circ}\mathrm{C} \right ) &= \left ( 36~\mathrm{g} \right ) \left ( 4.184~\mathrm{J~g^{-1}~^{\circ}C^{-1}} \right ) \left ( 38.75~^{\circ}\mathrm{C} - 22.4~^{\circ}\mathrm{C} \right ) \\[2ex] c_{s,\mathrm{rod}} \left ( 195.017~\mathrm{g~^{\circ}C^{-1}} \right ) &= 2462.70~\mathrm{J} \\[2ex] c_{s,\mathrm{rod}} &= 12.6~\mathrm{J~g^{-1}~^{\circ}C^{-1}} \end{align*}\]

Practice – Specific Heat

---

A small piece of silver with a temperature of 126 °C was placed into 63.0 g of water with a temperature of 22.3 °C. What is the mass of the silver if the equilibrated temperature is 26.8 °C? The specific heat of silver is 0.223 J g^–1 °C^–1.

Solution

\[\begin{align*} -q_{\mathrm{silver}} &= q_{\mathrm{water}} \\[2ex] -mc_{s,\mathrm{silver}}\Delta T &= mc_{l,\mathrm{water}}\Delta T \\[2ex] -m \left ( 0.223~\mathrm{J~g^{-1}~^{\circ}C^{-1}} \right ) \left ( 26.8~^{\circ}\mathrm{C} - 126~^{\circ}\mathrm{C} \right ) &= \left ( 63~\mathrm{g} \right ) \left ( 4.184~\mathrm{J~g^{-1}~^{\circ}C^{-1}} \right ) \left ( 26.8~^{\circ}\mathrm{C} - 22.3~^{\circ}\mathrm{C} \right ) \\[2ex] m \left ( 22.1216~\mathrm{J~g^{-1}}\right ) &= 1186.164~\mathrm{J} \\[2ex] &= 53.6~\mathrm{g} \end{align*}\]

Practice – Heat Transfer

---

What is the heat released by a 2.67 g piece of 98.0 °C lead that was added to a 25 mL sample of 22 °C water in an insulated cup calorimeter? The final temperature of the water was found to be 32.6 °C and some heat was absorbed by the calorimeter. The specific heat of lead is 0.128 J g^–1 K^–1.

Solution

Recall that a temperature interval in K is equivalent to a temperature interval in °C. Therefore, the specific heat of lead can be written as 0.128 J g^–1 °C^–1.

\[\begin{align*} -q_{\mathrm{lead}} &= mc_{s,\mathrm{lead}}\Delta T \\[2ex] &= \left ( 2.67~\mathrm{g}\right ) \left ( 0.128~\mathrm{J~g^{-1}~^{\circ}C^{-1}} \right ) \left ( 32.6~^{\circ}\mathrm{C} - 98.0~^{\circ}\mathrm{C} \right ) \\[2ex] &= 22.4~\mathrm{J} \end{align*}\]

Side note: The change in heat of the lead was –22.35 J.

Practice – Calorimetry

---

What is the equilibrated temperature in an insulated cup calorimeter if 50.89 g of 23.5 °C water had a 1.63 g of a 96.2 °C copper added to it? The specific heat of copper is 0.385 J g^–1 °C^–1.

Solution

This problem assumes that the reader has committed the specific heat of water (4.184 J g^–1 °C^–1) to memory.

\[\begin{align*} -q_{\mathrm{copper}} &= q_{\mathrm{water}} \\[2ex] -mc_{s,\mathrm{copper}}\Delta T &= mc_{l,\mathrm{water}}\Delta T \\[2ex] -\left (1.63~\mathrm{g}\right ) \left ( 0.385~\mathrm{J~g^{-1}~^{\circ}C^{-1}} \right ) \left ( \Delta T \right ) &= \left ( 50.89~\mathrm{g} \right ) \left ( 4.184~\mathrm{J~g^{-1}~^{\circ}C^{-1}} \right ) \left ( \Delta T \right ) \\[2ex] -0.62755~\mathrm{J~^{\circ}C^{-1}} \left ( T_f - T_{i,\mathrm{copper}} \right ) &= 212.92~\mathrm{J~^{\circ}C^{-1}} \left ( T_f - T_{i,\mathrm{water}} \right ) \\[2ex] \left ( -0.62755~\mathrm{J~^{\circ}C^{-1}} \times T_f \right ) - \left ( -0.62755~\mathrm{J~^{\circ}C^{-1}} \times 96.2~^{\circ}\mathrm{C} \right ) &= \left ( 212.92~\mathrm{J~^{\circ}C^{-1}} \times T_f \right ) - \left ( 212.92~\mathrm{J~^{\circ}C^{-1}} \times 23.5~^{\circ}\mathrm{C} \right ) \\[2ex] \left ( -0.62755~\mathrm{J~^{\circ}C^{-1}} \times T_f \right ) + \left ( 60.37031~\mathrm{J} \right ) &= \left ( 212.92~\mathrm{J~^{\circ}C^{-1}} \times T_f \right ) - \left ( 5003.62~\mathrm{J} \right ) \\[2ex] \left ( -0.62755~\mathrm{J~^{\circ}C^{-1}} \times T_f \right ) - \left ( 212.92~\mathrm{J~^{\circ}C^{-1}} \times T_f \right ) &= -5063.99~\mathrm{J}\\[2ex] -213.548~\mathrm{J~^{\circ}\mathrm{C}^{-1}} \times T_f &= -5063.99~\mathrm{J}\\[2ex] T_f &= 23.7~^{\circ}\mathrm{C} \end{align*}\]

Practice – Enthalpy

---

Given the following reactions
\[\begin{align*} \mathrm{2Al}(s) + \mathrm{6HCl}(aq) &\longrightarrow \mathrm{2AlCl_3}(aq) + \mathrm{3H_2}(g) \quad &&\Delta H = -1049~\mathrm{kJ} \\ \mathrm{HCl}(g) &\longrightarrow \mathrm{HCl}(aq) \quad &&\Delta H = -74.8~\mathrm{kJ} \\ \mathrm{H_2}(g) + \mathrm{Cl_2}(g) &\longrightarrow \mathrm{2HCl}(g) \quad &&\Delta H = -1845~\mathrm{kJ} \\ \mathrm{AlCl_3}(s) &\longrightarrow \mathrm{AlCl_3}(aq) \quad &&\Delta H = -323~\mathrm{kJ} \end{align*}\] calculate the ΔH (in kJ) for the following reaction:
\[2\mathrm{Al}(s) + \mathrm{3Cl_2}(g) \longrightarrow \mathrm{2AlCl_3}(s)\]

Solution

Transform and combine the given reference reactions to obtain the reaction of interest. Transform ΔH appropriately.

Steps:

1. Write the first reaction down as-is to get 2Al(s) on the left.
2. Multiply the second reaction by 6 to get 6HCl(aq) on the right.
3. Multiply the third reaction by 3 to get 3Cl₂(g) on the left.
4. Reverse and multiply the fourth reaction by 2 to get 2AlCl₃(s) on the right.

\[\begin{align*} \mathrm{2Al}(s) + \mathrm{6HCl}(aq) &\longrightarrow \mathrm{2AlCl_3}(aq) + \mathrm{3H_2}(g) \quad &&\Delta H = -1049~\mathrm{kJ} \\ \mathrm{6HCl}(g) &\longrightarrow \mathrm{6HCl}(aq) \quad &&\Delta H = -448.8~\mathrm{kJ} \\ \mathrm{3H_2}(g) + \mathrm{3Cl_2}(g) &\longrightarrow \mathrm{6HCl}(g) \quad &&\Delta H = -5535~\mathrm{kJ} \\ \mathrm{2AlCl_3}(aq) &\longrightarrow \mathrm{2AlCl_3}(s) \quad &&\Delta H = 646~\mathrm{kJ} \end{align*}\]

Adding the reactions together gives

\[\begin{align*} 2\mathrm{Al}(s) + \mathrm{3Cl_2}(g) \longrightarrow \mathrm{2AlCl_3}(s) \quad \Delta H = -6387~\mathrm{kJ} \end{align*}\]

Practice – Heat Transfer

---

What is the heat energy released when 2 moles of nitrogen react in the following reaction?
N₂(g) + 3H₂(g) → 2NH₃(g)     ΔH = –92.0 kJ mol^–1

Solution

\[\begin{align*} \Delta H &= \left ( \dfrac{-92.0~\mathrm{kJ}}{\mathrm{mol~N_2}} \right ) \left ( \dfrac{2~\mathrm{mol~N_2}}{} \right ) \\[2ex] &= -184~\mathrm{kJ} \end{align*}\]

Practice – Enthalpy of Reaction

---

What is the heat energy released when 0.50 moles of ammonia is formed according to the following reaction?
     N₂(g) + 3H₂(g) → 2NH₃(g)     ΔH = –92.0 kJ mol^–1

Solution

\[\begin{align*} \Delta H &= \left ( \dfrac{-92.0~\mathrm{kJ}}{2~\mathrm{mol~NH_3}} \right ) \left ( \dfrac{0.50~\mathrm{mol~NH_3}}{} \right ) \\[2ex] &= -23~\mathrm{kJ} \end{align*}\]

Practice – Heat Transfer

---

Find the ΔH (in kJ) when 100.0 mL of a 1 M solution of sodium hydroxide is reacted with 100.0 mL of a 1 M HCl solution in a calorimeter if the temperature rises from 21.0 °C to 34.6 °C. Assume the specific heat of the solution is 4.184 J g^–1 °C^–1 and the density of the solution is that of water (1 g mL^–1).

Solution

The mass of the solution is 200.0 g.

\[\begin{align*} q &= mc\Delta T \\[2ex] &= \left (200.0~\mathrm{g} \right ) \left (4.184~\mathrm{J~g^{-1}~^{\circ}C^{-1}} \right ) \left (34.6~^{\circ}\mathrm{C} - 21.0~^{\circ}\mathrm{C} \right ) \left (\dfrac{\mathrm{kJ}}{10^3~\mathrm{J}} \right ) \\[2ex] &= 11.38~\mathrm{kJ} \end{align*}\]

The solution temperature increased due to the solution absorbing 11.38 kJ of heat energy. Therefore, the reaction produced 11.38 kJ of energy (exothermic) making ΔH = –11.38 kJ.

\[\begin{align*} -q_{\mathrm{reaction}} &= q_{\mathrm{solution}} \\[2ex] q_{\mathrm{reaction}} &= -q_{\mathrm{solution}} \\[2ex] &= -11.38~\mathrm{kJ} \end{align*}\]

The final answer to the appropriate number of 3 significant figures is –11.4 kJ mol^–1.

Practice – Internal Energy

---

What is the change of internal energy of a system (in kJ) if 26.5 kJ of heat is evolved from the system and the surroundings do 562 kJ of work on the system?

Solution

The system evolves heat making q negative. Work is done on the system making w positive.

\[\begin{align*} \Delta U &= q + w \\[2ex] &= -26.5~\mathrm{kJ} + 562~\mathrm{kJ} \\[2ex] &= 536~\mathrm{kJ} \end{align*}\]

Practice – Internal Energy

---

What is the change in internal energy of a system that absorbs 36.8 kJ of heat and does 36.0 kJ of work on the surroundings?

Solution

The system absorbs heat making q positive. The system performs work on the surroundings making w negative.

\[\begin{align*} \Delta U &= q + w \\[2ex] &= 36.8~\mathrm{kJ} -36.0~\mathrm{kJ} \\[2ex] &= 0.8~\mathrm{kJ} \end{align*}\]

Practice – Kinetic Molecular Theory

---

If the velocity of an object decreases by half but the mass remains constant, how does the kinetic energy (KE) change?

A. KE is now 1/4th the original KE B. KE is now four times greater C. KE is now two times greater D. KE is now 1/2 the original KE

Solution

Answer: A

Recall the kinetic energy equation.

\[\mathrm{KE} = \dfrac{1}{2}mv^2\]

If v is one-half, then v² is one-fourth and KE is one-fourth of what it originally was.

Practice – Kinetic Molecular Theory

---

If the mass of an object is decreased by half but the velocity remains constant, how does kinetic energy (KE) change?

A. KE is now 1/4th the original KE B. KE is now four times greater C. KE is now two times greater D. KE is now 1/2 the original KE

Solution

Answer: D

Recall the kinetic energy equation.

\[\mathrm{KE} = \dfrac{1}{2}mv^2\]

If m is one-half, then KE is one-half of what it originally was.

Practice – Energy and Wavelength

---

If a photon emits light at a wavelength of 642.6 nm, what is the energy (in J) associated with that photon?

Solution

\[\begin{align*} E &= \dfrac{hc}{\lambda} \\[2ex] &= \dfrac{\left (6.626\times 10^{-34}~\mathrm{J~s} \right ) \left (2.998\times 10^{8}~\mathrm{m~s^{-1}}\right ) } {\left ( 642.6~\mathrm{nm} \right ) \left ( \dfrac{\mathrm{m}}{10^9~\mathrm{nm}} \right ) } \\[2ex] &= 3.09\times 10^{-19}~\mathrm{J} \end{align*}\]

Practice – Energy and Frequency

---

A photon has a frequency of 4.8×10¹⁸ Hz. What is the wavelength (in nm) of this light?

Solution

\[\begin{align*} c &= \lambda\nu \rightarrow \\[2ex] \lambda &= \dfrac{c}{\nu} \\[2ex] &= \left ( \dfrac{2.998\times 10^{8}~\mathrm{m~s^{-1}}} {4.8\times 10^{18}~\mathrm{s^{-1}}} \right ) \left ( \dfrac{10^9~\mathrm{nm}}{\mathrm{m}} \right ) \\[2ex] &= 0.062~\mathrm{nm} \end{align*}\]

Practice – Energy and Frequency

---

What is the energy (in J) associated with a photon that has a frequency of 2.10×10¹⁵ s^–1?

Solution

Find the wavelength of the photon.

\[\begin{align*} c &= \lambda\nu \rightarrow \\[2ex] \lambda &= \dfrac{c}{\nu} \\[2ex] &= \left ( \dfrac{2.998\times 10^{8}~\mathrm{m~s^{-1}}} {2.1\times 10^{15}~\mathrm{s^{-1}}} \right ) \\[2ex] &= 1.4276\times 10^{-7}~\mathrm{m} \end{align*}\]

Find the energy of the photon.

\[\begin{align*} E &= \dfrac{hc}{\lambda} \\[2ex] &= \dfrac{\left (6.626\times 10^{-34}~\mathrm{J~s} \right ) \left (2.998\times 10^{8}~\mathrm{m~s^{-1}}\right ) } {1.4276\times 10^{-7}~\mathrm{m}} \\[2ex] &= 1.39\times 10^{-18}~\mathrm{J} \end{align*}\]

Practice – Electron Transitions

---

An electron transitions from an n=5 energy state to an n=2 energy state. What is the energy (in J) associated with this transition in a hydrogen atom?

Solution

Find the wavelength using the Rydberg equation.

\[\begin{align*} \dfrac{1}{\lambda} &= R_{\infty} \left ( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right ) \\[2ex] \lambda &= \left [ R_{\infty} \left ( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right ) \right ]^{-1} \\[2ex] &= \left [ 1.097\times 10^{7}~\mathrm{m}^{-1} \left ( \dfrac{1}{2^2} - \dfrac{1}{5^2} \right ) \right ]^{-1} \\[2ex] &= 4.341\times 10^{-7}~\mathrm{m} \end{align*}\]

Find the energy of the photon.

\[\begin{align*} E &= \dfrac{hc}{\lambda} \\[2ex] &= \left ( \dfrac{\left (6.626\times 10^{-34}~\mathrm{J~s} \right ) \left (2.998\times 10^{8}~\mathrm{m~s^{-1}}\right ) } {4.341\times 10^{-7}~\mathrm{m}} \right )\\[2ex] &= 4.576\times 10^{-19}~\mathrm{J} \end{align*}\]

Practice – Electron Transitions

---

An electron transitions from an n=3 energy state to an n=2 energy state. What is the energy (in J) associated with this transition in a hydrogen atom?

Solution

Find the wavelength using the Rydberg equation.

\[\begin{align*} \dfrac{1}{\lambda} &= R_{\infty} \left ( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right ) \\[2ex] \lambda &= \left [ R_{\infty} \left ( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right ) \right ]^{-1} \\[2ex] &= \left [ 1.097\times 10^{7}~\mathrm{m}^{-1} \left ( \dfrac{1}{2^2} - \dfrac{1}{3^2} \right ) \right ]^{-1} \\[2ex] &= 6.563\times 10^{-7}~\mathrm{m} \end{align*}\]

Find the energy of the photon.

\[\begin{align*} E &= \dfrac{hc}{\lambda} \\[2ex] &= \left ( \dfrac{\left (6.626\times 10^{-34}~\mathrm{J~s} \right ) \left (2.998\times 10^{8}~\mathrm{m~s^{-1}}\right ) } {6.563\times 10^{-7}~\mathrm{m}} \right )\\[2ex] &= 3.027\times 10^{-19}~\mathrm{J} \end{align*}\]

Practice – Electron Configuration

---

Which of the following is the electron configuration for Mn?

1. 1s²2s²2p⁶3s²3p⁶3d⁷
2. 1s²2s²2p⁶3s²3p⁶4s²3d⁵
3. 4s²3d⁵
4. 1s²2s²2p⁶3s²3p⁶4s¹3d⁶

Solution

Answer: B

Practice – Electron Configuration

---

Which of the following is the electron configuration for Cu⁺?

1. 1s²2s²2p⁶3s²3p⁶3d¹⁰
2. 1s²2s²2p⁶3s²3p⁶4s¹3d⁹
3. 1s²2s²2p⁶3s²3p⁶4s²3d⁹
4. 1s²2s²2p⁶3s²3p⁶4s¹3d¹⁰

Solution

Answer: A

Practice – Electron Configuration

---

Which of the following is the complete electron configuration for O^2–?

1. [He]2s²2p⁵
2. 2s²2p⁶
3. 1s²2s¹2p⁶
4. 1s²2s²2p⁶

Solution

Answer: D

Practice – Electron Configuration

---

Which of the following is the correct electron configuration for Au?

1. [Kr]6s²4f¹⁴5d⁹
2. [Xe]6s¹4f¹⁴5d¹⁰
3. 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s²4d¹⁰5p⁶6s²4f¹⁴5d⁹
4. [Xe]6s²5f¹⁴5d⁹

Solution

Answer: B

Practice – Magnetism

---

Which of the following elements are diamagnetic?

1. lead
2. ruthenium
3. lithium
4. tin

Solution

Answer: A and D

Practice – Magnetism

---

Which of the following elements are paramagnetic?

1. Na⁺
2. Ca²⁺
3. Cr³⁺
4. O^2–

Solution

Answer: C

Practice – Quantum Numbers

---

Choose the set of quantum numbers (ordered as n, l, m_l, m_s) that is not possible.

1. 3, 1, 0, –1/2
2. 4, 0, 0, +1/2
3. 2, 3, 1, +1/2
4. 4, 3, –2, –1/2

Solution

Answer: C

Practice – Quantum Numbers

---

Choose the only set of quantum numbers (ordered as n, l, m_l, m_s) that is possible.

1. 6, 1, –1, +1/2
2. 3, 0, 1, –1/2
3. 2, 2, –1, +1/2
4. 1, 1, 0, –1/2

Solution

Answer: A

Practice – Quantum Numbers

---

Which of the following is a possible set of quantum numbers (ordered as n, l, m_l, m_s) for an electron found in the third period of a d orbital?

1. 3, 1, –1, +1/2
2. 4, 2, –1, –1/2
3. 3, 3, 0, –1/2
4. 3, 2, –2, +1/2

Solution

Answer: D

Practice – Electron Configuration

---

What is the number of valence electrons for Co?

1. 9
2. 27
3. 7
4. 2

Solution

Answer: A

Practice – Electron Configuration

---

Which of the following pairs is isoelectronic?

1. C^4–, Cl^–
2. O^2–, Na⁺
3. K⁺, Na⁺
4. O^2–, S^2–

Solution

Answer: B

Practice – Periodic Trends

---

As one moves left to right across the periodic table, how do the following atomic properties generally change?
     i. atomic radius
     ii. electronegativity
     iii. electron affinity

1. decreases, decreases, increases
2. increases, increases, decreases
3. increases, increases, increases
4. decreases, increases, increases

Solution

Answer: D

Practice – Periodic Trends

---

As one moves down the periodic table, how do the following atomic properties generally change?
     i. atomic radius
     ii. electronegativity
     iii. electron affinity

1. decreases, decreases, increases
2. increases, decreases, decreases
3. increases, increases, increases
4. decreases, increases, increases

Solution

Answer: B

Practice – Frequency

---

Tsunami waves travel through the deep ocean speeds of around 800 km h^–1 and have wavelengths of around 200 km. Find the frequency of a tsunami wave with this exact speed and wavelength.

1. 900 s^–1
2. 0.00111 Hz
3. 4.00 s^–1
4. 900 h^–1
5. 0.0144 daHz

Solution

Answer: B

While c = λν is typically used when characterizing light, the speed of light, c, can be replaced with the speed of a tsunami wave.

\[\begin{align*} \mathrm{speed} &= \lambda\nu \rightarrow \\[2ex] \nu &= \dfrac{\mathrm{speed}}{\lambda} \\[2ex] &= \dfrac{800~\mathrm{km~h^{-1}}}{200~\mathrm{km}}\\[2ex] &= 4.00~\mathrm{h^{-1}} \end{align*}\]

Since the computed frequency is not one of the given answers, convert the value into s^–1.

\[\begin{align*} \left ( \dfrac{4.00}{\mathrm{h}} \right ) \left ( \dfrac{1~\mathrm{h}}{60~\mathrm{min}} \right ) \left ( \dfrac{1~\mathrm{min}}{60~\mathrm{s}} \right ) = 0.00111~\mathrm{s^{-1}} \equiv 0.00111~\mathrm{Hz} \end{align*}\]

Practice – Calorimetry

---

Consider the following reaction:

\[\mathrm{HCl}(aq) + \mathrm{NaOH}(aq) \longrightarrow \mathrm{H_2O}(l) + \mathrm{NaCl}(aq) \qquad \Delta H = -56.2~\mathrm{kJ~mol^{-1}}\]

An unknown mass of HCl(aq) is neutralized. The overall solution has a mass of 0.826 kg and rises in temperature by 6.5 K. Assuming that the specific heat of the solution is the same as water, how many grams of HCl were neutralized?

1. 0.40
2. 15
3. 22
4. 400
5. not enough information

Solution

Answer: B

Recall that

\[\begin{align*} q &= mc_s\Delta T \end{align*}\]

Determine the heat absorbed by the solution. It is assumed that the reader has committed to memory the specific heat of water.

\[\begin{align*} q_{\mathrm{soln}} &= \left ( mc_{s} \Delta T \right )_{\mathrm{soln}} \\[2ex] &= \left ( 0.826~\mathrm{kg} \right ) \left ( \dfrac{10^3~\mathrm{g}}{\mathrm{kg}}\right ) \left ( 4.184~\mathrm{J~g^{-1}~^{\circ}\mathrm{C}^{-1}} \right ) \left ( 6.5~^{\circ}\mathrm{C} \right ) \\[2ex] &= 22464~\mathrm{J} \end{align*}\]

Therefore, the heat released by the reaction is

\[\begin{align*} q_{\mathrm{rxn}} &= -q_{\mathrm{soln}} \\[2ex] &= -22464~\mathrm{J} \end{align*}\]

Combine the heat released with the enthalpy of reaction to determine the mass of HCl that reacted.

\[\begin{align*} m_{\mathrm{HCl}} &= \left ( \dfrac{-22464~\mathrm{J}}{} \right ) \left ( \dfrac{\mathrm{kJ}}{10^3~\mathrm{J}} \right ) \left ( \dfrac{\mathrm{mol~HCl}}{-56.2~\mathrm{kJ}} \right ) \left ( \dfrac{36.46~\mathrm{g~HCl}}{\mathrm{mol~HCl}} \right ) \\[2ex] &= 15~\mathrm{g~HCl} \end{align*}\]

Practice – Electron Configuration

---

Which of the following would not have 2 valence electrons?

1. Be
2. Na^–
3. Sc⁺
4. Zn
5. all of these have 2 valence electrons

Solution

Answer: E

1. Be: [He]2s²
2. Na^–: [Ne]3s²
3. Sc⁺: [Ar]4s¹3d¹
4. Zn: [Ar]4s²3d¹⁰
5. all of these have 2 valence electrons

Practice – Heat

---

Bobert was out in his yard barefoot when he burnt his feet on one of his concrete stepping stones. He makes bad decisions and decided his best course of action was to cool this stepping stone by chucking it in his children’s kiddie pool. The stepping stone had a mass of 2.30 kg and was 44.4 °C when he burned his feet. The kiddie pool contained 28.0 L of water at 31.0 °C before the addition of the stone. What was the final temperature (in °C) of the water after the stone was tossed in and the system allowed to equilibrate?
   c_s,concrete = 0.880 J g^–1 °C^–1
   d_H2O = 1.00 g mL^–1

1. 0.0309
2. 31.2
3. 37.7
4. 43.7
5. 44.6

Solution

Answer: B

Convert all masses to grams.

\[\begin{align*} m_{\mathrm{concrete}} &= \left ( \dfrac{2.30~\mathrm{kg}}{} \right ) \left ( \dfrac{10^3~\mathrm{g}}{\mathrm{kg}} \right ) \\[2ex] &= 2300~\mathrm{g}\\[4ex] m_{\mathrm{water}} &= \left ( \dfrac{28.0~\mathrm{L}}{} \right ) \left ( \dfrac{10^3~\mathrm{mL}}{\mathrm{L}} \right ) \left ( \dfrac{1.00~\mathrm{g}}{\mathrm{L}} \right ) \\[2ex] &= 28000~\mathrm{g} \end{align*}\]

Solve for the final temperature. It is assumed that the reader has committed to memory the specific heat of water.

\[\begin{align*} q_{\mathrm{concrete}} &= -q_{\mathrm{water}} \\[2ex] (mc_s\Delta T)_{\mathrm{concrete}} &= -(mc_s\Delta T)_{\mathrm{water}} \\[2ex] \left [ \left ( 2300~\mathrm{g} \right ) \left ( 0.880~\mathrm{J~g^{-1}~^{\circ}\mathrm{C}} \right ) \left ( T_{\mathrm{f}} - 44.4~^{\circ}\mathrm{C} \right ) \right ]_{\mathrm{concrete}} &= - \left [ \left ( 28000~\mathrm{g} \right ) \left ( 4.184~\mathrm{J~g^{-1}~^{\circ}\mathrm{C}} \right ) \left ( T_{\mathrm{f}} - 31.0~^{\circ}\mathrm{C} \right ) \right ]_{\mathrm{water}} \\[2ex] T_{\mathrm{f}} &= 31.2~^{\circ}\mathrm{C} \end{align*}\]

Practice – Quantum Numbers

---

Which of the following sets of quantum numbers describes the highest energy electron in the ground state electron configuration of potassium (K)?

1. n=1, l=0, m_l=0, m_s=+1/2
2. n=3, l=2, m_l=1, m_s=+1/2
3. n=4, l=2, m_l=1, m_s=+1/2
4. n=4, l=0, m_l=0, m_s=+1/2
5. n=4, l=3, m_l=3, m_s=+1/2

Solution

Answer: D

The electron configuration for K is

\[[\mathrm{Ar}]4s^1\]

Core electrons will all have lower energy than the 4s¹ valence electron. For the valence electron, n=4 and l=0 since the atomic orbital is an s orbital. m_l can only be 0 and m_s can only be +/- 1/2.

Practice – Electron Configuration

---

Match the concept to its name.

ID	Name	ID	Concept
A	Pauli exclusion principle	X	Electrons are added to the lowest energy orbitals before moving to higher energy orbitals
B	Aufbau principle	Y	No two electrons can have the same four quantum numbers
C	Hund’s Rule	Z	The most stable arrangement of electrons maximizes the number of electrons with the same spin

1. A:X, B:Y, C:Z
2. A:Y, B:X, C:Z
3. A:Z, B:X, C:Y
4. A:Y, B:Z, C:X
5. A:X, B:Z, C:Y

Solution

Answer: B

Practice – Hess’s Law

---

Consider the following equations:

\[\begin{align*} \mathrm{2B}(s) + \mathrm{3H_2}(g) &\longrightarrow \mathrm{B_2H_6}(g) \quad &&\Delta H = 36~\mathrm{kJ~mol^{-1}} \\ \mathrm{2B}(s) + \frac{3}{2}\mathrm{O_2}(g) &\longrightarrow \mathrm{B_2O_3}(s) \quad &&\Delta H = -1273~\mathrm{kJ~mol^{-1}} \\ \mathrm{H_2}(g) + \frac{1}{2}\mathrm{O_2}(g) &\longrightarrow \mathrm{H_2O}(l) \quad &&\Delta H = -286~\mathrm{kJ~mol^{-1}} \\ \mathrm{H_2O}(l) &\longrightarrow \mathrm{H_2O}(g) \quad &&\Delta H = 44~\mathrm{kJ~mol^{-1}} \end{align*}\]

Calculate ΔH (in kJ mol^–1) for the combustion of borane given as

\[\mathrm{B_2H_6}(g) + \mathrm{3O_2}(g) \longrightarrow \mathrm{B_2O_3}(s) + \mathrm{3H_2O}(g)\]

1. –1463
2. –1151
3. –1963
4. –2035
5. –3612

Solution

Answer: D

Transform and combine the given reference reactions to obtain the reaction of interest. Transform ΔH appropriately.

Steps:

1. Reverse the first reaction to get B₂H₆ on the left.
2. Add the second reaction to get B₂O₃ on the right.
3. Add three times the third reaction to get rid of H₂ on the right.
4. Add three times the fourth reaction.

\[\begin{align*} \mathrm{B_2H_6}(g) &\longrightarrow \mathrm{2B}(s) + \mathrm{3H_2}(g) \quad &&\Delta H = -36~\mathrm{kJ~mol^{-1}} \\ \mathrm{2B}(s) + \frac{3}{2}\mathrm{O_2}(g) &\longrightarrow \mathrm{B_2O_3}(s) \quad &&\Delta H = -1273~\mathrm{kJ~mol^{-1}} \\ \mathrm{3H_2}(g) + \frac{3}{2}\mathrm{O_2}(g) &\longrightarrow \mathrm{3H_2O}(l) \quad &&\Delta H = -858~\mathrm{kJ~mol^{-1}} \\ \mathrm{3H_2O}(l) &\longrightarrow \mathrm{3H_2O}(g) \quad &&\Delta H = 132~\mathrm{kJ~mol^{-1}} \end{align*}\]

Adding the reactions together gives

\[\begin{align*} \mathrm{B_2H_6}(g) + \mathrm{3O_2}(g) \longrightarrow \mathrm{B_2O_3}(s) + \mathrm{3H_2O}(g) \qquad \Delta H = -2035~\mathrm{kJ~mol^{-1}} \end{align*}\]

Practice – Electron Configuration

---

Assume that a particular anion and a particular cation are isoelectronic. Which has the larger radius? Choose the most correct answer.

1. The anion has the larger ionic radius
2. The cation has the larger ionic radius
3. The ionic radii are the same because they are isoelectronic
4. Which is larger is dependent on the number of neutrons
5. There is no way to tell

Solution

Answer: A

Practice – Wavelength

---

A mole of photons has an energy of 292 kJ. What is the wavelength (in nm) of one photon?

1. 6.80×10^–31
2. 6.80×10^–22
3. 4.10×10^–7
4. 4.10×10²
5. 4.85×10^–19

Solution

Answer: D

Determine the energy of a single photon.

\[\begin{align*} \left ( \dfrac{292~\mathrm{kJ}}{\mathrm{mol}} \right ) \left ( \dfrac{\mathrm{mol}}{6.022\times 10^{23}~\mathrm{photons}} \right ) \left ( \dfrac{10^3~\mathrm{J}}{\mathrm{kJ}} \right ) = 4.85\times 10^{-19}~\mathrm{J~photon^{-1}} \end{align*}\]

Find the wavelength of the photon.

\[\begin{align*} E &= \dfrac{hc}{\lambda} \rightarrow \\[2ex] \lambda &= \dfrac{hc}{E} \\[2ex] &= \left ( \dfrac{(6.626\times 10^{24}~\mathrm{J~s})(2.998\times 10^{8}~\mathrm{m~s^{-1}})} {4.85\times 10^{-19}~\mathrm{J}}\right ) \left ( \dfrac{10^{9}~\mathrm{nm}}{\mathrm{m}} \right ) \\[2ex] &= 410~\mathrm{nm} \end{align*}\]

Practice – Electron Configuration

---

Give the ground state electron configuration for Se.

1. 1s²2s²2p⁶3s²3p⁶ 4s²4p⁴
2. 1s²2s²2p⁶3s²3p⁶ 4s²3d¹⁰4p⁴
3. 1s²2s²2p⁶3s²3p⁶ 4s²4d¹⁰4p⁴
4. 1s²2s²2p⁶3s²3p⁶ 4s²4p⁴4d¹⁰

Solution

Answer: B

Practice – Work

---

A system under 5.0 atm of constant pressure is compressed from 2.50 L to 0.14 L. What is the value of work for the system? (1 L atm = 101.3 J)

1. 1.2
2. 12
3. –1.2
4. –12
5. none of the above

Solution

Answer: A

Find the work performed in kJ.

\[\begin{align*} w &= -P\Delta V \\[2ex] &= -\left ( 5.0~\mathrm{atm} \right ) \left ( 0.14~\mathrm{L} - 2.50~\mathrm{L} \right ) \left ( \dfrac{101.3~\mathrm{J}}{\mathrm{L~atm}} \right ) \left ( \dfrac{\mathrm{kJ}}{10^3~\mathrm{J}} \right ) \\[2ex] &= 1.18~\mathrm{kJ} \end{align*}\]

Practice – Heat

---

175 mL of boiling 100 °C water (d ≈ 1.00 g mL^–1) is poured into a 23.00 °C ceramic mug. The mug and water come to a final temperature 24.00 °C. Assuming 100% thermal energy transfer, what is the heat capacity (in J °C^–1) of the mug?

1. 5.56×10⁴
2. –5.56×10⁴
3. 1.75×10⁴
4. –1.75×10⁴
5. not enough information; cannot be determined

Solution

Answer: A

Find the mass of the water.

\[\begin{align*} m_{\mathrm{water}} &= \left ( \dfrac{175~\mathrm{mL}}{} \right ) \left ( \dfrac{1.00~\mathrm{g}}{\mathrm{L}} \right ) \\[2ex] &= 175~\mathrm{g} \end{align*}\]

Find the heat capacity of the mug.

\[\begin{align*} q_{\mathrm{mug}} &= -q_{\mathrm{water}} \\[2ex] (C\Delta T)_{\mathrm{mug}} &= -(mc_s\Delta T)_{\mathrm{water}} \\[2ex] \left [ \left (C \right ) \left ( 24.00~^{\circ}\mathrm{C} - 23.00~^{\circ}\mathrm{C} \right ) \right ]_{\mathrm{mug}} &= - \left [ \left ( 175~\mathrm{g} \right ) \left ( 4.184~\mathrm{J~g^{-1}~^{\circ}\mathrm{C}} \right ) \left ( 24.00~^{\circ}\mathrm{C} - 100.00~^{\circ}\mathrm{C} \right ) \right ]_{\mathrm{water}} \\[2ex] C_{\mathrm{mug}} &= 5.56\times 10^{4}~\mathrm{J}~^{\circ}\mathrm{C}^{-1} \end{align*}\]

Practice – Work

---

25 mL of 0.45 M sulfuric acid and 15 mL of 1.20 M potassium hydroxide are combined in a small flask at constant pressure. How much heat (in kJ) is produced?

\[\mathrm{H_2SO_4}(aq) + \mathrm{2KOH}(aq) \longrightarrow \mathrm{K_2SO_4}(aq) + \mathrm{H_2O}(l) \qquad \Delta H = -104.2~\mathrm{kJ~mol^{-1}}\]

1. 1.17
2. 0.94
3. 1.88
4. 0.59
5. Heat cannot be directly related to the given question.

Solution

Answer: B

Determine the heat generated if each reactant reacted completely. The limiting reactant will give the smaller in magnitude energy.

H₂SO₄

\[\begin{align*} \left ( \dfrac{25~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right ) \left ( \dfrac{0.45~\mathrm{mol}}{\mathrm{L}} \right ) \left ( \dfrac{-104.2~\mathrm{kJ}}{\mathrm{mol}} \right ) = -1.17~\mathrm{kJ} \end{align*}\]

KOH

\[\begin{align*} \left ( \dfrac{15~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right ) \left ( \dfrac{1.20~\mathrm{mol}}{\mathrm{L}} \right ) \left ( \dfrac{-104.2~\mathrm{kJ}}{2~\mathrm{mol}} \right ) = -0.94~\mathrm{kJ} \end{align*}\]

KOH is limiting and the heat produced is 0.94 kJ.

Practice – Properties

---

Which of the following describes a chemical property?

1. Mercury(II) oxide is orange
2. Hydrogen is flammable
3. Mercury is a liquid
4. Calcium carbonate is brittle
5. Water has a density of 1.0 g mL^–1

Solution

Answer: B

The others are physical properties.

Practice – Units

---

Which of these is not a fundamental SI unit?

1. kilogram
2. second
3. liter
4. meter
5. ampere

Solution

Answer: C

Volume is a derived unit from the base SI unit of length.

Practice – Matter

---

Which statement about the phases of matter is correct?

1. Gases and liquids expand to fill the volume of their container
2. Solids and plasmas take the shape of their container
3. Solids and liquids have definite volumes that are nearly independent of pressure
4. The molecules in a gas are considered to be very close together
5. Liquids are generally considered to contain many electrically charged particles.

Solution

Answer: C

Practice – Accuracy and Precision

---

The density of gold is 19.32 g cm^–3. A student performs three separate measurements of the volume of a piece of gold weighing 37.5 g and obtains the following results:
     Trial 1: 3.5 mL
     Trial 2: 3.7 mL
     Trial 3: 3.4 mL

Which of the following bests describes these results?

1. Both accurate and precise
2. Accurate but not precise
3. Precise but not accurate
4. Extremely perfect
5. Very approximate

Solution

Answer: C

Determine the density of gold from the given trial information.

\[\begin{align*} d_{\mathrm{Au,Trial~1}} &= \dfrac{37.5~\mathrm{g}}{3.5~\mathrm{mL}} = 10.7~\mathrm{g~mL^{-1}} \rightarrow 11~\mathrm{g~mL^{-1}}\\[2ex] d_{\mathrm{Au,Trial~2}} &= \dfrac{37.5~\mathrm{g}}{3.7~\mathrm{mL}} = 10.1~\mathrm{g~mL^{-1}} \rightarrow 10~\mathrm{g~mL^{-1}}\\[2ex] d_{\mathrm{Au,Trial~3}} &= \dfrac{37.5~\mathrm{g}}{3.4~\mathrm{mL}} = 11.0~\mathrm{g~mL^{-1}} \rightarrow 11~\mathrm{g~mL^{-1}} \end{align*}\]

The determined values are relatively close together (precise) but not close to the actual density of gold (not accurate).

Practice – Metric System

---

Which of these values are equal?

I. 3×10¹¹ picograms II. 3×10⁹ micrograms III. 3×10^–1 grams IV. 3×10^–4 kilograms V. 3×10² megagrams

1. I and III
2. I, III, and IV
3. II and IV
4. IV and V
5. I, II, and III

Solution

Answer: B

Practice – Temperature

---

The title of American author Ray Bradbury’s dystopian science fiction novel, Fahrenheit 451, supposedly refers to the autoignition temperature of paper. What might this book have been called if Ray Bradbury had been British?

1. Celsius 233
2. Celsius 219
3. Celsius 780
4. Celsius 268
5. Celsius 754

Solution

Answer: A

\[\begin{align*} T_{^{\circ}\mathrm{C}} &= \dfrac{5}{9}\left ( ^\circ\mathrm{F} - 32 \right ) \\[2ex] &= \dfrac{5}{9}\left ( 451 - 32 \right ) \\[2ex] &= 233~\mathrm{^{\circ}C} \end{align*}\]

Practice – Significant Figures

---

Which of these numbers has the most significant figures?

1. 425,000
2. 3.760×10⁴
3. 0.00088
4. 808.00
5. 7.01×10^–9

Solution

Answer: D

Practice – Temperature

---

A sample of mercury was warmed from its freezing point (234 K) to room temperature (298 K) for a total change of 64 K. By how many °C was the sample warmed?

1. 64
2. 234
3. 337
4. –209
5. 96

Solution

Answer: A

A change in 1 K is equivalent to a change in 1 °C. Therefore, a change in 64 K is equivalent to a change in 64 °C.

Practice – Significant Figures

---

What is the correct answer (to the appropriate number of significant figures) for the following calculation
     (3.5 + 32.00) × (0.07 + 2.100)

1. 80
2. 77
3. 77.0
4. 77.03
5. 77.04

Solution

Answer: C

Practice – Dimensional Analysis

---

The density of air at sea level and 15 °C is 1.225 kg m^–3. How many kilograms would 5.00 mL of air weigh?

1. 6.12×10^–6
2. 4.08×10^–5
3. 6.12×10^–3
4. 2.45×10⁵
5. 4.08 kg

Solution

Answer: A

\[\begin{align*} \left ( \dfrac{5.00~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{cm^3}}{\mathrm{mL}} \right ) \left ( \dfrac{\mathrm{m}}{10^2~\mathrm{cm}} \right )^3 \left ( \dfrac{1.225~\mathrm{kg}}{\mathrm{m}^3} \right ) = 6.13 \times 10^{-6}~\mathrm{kg} \end{align*}\]

Practice – Elements

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Which of these statements is false?

1. Elements cannot be broken down into simpler substances
2. A homogeneous mixture can be separated using physical changes
3. Atoms are the smallest particles of an element that have the properties of that element
4. Compounds are pure substances
5. A molecule must consist of two or more atoms of different elements joined together by chemical bonds

Solution

Answer: E

Practice – Dimensional Analysis

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Sulfuric acid is one of the most widely used chemicals in the world and 1.2×10⁸ kg produced per day. Given that the density of sulfuric acid is 1.84 g mL^–1, how many milliliters of sulfuric acid are produced per second?

1. 7.5×10²
2. 4.5×10⁷
3. 2.6×10⁶
4. 7.5×10⁵
5. 2.6×10³

Solution

Answer: D

\[\begin{align*} \left ( \dfrac{1.2\times 10^{8}~\mathrm{kg}}{\mathrm{d}} \right ) \left ( \dfrac{10^3~\mathrm{g}}{\mathrm{kg}} \right ) \left ( \dfrac{\mathrm{mL}}{1.84~\mathrm{g}} \right ) \left ( \dfrac{\mathrm{d}}{24~\mathrm{h}} \right ) \left ( \dfrac{\mathrm{h}}{60~\mathrm{min}} \right ) \left ( \dfrac{\mathrm{min}}{60~\mathrm{s}} \right ) = 7.5\times 10^{5}~\mathrm{mL~s^{-1}} \end{align*}\]

Practice – Law

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Which of these describes an experiment whose results are consistent with the law of multiple proportions?

1. Several samples of iron(III) oxide were tested and all were 69.9% iron and 30.1% oxygen by mass
2. In two separate experiments, 1.000 g of iron were reacted with either 0.4297 g or 0.2865 g of oxygen to give two new compounds
3. When a sample of iron metal reacts with oxygen, the total mass of the system remains unchanged
4. A mixture of iron pellets and iron(III) oxide powder is separated using a magnet
5. An iron bar weighing 15.2 g was found to have a volume of 1.93 cm³

Solution

Answer: B

Practice – History

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Which experiment(s) proved that the “plum pudding” model of the atom was incorrect?

1. Rutherford’s gold foil experiment
2. J. J. Thompson’s cathode raw tube
3. Frederic Soddy’s experiments with isotopes
4. Millikan’s oil drop experiment
5. Roentgen’s discovery of X-rays

Solution

Answer: A

Practice – Atomic Structure

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A neutral atom of molybdenum has a mass number of 95. How many protons, neutrons, and electrons does it contain?

1. 42 protons, 95 neutrons, and 42 electrons
2. 42 protons, 53 neutrons, and 95 electrons
3. 42 protons, 53 neutrons, and 42 electrons
4. 42 protons, 53 neutrons, and 53 electrons
5. 95 protons, 95 neutrons, and 95 electrons

Solution

Answer: C

Practice – Atomic Structure

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Which of the following could describe an atom or ion with 33 electrons?

1. A selenium-79 atom
2. A gallium-69 cation
3. A germanium-72 anion
4. all of these
5. none of these

Solution

Answer: C

Practice – Structural Formulas

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The structural formulas for two different compounds (methyl ether and ethanol) are shown below.

Which of the following might not be true about these two compounds?

1. They have the same molecular formula
2. They have the same empirical formula
3. They are structural isomers
4. They have the same boiling point
5. They obey the law of constant composition

Solution

Answer: C

Practice – Periodic Table

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Which term refers to the columns in the periodic table?

1. groups
2. periods
3. series
4. classes
5. kingdoms

Solution

Answer: A

Practice – Periodic Table

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Which statement(s) about the elements are false?

1. All of the lanthanides and actinides are metals
2. Some of the halogens occur naturally as diatomic gases
3. Hydrogen is an alkali metal
4. 1. and (b)
5. 2. and (c)

Solution

Answer: C

Practice – Percent Abundance

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Boron has two stable isotopes. The most abundant (80.1%) is boron-11 with an atomic mass of 11.01 amu. What are the percent abundance and atomic mass of the other stable isotope?

1. 18.4%, 10.8 amu
2. 19.9%, 6.73 amu
3. 10.8%, 10.0 amu
4. 19.9%, 10.0 amu
5. 19.9%, 10.8 amu

Solution

Answer: D

The other stable isotope of boron has a percent abundance of

\[100\% - 80.1\% = 19.9\%\]

Determine the mass (in amu) of this isotope.

\[\begin{align*} 10.81 &= \left ( \dfrac{80.1}{100} \right ) \left ( 11.01 \right ) + \left ( \dfrac{19.9}{100} \right )x \\[2ex] x &= 10.0~\mathrm{amu} \end{align*}\]

Practice – Ions

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Which of the following shows an ion with its most common charge?

1. Al³⁺
2. Mg⁺
3. Br^2–
4. Se^5–
5. none of these

Solution

Answer: A

Practice – Ions

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Which of the following correctly names the ion?

1. NO₃^–, pernitrate
2. PO₄^3–, perphosphate
3. SO₄^2–, persulfate
4. ClO₄^–, perchlorate
5. CO₃^2–, percarbonate

Solution

Answer: D

Practice – Empirical Formula

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Predict the empirical formula of alumina, a mineral containing Al³⁺ cations and O₂^2– anions.

1. AlO₃
2. Al₂O
3. AlO
4. Al₂O₃
5. Al₃O₂

Solution

Answer: D

Practice – Empirical Formula

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What is the correct systematic name for the compound P₄O₁₀?

1. Tetraphosphate
2. Phosphorus(V) oxide
3. Phosphoric acid
4. Phosphorus(IV) oxide
5. Tetraphosphorus decaoxide

Solution

Answer: E

Practice – Empirical Formula

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What is the correct systematic name for the compound Fe(NO₃)₂?

1. Iron nitrate
2. Iron dinitrate
3. Iron(III) nitrate
4. Iron(II) nitrate
5. Iron(II) dinitrate

Solution

Answer: D

Practice – Reaction Yield

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Steam reforming of methane (CH₄) is the most important industrial process for the production of hydrogen gas (H₂) and is represented by the following balanced reaction:

\[\mathrm{H_2O}(g) + \mathrm{CH_4}(g) \longrightarrow \mathrm{CO}(g) + 3\mathrm{H_2}(g)\] The United States produces around 9.0×10⁶ tons of hydrogen gas per year using this process (1 ton = 907 kg). The yield for this process is 65%. How much methane (in kg) is consumed in this process per year?

1. 3.4×10¹⁰
2. 2.2×10¹⁰
3. 3.3×10¹³
4. 6.5×10¹⁰
5. 1.03.3×10¹¹

Solution

Answer: A

Determine the theoretical yield of the reaction.

\[\begin{align*} \%~\mathrm{yield} &= \dfrac{\mathrm{actual}}{\mathrm{theoretical}} \times 100 \rightarrow \\[2ex] \mathrm{theoretical} &= \dfrac{\mathrm{actual}\times 100}{\%~\mathrm{yield}}\\[2ex] &= \dfrac{(9.0\times 10^{6}~\mathrm{tons~H_2})(100)}{65}\\[2ex] &= 1.385\times 10^{7}~\mathrm{tons~H_2} \end{align*}\]

Convert tons of H₂ to mol H₂.

\[\begin{align*} n_{\mathrm{H_2}} &= \left ( \dfrac{1.385\times 10^{7}~\mathrm{tons~H_2}}{} \right ) \left ( \dfrac{\mathrm{907~kg}}{\mathrm{ton}} \right ) \left ( \dfrac{\mathrm{10^3~g}}{\mathrm{kg}} \right ) \left ( \dfrac{\mathrm{mol~H_2}}{\mathrm{2.016~g~H_2}} \right )\\[2ex] &= 6.2311\times 10^{12}~\mathrm{mol~H_2} \end{align*}\]

Using stoichiometry, convert mol H₂ to kg of CH₄.

\[\begin{align*} m_{\mathrm{CH_4}} &= \left ( \dfrac{6.2311~\mathrm{mol~H_2}}{} \right ) \left ( \dfrac{1~\mathrm{mol~CH_4}}{3~\mathrm{mol~H_2}} \right ) \left ( \dfrac{\mathrm{16.04~g~CH_4}}{\mathrm{mol~CH_4}} \right ) \left ( \dfrac{\mathrm{kg}}{\mathrm{10^3~g}} \right )\\[2ex] &= 3.33\times 10^{10}~\mathrm{kg~CH_4} \end{align*}\]

Practice – Reaction Yield

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In a fantasy universe, a student combines 23 mL of 0.114 M aluminum bicarbonate solution and 6.75 mL of 0.89 M rubidium hydroxide solution. Assuming a 100% yield, what mass (in g) of precipitate is formed?

1. 0.20
2. 0.156
3. 0.820
4. 0.468
5. 1.1

Solution

Answer: B

Write out a balanced chemical equation. Use solubility rules to determine the phases of each substance and to identify the precipitate of reaction (here Al(OH)₃).

\[\mathrm{Al(HCO_3)_3}(aq) + \mathrm{3RbOH}(aq) \longrightarrow \mathrm{Al(OH)_3}(s) + \mathrm{3RbHCO_3}(aq)\]

Determine the theoretical yield by treating each reactant as undergoing complete reaction.

For Al(HCO₃)₃:

\[\begin{align*} m_{\mathrm{Al(OH_3)}} &= \left ( \dfrac{23~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right ) \left ( \dfrac{0.114~\mathrm{mol}}{\mathrm{L}} \right ) \left ( \dfrac{1~\mathrm{mol~Al(OH)_3}}{1~\mathrm{mol~Al(HCO_3)_3}} \right ) \left ( \dfrac{78.00~\mathrm{g~Al(OH)_3}}{\mathrm{mol~Al(OH)_3}} \right ) \\[2ex] &= 0.20~\mathrm{g~Al(OH)_3} \end{align*}\]

For RbOH:

\[\begin{align*} m_{\mathrm{Al(OH_3)}} &= \left ( \dfrac{6.75~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right ) \left ( \dfrac{0.89~\mathrm{mol}}{\mathrm{L}} \right ) \left ( \dfrac{1~\mathrm{mol~Al(OH)_3}}{3~\mathrm{mol~RbOH}} \right ) \left ( \dfrac{78.00~\mathrm{g~Al(OH)_3}}{\mathrm{mol~Al(OH)_3}} \right ) \\[2ex] &= 0.156~\mathrm{g~Al(OH)_3} \end{align*}\]

RbOH is the limiting reactant and therefore, only 0.156 g Al(OH)₃ is formed.

Practice – Oxidation States

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Consider each of the compounds below. Which underlined element has the largest oxidation number?

     MgO₂
     CH₄
     F₂
     S^2–
     NH₄Cl

1. O in MgO₂
2. C in CH₄
3. F in F₂
4. S in S^2–
5. N in NH₄Cl

Solution

Answer: C

Practice – Oxidation States

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Solution A is made by dissolving 1.03 g K₃PO₄ in 1.000 L of water. A 5.00 mL portion of Sample A is diluted to 1.000 L to form Solution B. Finally, Solution C is made by diluting 7.00 mL of Solution B to 1.000 L. What is the molar concentration of K⁺ in solution C?

1. 4.90×10^–3
2. 2.45×10^–5
3. 1.72×10^–7
4. 1.47×10^–2
5. 5.09×10^–7

Solution

Answer: E

Determine the molar concentration of K₃PO₄ in Solution A.

\[\begin{align*} c_{\mathrm{K_3PO_4}} &= \left ( \dfrac{1.03~\mathrm{g}}{1.000~\mathrm{L}} \right ) \left ( \dfrac{\mathrm{mol}}{212.27~\mathrm{g}} \right ) \\[2ex] &= 0.0048523~M~\mathrm{K_3PO_4} \end{align*}\]

Determine the molar concentration of K₃PO₄ in Solution B.

\[\begin{align*} M_1V_1 &= M_2V_2 \\[2ex] M_2 &= \dfrac{M_1V_1}{V_2} \\[2ex] &= \dfrac{(0.0048523~M)(0.00500~\mathrm{L})}{1.000~\mathrm{L}} \\[2ex] &= 2.426\times 10^{-5}~M~\mathrm{K_3PO_4} \end{align*}\]

Determine the molar concentration of K₃PO₄ in Solution C.

\[\begin{align*} M_1V_1 &= M_2V_2 \\[2ex] M_2 &= \dfrac{M_1V_1}{V_2} \\[2ex] &= \dfrac{(2.425\times 10^{-5}~M)(0.00700~\mathrm{L})}{1.000~\mathrm{L}} \\[2ex] &= 1.6983\times 10^{-7}~M~\mathrm{K_3PO_4} \end{align*}\]

Use stoichiometry to determine the concentration of K⁺ ions in Solution C.

\[\begin{align*} c_{\mathrm{K^+}} &= \left ( \dfrac{1.6983\times 10^{-7}~M~\mathrm{K_3PO_4}}{} \right ) \left ( \dfrac{3~\mathrm{mol~K^+}}{\mathrm{mol~K_3PO_4}} \right ) \\[2ex] &= 5.09\times 10^{-7}~M~\mathrm{K^+} \end{align*}\]

Practice – Ionic Reactions

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Which of the following statements is incorrect?

1. AgNO₃, Hg₂(NO₃)₂, and Pb(NO₃)₂ are soluble
2. In the reaction between MgSO₄ and CaI₂, Mg²⁺ is a spectator ion
3. The net ionic equation for the reaction of H₂CO₃ and NaOH is H⁺ + OH^– → H₂O
4. Hydrofluoric acid is a weak acid
5. A proton and H⁺ refer to the same thing

Solution

Answer: C

Since H₂CO₃ is a weak acid, it will appear in the net ionic equation.

Practice – Oxidation-Reduction Reactions

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In a typical combustion reaction, what happens to oxygen?

1. It is oxidized
2. It is reduced
3. It is both oxidized and reduced
4. It is neither oxidized or reduced

Solution

Answer: B

In a combustion reaction, a hydrocarbon reacts with oxygen gas to form carbon dioxide and water. A general reaction scheme is given.

\[\mathrm{C}_x\mathrm{H}_y + \mathrm{O_2}(g) \longrightarrow \mathrm{CO_2} + \mathrm{H_2O}\] Each oxygen atom in O₂ has an oxidation state of 0 since oxygen exists as oxygen gas in its elemental form. The oxygens in carbon dioxide and water each have a –2 oxidation state. Since oxygen’s oxidation state goes from 0 → –2, oxygen is reduced.

Practice – Molecular Formula

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A sample of caffeine is found to consist of 17.77 g C, 1.86 g H, 10.04 g N, and 5.92 g O. Given that the molar mass of caffeine is about 194 g mol^–1, what is the molecular formula of caffeine?

1. C₄H₅N₂O
2. C₁₀HN₅O₃
3. C₁₉H₂N₁₁O₆
4. C₈H₁₀N₄O₂
5. C₁₇H₂N₁₀O₆

Solution

Answer: D

Convert each element to moles.

\[\begin{align*} n_{\mathrm{C}} &= \left ( \dfrac{17.77~\mathrm{g~C}}{} \right ) \left ( \dfrac{\mathrm{mol~C}}{12.01~\mathrm{g~C}} \right ) \\[2ex] &= 1.480~\mathrm{mol~C} \\[4ex] n_{\mathrm{H}} &= \left ( \dfrac{1.86~\mathrm{g~H}}{} \right ) \left ( \dfrac{\mathrm{mol~H}}{12.01~\mathrm{g~H}} \right ) \\[2ex] &= 1.85~\mathrm{mol~H} \\[4ex] n_{\mathrm{N}} &= \left ( \dfrac{10.04~\mathrm{g~N}}{} \right ) \left ( \dfrac{\mathrm{mol~N}}{12.01~\mathrm{g~N}} \right ) \\[2ex] &= 0.7166~\mathrm{mol~N} \\[4ex] n_{\mathrm{O}} &= \left ( \dfrac{5.92~\mathrm{g~O}}{} \right ) \left ( \dfrac{\mathrm{mol~O}}{12.01~\mathrm{g~O}} \right ) \\[2ex] &= 0.37~\mathrm{mol~O} \\[4ex] \end{align*}\]

Write a formula using the mole quantities of each element as subscripts. Divide each mole quantity by the smallest number to obtain a whole number. This gives the empirical formula.

\[\begin{align*} &\mathrm{C}_{1.480}\mathrm{H}_{1.85}\mathrm{N}_{0.7166}\mathrm{O}_{0.37} \rightarrow \\[1.5ex] &\mathrm{C}_{1.480/0.37}\mathrm{H}_{1.85/0.37}\mathrm{N}_{0.7166}\mathrm{O}_{0.37/0.37} \rightarrow \\[1.5ex] &\mathrm{C}_{4}\mathrm{H}_{5}\mathrm{N}_{2}\mathrm{O} \end{align*}\]

The molar mass for the empirical formula above is 97.10 g mol^–5. The molar mass of caffeine is 194 g mol^–1. Make a ratio of the two molar masses. The resulting number is what you multiply each subscript in the empirical formula with to get the molecular formula.

\[\dfrac{194~\mathrm{g~mol^{-1}}}{97.10~\mathrm{g~mol^{-1}}} = 2\]

Therefore,

\[\begin{align*} &\mathrm{C}_{4\times 2}\mathrm{H}_{5\times 2}\mathrm{N}_{2\times 2}\mathrm{O}_{1\times 2} \rightarrow \\[1.5ex] &\mathrm{C}_{8}\mathrm{H}_{10}\mathrm{N}_{4}\mathrm{O}_2 \end{align*}\]

Practice – Acid-Base Reaction

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A 40.25 mL solution of hydroiodic acid is found to require 19.20 mL of 3.14 M calcium hydroxide to neutralize it. What was the molar concentration of the hydroiodic acid solution?

1. 1.50
2. 3.00
3. 3.29
4. 6.58
5. 13.2

Solution

Answer: B

Write a balanced chemical equation.

\[2\mathrm{HI} + \mathrm{Ca(OH)_2} \longrightarrow \mathrm{CaI_2} + \mathrm{2H_2O}\]

Determine the moles of Ca(OH)₂ that was required for neutralization.

\[\begin{align*} n_{\mathrm{Ca(OH)_2}} &= \left ( \dfrac{19.20~\mathrm{mL}}{} \right ) \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right ) \left ( \dfrac{3.14~\mathrm{mol}}{\mathrm{L}} \right ) \\[2ex] &= 0.06029~\mathrm{mol} \end{align*}\]

Determine the moles of OH^– that was provided by Ca(OH)₂.

\[\begin{align*} n_{\mathrm{OH^-}} &= \left ( \dfrac{0.06029~\mathrm{mol~Ca(OH)_2}}{} \right ) \left ( \dfrac{2~\mathrm{mol~OH^-}}{\mathrm{mol~Ca(OH)_2}} \right ) \\[2ex] &= 0.1206~\mathrm{mol} \end{align*}\]

The neutralization reaction that occurs between a strong acid (such as HI which becomes H₃O⁺ in water) and strong base (OH^–) is

\[\mathrm{H_3O^+} + \mathrm{OH^-} \longrightarrow \mathrm{H_2O}\]

As seen, 1 mol of OH^– is required to neutralize one mol of strong acid. Therefore, the mol of HI that was neutralized in solution is

\[\begin{align*} n_{\mathrm{HI}} &= n_{\mathrm{OH^-}} \\[1.5ex] &= 0.1206~\mathrm{mol} \end{align*}\]

Determine the molar concentration of the original HI.

\[\begin{align*} c_{\mathrm{HI}} &= \dfrac{0.1206~\mathrm{mol}} {\left ( 40.25~\mathrm{mL} \right ) \left ( \dfrac{\mathrm{L}}{10^3~\mathrm{mL}} \right )} \\[2ex] &= 3.00~M \end{align*}\]

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"Chemistry I: Online Resource" was prepared by Eric Van Dornshuld.    |    Release: Alpha 1.0
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