1.7 Practice Problems
Attempt these problems as if they were real exam questions in an exam environment.
Only look up information if you get severely stuck. Never look at the solution until you have exhausted all efforts to solve the problem. Having to look up information or the solution should be an indicator that the previous layers (1–5) in the Structured Learning Approach have not been mastered.
If Mississippi State University has exactly 221 classrooms, and each classroom holds an estimated 150 students, how many students could be in class at MSU at the same time?
- 35,986
- 33,000
- 25,674
- 42.898
- 33.150
Solution
Answer: B
Concept: Unit conversion
\[\left ( \dfrac{221~\mathrm{classrooms}}{}\right ) \left ( \dfrac{150~\mathrm{students}}{1~\mathrm{classroom}}\right ) = 33,000~\mathrm{students}\]
How many significant figures should the final answer contain for the following calculation?
(9.284 + 2.6) * 5.31421 =
- 1
- 2
- 3
- 4
- 5
Solution
Answer: C
Concept: Significant figures
A chemical company owner finds it takes 1.5 hours to run a reaction that produces 15 grams of a product, at a profit of $10.00 per gram. If the reaction runs continuously for 24 hours per day, five days per week, for 25 weeks, approximately how much money (in $) does the chemical company stand to profit from this process alone?
- 1,080,000
- 300,000
- 3,750
- 162,000
- 2,300
Solution
Answer: B
Concept: Dimensional analysis
\[\begin{align*} \mathrm{profit} &= \left ( \dfrac{25~\mathrm{wk}}{} \right ) \left ( \dfrac{5~\mathrm{d}}{\mathrm{wk}} \right ) \left ( \dfrac{24~\mathrm{h}}{\mathrm{d}} \right ) \left ( \dfrac{15~\mathrm{g}}{1.5~\mathrm{h}} \right ) \left ( \dfrac{\$10.00}{\mathrm{g}} \right ) \\ &= \$300,000 \end{align*}\]
You have started to work in a synthesis lab, and you have made a new substance with a molar mass of 265.18 g mol–1. You make 26.3 grams of this substance. How many moles of this substance have you made?
- 6.65
- 0.0992
- 0.287
- 6,974.23
- 55.0
Solution
Answer: B
Concept: Dimensional analysis
\[\begin{align*} \mathrm{n} &= \left ( \dfrac{26.3~\mathrm{g}}{} \right ) \left ( \dfrac{\mathrm{mol}}{265.18~\mathrm{g}} \right ) &= 0.0992 \end{align*}\]
You work at a pawn shop and are told a lump of metal is solid platinum. You are skeptical, but since you took General Chemistry I, you have a pretty good idea of how to determine whether the metal is indeed platinum. You look up a table of metals and find the following densities for some common metals (shown below). Using the displacement method to determine the volume and a balance to obtain the mass, you record the following values:
Initial volume: 101.56 mL
Final volume after adding metal: 106.28 mL
Mass: 56.8 g
Is the lump of metal platinum? If not, what metal is it most likely?Metal Density (g mL–1) Mild steel
7.85
Silver
10.49
Aluminum
2.70
Palladium
12.02
Platinum
21.45
- No, silver
- No, aluminum
- No, palladium
- No, mild steel
- Yes, platinum
Solution
Answer: C
Concept: Density
Find the volume of the metal.
\[\begin{align*} V_{\mathrm{metal}} &= V_f - V_i \\ &= 106.28~\mathrm{mL} - 101.56~\mathrm{mL} \\ &= 4.72~\mathrm{mL} \end{align*}\]
Find the density of the metal lump using the given mass and calculated volume.
\[\begin{align*} d_{\mathrm{metal}} &= \dfrac{m}{V} \\[1.5ex] &= \dfrac{56.8~\mathrm{g}}{4.72~\mathrm{mL}} \\[1.5ex] &= 12.0~\mathrm {g~mL} \end{align*}\]
According to the table, palladium has a density of approximately 12 g mL–1.
Which statement BEST represents the meaning of a hypothesis?
- A scientific procedure undertaken to make a discovery or demonstrate a known fact.
- A proposed explanation made based on limited evidence as a starting point for further investigation.
- An estimate without sufficient information to be sure of being correct.
- A statement of fact, deduced from observation to the effect that a particular natural or scientific phenomenon always occurs if certain conditions are present.
- A person who is studying or has expert knowledge of one or more of the natural or physical sciences.
Solution
Answer: B
Concept: Scientific method
Ether has a density of 713 kg cm–3. What is the mass (in g) of a 1.5 mL sample of ether based on this data? (1 mL = 1 cm3)
- 1.07
- 1.55
- 3.55
- 6.02
- 0.155
Solution
Answer: A
Concept: Dimensional analysis
Recall that 1 m3 is 1×106 cm6.
Since 1 m = 100 cm,
(1 m)3 = (100 cm)3 →
13 m3 = 1003 cm3 →
1 m3 = 1×106 cm3.\[\begin{align*} m &= \left ( \dfrac{713~\mathrm{kg}}{\mathrm{m}^{3}} \right ) \left ( \dfrac{\mathrm{m}^{3}}{10^6~\mathrm{cm}^{3}} \right ) \left ( \dfrac{1~\mathrm{mL}}{\mathrm{cm}^{3}} \right ) \left ( \dfrac{10^3~\mathrm{g}}{\mathrm{kg}} \right ) \\ &= 1.07~\mathrm{g} \end{align*}\]