General Chemistry - Chapter 1 Quiz

Course

Unit 1

Chapter 1

General Chemistry

Molecular Structure and Properties

Introduction

This quiz is part of the MEEP curriculum on General Chemistry. Information about Project MEEP and other General Chemistry resources are available below.

Project MEEP

General Chemistry

Quiz Logistics

The quiz covers the following lectures and discussions.

Lecture 1 – The Essence of Chemistry
Lecture 2 – Units and Implications of Measurement
Lecture 3 – Phases, Properties, Particles, and the Periodic Table
Lecture 4 – Chemical Reactions and Empirical Analysis
(Implied) Discussion 1 – Chemical Nomenclature and Applications

The quiz has the following format.

Time Limit	Questions	Points
30 Minutes	20	25

The first 15 questions are worth 1 point each. The last 5 questions are worth 2 points each.

Only calculators and scratch papers are allowed for this quiz.

The quiz has the following grading scheme.

F	C	B	A
0	12	16	20

The numbers denote the lowest score possible to obtain the corresponding grade.

This section of the quiz is intentionally left blank.
Scroll down for the quiz!

You have 30 minutes to answer 20 questions.
Don’t click on the answers unless you finished the quiz!

Good luck!

Quiz Starts Here

Question 1

Below is a diagram of an ice-calorimeter used by French chemists Antoine Lavoisier and Pierre-Simon Laplace in the winter of 1792-1793. The calorimeter was used to determine the heat involved in various chemical changes.

The world's first ice-calorimeter, used in the winter of 1782–83, by Antoine Lavoisier and Pierre-Simon Laplace.

Which field of chemistry did those two chemists contribute greatly towards? (1 pt)

A. Electrochemistry
B. Thermochemistry
C. Nuclear chemistry
D. Phlogiston chemistry

Answer

Thermochemistry. See Lecture 1.

Question 2

In the 19th century, chemists further expanded the discipline with new discoveries. Which of the following scientists laid the cornerstone of nuclear chemistry? (1 pt)

A. Marie Skłodowska-Curie
B. Ernest Rutherford
C. Josiah Willard Gibbs
D. Dmitri Mendeleev

Answer

Marie Skłodowska-Curie. See Lecture 1.

Question 3

In the 19th century, chemists further expanded the discipline with new discoveries. Which of the following scientists essentially invented the periodic table? (1 pt)

A. Marie Skłodowska-Curie
B. Ernest Rutherford
C. Josiah Willard Gibbs
D. Dmitri Mendeleev

Answer

Dmitri Mendeleev. See Lecture 1.

Question 4

Which of the following conversions is correct? (1 pt)

A. 1 µm = 1,000 nm
B. 1 kg = 1,000,000 µg
C. 1 mmol = 1,000 nmol
D. 1 dm = 100 cm

Answer

1 µm = 1,000 nm

Here are the unit prefixes and their corresponding factors.

M (mega) – $10^{6}$
k (kilo) – $10^{3}$
d (deci) – $10^{-1}$
c (centi) – $10^{-2}$
m (milli) – $10^{-3}$
µ (micro) – $10^{-6}$
n (nano) – $10^{-9}$
p (pico) – $10^{-12}$

Using this chart, we can correct the incorrect options.

Incorrect	Correct
1 kg = 1,000,000 µg	1 kg = 1,000,000,000 µg
1 mmol = 1,000 nmol	1 mmol = 1,000,000 nmol
1 dm = 100 cm	1 dm = 10 cm

Questions 5-6 require the following information.

Suppose we have the following reaction:

$A + 2\ B \longrightarrow C + D$

Question 5

Assume you measured 3.51 moles of substance A. Theoretically, what is the least amount of substance B in moles required for substance A to react completely? (1 pt)

Answer

7.02 mol

Stoichiometric factors do not follow sig fig rules because they are viewed as constants, not measurements. So, the answer should have 3 sig figs.

(1) $\begin{equation*} 3.51 \text{mol of A} \times \frac{2 \text{ mol of B}}{1 \text{ mol of A}} = 7.02 \text{ mol of B} \end{equation*}$

Question 6

The 3.51 moles of substance A is measured to be 130.0 grams. What is the molar mass for substance A? (1 pt)

Answer

37.0 g/mol

We use the following formula.

(2) $\begin{equation*} M = \frac{m}{n} = \frac{130.0 \text{ g}}{3.51 \text{ mol}} = 37.0 \text{ g/mol} \end{equation*}$

We take 3 sig figs because 3.51 has 3 sig figs while 130.0 has 4 sig figs.

Questions 7-9 require the following information.

Our stomach acid is highly acidic. For example, it can react with amino acid glutamine described by the following reaction:

$\text{Glutamine} + HCl \rightarrow \text{Glutamate} + NH_4Cl$

The molar mass of those compounds are:

Glutamine: 146 g/mol

$HCl$ : 36.5 g/mol

Glutamate: 147 g/mol

$NH_4Cl$ : 53.5 g/mol

Question 7

Stomach acid has a high pH value. (1 pt)

True
False

Answer

False

Recall that pH is a measurement of how acidic a solution is. The lower the pH, the more acidic the solution is.

Question 8

Suppose a person ingested a glutamine tablet that contains 500 mg of glutamine. Assuming that all the glutamine reacted inside the person’s stomach, how many grams of glutamate is produced from the tablet? (Round to 3 sig figs) (1 pt)

Answer

0.503 grams

This is a bit tricky because it requires unit conversion.
For convenience, glutamine is abbreviated as Gln while glutamate is abbreviated as Glu.

Step 1: Convert units (mg) to SI (g).

(3) $\begin{equation*} 500 \text{ mg of Gln} \times \frac{1\text{ g}}{1000\text{ mg}} = 0.500 \text{ g of Gln} \end{equation*}$

Step 2: Convert mass of Gln into moles of Gln.

(4) $\begin{equation*} n_{Gln} = \frac{m}{M} = \frac{0.500 \text{ g of Gln}}{146 \text{ g/mol}} = 0.00342 \text{ mol of Gln} \end{equation*}$

Step 3: Stoichiometric manipulation from moles of Gln to moles of Glu.

(5) $\begin{equation*} n_{Glu} = 0.00342 \text{ mol of Gln} \times \frac{1 \text{ mol of Glu}}{1 \text{ mol of Gln}} = 0.00342 \text{ mol of Glu} \end{equation*}$

Step 4: Convert moles of Glu into mass of Glu.

(6) $\begin{equation*} m = nM = 0.00342 \text{ mol of Glu} \times 147 \text{ g/mol} = 0.503 \text{ g of Glu} \end{equation*}$

Question 9

You want to replicate this experiment in lab using the same 500 mg glutamine tablet and lab-made $HCl$ with a molar concentration of 6 mmol/L. What is the least amount (in mL) of 6 mmol/L $HCl$ needed to fully react with 500 mg glutamine? (round to 3 sig figs) (1 pt)

Answer

Both 570 or 571 are accepted depending on if we calculate the steps separately or together in one go.

Similarly, a lot of unit conversions are required here.

Step 1: Convert units (mg) into SI (g).
Step 2: Convert mass of Gln into moles of Gln.

(7) $\begin{equation*} 500 \text{ mg of Gln} \times \frac{1\text{ g}}{1000\text{ mg}} = 0.500 \text{ g of Gln} \end{equation*}$

(8) $\begin{equation*} n_{Gln} = \frac{m}{M} = \frac{0.500 \text{ g of Gln}}{146 \text{ g/mol}} = 0.00342 \text{ mol of Gln} \end{equation*}$

Step 3: Stoichiometric manipulation from moles of Gln to moles of HCl.
Since they have a 1:1 ratio according to the chemical equation, we have $0.00342 \text{ mol of HCl}$ .

Step 4a: Convert mol of HCl into mmol of HCl because the molarity is expressed in mmol/L.
Step 4b (alternative): Convert the molarity unit from mmol/L into mol/L.
We will demonstrate using Step 4a. Feel free to pursue Step 4b if you have time.

(9) $\begin{equation*} 0.00342 \text{ mol of HCl} \times \frac{1000\text{ mmol}}{1\text{ mol}} = 3.42 \text{ mmol of HCl} \end{equation*}$

Step 5: Convert mmol of HCl into L of HCl solution.

(10) $\begin{equation*} V = \frac{n}{c} = \frac{3.42 \text{ mmol of HCl}}{6 \text{ mmol/L}} = 0.570\text{ L of HCl solution} \end{equation*}$

Step 6: Convert L into mL per the prompt.

(11) $\begin{equation*} 0.570\text{ L of HCl solution} \times \frac{1000 \text{ mL}}{1 \text{ L}} = 570 \text{ mL of HCl solution} \end{equation*}$

Note:
If we go through Steps 1-6 one by one, we will end up with 570 mL.
If we write Steps 1-6 in one condensed equation, we will end up with 571 mL.

Questions 10-11 require the following information.

Ferroquine is a drug that is designed to treat malaria. It has a chemical formula of $C_{23}H_{24}ClFeN_3$ and a molar mass of 434 g/mol.

It can react with hydrogen peroxide to produce hydroxy radicals that can lead to bacterial death. Below is a simplified version of this reaction. The dot denotes a radical.

$Fe^{2+} + H_2O_2 \rightarrow Fe^{3+} + OH^- + \cdot OH$

Question 10

You measured 2.712 grams of ferroquine in lab. How many nitrogen atoms are in this sample of ferroquine? (1 pt)

Answer

$1.129 \times 10^{22}$

For convenience, ferroquine will be abbreviated as FQ.

This can be solved by the following equation.

(12) $\begin{equation*} \frac{2.712 \text{ g}}{434 \text{ g/mol}} \times \frac{3 \text{ mol of N}}{1 \text{ mol of FQ}} \times (6.02 \times 10^{23} \text{ particles per mol}) = 1.129 \times 10^{22} \text{ N atoms} \end{equation*}$

We take 4 sig figs because our only measurement in this prompt (2.712 grams) contains 4 sig figs.

Question 11

According to the reaction provided above, which substance is oxidized? (1 pt)

Iron (II) ion
Hydrogen peroxide
Iron (III) ion
Hydroxy radical

Answer

Iron (II) ion

A substance is oxidized when its oxidation number increases. In this case, $Fe^{2+}$ becomes $Fe^{3+}$ – its oxidation number increased from +2 to +3.

We usually only refer to reactants when talking about substances being oxidized or reduced.

Question 12

Which of the following is a physical property? (1 pt)

Reactivity
Oxidation state
Toxicity
Color

Answer

Color. See Lecture 3.

Question 13

The atomic number of silver is 47. Using this information, how many electrons are in a $Ag^+$ ion? (1 pt)

Answer

Since the atomic number of silver is 47, it means that silver has 47 protons.

In its atomic state, silver has 47 protons and 47 electrons.
However, since it has 1 positive charge, there should only be 46 electrons.

(+47 proton charge) + (-46 electron charge) = (1+ atom charge)

Question 14

Uranium-238 is radioactive and participates in a decay chain that can be described as the following simplified reaction.

(13) $\begin{equation*} ^{238}_{92}U \xrightarrow{\text{a complicated decay chain}}\ ^{206}_{82}Pb \end{equation*}$

During this decay chain, how many neutrons total would one 238-uranium atom lose to become 206-Pb? (1 pt)

Answer

Recall that the number at the top represents the number of protons + neutrons.
Recall that the number at the bottom represents the number of protons only.

From $^{238}_{92}U$ to $^{206}_{82}Pb$ , we know that:

It lost 10 protons during the process, because 92 – 82 = 10.
It lost 32 protons and neutrons during the process, because 238 – 206 = 32.
As a result, it lost 32 – 10 = 22 neutrons total.

Question 15

What is the oxidation state of iron in $Fe_3O_4$ ? (1 pt)

Answer

+8/3

It’s rare, but oxidation states can be a fraction.

(14) $\begin{equation*} 3\times \text{Fe} + 4 \times (-2) = 0 \Rightarrow \text{Fe} = +\frac{8}{3} \end{equation*}$

Questions 16-20 require the following information.

Cisplatin $PtCl_2(NH_3)_2$ is a medicine used in treating a variety of cancers, including testicular cancer, ovarian cancer, cervical cancer, bladder cancer, head and neck cancer, esophageal cancer, lung cancer, mesothelioma, brain tumors, and neuroblastoma. It works in part by binding to cancer cell DNA and inhibiting its replication, halting cancer cell replication.

Synthesis of cisplatin involves multiple steps. However, a simplified chemical equation is provided below.

(15) $\begin{equation*} \mathbf{c_1}\ K_2PtCl_4 + \mathbf{c_2}\ NH_3 \rightarrow \mathbf{c_3}\ PtCl_2(NH_3)_2 + \mathbf{c_4}\ KCl \end{equation*}$

Question 16

Potassium tetrachloroplatinate is one of the reactants used in cisplatin synthesis. What is the oxidation state of platinum in potassium tetrachloroplatinate? (2 pts)

Answer

Recall that Group 1 elements (alkali metals) have an oxidation state of +1.
Recall that Group 17 elements (halogens) have an oxidation state of -1.

Thus, K has an oxidation state of +1, while Cl has an oxidation state of -1.
We can calculate the rest.

Question 17

The synthesis of cisplatin is a redox reaction. (2 pts)

True
False

Answer

False.

It is not a redox reaction because no oxidation state is changed throughout the reaction.

Question 18

As indicated in the reaction equation, the chemical equation is not properly balanced. If the equation is properly balanced, what is the sum of all its stoichiometric factors ( $c_1 + c_2 + c_3 + c_4$ )? (2 pts)

Answer

This is the balanced equation.

(16) $\begin{equation*} K_2PtCl_4 + 2\ NH_3 \rightarrow PtCl_2(NH_3)_2 + 2\ KCl\end{equation*}$

Question 19

Suppose you measured 1.3 moles of $K_2PtCl_4$ and reacted it with a measured 2.2 moles of $NH_3$ . Which compound is the limiting reagent? (2 pts)

Potassium tetrachloroplatinate
Ammonia
Cisplatin
Potassium chloride

Answer

Ammonia

We can use our stoichiometric table here.
Here’s the balanced equation: $K_2PtCl_4 + 2\ NH_3 \rightarrow PtCl_2(NH_3)_2 + 2\ KCl$

If we have 1.3 moles of potassium tetrachloroplatinate, we need at least 2.6 moles of ammonia to fully react with it, which is larger than what is given. As a result, ammonia is the limiting reagent.

Question 20

After the reaction from Question 19, you yielded 1.0 moles of cisplatin. What is the percent yield of your reaction? (2 pts)

Answer

91%

As listed in Question 19, we have 1.3 moles of potassium tetrachloroplatinate reacting with 2.2 moles of ammonia. Since we know that ammonia is the limiting reagent, only 1.1 moles of potassium tetrachloroplatinate participate in the reaction.

Because potassium tetrachloroplatinate and cisplatin have a 1:1 stoichiometric factor ratio, the theoretical yield of cisplatin is 1.1 moles. We can calculate our percent yield now: $\frac{1.0 \text{ mol}}{1.1 \text{ mol}} \times 100\% = 91\%$

We take 2 sig figs because our measurements (1.3 moles, 1.0 moles, 2.2 moles) have 2 sig figs.

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General Chemistry – Chapter 1 Quiz

Quiz Logistics

Quiz Starts Here

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Question 9

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Question 19

Question 20