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Angelica Cadondon ([email protected])General Chemistry Peer Tutoringhttp://sites.uci.edu/gcptutoring/Chem 1B - Finkeldei

Midterm 1 Review KEY

Gas Laws and Stoichiometry

Below are variations of the Ideal Gas Equation for you to use:

● With initial and final states (remove other constants when needed):

● With density:

● With molar mass:

1. Sodium azide (NaN3) is used in some automobile air bags. The impact of a collision

triggers the decomposition of NaN3 as follows:

2NaN3(s) → 2Na(s) + 3N2(s)

The nitrogen gas produced quickly inflates the bag between the driver and the windshield and

dashboard. Calculate the volume of N2 generated at 85°C and 812 mmHg by the decomposition

of 50.0 g of NaN3.

Step 1) Use the mass of NaN3 to find the mols of N2.

50.0 g NaN3 x (1 mol/ 64.99 g/mol NaN3) = 0.769 mol NaN3 x (3 mol N2 / 2 mol NaN3) = 1.15

mol N2

Step 2) Use the ideal gas law to find volume.

V = nRT/P = [(1.15 mol)(0.08206 L·atm/K·mol)(358 K)] / 1.07 atm = 31.6 L

2. If 2.92 g of a gas occupies 1.00 L at 25°C and a pressure of 1.00 atm, what is the gas?

M = mRT/PV = [(2.92 g)(0.08206 L·atm/K·mol)(298 K)] / [(1.00 atm)(1.00 L)]


= 71.4 g

Because almost all gases are diatomic, 71.4 g is the mass of a diatomic gas (there is no noble gas

with a molar mass around 71.4 g).

71.4 g / 2 = 35.7 g = Cl → the gas is Cl2

3. Why don't gases strictly follow the ideal gas law? (fill in the blanks for each)

a. Real gas molecules take up volume and attractions exist between

molecules.

b. Since the molecules have an attraction, the pressure decreases because kinetic

energy decreases .

c. Since molecules have volume, at high pressure the molecules are forced together

so closely and cannot be forced together anymore, the volume appears higher

than it should be.

4. Using the balanced equation determine the number of O2 molecules in the visualization

before any reaction occurs.

C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(g)

a. 12

b. 6

c. 10

d. 9

e. 8


We know that we should have the same ratio of O2 molecules before and after the reaction. There

are 6 O molecules in the reactants and products of the balanced equation, so we can conclude

that the reactants in the container should contain the exact amount of O molecules as the

products. The products container has 18 O molecules, so the reactants must also have 18 O

molecules = 9 O2.

5. The following figure shows three chambers with equal volumes, connected by closed

valves. Chamber A contains argon gas and chamber B contains helium gas, with amounts

proportional to the number of atoms shown.

If the pressure in A is 760 Torr before the valves are opened, what is the pressure (in Torr) in the

chambers after the valves are opened?

a. 570 Torr

b. 326 Torr

c. 2280 Torr

d. 977 Torr

e. 591 Torr

Step 1) Write down given values and determine what remains constant and what changes.

● The chambers have equal volumes, which we can arbitrarily assign each as 1 L. Vi of

container A is then 1 L, but Vf changes once the valves are opened. Vf will then be 3 L

because all containers are now connected.

● The moles also change once the valves open. ni (moles of A) is 9 mols and nf (moles in A

and B) is 21 mols.


● The pressure changes once the valves open. Pi is 760 Torr and Pf is what we need to

find.

PiVi /ni = PfVf /nf → Pf = PiVi nf / niVf

= [(760 Torr)(1 L)(21 mol)] / [(9 mol)(3 L)] = 591 Torr

Partial Pressures and Root Mean Square Speed

6. A mixture of 50.0 g of O2 gas and 50.0 g of methane gas is placed in a container under a

total pressure of 600 mmHg. What is the partial pressure of the oxygen gas in the

mixture?

Step 1) Convert the mass of the gases to mols.

50.0 g O2 x (1 mol/ 32.0 g O2) = 1.563 mol O2

50.0 g O2 x (1 mol/ 16.04 g CH4) = 3.117 mol CH4

Step 2) Find the mol fraction and calculate partial pressure.

XO2 = 1.563 mol O2 / (1.563 mol O2 + 3.117 mol CH4) = 0.334

PO2 = (0.334)(600 mmHg) = 200 mmHg

7. Which molecules of the following gases will have the greatest root mean square speed?

a. CO2 at 1 atm and 273 K

b. CO at 0.1 atm and 273 K

c. N2 at 1 atm and 273 K

d. H2 at 0.5 atm and 273 K

e. All of the molecules have the same root mean square speed.

Firstly, notice that all molecules are at the same temperature. The only depending factor on Vrms

is the molar mass, and they have an inverse relationship according to the equation. Therefore, the

lightest molecule will have the highest Vrms.

8. Consider the following samples of gases at 25°C:


Which of the following is the correct order of the total pressure for samples A, B, and C?

a. C > A > B

b. A = B = C

c. A > B > C

d. C < A = B

e. A = C > B

The only information given is the temperature and number of moles. We cannot use the equation

for partial pressures to find the total pressure because partial pressures are not given. Therefore,

we will use the ideal gas law and examine the relationship between pressure and moles. Since

everything else remains constant (T, V, and R), we can see that pressure and the number of

moles have an equal relationship.

Container A and C have 7 moles while container B has 4. A and C have the most pressure.

Intermolecular Forces

9. What kind of attractive forces must be overcome in order to:

a. Melt H2O(s)? LDF, dipole-dipole, H-bonding

b. Boil molecular bromine? LDF

c. Melt NaCl(s)? Ionic bonds

10. Diethyl ether (left) has a boiling point of 34.5°C, and 1-butanol (right) has a boiling point

of 117°C:


Both of these compounds have the same numbers and types of atoms. Explain the difference in

their boiling points.

1-butanol has hydrogen bonding with O-H as a donor and O as an acceptor. Diethyl ether has no

hydrogen bonding.

11. Select all substances with a hydrogen bond acceptor atom that can participate in

hydrogen bonding with an appropriate hydrogen bond donor atom.

a. (ii), (v)

b. (i), (iii), (iv), (vi)

c. (i), (iii), (iv)

d. (i), (ii), (iii), (iv), (vi)

e. All

All three substances have an O, N, or F atom with available lone pairs for hydrogen bonding.

12. Which of the following molecules has the strongest dipole-dipole forces?

a. C2F4

b. N2

c. CH3I

d. BF3

e. NH3


N and H have the highest electronegativity difference and the shape of the molecule is not

symmetrical due to the lone pair. The N-H dipole moments are additive and the net dipole

moment points towards the lone pair.

Viscosity, Immiscibility, and Thermochemistry

13. Rank the viscosity (1 being highest) of for the following substances at the listed

temperatures.

3 1 2 4

Two factors shown here affect viscosity: temperature and intermolecular forces. The highest

viscosity is the second compound because of the low temperature and H-bonding capabilities.

Because the other three compounds are at the same temperature, they will be ranked by

intermolecular forces alone. The first compound exhibits dipole-dipole forces while the last

compound exhibits only LDF.

14. Which statement is NOT TRUE for why methanol, CH3OH, does dissolve well in water?

a. Solute-solvent interactions, which involve hydrogen bonding, are relatively

strong.

b. “Like dissolves like” holds true.

c. Solute-solvent interactions are similar in strength to original solute-solute

interactions.

d. Methanol makes strong covalent bonds to water when it dissolves.

When miscible liquids interact with each other, they do so through intermolecular forces.

Covalent bonds are not being made or broken.

15. Which of the following processes are exothermic?

a. The second ionization energy of Mg

b. The sublimation of Li


c. The breaking of the bond of I2

d. The formation of NaBr from its constituent elements in their standard state

a) Endothermic: If you recall, removing a second electron from a cation is highly

unfavorable. You would need to supply energy.

b) Endothermic: Sublimation is the process of a solid turning into a gas. Heat energy is

needed to break these intermolecular forces.

c) Endothermic: Energy needs to be supplied to break bonds. Forming bonds releases

energy.

d) Exothermic! Again, forming bonds releases energy. Atoms are much happier when they

are bonded and they release energy because it is easier and more stable to be in a

relationship.

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