Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 109

Chem 109, Dr. Paselk

Hour Exam I
 Name
Fall 1994

(100 pts)
 Lab

Answer Key

This examination is scheduled for fifty (50) minutes. At the end of this fifty minute period you will have five (5) minutes to place your examination at the front of the examination room.

Show Work for Credit!!!!!!

1. (12 points @ 2 points each). Answer the following in the spaces provided:

How many moles of NaBr are in 73 mL of a 1.500 M solution? 0.11 moles

What is the mass of 6.02 x 1020 atoms of gold? 0.198 g

How many P4 molecules are there in 6.0 g of phosphorus? 2.9 x 1022

How many molecules are there in 22.4 l of N2 (g) at 273 K under 1.00 atmosphere pressure? 6.02 x 1023

If equal volumes of each of the gases: H2, O2, N2, CH4, Ne filled the same sized balloons and a small hole is poked in each, which balloon will deflate last? O2

What is the % by mass of carbon in CO? 42.88 %

 

2. (8 points @ 2 points each) Briefly answer the following:

Name the compound with the formula: WO3? tungsten(VI) oxide

What is the value for the heat of formation for oxygen (O2) zero

What is the formula for cobalt (III) sulfate? Co2(SO4)3

What are the units for molarity? moles/L

3. (8 points @ 4 points each) Complete and/or balance with integral coefficients:

_4__ NH3 + __5__ O2 Æ __4__ NO + __6__ H2O

_2__ B + __6__ KOH Æ __2__ K3BO3 + __3__ H2

4. (8 points) The density of mercury is 13.53 g per mL at 25°C. What is its molar concentration under these conditions?

(1000 mL/L)(13.53 g) / (200.6 g/mole) = 67 M

5. (16 points) A 6.768 g sample of methanol (CH3OH) is ignited in a rigid 2.500 liter container into which 3.50 liters of oxygen at 2.017 atm and 23.0 °C has been introduced.

a. Write a balanced equation for the reaction, using integral coefficients.

2 CH3OH + 3 O2 Æ 2 CO2 + 4 H2O

b. Assuming the reaction goes to carbon dioxide and water products, calculate the number of liters of carbon dioxide produced at the above temperature and pressure.

Equation:  2 CH3OH + 3 O2 Æ 2 CO2 + 4 H2O
Stoichiometry (n or V): 2 : 3 : 2 : 4
Before reaction:  6.768g/32.04 g/mol = 0.2112 mole   (2.017 atm)(3.500 L)/(0.0821 L atm/mol K)(296.15 K) = 0.2904 mole   0   0
From the stoichiometry can see that oxygen is limiting (would need 3/2 0.2112 = 0.3168 mol O2) -
After reaction: 0.0176 mol   0 mol   0.1936 mol   0.3872 mol

nCO2 = 0.1936 mol

V = nRT/P = (0.1936 mol)(0.0821 L*atm/mol*K)(296.1 K) / 2.017 atm = 2.33 L

6. ( 15 points) A compound composed of mercury, nitrogen and oxygen is analyzed and found to contain 76.39% mercury and 5.33% nitrogen.

a. Find the empirical formula for this compound. Assume 100 g, then

(76.39 g)/(200.6 g/mol) = 0.3808 mol

(5.33 g)(14.01 g/mol) = 0.3804 mol

(18.28 g)(16.00 g/mol) = 1.143 mol

 Hg.38/.38N.38/.38O1.14/.38

= HgNO3

b. If the compound has a formula weight of approximately 500 g, find its "molecular" formula.

Formula weight = (200.6 + 14.01 + 3 (16.00)) g/mol = 262.6

500/262.6 = 1.9, so 2 formulae/molecule

Thus Hg2N2O6

c. Name this compound. (Hint: it is made up of only two types of ion)

Mercury(I) nitrate = Hg2(NO3)2

7. (15 points) Butane (C4H10) is a useful aerosol propellant since it is fairly innocuous environmentally. Consider a 925 mL aerosol can containing 7.25 g of butane at 23.0 °C. What will the pressure in this can become if it is heated to 690.0°C in a fire, assuming the can is rigid (there is no volume change)?

PV = nRT; P = nRT/V

MW = 4 (12.01) = 10 (1.008) = 58.12 g/mol

n = 7.25 g/58.12 g/mol = 0.1247 mol

T = 690 + 273.15 = 963 K

V = 0.925 L

P = (0.1247 mol)(0.0821 L atm/mol K)(963 K) / (0.925 L)

P = 10.7 atm

8. (8 points) A 0.95 kg cast iron pan is heated from room temperature (20.0 °C) to 125.0 °C on a gas stove. Assuming the heat capacity of the pan remains constant cover this temperature range, calculate the quantity of heat absorbed by the pan ( heat capacity of Fe = 25.1 J °C-1mol-1).

Heat = cnDT

=(25.1 J °C-1mol-1)(950g/55.85 g/mol)(105°C) = 4.5 x 104

9. (10 points)

a) What is the molarity of a solution made by dissolving 5.34 g of magnesium iodide (MgI2) in enough water to make up a volume of 500.0 mL.

MW = 24.30 + 2(126.9) = 278.1 g/mol

M = (5.34g/278.1 g/mol)/(0.5000 L) = 0.0384 M

b. How many mL of a 0.250 M solution of magnesium iodide will be needed to make up 250.0 mL of a solution which has an iodide concentration of 0.025 M?

(V)(0.250 M) = (250 mL)(0.025 M)

V = (250 mL)(0.025 M)/(0.250 M) = 25 mL


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© R A Paselk

Last modified 18 December 2006