2007 Solutions

1.               B

2.               C   2100g/(44g/mole) = 47.7 moles;   47.7 moles(32g/mole) = 1527 g

3.               D   P = nRT/V = (5/32)(8.31(-25+273)/0.3 = 1073 kPa

4.               B

5.               B   (snow= liquid water to solid is exothermic; because the reverse(melting) needs to absorb heat )

6.               C   (really -396 kJ/mole)

7.               A

8.               C

9.               D

10.            A

11.            D

12.            A

13.            C

14.            A

15.       PV = 90 (choose any two values of P and V)

If P = 12, then V= 90/12 = 7.5 mL

16.       1.27 g KClO3 (mol/ 122.5 g) = 0.010367 moles KClO3

but it's not a gas, so we need to to look at equation:

0.010367 moles KClO3(3 O2/2KClO3) = 0.01555 moles O2

R = PV/(nT) = 94.7 kPa(0.468 L)/(0.01555 moles)/(293 K) = 9.72 kPaL/(Kmole) : not an IDEAL gas

(question is not based on real data or experiment went awry becuase in reality oxygen does behave like an ideal gas!)

17.       n = PV/(RT) = 0.0134 moles

Molar mass = m/n = (51.96 -51.02)g/ 0.0134 moles = 70.2 g/mole = CHF3

18.       n = PV/(RT) = 0.00691 moles of H2

0.00691 moles of H2 (1 Mg/ 1 H2) = 0.00691 moles of Mg

0.00691 moles of Mg (24.3 g/mole) = 0.168 g

19.       Q = mcDT

= (225 + 375)g (4.19 J/gC)(30.0 C)

= 75420 J

n = CV = 1mole/L (0.225L) = 0.225 moles

DH = -Q = -75420 J

DH/n =-75420 J/0.225 moles = -335 kJ/mole

20.            Multiply equation(2) by 3 and double equation (1). Reverse equation(3).

Note equation (4) is no good because it does not have liquid water.

Add the three equations: +278 kJ + 3(-285 kJ) + -2(394kJ) = -1365 kJ

DH = -1365 kJ which is for 1 mole of C2H5OH

-5509 kJ (1mole/-1365 kJ)(46g/mole) = 184 g.

21.             Expt 1

Q = mcDT  = 566 J

DH = -Q = -566 J

n = 0.200/65.3 = 0.00306 moles

DH/n  = -566J / 0.00306 moles = -185 000 J = -185 k J/mole

Do the same for expt 2: DH/n  =          -91 .9 kJ/mole

Zn + 2HCl àZnCl2 + H2                                       DH = -185 kJ

ZnCl2 + H2O àZnO + 2 HCl                                 DH = 91.9 kJ (notice: we reversed equation and sign to cancel ZnCl2)

H2 + 0.5 O2 à H2O                                               DH = -286 kJ

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Zn + 0.5 O2 à ZnO                                                     DH = -379 kJ

22.                      use _Qlost = Q gained

c = 0.623 J/goC

23.       A                     CH4:                fewest bonds to break

B                     CH3OH            gaseous states react faster

24.       after 7 minutes, 0.244 moles HCl remain. There were originally 0.250.

So 0.006 moles of HCl reacted

Ratio is 2:1, so only 0.006 moles/2 = 0.003 moles of carbon dioxide formed in 7 minutes

Average rate = Dn/Dt = 0.003 moles/7 minutes = 4.29 X10-4 moles CO2/min

25.       A         curve 1 has no catalyst. Less is being made in the same given time.

B         Curve 2 has a catalyst. Ae is lower.

26.

 2 H2 2 NO N2 (don’t use liquid water) Initial 4/2 = 2M 2 M 0 Changing 0.40 *2 =0.80 0.80 0.40 Equilibrium 2 – 0.80 = 1.2 1.2 0.40

K = [N2]/([H2]2[NO]2

= 0.19

27.

 HCOOH H2O H3O+ HCOO- Initial 0.10 - 0 0 Changing x - x x Equilibrium 0.10 - x - x x

KA = x(x)/(0.10-x) = 1.77 X 10-4

x = 0.004119567520

pH = -log(H+) = -log(0.004119567520) = 2.39

28.       A)        aluminum is oxidized

B)      2 Cu+2  + 6e-    à  3 Cu                       0.34 V

2 Al     à2 Al+3 + 6e                          1.66 V

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2 Cu+2 +  2 Al à2 Al+3 + 3 Cu           2.00 V

c)         2.00 V

d)         Cu+2 at the reduction half cell (cathode)

29.       a)         2 Ag+   +          2e à 2 Ag                   0.80 V

Cu à Cu+2       + 2e-                            -0.34 V

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2 Ag+ +  Cu      à Cu+2 + 2 Ag            0.46 V