Honors Chemistry

Answers to Chapter 5 Study Questions

1.  a)  STP:  PT = 1 atm; T = 273 K;  ;  nT =  0.039 + 0.010 + 0.001 = 0.050 moles

=  0.20 atm

b)  V  =  ?;  STP:  T = 273 K, PT = 1 atm;  nT = 0.050 mol;  PV = nRT

=  1.1 L

2.  a)  PT  = PH2 + PH2O; Find PH2O in Table from lab report; at 19°C, PH2O = 16 mmHg

PH2  =  PT  -  PH2O  =  756 - 16  =  740. mmHg

b)    =  0.974 atm

3.  V1  =  600. cm3;  T1  =  25°C  =  298 K;  P1  =  750. mmHg

V2  =  480. cm3;  T2  =  41°C  =  314 K;  P2  =  ?

4.    =  0.178 g/L

5.   a)  2 C4H10(g)  +  13 O2(g)  ®   10 H2O(l)  +  8 CO2(g)

b)     =   3.2 L O2

c)   = 17.9 L CO2

d)  = 7.5 x 1023 molecules

6.   n = 1 mole;  T  =  68°C = 341 K;  P  =  2.00 atm;  V  =  ?

14.0 L

7.    =  11.2 L

= 3.93 g/L

9.  Find molar volume at 710 mmHg and 36°C and then use conversion factors:

T = 36 + 273 = 309 K;   = 0.934 atm

27.1 L

=  3.70 L O2

10.  ;  so use PV = nRT to calculate n;  T = 29 + 273 = 302 K;  P = 1 atm

=  28.0 g/mole

11. = 0.341 mol CO2;    = 0.750 mole CH4;

= 15.7 L

12. PH2= PT - PH2O; At 22°C, PH2O = 20 mm Hg; PH2 = 750 mmHg - 20 mmHg = 730 mmHg

= 0.960 atm;  T = 22 + 273 = 295 K;

=  0.119 mol H2 =  n

3.00 L