Page 317 - Theory and Problems of BEGINNING CHEMISTRY

306 SCIENTIFIC CALCULATIONS [APP.

A.17. Solve each of the following equations for the indicated variable, given the other values listed.

Solve Solve
Equation for: Given Equation for: Given

(a) PV = nRT P n = 1.50 mol (e) πV = nRT π V = 2.50 L
0.0821 L·atm
R = n = 0.0300 mol
(mol·K)
T = 305 K T = 298 K
0.0821 L·atm
V = 4.00 L R =
(mol·K)
(b) M = n/V V M = 1.65 mol/L ( f ) P 1 V 1 = P 2 V 2 P 2 V 1 = 242 mL
n = 0.725 mol V 2 = 505 mL
P 1 = 0.989 atm
(c) d = m/V V d = 2.23 g/mL (g) E = hv v E = 6.76 × 10 −19 J
m = 133 g h = 6.63 × 10 −34 J·s
P 1 P 2
◦
(d) = T 1 T 2 = 273 K (h)
t = k b m k b
t = 0.400 C
T 1 T 2
P 1 = 888 torr m = 0.250 m
P 2 = 777 torr

nRT (1.50 mol)[0.0821 L·atm/(mol·K)](305 K)
Ans. (a) P = = = 9.39 atm
V 4.00 L
n 1L
(b) V = = 0.725 mol = 0.439 L
M 1.65 mol
m 1mL
(c) V = = 133 g = 59.6mL
d 2.23 g
P 1 T 2 (888 torr)(273 K)
(d) T 1 = = = 312 K
P 2 777 torr
nRT (0.0300 mol)[0.0821 L·atm/(mol·K)](298 K)
(e) π = = = 0.294 atm
V 2.50 L
P 1 V 1 (0.989 atm)(242 mL)
( f ) P 2 = = = 0.474 atm
V 2 505 mL
15
(g) v = E/h = (6.76 × 10 −19 J)/(6.63 × 10 −34 J·s) = 1.02 × 10 /s
(h) k b =
t/m = (0.400 C)/(0.250 m) = 1.60 C/m
◦
◦

CALCULATOR MATHEMATICS
2
2
A.18. Note the difference between (5.71) and 5.71 × 10 . Which calculator operation key should be used for
entering each of these?
2
2
Ans. For (5.71) , the x 2 key should be pressed after entering 5.71, giving 32.6. For 5.71 × 10 , the EXP or
EE key followed by 2 should be pressed after entering 5.71, giving 571.
A.19. Calculate the reciprocal of 4.00 × 10 −8 m.
7
7
Ans. 2.50 × 10 m −1 = 2.50 × 10 /m
Note that the unit for the reciprocal is different from the unit for the given value.