11. Which of the following (with specific heat capacity, C, provided) would show
ID: 579560 • Letter: 1
Question
11. Which of the following (with specific heat capacity, C, provided) would show the smallest temperature change upon gaining 200.0 J of heat? a. 50.0 g Al, C 0.903 J/g"C b. 50.0 g Cu, Ccu 0.385 /g"C c. 25.0 g Au, CA 0.128 J/g c d, 25.0 g Ag, CAg = 0.235 J/g°C e. 25.0g H,0, Cizo 4.184 /g c 12. Calculate the change in internal energy (AE) for a system that is absorbing 35.8 k of heat and is expanding from 8.00 to 24.0 L in volume at 1.00 atm. (Remember that 101.31 = 1 L-atm) a. +51.8 k b. -15.8 k C. -16.6 k d. -29.3 k e. +34.2 kl 13. Suppose, at 298 K, E°= 0.011V for the following redox reaction: 2N20(g) + 302(g) 2N,04(g) . 2418 Calculate E at 298 K when PN20 0.15 atm, Po2 0.050 atm, and P20s 1.85 atm. Pick closest answer. (Hint: Find the number of electrons transferred first) a. 0.01 v b. 0.08 V C. -0.06 V d. -0.04 V e. .020VExplanation / Answer
Q11
smallest temperature change --> must be the LARGEST heat capacity
heat capacity = m*Cp
from the list:
a)
50*0.903 = 45.15
b)
no need, since Cp is lower than Aluminium
c)
no need, since both m, Cp are lower
d)
no need, since m, Cp are bot lower
E)
m*C = 25*4.184 = 104.6
therefore
from all, choose water system, since Cp >> Cpmetals
therefore, a small amount of water will not change drastically
Q12
Q - W = dU
Q = +35.8 kJ since absorbing
Expansion is W = opsitive
W = -P*(Vf-Vi)
W = -1*(24-8) = -16
W = -16*101.3 = -1620.8 J = -1.62 kJ
dU = 35.8 - 1.62
dU = 34.2 kJ
q13
Apply
Ecell = E° - 0.0592/n*ln(Q)
n = 12 e- since O2 = 4 e- then 3O2 = 3x4 = 12 e-
Ecell = 0.011 - 0.0592/12*ln((1.85^2)/((0.15^2)*0.05))
Ecell = -0.02
best answer is E
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