1) 1.64 mol of nickel at 150.13C is placed in 1.00 L of water at 25.09C. The fin
ID: 822293 • Letter: 1
Question
1) 1.64 mol of nickel at 150.13C is placed in 1.00 L of water at 25.09C. The final temperature of the nickel-water mixture is 26.34C. what is the specific heat of nicker?
a)0.349 J/K-1 mol-1
b) 0.554 J/K-1 mol-1
c) 2.28 J/K-1 mol-1
d) 25.7 J/K-1 mol-1
e) 31.6 J/K-1 mol-1
2) The standard molar entropy of NO is 211 J/K-1 mol-1. What is the standard entropy of 3 moles of NO at 10 bar?
a) 202 J/K-1 mol
b) 576 J/K-1
c) 606 J/K-1 mol-1
d) 624 J/K-1
e) 633 J/K-1
PLEASE HELP . THANKYOU IN ADVANCE. I WILL RAISE POINTS IF YOU FEEL IT IS NECESSARY THANK YOU FOR YOUR HELP
Explanation / Answer
Heat lost by nickel = heat gined by water (principle of calorimetry)
C = specific heat
moles_nickel * C_nickel * change in temp_nickel = mass_water * C_water * change in temp_water
So,
C_nickel = (mass_water * C_water * change in temp_water) / (moles_nickel * change in temp_nickel)
mass_water = 1 kg
C_water = 4186 JK^-1kg^-1
moles of Nickel = 1.64
Chane in temp is given in the problem
Just plug in the value (refer to the first answer)
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