A. A total of 2.00 mol of a compound is allowed to react with water in a foam co
ID: 487902 • Letter: A
Question
A. A total of 2.00 mol of a compound is allowed to react with water in a foam coffee cup and the reaction produces 190 g of solution. The reaction caused the temperature of the solution to rise from 21.00 to 24.70 C. What is the enthalpy of this reaction? Assume that no heat is lost to the surroundings or to the coffee cup itself and that the specific heat of the solution is the same as that of pure water. Enter your answer in kilojoules per mole of compound to three significant figures.
B. Calculate the heat required to raise the temperature of 52.5 g of gold from 56.3 °C to 98.4 °C
C. Calculate the specific heat capacity for a 15.3-g sample of gold that absorbs 87.2 J when its temperature increases from 35.0 °C to 79.2 °C.
D. A 68-g sample of sodium is at an initial temperature of 42 °C. If 1840. joules of heat are applied to the sample, what is the final temperature of the sodium?
E. Calculate the heat required to raise the temperature of 75.1 g of mercury from 31.7 °C to 53.8 °C. The specific heat capacity of mercury is 0.14 J/(g °C).
Explanation / Answer
A. Heat required to raise the temperature,
q = mcdeltat
m = 190g = 0.19kg
c=4,18J/kgC
delta t = 24.70 - 21.00 = 3.7
q = 0.19kg*4.18kJ/kgC*3.7C
q = 2.94kJ
2 mol of the compound is used to make the solution,
thus enthalpy = 2.94/2kJ/mol = 1.47kJ/mol
B.m = 52.5g
delta t = 98.4C-56.3C = 42.1C
c for gold = 0.129J/gC
so,q = mcdeltat
q = 52.5g*0.129J/gC*42.1C
q= 285.12J = 0.285kJ
C.m = 15,3g
q=87.2J
delta t = 79.2C-35.0C
delta t = 44.2C
q = mcdeltat
c = q/m.delta t
c= 87.2J/15,3g*44.2C
c = 0.129J/gC
D.m = 68g
Ti=42C
q=1840J
c for sodium = 1.23J/gC
q = mc(Tf-Ti)
q/mc = Tf-Ti
1840J/68g*1.23J/gc
22C=Tf-42C
Tf = 64C
E.
m=75.1g
c = 0.14J/gC
delta t = 53.8C - 31.7C = 22.1C
q = 75.1g*0.14J/gC*22.1C
q = 232.36J
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