What mass of ammonium nitrate is required to lower the temperature of 400 mL of
water from 25 oC to 1oC? The Hsoln for ammonium nitrate is 34 kJ/mol.
1
Expert's answer
2018-02-15T07:15:45-0500
400 ml = 400 g of water ΔT = 25 – 1 = 24 o C C = 4.18 J/g∙C Q = cmΔT = 4.18 J/g∙C x 400 g x 24 C = 40128 J = 40.128 kJ 40.128 kJ / 34 kJ/mol = 1.18 mol of ammonium nitrate required 80.043 g/mol x 1.18 mol = 94.45074 g. Answer: 94.45074 g.
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Comments
Assignment Expert
15.02.18, 15:10
Dear Katelyn, it just influnces on the sign of Q, which means energy
releasing or consumption. Since it is irrelevant in this problem, one
can use any ΔT definition.
Katelyn
15.02.18, 14:40
Wouldn’t deltaT be 1-25= -24? Since deltaT is tfinal-tinitial?
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Comments
Dear Katelyn, it just influnces on the sign of Q, which means energy releasing or consumption. Since it is irrelevant in this problem, one can use any ΔT definition.
Wouldn’t deltaT be 1-25= -24? Since deltaT is tfinal-tinitial?
Leave a comment