In a calorimeter, 100 g of ice melts at 0oc. the enthalpy of fusion of the ice is 334 j/g. how much heat was absorbed?
question 21 options:

33.4 kj

334 j

334 kj

33.4 j

question 22 (2 points)

in a calorimeter, 1.0 kg of ice melts at 0oc. the enthalpy of fusion of the ice is 334 j/g. how much heat was absorbed?
question 22 options:

334 kj

33.4 j

334 j

33.4 kj

question 23 (2 points)

in a calorimeter, 10.0 g of ice melts at 0oc. the enthalpy of fusion of the ice is 334 j/g. how much heat was absorbed?
question 23 options:

334 kj

0.33 kj

3.34 kj

33.4 kj

4.13 unit assessment: chemical thermodynamics, part 1 pool 13
question 24 (2 points)

what is the change in enthalpy when 250 g of water vapor condenses at 100oc? (δhv = 40.67 kj/mol)
question 24 options:

565 kj

–565 kj

291 kj

–291 kj
question 25 (2 points)

what is the change in enthalpy when 180 g of water vapor condenses at 100oc? (δhv = 40.67 kj/mol)
question 25 options:

565 kj

–565 kj

407 kj

–407 kj
question 26 (2 points)

what is the change in enthalpy when 250 g of water evaporates at 100oc? (δhv = 40.67 kj/mol)
question 26 options:

291 kj

565 kj

–291 kj

–565 kj

For questions 21 - 23 we use this formula: Q = ( m ) ( L sub F )

21: Given: 100 g of ice melts at 0oC. The enthalpy of fusion of the ice is 334 J/g
Q = (100 g) (334 J/g) = 33,400 J
33,400 J * 1kJ/1000J = 33.4 kJ

22. Given:  1.0 kg of ice melts at 0oC. The enthalpy of fusion of the ice is 334 J/g
1.0 kg * 1000 g/kg = 1000 g
Q = (1000 g) (334 J/g) = 334,000 J
334,000 J * 1kJ/1000J = 334 kJ

23. Given: 10.0 g of ice melts at 0oC. The enthalpy of fusion of the ice is 334 J/g
Q = (10 g) (334 J/g) = 3340 J
3340 J * 1kJ/1000J = 3.34 kJ

For questions 24-26, we use Q = m x Hv / M(H2O)   where: M(H20) = 18 g/mol

24. Given: change in enthalpy when 250 g of water vapor condenses at 100oC? (ΔHv = 40.67 kJ/mol)
Q = 250 g * (40.67 kJ/mol / 18 g/mol)
Q = 250 g * 40.67 kJ/mol * 1 mol/18g
Q = 250 * 40.67 kJ / 18
Q = 564.86 kJ or 565 kJ

25. Given: change in enthalpy when 180 g of water vapor condenses at 100oC? (ΔHv = 40.67 kJ/mol)
Q = 180 g * (40.67 kJ/mol / 18 g/mol)
Q = 180 g * 40.67 kJ/mol * 1mol/18g
Q = 180 * 40.67 kJ / 18
Q = 406.7 kJ or 407 kJ

26. Given: change in enthalpy when 250 g of water evaporates at 100oC? (ΔHv = 40.67 kJ/mol)
Q = 250 g * (40.67 kJ/mol / 18 g/mol)
Q = 250 g * 40.67 kJ/mol * 1mol/18g
Q = 250 * 40.67 kJ  / 18
Q = 564.86 kJ or 565 kJ

1. o2(g)as a product

The total number of electrons is 14