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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

Verified Answer :

taskmasters

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

Resulatant values of absorbed heat for the given question are 33.4 kJ, 334 kJ, 3.3 kJ, 565 kJ, 407 kJ and 565 kJ.

How do we calculate absorbed heat?

The amount of heat abosorbed in any reaction will be calculated by using the below equation as:

Q = m(ΔHf), where

m = mass of substance

ΔHf = enthalpy of fusion

  • Amount of heat absorbed by 100g of ice is:

Q = (100 g)(334 J/g) = 33,400 J = 33.4 kJ

  • Amount of heat absorbed by 1kg or 1000g of ice is:

Q = (1000 g)(334 J/g) = 334,000 J = 334 kJ

  • Amount of heat absorbed by 10g of ice is:

Q = (10 g)(334 J/g) = 3340 J = 3.3 kJ

The amount of heat abosorbed in any reaction will be calculated by using the below equation as:

Q = m x Hv / M(H₂O)

  • Amount of heat for 250g of water vapor condenses:

Q = (250 g)(40.67 kJ/mol / 18 g/mol) = 564.86 kJ or 565 kJ

  • Amount of heat for 180g of water vapor condenses:

Q = (180 g)(40.67 kJ/mol / 18 g/mol) = 406.7 kJ or 407 kJ

  • Amount of heat for 250g of water evaporates:

Q = (250 g)(40.67 kJ/mol / 18 g/mol) = 564.86 kJ or 565 kJ

Hence required values are discussed above.

To know more about absorbed heat, visit the below link:

https://brainly.com/question/8828503

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