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    Principles of Chemistry
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    Exam 11: Liquids, Solids, and Intermolecular Forces
  5. Question
    The Enthalpy Change for Converting 1
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The Enthalpy Change for Converting 1

Question 30

Question 30

Multiple Choice

The enthalpy change for converting 1.00 mol of ice at -50.0°C to water at 70.0°C is The enthalpy change for converting 1.00 mol of ice at -50.0°C to water at 70.0°C is   The specific heats of ice,water,and steam are     And   Respectively.For H<sub>2</sub>O,∆H<sub>fus </sub>= 6.01 kJ/mol,and ∆H<sub>vap </sub>= 40.67 kJ/mol. A) 12.28 B) 6.41 C) 13.16 D) 7154 E) 9.40
The specific heats of ice,water,and steam are
The enthalpy change for converting 1.00 mol of ice at -50.0°C to water at 70.0°C is   The specific heats of ice,water,and steam are     And   Respectively.For H<sub>2</sub>O,∆H<sub>fus </sub>= 6.01 kJ/mol,and ∆H<sub>vap </sub>= 40.67 kJ/mol. A) 12.28 B) 6.41 C) 13.16 D) 7154 E) 9.40
The enthalpy change for converting 1.00 mol of ice at -50.0°C to water at 70.0°C is   The specific heats of ice,water,and steam are     And   Respectively.For H<sub>2</sub>O,∆H<sub>fus </sub>= 6.01 kJ/mol,and ∆H<sub>vap </sub>= 40.67 kJ/mol. A) 12.28 B) 6.41 C) 13.16 D) 7154 E) 9.40
And
The enthalpy change for converting 1.00 mol of ice at -50.0°C to water at 70.0°C is   The specific heats of ice,water,and steam are     And   Respectively.For H<sub>2</sub>O,∆H<sub>fus </sub>= 6.01 kJ/mol,and ∆H<sub>vap </sub>= 40.67 kJ/mol. A) 12.28 B) 6.41 C) 13.16 D) 7154 E) 9.40
Respectively.For H2O,∆Hfus = 6.01 kJ/mol,and ∆Hvap = 40.67 kJ/mol.


A) 12.28
B) 6.41
C) 13.16
D) 7154
E) 9.40

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