What is the delta H when 72.0 grams H2O condenses at 100.00C? Here are some constants that you MAY need. specific

Question

What is the delta H when 72.0 grams H2O condenses at 100.00C?

Here are some constants that you MAY need.

specific heats heat of fusion heat of vaporization

H2O(s) = 2.1 J/g0C 6.01 kJ/mole 40.7 kJ/mole

H2O(L) = 4.18 J/g0C

H2O(g) = 1.7 J/g0C

2930 kJ

163 kJ

-163 kJ

-2930 kJ

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Sapo 2 months 2021-07-22T15:48:17+00:00 1 Answers 7 views 0

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    2021-07-22T15:50:12+00:00

    Answer: The value of \Delta H is -163 kJZ

    Explanation:

    The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

    \text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}} ……(1)

    We are given:

    Given mass of water = 72.0 g

    Molar mass of water = 18 g/mol

    Putting values in equation 1, we get:

    \text{Moles of water}=\frac{72.0g}{18g/mol}\\\\\text{Moles of water}=4mol

    Calculating the heat released for the condensation process:

    \Delta H=n\times \Delta H_{(vap)} ……(2)

    where,

    \Delta H = amount of heat released

    n = number of moles of water = 4 moles

    \Delta H_{(vap)} = specific heat of vaporization = -40.7 kJ/mol

    Negative sign represents the amount of heat released.

    Putting values in equation 2:

    \Delta H=4mol\times (-40.7kJ/mol)=-163kJ

    Hence, the value of \Delta H is -163 kJ

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