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aniked [119]
2 years ago
14

The enthalpy of vaporization of liquid water at 100°C is 2257 kJ/kg. Determine the enthalpy for apordato of iuod eeat capacity o

f liquid water is 4 154 kJl(kg °C) and for water vapor is 1.859 kJ/(kg °C)
Chemistry
1 answer:
igomit [66]2 years ago
3 0

Explanation:

The given data is as follows.

         T_{1} = 100^{o}C,       T_{2} = 10^{o}C

       \Delta H_{vap1} = 2257 kJ/kg,     \Delta H_{vap2} = ?

For water, C_{p} = 4.184 kJ/kg ^{o}C

Formula to calculate heat of vaporization is as follows.

  \Delta H_{vap1} - \Delta H_{vap2} = C_{p}(T_{1} - T_{2})

Hence, putting the values into the above formula as follows.

\Delta H_{vap1} - \Delta H_{vap2} = C_{p}(T_{1} - T_{2})

2257 kJ/kg - \Delta H_{vap2} = 4.184 kJ/kg ^{o}C (100 - 10)^{o}C

            \Delta H_{vap2} = 2257 kJ/kg - 376.56 kJ/kg

                                       = 1880.44 kJ/kg

Thus, we can conclude that enthalpy of liquid water at 10^{o}C is 1880.44 kJ/kg.

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The expression for enthalpy change is,

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Putting the values we get :

\Delta H=[2\times H_f{Fe}+3\times H_f{H_2O}]-[1\times H_f{Fe_2O_3}+3\times H_f{H_2}]

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2 years ago
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How many moles are in 5.0 x 10^25 atoms of iron?
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Hello!

To find the number of moles that are in the given amount, we need to divide the total number of atoms by Avogadro's number, which is 1 mole is equal to 6.02 x 10^23 atoms.

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7 0
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