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ANEK [815]
3 years ago
5

Conversion of 0.702 g of liquid water to steam at 100.0°C requires 1.56 kJ of heat. Calculate the molar enthalpy of evaporation

of water at 100.0°C.
Chemistry
1 answer:
kifflom [539]3 years ago
7 0

Answer:

The molar enthalpy of evaporation of water at 100.0°C is 40 kJ

Explanation:

Calculation of the moles of H_2O as:-

Mass = 0.702 g

Molar mass of H_2O = 18 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{0.702\ g}{18\ g/mol}

Moles= 0.039\ mol

Given that:- \Delta H=+1.56\ kJ

It means that when 0.039 moles of water undergoes evaporation, 1.56 kJ of energy is used

Also,

When 1 mole of water undergoes evaporation, \frac{1.56}{0.039}\ kJ of energy is used

<u>The molar enthalpy of evaporation of water at 100.0°C is 40 kJ</u>

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The independent variable has control and affects the
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Answer:

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3 0
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Read 2 more answers
A solution contains 0.036 M Cu2+ and 0.044 M Fe2+. A solution containing sulfide ions is added to selectively precipitate one of
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Answer:

The precipitate is CuS.

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Explanation:

<u>Step 1: </u>Data given

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Ksp (CuS) = 1.3 × 10-36

Ksp (FeS) = 6.3 × 10-18

Step 2:  Calculate precipitate

CuS → Cu^2+ + S^2-         Ksp= 1.3*10^-36

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Calculate the minimum of amount needed to form precipitates:

Q=Ksp

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[S2-]= 3.61*10^-35 M

<u>For Iron</u>  we have: Ksp=[Fe2+]*[S2-]

Ksp(FeS) = 6.3*10^-18 = 0.044M*[S2-]

[S2-]= 1.43*10^-16 M

CuS will form precipitates before FeS., because only 3.61*10^-35 M Sulfur Ions are needed for CuS. For FeS we need 1.43*10^-16 M Sulfur Ions which is much larger.

The precipitate is CuS.

Sulfide will precipitate at  [S2-]= 3.61*10^-35 M

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3 years ago
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