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kow [346]
3 years ago
10

Calculate the enthalpy change, ΔH, for the process in which 31.6 g of water is converted from liquid at 3.2 ∘C to vapor at 25.0°

C . For water, ΔHvap = 44.0 kJ/mol at 25.0°C and Cs = 4.18 J/(g*°C) for H2O(l).
Chemistry
1 answer:
amm18123 years ago
5 0

Answer:

Enthalpy change = 74.36 kJ

Explanation:

Enthalpy change is defines as the heat absorbed or evolved during a chemical reaction at constant pressure and volume

Reaction:

H2O(l) --> H2O(g)

DHr = mCsDT

Where m is the mass of water

Cs is the specific heat capacity

DT is the temperature difference

= 31.6 * 4.18 * (25 - 3.2)

= 2879.518 J

DHvap = 44kJ/mol

Moles of water = mass/molecular weight

Molecular weight = (1*2) + 16

= 18g/mol

Moles of water = 31.6/18

= 1.756 moles

DHvap = 44 * 1.756

= 77.244kJ

DH = DHproduct - DHreactant

DHproduct = DHvap = 77.24kJ

DHreactant = DHr = 2879.518J = 2.880kJ

= 77.24 - 2.880

DH = 74.36kJ

Enthalpy change = 74.36 kJ

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3 years ago
The concentration of CI ion in a sample of H,0 is 15.0 ppm. What mass of CI ion is present in 240.0 mL of H,0, which has a densi
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Answer:

Mass of solute = 0.0036 g

Explanation:

Given data:

Concentration of Cl⁻ = 15.0 ppm

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Formula:

ppm = mass of solute / mass of sample ×1,000,000

by putting values,

15.0 ppm = (mass of solute / 240 g) ×1,000,000

Mass of solute = 15.0 ppm ×  240 g / 1,000,000

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1 year ago
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Thus as shown in the balanced equation two moles of Fe2O3 are formed when 0.5 moles of O2 reacted with mixture of FeO and Fe3O4

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