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Yuki888 [10]
3 years ago
11

How much heat is evolved in converting 1.00 mol of steam at 140.0 ?C to ice at -45.0 ?C? The heat capacity of steam is 2.01 J/(g

??C) and of ice is 2.09 J/(g??C).
Chemistry
1 answer:
Serhud [2]3 years ago
5 0
The chemical formula for steam is H2O(g)

mass of steam = mole of steam * molar mass of H2O
= 1 * (1*2+16) = 18 g

by E = m * c * change in temperature,
From 140C steam to 100C steam
18 * 2.01 * (140-100) = 1447.2J

latent heat of vaporization of water = 2.26x10^6 J/kg* 0.018 = 40680 J

From 100C water to 0C water
18 * 4.2 * (100-0) = 7560 J

latent heat of fusion of ice = 334 J/g * 18 g = 6012 J

From 0C ice to -45C ice
18 * 2.09 * (45-0)
= 1692.9 J

1447.2 + 40680 + 7560 + 6012 + 1692.9 = 57392.1 J
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Use the half-reaction method to balance the equation for the conversion of ethanol to acetic acid in acid solution:CH₃CH₂OH + Cr
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Balanced chemical equation is 3CH₃CH₂OH + 2Cr₂O₇²⁻ + 16H⁺ → 3CH₃COOH + 4Cr³⁺ + 11H₂O.

<h3>What is Balanced Chemical Equation ?</h3>

The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.

The chemical equation

CH₃CH₂OH + Cr₂O₇²⁻ → CH₃COOH + Cr³⁺

First assign the oxidation number for each atom in the equation.

\overset{+3}{C}\overset{+1}{H_3} \overset{-1}{C} \overset{+1}{H_2} \overset{-2}{O} \overset{+1}{H} + \overset{+6}{Cr_2} \overset{-2}{O_7} + \overset{+1}{H} \rightarrow \overset{+3}{C}\overset{+1}{H_3} \overset{+3}{C}\overset{-2}{O} \overset{-2}{O} \overset{+1}{H} + \overset{+3}{Cr}

Oxidation: C₂H₆O → C₂H₄O₂ + 4e⁻

Reduction: Cr₂O₇ + 6e⁻  → 2Cr⁺³

Now, balance the charge

Oxidation: C₂H₆O → C₂H₄O₂ + 4e⁻ + 4H⁺

Reduction: Cr₂O₇ + 6e⁻ + 14H⁺ → 2Cr⁺³

Now balance the oxygen atoms

Oxidation: C₂H₆O + H₂O → C₂H₄O₂ + 4e⁻ + 4H⁺

Reduction: Cr₂O₇ + 6e⁻ + 14H⁺ → 2Cr⁺³ + 7H₂O

Now, make electron gain equivalent to lost

Oxidation: C₂H₆O + H₂O → C₂H₄O₂ + 4e⁻ + 4H⁺ }  × 3

Reduction: Cr₂O₇ + 6e⁻ + 14H⁺ → 2Cr⁺³ + 7H₂O } × 2

Now,

Oxidation: 3C₂H₆O + 3H₂O → 3C₂H₄O₂ + 12e⁻ + 12H⁺

Reduction: 2Cr₂O₇ + 12e⁻ + 28H⁺ → 4Cr⁺³ + 14H₂O

Now, add the both equations

3CH₃CH₂OH + 2Cr₂O₇²⁻ + 16H⁺ → 3CH₃COOH + 4Cr³⁺ + 11H₂O

Thus from the above conclusion we can say that the balanced chemical equation is 3CH₃CH₂OH + 2Cr₂O₇²⁻ + 16H⁺ → 3CH₃COOH + 4Cr³⁺ + 11H₂O.

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D) energy required to remove a valence electron

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