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Diano4ka-milaya [45]
3 years ago
5

10. On the basis of the general solubility rules given

Chemistry
1 answer:
Crank3 years ago
3 0
I think it’s c but I could be wrong
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Complete and balance the following equation: NO2−(aq)+Cr2O72−(aq)→Cr3+(aq)+NO3−(aq)(acidic solution) Express your answer as a ne
Georgia [21]

Answer:

Part A: Cr₂O₇²⁻(aq) +  8H+ + 3NO₂⁻(aq) → 2Cr³⁺(aq) + 3NO₃⁻(aq)  + 4H₂O.

Part B: 2HNO₃(aq) + 2S(s) + H₂O(l) → N₂O(g) + 2H₂SO₃(aq).

Part C: 2Cr₂O₇²⁻(aq) + 3CH₃OH(aq) + 16H⁺ → 4Cr³⁺(aq) + 3HCO₂H(aq) + 11H₂O.

Part D: 4BrO₃⁻(aq) + 5N₂H₄(aq) + 4H⁺(aq) → 2Br₂(I) + 5N₂(g) + 12H₂O(aq).

Part E: NO₂⁻(aq) + 2Al(s) + OH⁻(aq) + H₂O(aq) → NH₃(aq) + 2AlO₂⁻(aq).

Part F: H₂O₂(aq) + ClO₂(aq) → ClO₂⁻(aq) + O₂(g) + 2H⁺(aq).

Explanation:

<em>Part A: Complete and balance the following equation: NO₂⁻(aq) + Cr₂O₇²⁻(aq) → Cr³⁺(aq) + NO₃⁻(aq) (acidic solution). Express your answer as a net chemical equation including phases.</em>

  • To balance and write the net chemical equation, we should write the two-half reactions:
  • The two half reactions are:

The reduction reaction: Cr₂O₇²⁻(aq) +  14H⁺ + 6e → 2Cr³⁺(aq) + 7H₂O.

The oxidation reaction: H₂O(aq) + NO₂⁻(aq) → NO₃⁻(aq) + 2H⁺(aq) + 2e.

  • Before adding the reactions number of electron gained must be equal to number of electrons lost. we multiply the reduction reaction by 1 (be the same) and the oxidation reaction by 3 (3H₂O(aq) + 3NO₂⁻(aq) → 3NO₃⁻(aq) + 6H⁺(aq) + 6e) to equalize the no. of electrons in the two-half reactions.

  • Add up both reactions:

Cr₂O₇²⁻(aq) +  14H⁺ + 6e → 2Cr³⁺(aq) + 7H₂O.

3H₂O(aq) + 3NO₂⁻(aq) → 3NO₃⁻(aq) + 6H⁺(aq) + 6e.

  • So, the net redox reaction will be:

<em>Cr₂O₇²⁻(aq) +  8H⁺ + 3NO₂⁻(aq) → 2Cr³⁺(aq) + 3NO₃⁻(aq)  + 4H₂O.</em>

<em>Part B: Complete and balance the following equation: S(s) + HNO₃(aq) → H₂SO₃(aq) + N₂O(g) (acidic solution) Express your answer as a net chemical equation including phases.</em>

  • To balance and write the net chemical equation, we should write the two-half reactions:
  • The two half reactions are:

The oxidation reaction: S(s) + 3H₂O(l) → H₂SO₃(aq) + 4H⁺(aq) + 4e.

The reduction reaction: 2HNO₃(aq) + 8H⁺(aq) + 8e → N₂O(g) + 5H₂O(l).

  • Before adding the reactions number of electron gained must be equal to number of electrons lost. we multiply the oxidation reaction by 2 (2S(s) + 6H₂O(l) → 2H₂SO₃(aq) + 8H⁺(aq) + 8e) and the reduction reaction by 1 (be the same) to equalize the no. of electrons in the two-half reactions.

  • Add up both reactions:

2S(s) + 6H₂O(l) → 2H₂SO₃(aq) + 8H⁺(aq) + 8e.

2HNO₃(aq) + 8H⁺(aq) + 8e → N₂O(g) + 5H₂O(l).

So, the net redox reaction will be:

<em>2HNO₃(aq) + 2S(s) + H₂O(l) → N₂O(g) + 2H₂SO₃(aq).</em>

<em>Part C: Complete and balance the following equation: Cr₂O₇²⁻(aq) + CH₃OH(aq) → HCO₂H(aq) + Cr³⁺(aq) (acidic solution), Express your answer as a net chemical equation including phases.</em>

  • To balance and write the net chemical equation, we should write the two-half reactions:
  • The two half reactions are:

The reduction reaction: Cr₂O₇²⁻(aq) +  14H⁺ + 6e → 2Cr³⁺(aq) + 7H₂O.

The oxidation reaction: CH₃OH(aq) + H₂O(aq) → HCO₂H(aq) + 4H⁺(aq) + 4e.

  • Before adding the reactions number of electron gained must be equal to number of electrons lost. we multiply the reduction reaction by 2 (2Cr₂O₇²⁻(aq) +  28H⁺ + 12e → 4Cr³⁺(aq) + 14H₂O) and the reduction reaction by 3 (3CH₃OH(aq) + 3H₂O(aq) → 3HCO₂H(aq) + 12H⁺(aq) + 12e) to equalize the no. of electrons in the two-half reactions.

  • Add up both reactions:

2Cr₂O₇²⁻(aq) +  28H⁺ + 12e → 4Cr³⁺(aq) + 14H₂O.

3CH₃OH(aq) + 3H₂O(aq) → 3HCO₂H(aq) + 12H⁺(aq) + 12e.

So, the net redox reaction will be:

<em>2Cr₂O₇²⁻(aq) + 3CH₃OH(aq) + 16H⁺ → 4Cr³⁺(aq) + 3HCO₂H(aq) + 11H₂O.</em>

<em>Part D: Complete and balance the following equation: BrO₃⁻(aq) + N₂H₄(aq) → Br₂(l) + N₂(g)(acidic solution), Express your answer as a net chemical equation including phases.</em>

  • To balance and write the net chemical equation, we should write the two-half reactions:
  • The two half reactions are:

The oxidation reaction: N₂H₄(aq) → N₂(g) + 4e + 4H⁺(aq).

The reduction reaction: 2BrO₃⁻(aq) + 10e + 12H⁺(aq) → Br₂(I) + 6H₂O(aq).

  • Before adding the reactions number of electron gained must be equal to number of electrons lost. we multiply the oxidation reaction by 5 (5N₂H₄(aq) → 5N₂(g) + 20e + 20H⁺(aq)) and the reduction reaction by 2 (4BrO₃⁻(aq) + 20e + 24H⁺(aq) → 2Br₂(I) + 12H₂O(aq)) to equalize the no. of electrons in the two-half reactions.

  • Add up both reactions:

5N₂H₄(aq) → 5N₂(g) + 20e + 20H⁺(aq).

4BrO₃⁻(aq) + 20e + 24H⁺(aq) → 2Br₂(I) + 12H₂O(aq).

So, the net redox reaction will be:

<em>4BrO₃⁻(aq) + 5N₂H₄(aq) + 4H⁺(aq) → 2Br₂(I) + 5N₂(g) + 12H₂O(aq).</em>

<em />

<em>Very Important Note:</em>

<em>Due to the answer exceeds 5000 character, kindly find the answer of part E and F are in the attached word file with also other prats.</em>

Download docx
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The correct answer is actually  

B) energy that flows from warmer objects to cooler objects.

because temperature is a measure of the average amount of energy possessed by an object due to the random motions of its particles.  Heat is the energy that flows from warmer objects to cooler objects. Heat cannot flow in the opposite direction.

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