Predict the products of each reaction: part a part complete hclo4(aq)+fe2o3(s)→ express your answer as a balanced chemical equation.

Predict the products of each reaction: part a part complete hclo4(aq)+fe2o3(s)→ express your answer as a balanced chemical equation. identify all of the phases in your answer. 6hclo4(aq)+fe2o3(s)→2fe(clo4)3(aq)+ 3h2o(l) previous answers correct part b h2so4(aq)+sr(s)→ express your answer as a balanced chemical equation. identify all of the phases in your answer. request answer part c h3po4(aq)+koh(aq)→ express your answer as a balanced chemical equation. identify all of the phases in your answer. nothing request answer provide feedback correct. no additional followup.

Answers

Part A: 6HClO₄(aq) + Fe₂O₃(s) → 2Fe(ClO₄)₃(aq) + 3H₂O(l).

Part B: H₂SO₄(aq) + Sr(s) → SrSO₄(s) + H₂(g).

Part C: H₃PO₄(aq) + 3KOH(aq) → K₃PO₄(aq) + 3H₂O(l).

Explanation:

Part A: complete HClO₄(aq) + Fe₂O₃(s)→ Express your answer as a balanced chemical equation. Identify all of the phases in your answer.

The balanced chemical equation is:

6HClO₄(aq) + Fe₂O₃(s) → 2Fe(ClO₄)₃(aq) + 3H₂O(l).

It is clear that 6 mol of HClO₄ (in aqueous phase) react with 1 mol of Fe₂O₃ (in solid phase) to produce 2 mol of Fe(ClO₄)₃ (in aqueous phase) and 3 mol of H₂O (in liquid phase).

Part B: H₂SO₄(aq) + Sr(s) → Express your answer as a balanced chemical equation. Identify all of the phases in your answer.

The balanced chemical equation is:

H₂SO₄(aq) + Sr(s) → SrSO₄(s) + H₂(g).

It is clear that 1 mol of H₂SO₄ (in aqueous phase) reacts with 1 mol of Sr (in solid phase) to produce 1 mol of SrSO₄ (in solid phase) and 1 mol of H₂ (in gas phase).

Part C: H₃PO₄(aq) + KOH(aq) → Express your answer as a balanced chemical equation.

The balanced chemical equation is:

H₃PO₄(aq) + 3KOH(aq) → K₃PO₄(aq) + 3H₂O(l).

It is clear that 1 mol of H₃PO₄ (in aqueous phase) react with 3 mol of KOH (in aqueous phase) to produce 1 mol of K₃PO₄ (in aqueous phase) and 3 mol of H₂O (in liquid phase).

1) The answer would be x=0.7 or x=frac{7}{10}

2) Your answer would be C. 2k+3=5.4

3) Your answer would be x=124.16 where the 6 is repeating or x=frac{745}{6}

4) frac{2a}{7}+frac{6}{7} =frac{5}{7}, frac{4a}{7} +frac{3}{7}=frac{1}{7}

5) I am sorry, I do not understand number 5.

6) Your answer would be answer choice B. x=1.1 satisfies the equation because 7-1.5=5.5 and 1.1*5=5.5.

7) None of those equations is equal to the answer.

8) None of those equations is equal to the answer.

I hope this helps.

answers in order C,A,B,D,because the reactants only partially ionize in the solution,none of these( SO4 2- is a strong one),none of these( all are weak),same(beacause both Ka and Kb values are equal in those subtances),di,mono,tri,mono,mono,mono,di,di,mono,Na2Be(OH)4(aq)...

Explanation:

this quiz doesnt worth my time...but have a nice day.

Explanation:

Part A : LiCl(aq) + AgNO₃(aq)→

Chemical equation:

LiCl(aq) + AgNO₃(aq)  →  AgCl(s) + LiNO₃(aq)

Ionic equation:

Li⁺(aq)  + Cl⁻(aq) + Ag⁺(aq) + NO₃⁻(aq)  →  AgCl(s) + Li⁻(aq)  + NO⁻₃(aq)

Net ionic equation:

Cl⁻(aq) + Ag⁺(aq) →  AgCl(s)

C = H2SO4(aq)+Li2SO3(aq)→

Chemical equation:

H₂SO₄(aq) + Li₂SO₃(aq)  →  Li₂SO₄(aq) + SO₂(g) + H₂O(l)

Ionic equation:

2H⁺(aq)  + SO²⁻₄(aq) + 2Li⁺(aq)  + SO₃²⁻(aq)  →  2Li⁺ (aq) + SO₄²⁻(aq) + SO₂(g) + H₂O(l)

Net ionic equation:

2H⁺ + SO₃²⁻(aq)  →  SO₂(g) + H₂O(l)

Part E: HClO4(aq)+Ca(OH)2(aq)→

Chemical equation:

HClO₄(aq) + Ca(OH)₂(aq)  →  Ca(ClO₄)₂ (aq) + H₂O(l)

Balanced Chemical equation:

2HClO₄(aq) + Ca(OH)₂(aq)  →  Ca(ClO₄)₂ (aq) + 2H₂O(l)

Ionic equation:

2H⁺(aq) + 2ClO⁻₄(aq) + Ca²⁺(aq) + (OH)²⁻₂(aq)  →  Ca²⁺(aq) +(ClO₄)²⁻₂ (aq) + 2H₂O(l)

Net ionic equation:

2H⁺(aq) + (OH)²⁻₂(aq)  →  2H₂O(l)

Part F: Cr(NO3)3(aq)+LiOH(aq)→

Chemical equation:

Cr(NO₃)₃(aq) + LiOH (aq)  →   LiNO₃(aq) + Cr(OH)₃(s)

Balanced chemical equation;

Cr(NO₃)₃(aq) + 3LiOH (aq)  →   3LiNO₃(aq) + Cr(OH)₃(s)

Ionic equation:

Cr³⁺(aq) + 3NO₃⁻(aq) + 3Li⁺(aq) + 3OH⁻ (aq)  →   3Li⁺(aq) + 3NO⁻₃(aq) + Cr(OH)₃(s)

Net ionic equation:

Cr³⁺(aq) +  3OH⁻ (aq)  →    Cr(OH)₃(s)

Part H: HCl(aq)+Hg2(NO3)2(aq)→

Chemical equation:

HCl (aq) + Hg₂(NO₃)₂(aq)  → Hg₂Cl₂ (s) + HNO₃(aq)

Balanced chemical equation:

2HCl (aq) + Hg₂(NO₃)₂(aq)  → Hg₂Cl₂ (s) + 2HNO₃(aq)

Ionic equation;

2H⁺(aq) + 2Cl⁻ (aq) + 2Hg⁺(aq) + 2NO₃⁻(aq)  → Hg₂Cl₂ (s) + 2H⁺(aq) + 2NO⁻₃(aq)

Net ionic equation:

2Cl⁻ (aq) + 2Hg⁺(aq)   → Hg₂Cl₂ (s)

Option 1, Option 3, Option 4, Option 5

Step-by-step explanation:

Here two equations are given,

x + y = 7    (1)

2x + y = 5  (2)

Check option 1

Multiply the first equation by -1 and add the equations together.

-x - y = -7

2x+y = 5    correct

Similarly,

Option 1, Option 3, Option 4 and Option 5 are correct.

That's the final answer.

I hove it will helps you.

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:

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.

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:

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

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.

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:

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

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.

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:

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

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.

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:

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

Very Important Note:

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.

this looks hard

Step-by-step explanation:

y=4x+0  I hope that's what you were looking for

Step-by-step explanation:

All i know is that
1. x=0.7
2. C. 2k+3=5.4
3. x=124.1666667
6. A. 
7. D.
Hope this helps!! If you need anything else just ask! 🙂

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