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

All i know is that1. x=0.72. C. 2k+3=5.43. x=124.16666676. A. 7. D.Hope this helps!! If you need anything else just ask! ?

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=2) Your answer would be C. 2k+3=5.43) Your answer would be x=124.16 where the 6 is repeating or x=4) , 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 5Step-by-step explanation:Here two equations are given, x + y = 7    (1)2x + y = 5  (2)Check option 1Multiply the first equation by -1 and add the equations together.-x - y = -72x+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 hardStep-by-step explanation:

y=4x+0  I hope that's what you were looking forStep-by-step explanation: