Chemistry, 20.12.2019 11:31 chem64

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. request answer part b complete and balance the following equation: s(s)+hno3(aq)→h2so3(aq)+n2o(g)(acid ic solution) express your answer as a net chemical equation including phases. nothing request answer 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. nothing request answer part d complete and balance the following equation: bro3−(aq)+n2h4(aq)→br2(l)+n2(g)(aci dic solution) express your answer as a net chemical equation including phases. nothing request answer part e complete and balance the following equation: no2−(aq)+al(s)→nh4+(aq)+alo2−(aq)(b asic solution) express your answer as a net chemical equation including phases. nothing request answer part f complete and balance the following equation: h2o2(aq)+clo2(aq)→clo2−(aq)+o2(g)(b asic solution) express your answer as a net chemical equation including phases. nothing

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Answer from: emmybear103002

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.

Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

$\delta h_{reaction}\times \text{moles of no}=-311.85kj/mol\times 0.49 mol=-152.80kj/mol$

Answer from: Quest

Hydrogen is the most abundant element on the periodic table..hope it .

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Complete and balance the following equation: no2−(aq)+cr2o72−(aq)→cr3+(aq)+no3−( aq)(acidic solutio...

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