Chemistry, 21.09.2019 16:30 codie1103

2. balance each of the following reactions: a) as(s) + cl, (g) → ascl (8) b) cu(s) + hno,(aq) → cu2+ (aq) + no(g) (in acidic solution) c) mno; (aq) + i-(aq) → mn? "(aq) + 1,(s) (in acidic solution)

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Answer from: ykluhredd

a) 2 As(s) + Cl₂ (g) → 2 AsCl (g)

b) Cu(s) + 2 H⁺ + 2 HNO₂(aq) → Cu²⁺ + 2 NO(g) + 2 H₂O

c) 2 I⁻ + 4H⁺ + MnO₂ (aq) → I₂(s) + Mn²⁺ + 2 H₂O(aq)

Explanation:

a) As(s) + Cl₂ (g) → AsCl (g)

2 As(s) + Cl₂ (g) → 2 AsCl (g)

This reaction is balanced yet

b) Cu(s) + HNO₂ (aq) → Cu²⁺ (aq) + NO(g)

To balance a redox reaction the first step is to know which atom is oxidated and which reduced. To know it. We need to obtain the oxidation number for all atoms. Almost always, oxidation numbers of H is +1 and for O -2.

For Cu: The oxidation number of a pure element is 0.

HNO₂: 2×O: -4, H: +1 → N: +3

NO: O: -2 → N: +2

Thus, oxidation numbers are:

Cu(s) + HNO₂ (aq) → Cu²⁺ (aq) + NO(g)

0 +3 +2 +2

The atoms which increase oxidation number are oxidated and the atoms which decrease oxidation number are reduced. Thus, Cu is oxidated and N is reduced.

The next step is separate half-reactions, thus:

HNO₂ (aq) → NO(g) Reduction

Cu(s) → Cu²⁺ (aq) Oxidation

Then, we should balance, first, elements differents of oxygen and hydrogen: They are balanced yetOxygens must be balanced with H₂O and hydrogens with H⁺ (Because is acidic solution):

Cu(s) → Cu²⁺ (aq) Oxidation

H⁺ + HNO₂ (aq) → NO(g) + H₂O Reduction

After this, we must balance charges with electrons:

Cu(s) → Cu²⁺ (aq) + 2e⁻ Oxidation

1e⁻ + H⁺ + HNO₂ (aq) → NO(g) + H₂O Reduction

Electron number must be the same for oxidation and reduction

Cu(s) → Cu²⁺ (aq) + 2e⁻ Oxidation

2× (1e⁻ + H⁺ + HNO₂ (aq) → NO(g) + H₂O) Reduction

Last, we should sum half-reactions and cancel out common compounds:

Cu(s) + 2e⁻ + 2H⁺ + HNO₂ (aq) → Cu²⁺ + 2e⁻ + NO(g) + H₂O

Cu(s) + 2 H⁺ + 2 HNO₂(aq) → Cu²⁺ + 2 NO(g) + 2 H₂O

c) MnO₂ (aq) + I⁻ (aq) → Mn²⁺ (aq) + I₂(s)

Oxidation number:

MnO₂ (aq) + I⁻ (aq) → Mn²⁺ (aq) + I₂(s)

+4 -1 +2 0

Half reaction:

I⁻ (aq) → I₂(s) Oxidation

MnO₂ (aq) → Mn²⁺ (aq) Reduction

Then, we should balance, first, elements differents of oxygen and hydrogen.

2 I⁻ (aq) → I₂(s) Oxidation

MnO₂ (aq) → Mn²⁺ (aq) Reduction

Oxygens must be balanced with H₂O and hydrogens with H⁺ (Because is acidic solution):

2 I⁻ (aq) → I₂(s) Oxidation

4H⁺ + MnO₂ (aq) → Mn²⁺ + 2 H₂O(aq) Reduction

Charge balance:

2 I⁻ (aq) → I₂(s) + 2e⁻ Oxidation

2e⁻ + 4H⁺ + MnO₂ (aq) → Mn²⁺ + 2 H₂O(aq) Reduction

Last, we should sum half-reactions and cancel out common compounds:

2 I⁻ + 4H⁺ + MnO₂ (aq) → I₂(s) + Mn²⁺ + 2 H₂O(aq)

I hope it helps!

Answer from: Quest

120,630,890

step-by-step explanation:

110+10=120,620+10=630,880+10=890

Answer from: Quest

when the volume of gas is changed from 717.73 ml to 852 ml the temperature will change from 315 c to 452 c

calculation

this is calculated using the charles law formula

that is, v1/t1 = v2/t2

v1=?

t1 = 315 c into kelvin = 315 +273 = 588 k

v2 = 852 ml

t2 = 425c + 273 =698 k

make v1 the subject of the formula ( by multiplying both side by t1)

v1=( v2 xt1) / t2

v1= (852 ml x 588 k)/ 698 k= 717.73 ml

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