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3 years ago

How much energy is needed to change 44 grams of ice at -8° to steam at 107°?

1 answer:

5 0

Q=m(c∆t +heat of fusion + heat of evaporation)

m= 44g
c= 4.186 J/g.C
∆t= 107-(-8) =115 C
heat of fusion= 333.55 J/g
heat of evaporation=2260 J/g

Q=44(4.186*115 + 333.55 + 2260)
Q= 135297.36 J

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Balance the following redox equations by the half-reaction method:

frozen [14]

Answer:

a) Mn^2+ + 2OH- + H2O2 → MnO2 + 2H2O

b) 2Bi(OH)3 + 3SnO2^2- → 2Bi + 3SnO3^2- + 3H2O

c) Cr2O7^2- + 3C2O4^2- + 14H+ → 2Cr^3+ +7H2O + 6CO2

d) 2ClO3- + 2Cl- + 4H+ → Cl2 + 2H2O +2ClO2

e) 5 BiO3^- + 14 H+ + 2Mn^2+ → 5Bi^3+ + 7H2O + 2MnO4^-

Explanation:

(a) Mn2 + H2O2 → MnO2 + H2O (in basic solution)

Step 1: The half reactions

Oxidation: Mn2+ + 4OH- → MnO2 + 2H2O + 2e-

Reduction: H2O2 + 2e- + 2H2O → 2H2O + 2OH-

Step 2: Sum of both half reactions

Mn2+ + 4OH- + H2O2 → MnO2 + 2H2O + 2OH-

Step 3: the netto reaction

Mn^2+ + 2OH- + H2O2 → MnO2 + 2H2O

(b) Bi(OH)3 + SnO2^2- → SnO3^2- + Bi (in basic solution)

Step 1: The half reactions

Reduction: Bi(OH)3 + 3e- → Bi

Oxidation : Sno2^2- → SnO3^2- +2e-

Step 2: Balance the half reactions

2* (Bi(OH)3 + 3e- → Bi + 3OH-)

3* (Sno2^2- +2OH- → SnO3^2- +2e- + H2O)

Step 3: The netto reaction

2Bi(OH)3 + 3SnO2^2- → 2Bi + 3SnO3^2- + 3H2O

(c) Cr2O7^2- + C2O4^2- → Cr^3+ + CO2 (in acidic solution)

Step 1: The half reactions

Reduction: Cr2O7^2- + 6e- → 2Cr+

Oxidation : C2O4^2- → 2CO2 + 2e-

Step 2: Balance the half reactions

Cr2O7^2- + 6e- +14H+ → 2Cr+ +7H2O

3*(C2O4^2- → 2CO2 + 2e-)

Step 3: The netto reaction

Cr2O7^2- + 3C2O4^2- + 14H+ → 2Cr^3+ +7H2O + 6CO2

(d) ClO3^- + Cl^− → Cl^2 + ClO^2 (in acidic solution)

Step 1: The half reactions

Reduction: 2 ClO3^- + 10e- → Cl2

ClO3^- + e- → ClO2

2 Cl- + 2ClO3^- +8e- →2Cl2

Oxidation: 2Cl- → Cl2 + 2e-

Cl- → ClO2 + 5e-

Cl- +ClO3^- → 2ClO2 + 4e-

Step 2: Balance the reactions

2Cl- + 2ClO3^- + 8e- + 12H+ → 2Cl2 + 6H2O

2* (Cl- + ClO3^- + H2O → 2ClO2 + 4e- + 2 H+)

Step 3: The netto reaction

2ClO3^- + 2Cl- + 4H+ → Cl2 + 2H2O +2ClO2

(e) Mn^2 + BiO3^− → Bi^3 + MnO^4− (in acidic solution)

Step 1: The half reactions

Reduction: BiO3^- + 2e- → Bi^3+

Oxidation : Mn^2+ → MnO4^- +5e-

Step 2: Balanced the reactions

5* ( BiO3^- + 2e- + 6H+ → Bi^3+ + 3H2O)

2* ( Mn^2+ + 4H2O →MnO4^- + 5e- + 8H+)

Step 3: The netto reaction

5 BiO3^- + 14 H+ + 2Mn^2+ → 5Bi^3+ + 7H2O + 2MnO4^-

6 0

4 years ago