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Irina18 [472]
3 years ago
14

Complete and balance the following redox reaction in basic solution

Chemistry
1 answer:
inn [45]3 years ago
8 0

Answer:

balanced in ACID not BASE

Cr2O7^2-(aq) +3Hg(l) +14 H^1+ ----> 3Hg^2+ + 2Cr^3+(aq) + 7H2O

Answer

Cr2O7^2-(aq) +3Hg(l) +14 H^1+ ----> 3Hg^2+ + 2Cr^3+(aq) + 7H2O

Explanation:

Cr2O7^2-(aq) + Hg(l) ----> Hg^2+(aqH) + Cr^3+(aq)

add H^1+ (acid) to capture the O and make 7 water molecules

Cr2O7^2-(aq) + Hg(l) + H^1+ ----> Hg^2+(aqH) + Cr^3+(aq) + 7H2O

Cr goes from +6 to +3 by gaining 3 e

Hg goes from 0 to +2 by losing 2 e

we need 3 Hg for every 2 Cr

so

Cr2O7^2-(aq) +3Hg(l) +14 H^1+ ----> 3Hg^2+ + 2Cr^3+(aq) + 7H2O

2 Cr on the right and left

Net 12 positive charges on the right and the left

3 Hg on the right and left

14 H on the right and left

the equation is balanced

we cannot balance the equation in a basic solution with OH^1-

we have plenty of O in the dichromate ion. we need to convert it to water which take free H^1+ from the acid

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From the data below, calculate the total heat (in j) needed to convert 0.782 mol of gaseous ethanol at 300.0°c and 1 atm to liqu
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Answer:

You must remove \text{50.6 kJ} .

Explanation:

There are three heat transfers in this process:

Total heat = cool the vapour + condense the vapour + cool the liquid  

       q          =           q₁            +                q₂                   +           q₃

       q          =       nC₁ΔT₁        +          nΔHcond             +        nC₂ΔT₂

Let's calculate these heat transfers separately.

Data:

You don't give "the data below", so I will use my best estimates from the NIST Chemistry WebBook. You can later substitute your own values.

C₁ = specific heat capacity of vapour = 90 J·K⁻¹mol⁻¹

C₂ = specific heat capacity of liquid   = 115 J·K⁻¹mol⁻¹

ΔHcond = -38.56 kJ·mol⁻¹

Tmax = 300   °C

  b.p. =   78.4 °C

Tmin =   25.0 °C

n = 0.782 mol

Calculations:

ΔT₁ = 78.4 - 300 = -221.6 K

q₁ = 0.782 × 90 × (-221.6) = -15 600 J = -15.60 kJ

q₂ = 0.782 × (-38.56) = -30.15 kJ

ΔT = 25.0 - 78.4 = -53.4 K

q₃ = 0.782 × 115 × (-53.4) = -4802 J = 4.802 kJ

q = -15.60 - 53.4 - 4.802 = -50.6 kJ

You must remove \text{50.6 kJ} of heat to convert the vapour to a gas.

8 0
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