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crimeas [40]
3 years ago
12

which of these equations are balance a) 3 H2SO + 2AI = AI2(SO4)3 3H2 b) 2KCI + Pb(NO3)2 = 2KNO3 + PbCI2

Chemistry
1 answer:
mihalych1998 [28]3 years ago
6 0

Both equations are balanced.

a) <em>3H₂SO₄ + 2Al ⟶ Al₂(SO₄)₃ + 3H₂ </em>

<u>On the left</u> <u>On the right</u>

      6H                6H

      3S                 3S

    12O               12O

     2Al                2Al

b) <em>2KCl + Pb(NO₃)₂ ⟶ 2KNO₃ + PbCl₂ </em>

<u>On the left</u>   <u>On the right </u>

      2K                 2K

      2Cl               2Cl

      1Pb                1Pb

     2N                 2N

     6O                 6O

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Answer:

See Explanation

Explanation:

In thermodynamics theory the Free Energy (ΔG) of a chemical system is described by the expression ΔG = ΔG° + RTlnQ. When chemical system is at equilibrium ΔG = 0. Substituting into the system expression gives ...

0 = ΔG° + RTlnKc, which rearranges to ΔG° = - RTlnKc.  ΔG° in electrochemical terms gives ΔG° = - nFE°, where n = charge transfer, F = Faraday Constant = 96,500 amp·sec and E° = Standard Reduction Potential of the electrochemical system of interest.

Substituting into the ΔG° expression above gives

-nFE°(cell) = -RTlnKc => E°(cell) = (-RT/-nF)lnKc = (2.303·R·T/n·F)logKc

=> E°(cell) = (0.0592/n)logKc = E°(Reduction) - E°(Oxidation)

Application example:

Calculate the Kc value for a Zinc/Copper electrochemical cell.

Zn° => Zn⁺² + 2e⁻  ;    E°(Zn) = -0.76 volt  

Cu° => Cu⁺² + 2e⁻ ;    E°(Cu) =  0.34 volt

By natural process, charge transfer occurs from the more negative reduction potential to the more positive reduction potential.

That is,

           Zn° => Zn⁺² + 2e⁻ (Oxidation Rxn)

Cu⁺² + 2e⁻ => Cu°             (Reduction Rxn)

E°(Zn/Cu) = (0.0592/n)logKc

= (0.0592/2)logKc = E°(Cu) - E°(Zn) = 0.34v - (-0.76v) = 1.10v

=> logKc = 2(1.10)/0.0592 = 37.2

=> Kc = 10³⁷°² = 1.45 x 10³⁷

3 0
3 years ago
Cylinder of air at 1.5 atm of pressure is kept at room temperature while a piston compresses the air from 40 l down to 10 ml. wh
djyliett [7]

The new pressure, P₂ is 6000 atm.

<h3>Calculation:</h3>

Given,

P₁ = 1.5 atm

V₁ = 40 L = 40,000 mL

V₂ = 10 mL

To calculate,

P₂ =?

Boyle's law is applied here.

According to Boyle's law, at constant temperature, a gas's volume changes inversely with applied pressure.

                                        PV = constant

Therefore,

P₁V₁ = P₂V₂

Put the above values in the equation,

1.5 × 40,000 = P₂ × 10

P₂  = 1.5 × 4000

P₂  = 6000 atm

Therefore, the new pressure, P₂ is 6000 atm.

Learn more about Boyle's law here:

brainly.com/question/23715689

#SPJ4

8 0
2 years ago
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