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LenKa [72]
3 years ago
15

Given the concentrations, calculate the equilibrium constant for this reaction: I2(g) +Cl2(g)⇌2ICl(g) At equilibrium, the mola

r concentrations for reactants and products are found to be [I2]=0.50 M, [Cl2]=0.60 M, and [ICl]=5.0 M. What is the equilibrium constant (Kc) for this reaction?
Chemistry
1 answer:
lawyer [7]3 years ago
6 0

Answer:

use free science lessons

Explanation:

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A galvanic cell whose cell reaction is 2Fe3+(aq) + Zn(s) → 2Fe2+(aq) + Zn2+(aq) has a cell potential of 0.72V. What is the maxim
sesenic [268]

Answer:

138.96kJ is the maximum electrical work

Explanation:

The maximum electrical work that can be obtained from a cell is obtained from the equation:

W = -nFE

<em>Where W is work in Joules,</em>

<em>n are moles of electrons = 2mol e- because half-reaction of Zn is:</em>

Zn(s) → Zn²⁺(aq) + 2e⁻

F is faraday constant = 96500Coulombs/mol

E is cell potential = 0.72V

Replacing:

W = -2mol*96500Coulombs/mol*0.72V

W = - 138960J =

<h3>138.96kJ is the maximum electrical work</h3>

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2 years ago
If a balloon initially has a volume of 4.0 liters and a temperature of 10 degrees Celsius, what will the volume of the balloon b
Anika [276]
The answer is D. 8.1L
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Which chemical property makes a substance suitable as a fuel?
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3 years ago
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Fe + O2 → Fe2O3

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Hope this will help u mate :)

4 0
3 years ago
The rate constant for the reaction 3A equals 4b is 6.00×10 how long will it take the concentration of a to drop from 0.75 to 0.2
Lelu [443]

This question is incomplete, the complete question is;

The rate constant for the reaction 3A equals 4B is 6.00 × 10⁻³ L.mol⁻¹min⁻¹.

how long will it take the concentration of A to drop from 0.75 to 0.25M ?

from the unit of the rate constant we know it is a second reaction order

OPTIONS

a) 2.2×10^−3 min

b) 5.5×10^−3 min

c) 180 min

d) 440 min

e) 5.0×10^2 min

Answer:

it will take 440 min for the concentration of A to drop from 0.75 to 0.25M

Option d) 440 min is the correct answer

Explanation:

Given that;

Rate constant K =  6.00 × 10⁻³ L.mol⁻¹min⁻¹

3A → 4B

given that it is a second reaction order;

k = 1/t [ 1/A - 1/A₀]

kt = [ 1/A - 1/A₀]

t = [ 1/A - 1/A₀] / k

K is the rate constant(6.00 × 10⁻³)

A₀ is initial concentration( 0.75 )

A is final concentration(0.25)

t is time required = ?

so we substitute our values into the equation

t = [ (1/0.25) - (1/0.75)] / (6.00 × 10⁻³)

t = 2.6666 / (6.00 × 10⁻³)

t = 444.34 ≈ 440 min     {significant figures}

Therefore it will take 440 min for the concentration of A to drop from 0.75 to 0.25M

Option d) 440 min is the correct answer

6 0
3 years ago
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