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3 years ago

What is the value of the equilibrium constant for this redox reaction?

2 answers:

8 0

Correct answer: Option D, <span>K = 5.04 × 10^52
</span>
Reason:
We know that,
Ecell = $\frac{0.0592}{n}log(K)$ ,
where n = number of electrons = 2 (in present case)
K = equilibrium constant.

Also, Ecell = <span>+1.56 v

Therefore, 1.56 = </span> $\frac{0.0592}{2}log(K)$
Therefore, log (K) = 52.703
Therefore, K = 5.04 X 10^52

nignag [31]3 years ago

3 0

D for all the plato homies

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3 years ago

Read 2 more answers

A sample of hydrogen was collected by water displacement at 23.0°C and an atmospheric pressure of 735 mmHg. Its volume is 568 mL

ziro4ka [17]

Answer:

V = 552 mL or 0.552 L

Explanation:

First, we need to calculate the number of moles of H2 using the ideal gas equation which is:

PV = nRT

Solving for n:

n = PV / RT

Where:

P = Pressure

V = Volume

R = Gas constant (0.082 L atm / K mol)

T = Temperature in K

Let's convert first both pressure in atm, remember that 1 atm = 760 mmHg

P = 735 / 760 = 0.967 atm

Pwater = 21 / 760 = 0.028 atm

Finally temperature to Kelvin:

T = 23 + 273.15 = 296.15 K

Now, at first the hydrogen was collected by water displacement so pressure is:

P = 0.967 - 0.028 = 0.939 atm

Now the moles of hydrogen:

n = 0.939 * 0.568 / 0.082 * 296.15

n = 0.022 moles

Now that we have the moles, let's calculate the volume when the pressure is 735 mmHg

V = nRT/P

V = 0.022 * 0.082 * 296.15 / 0.967

V = 0.552 L or 552 mL

This is the volume that hydrogen occupies.

6 0

3 years ago