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3 years ago

How long will a current of 0.995 A need to be passed through water (containing H2SO4) for 5.00 L of O2 to be produced at STP

Chemistry

1 answer:

DIA [1.3K]3 years ago

3 0

Answer:

24 hours

Explanation:

The computation is shown below:

The needed mole of $O_2$ is

= 5 ÷22.4 = n

Also 1 mole of $O_2$ required four electric charge

Now the charge needed is

= n × 4 × 96,500 C

= 4 × 96,500 × 5 c ÷ 22.4

= 86160.714 C

Now

q = i t

t = q ÷ i

= 86160.714 C ÷ 0.995

= 86593.7 seconds

= 24 hours

Hence, the correct option is A.

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Explanation:

As we can see, we have been a particular reaction and Energy values as well.

ΔG° of reaction = -30.5 kJ/mol

Temperature = 37°C.

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The first step is to calculate the equilibrium constant for the reaction:

Equilibrium Constant K = $\frac{[HPO4-2] x [ADP]}{ATP}$

And we have values given for these quantities in the biological cell:

[HP04-2] = 2.1 x $10^{-3}$ M

[ATP] = 1.2 x $10^{-2}$ M

[ADP] = 8.4 x $10^{-3}$ M

Let's plug in these values in the above equation for equilibrium constant:

K = $\frac{[2.1x10^{-3}] x [8.4x10^{-3}] }{[1.2 x 10^{-2}] }$

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Now, we have to calculate the ΔG° of reaction for the biological cell:

But first we have to convert the temperature in Kelvin scale.

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Temp = 37 + 273

Temp = 310 K

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Where 8.314 = value of Gas Constant

ΔG° of reaction = (-30.5 x $10^{3}$ ) + (-16810.68)

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