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Nostrana [21]
3 years ago
11

A 3.0-L sample of helium was placed in container fitted with a porous membrane. Half of the helium effused through the membrane

in 24 hours. A 3.0-L sample of oxygen was placed in an identical container. How many hours will it take for half of the oxygen to effuse through the membrane?
Chemistry
1 answer:
zhenek [66]3 years ago
3 0

Explanation:

It is known that rate of effusion of gases are inversely proportional to the square root of their molar masses.

And, half of the helium (1.5 L) effused in 24 hour. So, the rate of effusion of He gas is calculated as follows.

          \frac{1.5 L}{24 hr}

            = 0.0625 L/hr

As, molar mass of He is 4 g/mol  and molar mass of O_{2} is 32 g/ mol.

Now,

   \frac{\text{Rate of He}}{\text{Rate of Oxygen}} = \sqrt{\frac{32}{4}

                               = 2.83

or, rate of O_{2} = \frac{\text{Rate of He}}{2.83}

                       = \frac{0.0625 L/hr}{2.83}

          Rate of O_{2} = 0.022 L/hr.

This means that 0.022 L of O_{2} gas effuses in 1 hr

So, time taken for the effusion of 1.5 L of O_{2} gas is calculated as follows.

         \frac{1.5 L}{0.022 L/hr}

                = 68.18 hour

Thus, we can conclude that 68.18 hours will it take for half of the oxygen to effuse through the membrane.

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<u>Answer:</u> The cell potential of the cell is +0.118 V

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<u>Reduction half reaction (cathode):</u>  Ni^{2+}+2e^-\rightarrow Ni(s)

In this case, the cathode and anode both are same. So, E^o_{cell} will be equal to zero.

To calculate cell potential of the cell, we use the equation given by Nernst, which is:

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