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Tresset [83]
3 years ago
14

In a mixture of He, O, and N, gases He exerts a partial pressure of 15.22 mm Hg and

Chemistry
1 answer:
gayaneshka [121]3 years ago
7 0

Answer:

P(N)  = 38.48 mmHg

Explanation:

Given data:

Partial pressure of He = 15.22 mmHg

Partial pressure of O = 35.21 mmHg

Partial pressure of N = ?

Total pressure = 88.91 mmHg

Solution:

According to Dalton law of partial pressure,

The total pressure inside container is equal to the sum of partial pressures of individual gases present in container.

Mathematical expression:

P(total) = P₁ + P₂ + P₃+ ............+Pₙ

Now we will solve this  problem by using this law.

P(total) = P(He) + P(O) + P(N)

88.91 mmHg = 15.22 mmHg + 35.21 mmHg + P(N)

88.91 mmHg = 50.43 mmHg + P(N)

P(N)  = 88.91 mmHg  - 50.43 mmHg

P(N)  = 38.48 mmHg

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Explanation :

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Initial pressure     8.00      5.00            0

At eqm.               (8.00-x) (5.00-x)        2x

The expression of equilibrium constant K_p for the reaction will be:

K_p=\frac{(p_{NO})^2}{(p_{N_2})(p_{O_2})}

Now put all the values in this expression, we get :

0.0025=\frac{(2x)^2}{(8.00-x)\times (5.00-x)}

By solving the terms, we get:

x=0.15atm

The equilibrium partial pressure of N_2 = (8.00 - x) = (8.00 - 0.15) = 7.8 atm

Therefore, the equilibrium partial pressure of N_2 is 7.8 atm.

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Since the half-life of 235U (7. 13 x 108 years) is less than that of 238U (4.51 x 109 years), the isotopic abundance of 235U has
Ymorist [56]

Answer:

\mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}

Explanation:

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The Half-life of ^{235}U = 7.13 \times 10^8 \ years is less than that of ^{238} U = 4.51 \times 10^9 \ years

Although we are not given any value about the present weight of ^{235}U.

So, consider the present weight in the percentage of ^{235}U to be  y%

Then, the time elapsed to get the present weight of ^{235}U = t_1

Therefore;

N_1 = N_o e^{-\lambda \ t_1}

here;

N_1 = Number of radioactive atoms relating to the weight of y of ^{235}U

Thus:

In( \dfrac{N_1}{N_o}) = - \lambda t_1

In( \dfrac{N_o}{N_1}) =  \lambda t_1 --- (1)

However, Suppose the time elapsed from the initial stage to arrive at the weight of the percentage of ^{235}U to be = t_2

Then:

In( \dfrac{N_o}{N_2}) =  \lambda t_2  ---- (2)

here;

N_2 =  Number of radioactive atoms of ^{235}U relating to 3.0 a/o weight

Now, equating  equation (1) and (2) together, we have:

In( \dfrac{N_o}{N_1}) -In( \dfrac{N_o}{N_2}) =  \lambda( t_1-t_2)

replacing the half-life of ^{235}U = 7.13 \times 10^8 \ years

In( \dfrac{N_2}{N_1})  = \dfrac{In 2}{7.13 \times 10^9}( t_1-t_2)      ( since \lambda = \dfrac{In 2}{t_{1/2}} )

∴

\mathtt{In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2}= t_1-t_2}

The time elapsed signifies how long the isotopic abundance of 235U equal to 3.0 a/o

Thus, The time elapsed is  \mathtt{ t_1-t_2= In(\dfrac{3}{y}) \times \dfrac{7.13 \times 10^8}{In2} \ years}

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