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Nataly [62]
4 years ago
5

A tank contains 50 kg of salt and 1000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drain

s from the tank at the rate 3 L/min. (a) What is the amount of salt in the tank initially
Chemistry
1 answer:
Studentka2010 [4]4 years ago
7 0

Answer : The amount of salt present in the tank initially are, 50 kg

Explanation : Given,

Mass of salt in the tank = 50 kg

Mass of water in the tank = 1000 L

Rate of flow of water in the tank = 6 L/min

Rate of flow of solution in the tank = 3 L/min

Now we have to determine the amount of salt present in the tank initially.

As per question, there are 50 kg of salt present in the tank initially in 1000 L of water.

Thus, the amount of salt present in the tank initially are, 50 kg

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Answer:

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

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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}} )

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\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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