2 years ago

St2 Second Laws

Chemistry

1 answer:

jeyben [28]2 years ago

7 0

Answer: b. 606

Explanation:

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weeeeeb [17]

Answer:

Therefore, the rate of change in the amount of salt is $\frac{dx}{dt} =( 5c}{ - \frac{x }{20})$

$\frac{grams }{min}$

Explanation:

Given:

Initial volume of water $V = 35$ lit

Flowing rate = 5 $\frac{Lit}{min}$

The rate of change in the amount of salt is given by,

$\frac{dx}{dt} =$ ( Rate of salt enters tank - rate of sat leaves tank )

Since tank is initially filled with water so we write that,

$x(0) = 0$

Let amount of salt in the solution is $c$ ,

$\frac{dx}{dt} = \frac{5c}{1 } - \frac{x(t) \times 5}{100}$

$\frac{dx}{dt} =( 5c}{ - \frac{x }{20})$ $\frac{grams}{min}$

Therefore, the rate of change in the amount of salt is $\frac{dx}{dt} =( 5c}{ - \frac{x }{20})$

$\frac{grams }{min}$

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3 years ago