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3 years ago

3. Your company makes 1000 boxes of 500 g baking soda per day.

Chemistry

1 answer:

inessss [21]3 years ago

5 0

500,000 g of baking soda is present in 1000 boxes of 500 g baking soda boxes.

Answer:

Option C.

Explanation:

As 500 g of baking soda is taken in each box of that company. The total weight of baking soda in all the boxes can be determined by adding the weights of each box. This is possible only when the number of boxes is less. But if the number of boxes are large, then we can determine the total weight of baking soda by multiplying the number of boxes with the weight in each box.

So in this case, 1000 boxes are present and in that 500 g of baking soda are present in each box.

So total grams of baking soda will be 1000 * 500 = 5,00,000 g.

Thus, 500,000 g of baking soda is present in 1000 boxes of 500 g baking soda boxes.

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<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>

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given amount of salt at time t is A(t)

initial amount of salt =300 gm =0.3kg

=>A(0)=0.3

rate of salt inflow =5*0.4= 2 kg/min

rate of salt out flow =5*A/(200)=A/40

rate of change of salt at time t , dA/dt= rate of salt inflow- ratew of salt outflow

$dA/dt=2-(A/40)\\\\dA=2dt-(A/40)dt\\\\dA+(A/40)dt=2dt$

integrating factor

$=e^{\int\limits (1/40) \, dt}$

integrating factor $=e^{(1/40)t}$

multiply on both sides by $=e^{(1/40)t}$

$dAe^{(1/40)t}+(A/40)e^{(1/40)t} dt =2e^{(1/40)t}t\\\\(Ae^{(1/40)t})=2e^{(1/40)t}t$

integrate on both sides

$\int\limits(Ae^{(1/40)t})=\int\limits2e^{(1/40)t}dt\\\\(Ae^{(1/40)t})=2*40e^{(1/40)t}+C\\\\A=80+(C/e^{(1/40)t})\\\\A(0)=0.3\\\\0.3=80+(C/e^{(1/40)t}^*^0)\\\\0.3=80+(C/1)\\\\C=0.3-80\\\\C=-79.7\\\\A(t)=80-(79.7/e^{(1/40)t})$

after long period of time means t - > ∞

${t \to \infty}\\\\ \lim_{t \to \infty} A_t \\\\ \lim_{t \to \infty} (80)-(79/{e^{(1/40)t}}\\\\=80-(0)\\\\=80$

<h3>Therefore, after long period of time 80kg of salt will remain in tank</h3>

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