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3 years ago

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Chemistry

1 answer:

Art [367]3 years ago

6 0

Answer:

1) saturated

2) unsaturated

3) saturated

4) unsaturated

5) saturated

6) saturated

7) saturated

8) unsaturated

9) saturated

10) saturated

40g of saturated strawberry powder in 100 ml of milk is more soluble.

Explanation:

We define a saturated solution as a solution that holds just as much solute as it can normally hold at a given temperature. Hence, a saturated solution. is unable to dissolve more solute and undissolved solutes begins to appear or gasses are given off.

An unsaturated solute contains less solute than it can normally hold at that particular temperature hence it can still dissolve more solute.

In the closure, we must note that the greater the volume of solvent, the greater the solubility of the solute. Hence it follows that; 40g of saturated strawberry powder in 100 ml of milk is more soluble.

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For parts a & b below, derive only the initial value problem set up.

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Answer:

Explanation:

From the given information:

Let y(t) be the amount of sugar in tank A at any time.

Then, the rate of change of sugar in the tank is given by:

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$So; \\ \\ \\ \dfrac{dy}{dt} = 0 - \dfrac{y}{100} \\ \\ \implies \dfrac{dy}{y }= -\dfrac{dt}{100 } \\ \\ \implies dx|y| = -\dfrac{t}{100}+C \\ \\ \implies e*{In|y|}= e^{-\dfrac{t}{100}+C} \\ \\ \implies y = Ce^{-\dfrac{t}{100}}$

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Now; y(0) = 25

25 = Ce⁰

C = 25

So; y(t) = 25 $e^{-\dfrac{t}{100}$

Thus, the amount of sugar at any time t = $\mathbf{25 e^{^{-\dfrac{t}{100}}}}$

B) For tank B :

$\dfrac{dy}{dt}= 1 - \dfrac{y}{50 } \\ \\ \dfrac{dy}{dt}= \dfrac{50-y}{50} \\ \\ \dfrac{dy}{50-y }= \dfrac{dt}{50}$

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0.236 gm of Metal X reacts with O2 forming 0.436 of the compound X2O3 what is the gram atomic mass of x?

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Explanation:

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$No. of moles = \frac{mass}{molarmass}\\= \frac{10g}{40 g/mol}\\= 0.25 mol$

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