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alexandr1967 [171]
3 years ago
8

For each described change, determine the generally expected impact on a salt's solubility. LABEL No solubility change, SOLUBILIT

Y DECREASE OR SOLUBILITY INCREASE
A- Cooling a saturated solution by 10 oC
B- Placing a saturated solution in a hot water bath
C- Doubling the amount of salt added to a saturated solution
Chemistry
1 answer:
Anna [14]3 years ago
4 0

Answer:

A- Solubility decrease.

B- Solubility increase.

C- No solubility change.

Explanation:

Hello!

In this case, since the solubility of salt stands for the maximum amount of salt that can be added to a specific mass of water, usually 100 g; we need to take into account that for table salt in aqueous solution, the higher the temperature the larger the solubility and the lower the temperature the smaller the solubility; it means that more salt is dissolved in the same mass of water at higher temperatures and vice versa. Therefore, A- would decrease the solubility as the solution is cooled down and B- would increase the solubility as the solution is heated up.

Moreover, since the mass of water is assumed to remain the same, adding more salt do not affect the solubility but increase the degree of saturation of the solution up to supersaturated, yet the solubility remains unchanged.

Best regards!

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A mixture containing 20 mole % butane, 35 mole % pentane and rest
notka56 [123]

Answer:

2.5 % butane, 42.2 % pentane and 55.3 % hexane

Explanation:

Hello,

In this case, the mass balance for each substance is given by:

Butane:z_bF=y_bD+x_bB\\\\Pentane: z_pF=y_pD+x_pB\\\\Hexane: z_hF=y_hD+x_hB

Whereas y accounts for the fractions at the outlet distillate and x for the fractions at the outlet bottoms. Moreover, with the 90 % recovery of butane, we can write:

0.9=\frac{y_bD}{z_bF}

So we can compute the product of the molar fraction of butane at the distillate by total distillate flow by assuming a 100-mol feed:

y_bD=0.9*z_bF=0.9*0.2*100mol=18mol

The total distillate flow:

y_bD=18mol\\\\D=\frac{18mol}{0.95} =18.95mol

And the total bottoms flow:

F=D+B\\\\B=F-D=100mol-18.95mol=81.05mol

Next, by using the mass balance of butane, we compute the molar fraction of butane at the bottoms:

x_b=\frac{z_bF-y_bD}{B} =\frac{0.2*100mol-18mol}{81.05} =0.025

Then, the molar fraction of pentane and hexane:

x_p=\frac{z_pF-y_pD}{B} =\frac{0.35*100mol-0.04*18.95mol}{81.05} =0.422

x_h=\frac{z_hF-y_hD}{B} =\frac{(1-0.2-0.35)*100mol-(1-0.95-0.04)*18.95mol}{81.05} =0.553

Therefore, the molar composition of the bottom product is 2.5 % butane, 42.2 % pentane and 55.3 % hexane.

NOTE: notice the result is independent of the value of the assumed feed, it means that no matter the basis, the compositions will be the same for the same recovery of butane at the feed, only the flows will change.

Regards.

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