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Ivan
3 years ago
13

A chemistry student weighs out 0.027 kg of an unknown solid compound X and adds it to 550. mL of distilled water at 30.° C. Afte

r 10 minutes of stirring, only some of the X has dissolved. The student drains off the solution, then washes, dries and weighs the X that did not dissolve. It weighs 0.008 kg.
1) Using the information above, can you calculate the solubility of X?
2) If so, calculate it. Remember to use the correct significant digits and units.
Chemistry
1 answer:
QveST [7]3 years ago
5 0

Answer:

1. Yes

2.The solubility of X is 34.55g/L

Explanation:

Solubility of  solute refers to how readily a solute will dissolve in a solvent at a particular temperature. Its the amount of moles or grams required to saturate 1dm^{3} or 1 Litre of water.

From the problem, when the liquid was drained off and amount of X which didn't dissolve was measured, it weighed 0.008kg, this means out of 0.027kg, 0.027-0.008 actually dissolved

= 0.019kg*1000 = 19g.

if 19g is required to saturate 550mL at 30°C,

then\frac{19*1000}{550} will saturate 1L

= 34.545g  will saturate 1Litre

The solubility thus is 34.55g/L

 

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4. How many grams of ammonium carbonate are needed to decompose in order to produce
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Answer:

14.23g of (NH4)2CO3

Explanation:

We'll begin by writing the balanced equation for the reaction.

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Next,, we shall determine the mass of (NH4)2CO3 that decomposed and the mass of CO2 produced from the balanced equation. This is illustrated below:

Molar mass of (NH4)2CO3 = 2[14+(4x1)] + 12 + (16x3)

= 2[14 +4] + 12 + 48

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Mass of (NH4)2CO3 from the balanced equation = 1 x 96 = 96g

Molar mass of CO2 = 12 + (2x16) = 44g/mol

Mass of CO2 from the balanced equation = 1 x 44 = 44g.

Summary:

From the balanced equation above,

96g of (NH4)2CO3 decomposed to produce 44g of CO2.

Finally, we can determine the mass of (NH4)2CO3 that decomposed to produce 6.52g of CO2 as follow:

From the balanced equation above,

96g of (NH4)2CO3 decomposed to produce 44g of CO2.

Therefore, Xg of (NH4)2CO3 will decompose to produce 6.52g of CO2 i.e

Xg of (NH4)2CO3 = (96 x 6.52)/44

Xg of (NH4)2CO3 = 14.23g

Therefore, 14.23g of (NH4)2CO3 is needed to produce 6.52g of CO2.

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