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Margarita [4]
3 years ago
15

4. The solubility of acetanilide in hot water is 5.5 g / 100 mL at 100oC and its solubility in cold water is 0.53 g / 100 mL at

0oC. What would be the maximum theoretical percent recovery from the crystallization of 2.5 g of acetanilide from 50 mL of water? Show the calculation.
Chemistry
1 answer:
Triss [41]3 years ago
7 0

Answer:

89,4%

Explanation:

If you have a solutio of 2,5g of acetanilide in 50mL of water and you warm this solution to 100°C you will dissolve all acetanilide because the maximum solubility in 50mL will be:

5,5g / 100mL → 2,75g / 50mL.

Then, if you cold the water to 0°C the solubility in 50mL will be:

0,53g / 100mL → 0,265g / 50mL.

That means you will precipitate:

2,5g - 0,265g = <em>2,235g of acetanilide</em>

The theoretical percent recovery will be:

2,2365g / 2,5g ×100 = <em>89,4%</em>

<em></em>

I hope it helps!

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What is the mass percentage of O in C₂H₄O₂? Provide an answer to two decimal places.
nasty-shy [4]

Answer:

53.29%

Explanation:

The molar mass of C2H4O2 is 60.05g and the 2 O's are 32.00g

so 32.00/60.05= 0.53288925895

and that as a decimal rounded to the nearest hundredths is 53.29%

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How many molecules are present in 2.00 mol of KCl?
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Answer:

The question specified 2 moles of potassium chloride. Clearly, there are 2 moles of chloride ions. How many moles of chloride ions in 95.2 g

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Which statement accurately describes part of the dissolving process of a polar solute in water? Solute molecules repel water mol
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Its water molecules surround solute molecules

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Which crystalline solds generally have the highest melting and boiling points?
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Metallic (w) with a boiling point of 566”
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Gaseous methane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Supposed 0.481 g of methane
Shalnov [3]

Answer:

1.08g

Explanation:  

Like all other hydrocarbons, methane burns in oxygen to form carbon iv oxide and water. The chemical equation of this equation is shown below;

  CH4 (g)  +  2O2  (g) ----------->  2H2O (l)   +   CO2 (g)

From the reaction equation, we can see that one mole of methane gave 2  moles of water. This is the theoretical yield. We need to note the actual yield.

To get the actual yield, we get the number of moles of methane reacted. To get this, we divide the mass of methane by  the molar mass of methane. The molar mass of methane is 16g/mol. The number of moles is thus 0.481/16 = 0.03 moles.

Since 1 mole methane gave 2 moles of water, this shows that 0.06 moles of water were produced. The mass of water thus produced is 0.06 multiplied by the molar mass of water. The molar mass of water is 18g/mol. The mass produced is 0.06 * 18 = 1.08g

Now, we do same for the mass of oxygen. From the reaction equation, 2 moles of oxygen produced two moles of water. Hence, we can see from here that the number of moles here are equal. We then proceed to calculate the actual number of moles of oxygen produced. This is the mass of the oxygen divided by the molar mass of molecular oxygen. The molar mass of molecular oxygen is 32g/mol. The number of moles thus produced is 0.54/32 = 0.016875 mole. The number of moles are equal, this means that the number of moles of oxygen produced is also 0.016875

Now, to get the mass of water produced, we multiply the number of moles by the molar mass of water. The molar mass of water is 18g/mol.

The mass is thus, 0.016875 * 18 = 0.30375g

1.08g is higher and thus is the maximum mass

3 0
3 years ago
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