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Volgvan
2 years ago
11

The solubility of acetanilide in hot water (5.5 g/100 ml at 100 C) is not very great, and it has some solubility in cold water (

0.53 g/100 ml at 0 C). What would be the maximum theoretical percent recovery from the crystallization of 4.0 g of acetanilide from 100 ml of water (assuming the solution is chilled to 0 C for filtration). Show your calculations.
Chemistry
2 answers:
sashaice [31]2 years ago
5 0

To answer the problem given:

|0.53−4.0| / 4.0 * 100%

= 3.47 / 4.0 * 100%

= 87%

The maximum theoretical percent recovery from the crystallization of 4.0 g of acetanilide from 100 ml of water is 87%. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

son4ous [18]2 years ago
4 0

Answer:

Maximum theoretical percent recovery of acetanilide is 86.8%

Explanation:

From the given information:

Solubility of acetanilide in 100 ml water at 100 C = 5.5 g

Solubility of acetanilide in 100 ml water at 0 C = 0.53 g

When 4.0 g of acetanilide is dissolved in 100 ml of water, all of it will go into solution. As the solution is cooled to 0 C only 0.53 g of it will dissolve.

Amount of acetanilide undissolved = 4.0 - 0.53 = 3.47 g\\\\Percent\ recovery = \frac{3.47g}{4.00} *100=86.75% (86.8%\ approx)

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0.45 g of hydrated sodium carbonate crystals were heated until 3.87 of anhydrous power remained.
snow_tiger [21]

Formula of hydrated sodium carbonate : Na₂CO₃.10H₂O, so moles of water in one mole of hydrated salt = 10

<h3>Further explanation</h3>

Hydrate is a compound that binds water (H₂O), usually in the form of crystals/ solids

If these compounds are dissolved in water or heated, the hydrates can decompose:

Example: X.YH₂O (s) → X (aq) + YH₂O (l)

The formula for the hydrated compound contains: YH2O

The mole ratio shows the ratio of the coefficients of the hydrate compound

10.45 hydrated sodium carbonate(Na₂CO₃.xH₂O) were heated until 3.87 of 3.87of anhydrous (Na₂CO₃) remained, so

mass H₂O released :

\tt 10.45-3.87=6.58~g

mass Na₂CO₃ = 3.87 g

mol ratio Na₂CO₃(MW= g/mol) : H₂O(MW=18 g/mol) =

\tt \dfrac{3.87}{105,9888}\div \dfrac{6.58}{18}=0.0365\div 0.3655=1\div 10

6 0
2 years ago
How many grams of NH3 can be prepared from 85.5 grams of N2 and 17.3 grams of H2 ?
Tcecarenko [31]
N2 + 3H2 ---> 2NH3

mass of N2 = 28g
mass of H2 = 2g
mass of NH3 = 17g

according to the reaction:
28g N2----------------- 3*2g H2
85,5g N2-------------------- x
x = 18,32g H2 >>>  so, nitrogen is excess

according to the reaction:
2*3g H2---------------------- 2*17g NH3
17,3g H2 ------------------------- x
x = 98,03g NH3

<u>answer: 98,03g of NH3</u>
4 0
2 years ago
Determine whether or not the mixing of each of the two solutions indicated below will result in a buffer.
WARRIOR [948]
Part A

75.0 mL of 0.10 M HF; 55.0 mL of 0.15 M NaF

This combination will form a buffer.

Explanation

Here, weak acid HF and its conjugate base F- is available in the solution

Part B

150.0 mL of 0.10 M HF; 135.0 mL of 0.175 M HCl

This combination cannot form a buffer.

Explanation

Here, moles of HF = 0.15 x 0.1 = 0.015 moles

Moles of HCl = 0.135 x 0.175 = 0.023

Since HCl is a strong acid and the number of HCl is higher than HF. This prevents the dissociation of HF and the conjugate base F- will not be available in the solution

Part C

165.0 mL of 0.10 M HF; 135.0 mL of 0.050 M KOH

This combination will form a buffer.

Explanation

Moles of HF = 0.165 x 0.1 = 0.0165 moles

Moles of KOH = 0.135 x 0.05 = 0.00675 moles

Moles of KOH is not sufficient for the complete neutralization of HF. Thus weak acid HF and its conjugate base F- is available in the solution and form a buffer

Part D

125.0 mL of 0.15 M CH3NH2; 120.0 mL of 0.25 M CH3NH3Cl

This combination will form a buffer

Explanation

Here, weak acid CH3NH3+ and its conjugate base CH3NH2 is available in the solution and form a buffer

Part E

105.0 mL of 0.15 M CH3NH2; 95.0 mL of 0.10 M HCl

This combination will form a buffer

Explanation

Moles of CH3NH2 = 0.105 x 0.15 = 0.01575 moles

Moles of HCl = 0.095 x 0.1 = 0.0095 moles

Thus the HCl completely reacts with CH3NH2 and converts a part of the CH3NH2 to CH3NH3+. This results weak acid CH3NH3+ and its conjugate base CH3NH2 is in the solution and form a buffer
5 0
2 years ago
25. In the investigation shown the changes in heat of the copper is greater than the change in the heat of the water. What error
forsale [732]

Answer:

Heat lost to the surroundings

Heat lost to the thermometer

Explanation:

All changes in heat, or energy, can be explained. Many of the reactions or changes we see in the world involve the conversion of energy. For example as we heat up a substance (eg. water), the amount of energy we put in should give us an exact temperature. However, this is a "perfect world" scenario, and does not occur in real life. Whenever heat is added to a substance like water, we always need to account for the energy that is going to be lost. For example, heat lost to evaporation or even the effect of measuring the temperature with a thermometer (the introduction of anything including a thermometer will affect the temperature).

3 0
2 years ago
A student heats a sample of Copper (II) sulfate in a crucible and records the data shown in the table. What is the complete form
liberstina [14]

Explanation:

Copper (II) sulfate is usually present as a hydrous state, which is of the form CuSO4 * nH2O, where n is a whole number.

Mass of sample (CuSO4 * nH2O)

= 152.00g - 128.10g = 23.90g.

Mass of water loss during heating

= 152.00g - 147.60g = 4.40g.

Molar mass of H2O = 18g/mol

Moles of H2O in sample

= 4.40g / (18g/mol) = 0.244mol.

Mass of anhydrous sample (CuSO4)

= 23.90g - 4.40g = 19.50g

Molar mass of CuSO4 = 159.61g/mol

Moles of CuSO4 in sample

= 19.50g / (159.61g/mol) = 0.122mol.

Since mole ratio of CuSO4 to H2O

= 0.122mol : 0.244mol = 1:2, n = 2.

Hence we have CuSO4 * 2H2O.

6 0
2 years ago
Read 2 more answers
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