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xeze [42]
2 years ago
12

A mixture of two compounds, A and B, was separated by extraction. After the compounds were dried, their masses were found to be:

119 mg of compound A and 97 mg of compound B. Both compounds were recrystallized and weighed again. After recrystallization, the mass of compound A was 83 mg and the mass of compound B was 79 mg. Calculate the percent recovery from recrystallization for both compounds.
Chemistry
1 answer:
Arlecino [84]2 years ago
3 0

Answer:

The percent recovery from re crystallization for both compounds A and B is 69.745 and 81.44 % respectively.

Explanation:

Mass of compound A in a mixture  = 119 mg

Mass of compound A after re-crystallization = 83 mg

Percent recovery from re-crystallization :

\frac{\text{Mass after re-crystallization}}{\text{Mass before re-crystallization}}\times 100

Percent recovery of compound A:

\frac{83 mg}{119 mg}\times 100=69.74\%

Mass of compound B in a mixture  = 97 mg

Mass of compound B after re-crystallization = 79 mg

Percent recovery of compound B:

\frac{79 mg}{97 mg}\times 100=81.44\%

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Kc for the reaction N2O4 <=> 2NO2 is 0.619 at 45 degrees C If 50.0g of N2O4 is introduced into an empty 2.10L container, w
Nadya [2.5K]

Answer:

p(N2O4) = 0.318 atm

p(NO2) = 7.17 atm

Explanation:

Step 1: Data given

Kc = 0.619

Temperature = 45.0 °C

Mass of N2O4 = 50.0 grams

Volume = 2.10 L

Molar mass N2O4 = 92.01 g/mol

Step 2: The balanced equation

N2O4 ⇔ 2NO2

Step 3: Calculate moles N2O4

Moles N2O4 = 50.0 grams / 92.01 g/mol

Moles N2O4 = 0.543 moles

Step 4: The initial concentration

[N2O4] = 0.543 moles/2.10 L = 0.259 M

[NO2]= 0 M

Step 5: Calculate concentration at the equilibrium

For 1 mol N2O4 we'll have 2 moles NO2

[N2O4] = (0.259 -x)M

[NO2]= 2x

Step 6: Calculate Kc

Kc = 0.619=  [NO2]² / [N2O4]

0.619 = (2x)² / (0.259-x)

0.619 = 4x² / (0.259 -x)

x = 0.1373  

Step 7: Calculate concentrations

[N2O4] = (0.259 -x)M = 0.1217 M

[NO2]= 2x = 0.2746 M

Step 8: The moles

Moles = molarity * volume

Moles N2O4 = 0.1217 M * 2.10  = 0.0256 moles

Moles NO2 = 0.2746 M * 2.10 = 0.577 moles

Step 9: Calculate partial pressure

p*V = n*R*T

⇒ with p = the partial pressure

⇒ with V = the volume = 2.10 L

⇒ with n = the number of moles

⇒ with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature = 45 °C = 318 K

p = (nRT)/V

p(N2O4) = (0.0256 *0.08206 * 318)/ 2.10

p(N2O4) = 0.318 atm

p(NO2) = (0.577 *0.08206 * 318)/ 2.10

p(NO2) = 7.17 atm

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