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xeze [42]
3 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]3 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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You have a mixture of the gases Cl₂ and O₂ along with some N₂ in a container at STP. The moles of Cl₂ and O₂ in the mixture are
Harman [31]

The mole fraction of N₂ in the mixture, given the data is 0.225

<h3>How to determine the mass of the mixture</h3>
  • Density of mixture = 2.063 g/L
  • Volume (at STP) = 22.4 L
  • Mass of mixture =?

Density = mass / volume

Cross multiply

Mass = Density × volume

Mass of mixture = 2.063 × 22.4

Mass of mixture = 46.2112 g

<h3>How to determine the mole fraction of N₂</h3>

We'll begin by calculating the mole of N₂ in the mixture. This can be obtained as follow:

  • Mass of mixture = 46.2112 g
  • Let the mole of Cl₂ = y
  • Let the mole of O₂ = Mole of Cl₂ = y
  • Total mole = 1 mole
  • Let the mole of N₂ = x =?

Total mole = Mole of N₂ + Mole of Cl₂ + Mole of O₂

1 = x + y + y

1 = x + 2y

2y = 1 - x

y = (1 - x) / 2

Mass of mixture = mass of N₂ + mass of Cl₂ + mass of O₂

Recall

mass = mole × molar mass

Thus,

Mass of mixture = (mole of N₂ × molar mass) + (mole of Cl₂ × molar mass) + (mole of O₂ × molar mass)

46.2112 = 28x + 71y + 32y

But

y = (1 - x) / 2

46.2112 = 28x + 71(1 - x) / 2 + 32(1 - x) / 2

Multiply through by 2

92.4224 = 56x + 71(1 - x) + 32(1 - x)

92.4224 = 56x + 71 - 71x + 32 - 32x

Collect like terms

92.4224 - 71 - 32 = 56x - 71x - 32x

-10.5776 = -47x

Divide both sides by -47

x = -10.5776 / -47

x = 0.225 mole

Finally, we can determine the mole fraction of N₂. This can be obtained as follow:

  • Mole of N₂ = x = 0.225 mole
  • Total mole = 1 mole
  • Mole fraction of N₂ = ?

Mole fraction = mole / total mole

Mole fraction of N₂ = 0.225 / 1

Mole fraction of N₂ = 0.225

Learn more about mole fraction:

brainly.com/question/2769009

#SPJ1

8 0
1 year ago
TRUE or FALSE: Units in the Sl system include feet, pounds, and<br> degrees.<br><br> True<br> False
luda_lava [24]

Answer:

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4 0
3 years ago
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Some COCl2 is placed in a sealed flask and heated to 756 K. When equilibrium is reached, the flask is found to contain COCl2 (7.
o-na [289]

Answer:

9.044\times 10^{-3} is the value of the equilibrium constant for this reaction at 756 K.

Explanation:

COCl_2\rightleftharpoons CO+Cl_2

Equilibrium concentration of COCl_2

[COCl_2]=7.40\times 10^{-4} M

Equilibrium concentration of CO

[CO]=3.76\times 10^{-2} M

Equilibrium concentration of Cl_2

[Cl_2]=1.78\times 10^{-4} M

The expression of an equilibrium constant can be written as;

K_c=\frac{[CO][Cl_2]}{[COCl_2]}

=\frac{3.76\times 10^{-2}\times 1.78\times 10^{-4}}{7.40\times 10^{-4}}

K_c=9.044\times 10^{-3}

9.044\times 10^{-3} is the value of the equilibrium constant for this reaction at 756 K.

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3 years ago
The fermentation of glucose (C6H12O6) produces ethyl alcohol (C2H5OH) and CO2:
Jlenok [28]
To answer your question I will use dimensional analysis, which is used by cancelling out the units. I will also use the balanced equation provided as a conversion factor.

A) First start out with the 0.300 mol of C6H12O6...
0.300 mol C6H12O6 * (2 mol CO2 / 1 mol C6H12O6) = 0.600 mol CO2

*The significant figures (sig figs) at still three, the 2 is a conversion counting number and does not count*

B) First change 2.00 g of C2H5OH to moles of C2H5OH...
The molecular mass of C2H5OH is...
2(12.01 g/mol) + 5(1.008 g/mol) + 16.00 g/mol + 1.008 g/mol = 46.07 g/mol
This can be used as a conversion factor to change grams to moles. 
2.00 g C2H5OH * (1 mol C2H5OH / 46.07 g C2H5OH) = 0.0434 mol C2H5OH

Second, you can change the moles of C2H5OH to moles of C6H12O6..
0.0434 mol C2H5OH * (1 mol C6H12O6 / 2 mol C6H12O6) = 0.0217 mol C6H12O6

Third, change moles of C6H12O6 to grams...
MM = 6(12.01 g/mol) + 12(1.008 g/mol) + 6(16.00 g/mol) = 180.16 g/mol
0.0217 mol C6H12O6 * (180.16 g C6H12O6 / 1 mol C6H12O6) = 3.91 g C6H12O6

C) Now I am going to put it all into one long dimensional analysis problem.
MM of CO2 = 44.01 g/mol
MM of C2H5OH = 46.07 g/mol

2.00 g C2H5OH * (1 mol C2H5OH / 46.07 g C2H5OH) * (2 mol CO2 / 2 mol C2H5OH) * (44.01 g CO2 / 1 mol CO2) = 1.91 g CO2

I hope this helped and I am sorry that I talked to much, I just didn't want to miss anything!
3 0
3 years ago
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Maurinko [17]
LD50 is defined as the lethal dose 50% which describes the amount of material required to kill 50% of the testing population. It is given in units of mg of chemical per kg of bodyweight of the recipient. 

Comparing hydrogen peroxide and acetic acid, we see that peroxide has a lower LD50 of 900 mg/kg, with acetic acid having LD50 = 3310 mg/kg. When comparing LD50 values, the smaller value will be the more toxic compound. What this means is that in this case, a smaller amount of peroxide is required to kill 50% of the testing population compared to acetic acid.

Therefore, 3% hydrogen peroxide is more hazardous to consume.
7 0
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