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xeze [42]
3 years ago
12

You have a solution of 600 mg of caffeine dissolved in 100 mL of water. The partition coefficient for aqueous caffeine extracted

with dichloromethane (DCM, CH2Cl2) is 4.6. Show all calculations for the following (the equation is in your lecture notes): a.How many mgs caffeine would be extracted from the 100 mL of water containing 600 mg of caffeine with one portion of 60 mL of dichloromethane

Chemistry
2 answers:
klio [65]3 years ago
4 0

Answer:

159 mg caffeine is being extracted in 60 mL dichloromethane

Explanation:

Given that:

mass of caffeine in 100 mL of water =  600 mg

Volume of the water = 100 mL

Partition co-efficient (K) = 4.6

mass of caffeine extracted = ??? (unknown)

The portion of the DCM = 60 mL

Partial co-efficient (K) = \frac{C_1}{C_2}

where; C_1= solubility of compound in the organic solvent and C_2 = solubility in aqueous water.

So; we can represent our data as:

K=(\frac{A_{(g)}}{60mL} ) ÷ (\frac{B_{(mg)}}{100mL} )

Since one part of the portion is A and the other part is B

A+B = 60 mL

A+B = 0.60

A= 0.60 - B

4.6=(\frac{0.6-B(mg)}{60mL} ) ÷ (\frac{B_{(mg)}}{100mL})

4.6 = \frac{(\frac{0.6-B(mg)}{60mL} )}{(\frac{B_{(mg)}}{100mL})}

4.6 × (\frac{B_{(mg)}}{100mL}) = (\frac{0.6-B(mg)}{60mL} )

4.6 B *\frac{60}{100} = 0.6 - B

2.76 B = 0.6 - B

2.76 + B = 0.6

3.76 B = 0.6

B = \frac{0.6}{3.76}

B = 0.159 g

B = 159 mg

∴ 159 mg caffeine is being extracted from the 100 mL of water containing 600 mg of caffeine with one portion of in 60 mL dichloromethane.

marissa [1.9K]3 years ago
4 0

Explanation:

Below is an attachment containing the solution.

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emmasim [6.3K]

Hey there!:

Mass = 17.92 g

Volume = 2.00 mL

Density =  ??

Therefore:

D = m / V

D = 17.92 / 2.00

D = 8.96 g/mL

Hope this helps!

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3 years ago
Which structure is responsible for bringing in the amino acids? (Points : 1)
Rainbow [258]
Well the answer to number 1 is definitely B
 
number 2 is B

number 3 is D

Number 4 is C

Hope this helps:)



5 0
3 years ago
Read 2 more answers
What is the pH of a solution that is 0.40 M NaBrO and 0.50 M HBrO (hypobromous acid) (Ka for HBrO = 2.3 x 10^-9)
Pie

Answer

pH=8.5414

Procedure

The Henderson–Hasselbalch equation relates the pH of a chemical solution of a weak acid to the numerical value of the acid dissociation constant, Kₐ. In this equation, [HA] and [A⁻] refer to the equilibrium concentrations of the conjugate acid-base pair used to create the buffer solution.

pH = pKa + log₁₀ ([A⁻] / [HA])

Where

pH = acidity of a buffer solution

pKa = negative logarithm of Ka

Ka =acid disassociation constant

[HA]= concentration of an acid

[A⁻]= concentration of conjugate base

First, calculate the pKa

pKa=-log₁₀(Ka)= 8.6383

Then use the equation to get the pH (in this case the acid is HBrO)

pH=8.6383+\log_{10}(\frac{0.40\text{ M}}{0.50\text{ M}})=8.5414

8 0
1 year ago
Magma can partially crystallize at depth and then rise to shallow depths where the remaining magma solidifies. The early-formed
Doss [256]
<h2>Phenocrysts and Porphyritic texture </h2>

Explanation:

  • The early formed crystals are of phenocrysts and the texture of these crystals is porphyritic texture.  
  • This crystallization occurs when early-forming plagioclase crystals which are rich in calcium start coating with plagioclase crystals which are rich in sodium.
  • On cooling, the magma is then processed in a volcanic eruption, after the eruption the liquid which is left behind will start cooling and forms a porphyritic texture.  

5 0
3 years ago
What is the formula weight (amu) of the molecule H2O? Use atomic masses of H and O as 1.008 amu and 16.00 amu respectively. Repo
vaieri [72.5K]

Answer:

Formula weight of H₂O molecule is  18.02 amu.

Explanation:

Given data:

Formula weight of H₂O = ?

Atomic mass of H = 1.008 amu

Atomic mass of O = 16.00 amu

Solution:

Formula weight:

"It is the sum of all the atomic weight of atoms present in given formula"

Formula weight of H₂O = 2×1.008 amu + 1×16.00 amu

Formula weight of H₂O = 18.02 amu

Thus, formula weight of H₂O molecule is  18.02 amu.

6 0
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