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Ad libitum [116K]
3 years ago
8

Why is performing extraction with several small portions of a solvent more officient than a single extraction with the same tota

l volume of the solvent?
Chemistry
1 answer:
elena55 [62]3 years ago
6 0

With various extractions the amount of material left in the trash will be lower, ergo the extraction will be more perfect. Various extractions with fewer amounts of solvent are more efficient than a single extraction with a huge amount of solvent.

<u>Explanation:</u>

Surely multiple extractions are better than the single large extraction. Because extraction is about maximizing outside field communication between the two solvents, and you easily get more surface area contact with fewer amounts.

You can merge two smaller portions quicker and more completely than with large portions.

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3 years ago
At 25 °C, how many dissociated OH– ions are there in 1243 mL of an aqueous solution whose pH is 2.07?
coldgirl [10]

<u>Answer:</u> The number of OH^- ions dissociated are 8.57\times 10^{11}

<u>Explanation:</u>

We are given:

pH = 2.07

Calculating the value of pOH by using equation, we get:

2.07+pOH=14\\\\pOH=14-2.07=11.93

To calculate hydroxide ion concentration, we use the equation to calculate pOH of the solution, which is:

pOH=-\log[OH^-]

We are given:

pOH = 11.93

Putting values in above equation, we get:

11.93=-\log[OH^-]

[OH^-]=10^{-11.93}=1.17\times 10^{-12}M

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Molarity of solution = 1.17\times 10^{-12}M

Volume of solution = 1243 mL = 1.243 L  (Conversion factor: 1 L = 1000 mL)

Putting values in above equation, we get:

1.17\times 10^{-12}M=\frac{\text{Moles of }OH^-}{1.243L}\\\\\text{Moles of }OH^-=(1.17\times 10^{-12}mol/L\times 1.243L)=1.424\times 10^{-12}mol

According to mole concept:

1 mole of a compound contains 6.022\times 10^{23} number of particles

So, 1.424\times 10^{-12}mol number of OH^- will contain = (1.424\times 10^{-12}\times 6.022\times 10^{23})=8.57\times 10^{11} number of ions

Hence, the number of OH^- ions dissociated are 8.57\times 10^{11}

3 0
4 years ago
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