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mestny [16]
4 years ago
8

What volume of a 4.5 m solution of phosphate is necessary to create a 80.0 ml stock of 2.0 m phosphate solution?

Chemistry
1 answer:
11Alexandr11 [23.1K]4 years ago
3 0

Answer: V1 = 45.6 ml

Explanation:

This is a dilution process whereby the concentration of phosphate in the solution is being reduced from 4.5m to 2m. Meanwhile. In a concentration process, the concentration of the phosphate solution will be increased. It can be achieved by removing solvents.

The dilution equation will be used. The equation is as follows:

M1V1 = M2V2 where

V1 = Initial volume of phosphate solution.

M1 = Initial concentration of phosphate solution.

V2 =Final volume of phosphate solution.

M2 = Final concentration of phosphate solution .

From the information given,

M1 = 4.5

V1 = ?

M2 = 2.0

V2 = 80

4.5 × V1 = 2×80

4.5V1 = 160

V1 = 160/4.5

= 35.6 ml

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Answer:

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Explanation:

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~Grams of KCIO3 are canceled out.~  

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1 mol of KCIO3 equals 2 mols of KCIO3 -- 2 moles from the equation  

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3 years ago
The net ionic equation for formation of an aqueous solution of nii2 accompanied by evolution of co2 gas via mixing solid nico3 a
Ann [662]

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Calculate the feed ratio of adipic acid and hexamethylene diamine that should be employed to obtain a polyamide of approximately
artcher [175]

Answer:

r= 0.9949 (For 15,000)

r=0.995 (For 19,000)

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3 years ago
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