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Katyanochek1 [597]
3 years ago
12

No links.

Chemistry
1 answer:
Gennadij [26K]3 years ago
5 0

Explanation:

first find the value of Ph

from Ph= -log[H+]

ph=-log[1.7x10^-8]=7.769

find Poh from 14=ph+poh

14=7.769+poh

Poh=6.231 part a

for part b the same as part a from the above procedure

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AleksAgata [21]

Explanation:

(w/w) % : The percentage mass or fraction of mass of the of solute present in total mass of the solution.

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2) 250 g of 0.500% (w/w) NaBr

Mass of solution = 250 g

Mass of solute = x

Required w/w % of solution = 0.500%

0.500\%=\frac{x}{250 g}\times 100

x=\frac{0.500\times 250 g}{100}=1.25 g

1.25 grams of solute needed to make 250 g of 0.500% (w/w) NaBr

3) 500 g of 1.25% (w/w) glucose

Mass of solution = 500 g

Mass of solute = x

Required w/w % of solution = 1.25%

1.25\%=\frac{x}{500 g}\times 100

x=\frac{1.25\times 500 g}{100}=6.25 g

6.25 grams of solute needed to make 500 g of 1.25% (w/w) (glucose)

4) 750 g of 2.00% (w/w) sulfuric acid.

Mass of solution = 750 g

Mass of solute = x

Required w/w % of solution = 2.00%

2.00\%=\frac{x}{750 g}\times 100

x=\frac{2.00\times 750 g}{100}=15.0 g

15.0 grams of solute needed to make 750 g of 2.00% (w/w) sulfuric acid.

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