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Anit [1.1K]
2 years ago
11

A 609 ml sample of naoh has a ph of 13.5292. if 594 ml of distilled water was added to the initial naoh solution, what would the

new ph of the solution be g
Chemistry
1 answer:
Delvig [45]2 years ago
8 0

The new pH of the solution after the addition of water would be 13.22.

<h3>Total volume of the solution</h3>

V(total) = 609 ml + 594 ml

V(total) = 1203 ml

<h3>Concentration of water </h3>

C = (594 ml)/(1203 ml)

C = 0.493

pOH = -log(0.493)

pOH = 0.31

New pH of the solution = 13.5292 - 0.31 = 13.22

Learn more about pH here: brainly.com/question/13557815

#SPJ1

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Covert from scientific notation<br> 1.3 x 10¹²=
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Answer:

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Explanation:

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We want to convert this from scientific notation.

Tip: in scientific notation the exponent tells you how many place you move the decimal point over. If the exponent is negative you move the decimal point to the left. Ex. For, 4.1 x 10^-8, we would move the decimal point over 8 times to the left to get .00000041. When the exponent is positive we move over to the right. Ex. For, 7.6 x 10^7 we would move the decimal point over 7 times to the right to get 76,000,000

So to convert 1.3 x 10^12 we simply move over the decimal point over 12 times to the right.

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2 years ago
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3 years ago
Gaseous hydrogen iodide is placed in a closed container at 425∘C, where it partially decomposes to hydrogen and iodine: 2HI(g)⇌H
murzikaleks [220]

Answer:

0.0184

Explanation:

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