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2 years ago

If the poh of vinegar is 9.45, what is its [oh− ]? round to the nearest hundredth. × 10n m n =

Mathematics

1 answer:

Alinara [238K]2 years ago

7 0

The [OH] of vinegar to the nearest hundredth is 2.82 * 10^-5

<h3>How to calculate the [OH] of a substance:</h3>

The sum of the pH and pOH of a substance is 14. Mathematically

[OH−] = Antilog (pOH)

Given the following parameters

[OH−] = Antilog (9.45)

Substitute

[OH−] = 2.82 * 10^-5

Hence the [OH] of vinegar to the nearest hundredth is 2.82 * 10^-5

Learn more on pH and pOH here:brainly.com/question/13557815

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I believe its answer is D.
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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

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Answer:

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3 years ago

Tan(x+pi) + 2sin(x+pi)=0 can you help prove this and make the left side equal the right?

vladimir1956 [14]

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To see why:

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The angle sum identity for $\sin x$ says that

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$\tan(x+\pi)+2\sin(x+\pi)=\tan x-2\sin x$

By definition of $\tan x$ ,

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$\sin x=0$ whenever $x$ is a multiple of $\pi$ , i.e. $x=n\pi$ for any integer $n$ .

If $0\le x$ , then $\cos x=\dfrac12$ is true for $x=\dfrac\pi3$ and $x=\dfrac{5\pi}3$ . Then to account for all other possible values, we add a multiple of $2\pi$ , so that $x=\dfrac\pi3+2n\pi$ or $x=\dfrac{5\pi}3+2n\pi$ for integers $n$ .

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3 years ago

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Natali [406]

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3 years ago

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