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Yuki888 [10]
3 years ago
14

Calculate the OH− concentration after 53 mL of the 0.100 M KOH has been added to 25.0 mL of 0.200 M HBr. Assume additive vol- um

es.
Answer in units of M.
Chemistry
1 answer:
Andre45 [30]3 years ago
8 0

Answer:

\large \boxed{\text{0.0038 mol/L}}

Explanation:

1. Calculate the initial moles of acid and base

\text{moles of acid} = \text{0.0250 L} \times \dfrac{\text{0.200 mol}}{\text{1 L}} = \text{0.005 00 mol}\\\\\text{moles of base} = \text{0.053 L} \times \dfrac{\text{0.100 mol}}{\text{1 L}} = \text{0.0053 mol}

2. Calculate the moles remaining after the reaction

                   OH⁻     +     H₃O⁺ ⟶ 2H₂O

I/mol:      0.0053       0.005 00

C/mol:    -0.00500   -0.005 00

E/mol:      0.0003              0

We have an excess of 0.0003 mol of base.

3. Calculate the concentration of OH⁻

Total volume = 53 mL + 25.0 mL = 78 mL = 0.078 L

\text{[OH}^{-}] = \dfrac{\text{0.0003 mol}}{\text{0.078 L}} = \textbf{0.0038 mol/L}\\\\\text{The final concentration of OH$^{-}$ is $\large \boxed{\textbf{0.0038 mol/L}}$}

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Calculate the number of moles in 200 grams of Sodium hydroxide​
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Answer:

The answer is 5 mol

Explanation:

Number Of Moles = Mass Of Substance ÷ Mass Of One Mole

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23 + 16 + 1

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Number of Mole = 200/40 = 5 mol

Thus, The number of Moles in 200 grams of Sodium hydroxide is 5 mol

<u>-TheUnknownScientist 72</u>

6 0
3 years ago
If a mixture of gases contained 78% nitrogen at a pressure of 984 torr and 22% carbon dioxide at 345 torr, what is the total pre
AysviL [449]

Answer:

P(total) = 1329 torr

Explanation:

Given data:

Pressure of nitrogen = 984 torr

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According to the this law,

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Mathematical expression,

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P(total) = 1329 torr

8 0
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Natalija [7]

Answer:

C. L mol-1 s-1.

Explanation:

Hello,

In this case, a rate law has the following general form:

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In such a way, since in this case x equals two (second-order rate), the units of k turns out:

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