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leva [86]
3 years ago
5

What volume would be needed to prepare 375 mL of a .45 M CaCl2 using only a solution of 1.0 M CaCl2 and water?

Chemistry
1 answer:
Neporo4naja [7]3 years ago
8 0

Answer:

168.75 ml

Explanation:

M1V1=M2V2

375ml*.45M=1M*V2

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Its formula :)

Explanation:

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Fireworks, KNO3<br> Give only the names of the elements alphabetically, separating them with commas.
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The answer will be K, N, O.

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4 years ago
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PH CHEM, PLEASE HELP QUICK! NO LINKS/VIRUSES PLEASE!
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Answer:

The pH of the solution is 11.48.

Explanation:

The reaction between NaOH and HCl is:

NaOH  +  HCl  →  H₂O  +  NaCl

From the reaction of 3.60x10⁻³ moles of NaOH and 5.95x10⁻⁴ moles of HCl we have that all the HCl will react and some of NaOH will be leftover:

n_{NaOH}} = n_{i_{NaOH}} - n_{HCl} = 3.60 \cdot 10^{-3} moles - 5.95 \cdot 10^{-4} moles = 3.01 \cdot 10^{-3} moles

Now, we need to find the concentration of the OH⁻ ions.

[OH^{-}] = \frac{n_{NaOH}}{V}

Where V is the volume of the solution = 1.00 L                

[OH^{-}] = \frac{n_{NaOH}}{V} = \frac{3.01 \cdot 10^{-3} moles}{1.00 L} = 3.01 \cdot 10^{-3} mol/L

Finally, we can calculate the pH of the solution as follows:

pOH = -log([OH^{-}]) = -log(3.01 \cdot 10^{-3}) = 2.52

pH + pOH = 14

pH = 14 - pOH = 14 - 2.52 = 11.48

Therefore, the pH of the solution is 11.48.

I hope it helps you!

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3 years ago
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