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ruslelena [56]
3 years ago
11

Write the balanced equation. Then calculate the volume of 0.65 M HCl required to completely neutralize 400.0 ml of 0.88 M KOH.

Chemistry
1 answer:
Misha Larkins [42]3 years ago
8 0
Hello!

The balanced equation for the neutralization of KOH is the following:

HCl(aq) + KOH(aq) → KCl(aq) + H₂O(l)

To calculate the volume of HCl required, we can apply the following equation:

 moles HCl = moles KOH \\  \\ cHCl*vHCl=cKOH*vKOH \\  \\ vHCl= \frac{cKOH*vKOH}{cHCl}= \frac{400 mL*0,88M}{0,65M}=  541,54mL

So, the required volume of HCl is 541,54 mL

Have a nice day!
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Answer:

<u><em>Arrhenius Acid:</em></u>

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Since H₂SO₄ have a proton (H⁺ ion) and it can donate it to be made a sulphate ion, So it is an Arrhenius acid.

See the following reaction =>

<u><em>H₂SO₄ + H₂O => HSO₄ + H₃O⁺</em></u>

<u><em>Arrhenius Base:</em></u>

An Arrhenius base is a a proton acceptor.

KOH accepts the proton to to made to KOH₂ and a proton acceptor.

See the following reaction =>

<u><em>KOH + H₂o => KOH₂ + OH⁻</em></u>

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3 years ago
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Answer:

–36 KJ.

Explanation:

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This can be obtained as illustrated below:

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Therefore, –36 KJ of energy is associated with 1.5 moles of D.

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