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True [87]
2 years ago
11

21.10g of NaOH and Ba3(OH)2 mixture is dissolved water to prepare 1.0dm³ Solution. To neutralize 25.OO mL of this solution needs

0.5 moldm-³ HCl 15.00mL. calculate the percentage of NaOH by mass in the mixture.​
Chemistry
1 answer:
alex41 [277]2 years ago
8 0

From the equation of te reaction, we know that the mass percent  of NaOH in the mixture is 1.4%.

<h3>What is neutralization?</h3>

Neutralization is a reaction that occurs between an acid and a base to yield salt and water only.

In tis case, the reaction of the NaOH and HCl occurs as follows; NaOH + HCl ----> NaCl + H2O

Number of moles of HCl reacted = 15/1000 * 0.5 moldm-³ = 0.0075 moles

Since the reaction is 1:1, 0.0075 moles of NaOH reacted.

Mass of NaOH =  0.0075 moles of NaOH * 40 g/mol = 0.3 g

Percent of NaOH = 0.3 g/21.10g * 100/1 = 1.4%

Learn more about percent concentration: brainly.com/question/202460

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Answer:

The answers are in the explanation

Explanation:

A buffer is the mixture of a weak acid with its conjugate base or vice versa. Thus:

<em>1)</em> Mixing 100.0 mL of 0.1 M HF with 100.0 mL of 0.05 M mol KF. <em>Will </em>result in a buffer because HF is a weak acid and KF is its conjugate base.

<em>2)</em> Mixing 100.0 mL of 0.1 M NH₃ with 100.0 mL of 0.1 M NH₄Br. <em>Will not </em>result in a buffer because NH₃ is a strong base.

<em>3) </em>Mixing 100.0 mL of 0.1 M HCN with 100.0 mL of 0.05 M KOH. <em>Will </em>result in a buffer because HCN is a weak acid and its reaction with KOH will produce CN⁻ that is its conjugate base.

<em>4)</em> Mixing 100.0 mL of 0.1 M HCl with 100.0 mL of 0.1 M KCl <em>Will not </em>result in a buffer because HCl is a strong acid.

<em>5)</em> Mixing 100.0 mL of 0.1 M HCN with 100.0 mL of 0.1 M KOH <em>Will not </em>result in a buffer because each HCN will react with KOH producing CN⁻, that means that you will have just CN⁻ (Conjugate base) without HCN (Weak acid).

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