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OLga [1]
3 years ago
10

A student has a 0.00124 M HCl (aq) solution and she titrates 25.00 mL of this solution against an unknown potassium hydroxide so

lution. The acid requires 44.25 mL of base to reach the end point, what is the molarity of the KOH (aq) solution?
Chemistry
1 answer:
malfutka [58]3 years ago
5 0

Answer:

The Molarity of KOH is

7,01.10^{-4}M

Explanation:

The endpoint indicates the volume necessary to neutralize the moles of acid.

In other words, the point at which the moles of both solutions are the same.

M_{(HCl)}V_{HCl}=n\\ \\M_{(KOH)}V_{KOH}=n

we match these equations and find the concentration of KOH

M_{(HCl)}. V_{(HCl)} =M_{(KOH)} .V_{(KOH)}\\ \\M_{(KOH)}=\frac{M_{(HCl)}. V_{(HCl)}}{V_{(KOH)}} \\\\M_{(KOH) =\frac{(25mL)(0,00124m)}{(44,25mL)}\\\\

M_{(KOH)}=7,01.10^{-4}

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1 mole of nitrogen converts to 0.5  moles of ammonium nitrate

the conversation factor is 0.5

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Convert the following with the correct number of significant figures:
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Answer:

1.78 × 10⁹ μg

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We have to convert 1.78 kg to μg.

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We will use the conversion factor 1 kg = 10³ g.

1.78 kg × 10³ g/1 kg = 1.78 × 10³ g

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When the same species undergoes both oxidation and reduction in a single redox reaction, this is referred to as a disproportionation. Therefore, divide it into two equal reactions.

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and do the usual changes

First, balance the two half reactions:

3. NO2 +H2O →NO^−3 + 2 H^+ + e−

4. NO2 +2 H^+ + 2e− → NO + H2O

Now multiply one or both half-reactions to ensure that each has the same number of electrons. Here, Eqn (3) x 2 results in each half-reaction having two electrons:

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Now add Eqn 4 and 5 (the electrons now cancel each other):

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and cancel terms that’s common to both sides:

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brainly.com/question/26227625

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