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77julia77 [94]
2 years ago
8

If it takes 25.0 mL of 0.050 M HCl to neutralize 34.5 mL of NaOH solution, what is the molarity of the NaOH solution?

Chemistry
1 answer:
Serhud [2]2 years ago
8 0

The molarity of the NaOH solution is 0.036 M

First, we will write the balanced chemical equation for the reaction

The balanced equation of reaction between HCl and NaOH is

HCl + NaOH → NaCl + H₂O

In the balanced chemical equation above,

It means 1 mole of HCl is required to completely neutralize 1 mole of NaOH

To determine the molarity of the NaOH solution,

We will determine the number of moles of HCl that reacted

From the question,

Volume of HCl = 25.0 mL = 0.025 L

Concentration of HCl = 0.050 M

From the formula

Number of moles = Concentration × Volume

Then,

Number of moles of HCl = 0.050 × 0.025

Number of moles of HCl = 0.00125 moles

This means 0.00125 moles of HCl reacted with 0.00125 moles of NaOH

Now, to determine the molarity (that is concentration) of NaOH,

Number of moles of NaOH = 0.00125 moles

Volume of NaOH = 34.5 mL = 0.0345 L

From the formula

Number of moles = Concentration × Volume

Then,

Concentration = \frac{Number \ of \ moles }{Volume}

Concentration \ of \ NaOH = \frac{0.00125}{0.0345} \\

∴ Concentration of NaOH = 0.03623 M

Concentration of NaOH = 0.036 M

Hence, the molarity of the NaOH solution is 0.036 M

Learn more here: brainly.com/question/25008908

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The half-life of nitrogen-13 is 10.0 minutes. if you begin with 53.3 mg of this isotope, what mass remains after 25.9 minutes ha
zimovet [89]

Hello!

The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.

We have the following data:

mo (initial mass) = 53.3 mg

m (final mass after time T) = ? (in mg)

x (number of periods elapsed) = ?

P (Half-life) = 10.0 minutes

T (Elapsed time for sample reduction) = 25.9 minutes

Let's find the number of periods elapsed (x), let us see:

T = x*P

25.9 = x*10.0

25.9 = 10.0\:x

10.0\:x = 25.9

x = \dfrac{25.9}{10.0}

\boxed{x = 2.59}

Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:

m =  \dfrac{m_o}{2^x}

m =  \dfrac{53.3}{2^{2.59}}

m \approx \dfrac{53.3}{6.021}

\boxed{\boxed{m \approx 8.85\:mg}}\end{array}}\qquad\checkmark

I Hope this helps, greetings ... DexteR! =)

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3 years ago
Consider two separate aqueous solutions: one of a weak acid HA and one of HCl. Assuming you started with 10 molecules of HA in o
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Answer:

The answer is given below:

Explanation:

Strong acids have the ability to dissociate quickly as compared to the weak acids which dissociates slowly and can do that up to a certain level.

HCl is a strong acid and HA as stated in the statement is a weak acid.

So, in the beaker which contains HCl will have:

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In the beaker with HA will have:

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