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Anna35 [415]
2 years ago
9

If 3.00 liters of 2M HCl neutralizes 1.00 liter of NaOH, what is the molarity of the NaOH?

Chemistry
1 answer:
PilotLPTM [1.2K]2 years ago
7 0

Answer:

M_{base}=6M

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to realize this is a question about titration, which base solved by knowing that the HCl reacts with the NaOH in a 1:1 mole ratio, and therefore, we can write the following:

M_{acid}V_{acid}=M_{base}V_{base}

Thus, we solve for the molarity of the base, NaOH, as shown below:

M_{base}=\frac{M_{acid}V_{acid}}{V_{base}} \\\\M_{base}=\frac{3L*2M}{1.00L}\\\\M_{base}=6M

Regards!

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The answers would be:

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In general, these are the best answers. The solute is what is being dissolved and the solvent is what dissolves. A solvent comes in greater amounts in a solution and it is the dissolving agent.

For example, sugar and water.

To make a sugar water solution, you will need to dissolve sugar in water. Sugar is the solute in this case because it is what is being dissolved. The water is the solvent, because it dissolves the sugar.

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Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of bromine at 10.0^oC = ?

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T_1 = temperature of propane = 10.0^oC=273+10.0=283.0K

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R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

\ln (\frac{1atm}{P_1})=\frac{30910J/mole}{8.314J/K.mole}\times (\frac{1}{283.0K}-\frac{1}{331.8K})

P_1=0.1448atm

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Answer:

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Explanation:

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