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Levart [38]
3 years ago
13

Calculate hydrochloric acid (umol)in 200 ul of a 0.5173Msolution of acid?

Chemistry
1 answer:
strojnjashka [21]3 years ago
7 0

<u>Answer:</u> The moles of hydrochloric acid is 1.0346\times 10^{-4}\mu mol

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Or,

\text{Molarity of the solution}=\frac{\text{Micro moles of solute}\times 10^6}{\text{Volume of solution (in }\mu L)}}

We are given:

Molarity of solution = 0.5173 M

Volume of solution = 200\mu L

Putting values in above equation, we get:

0.5173M=\frac{\text{Micro moles of HCl}\times 10^6}{200\mu L}\\\\\text{Micro moles of HCl}=1.0346\times 10^{-4}\mu mol

Hence, the moles of hydrochloric acid is 1.0346\times 10^{-4}\mu mol

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Answer:

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Explanation:

Hello there!

1) In this case, for these calorimetry problems, we can realize that since the temperature increases the reaction is exothermic because it is releasing heat to solution, that is why the temperature goes from 22.0 °C to 28.6 °C.

2) Now, for the total heat released by the reaction, we first need to assume that all of it is absorbed by the solution since it is possible to assume that the calorimeter is perfectly isolated. In such a way, it is also valid to assume that the specific heat of the solution is 4.184 J/(g°C) as it is mostly water, therefore, the heat released by the reaction is:

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\Delta _rH=\frac{Q_{rxn}}{n_{LiCl}} \\\\\Delta _rH=\frac{-8580J}{10.7g*\frac{1mol}{150.91g} }*\frac{1kJ}{1000J}  \\\\\Delta _rH=-121.0kJ/mol

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