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Daniel [21]
3 years ago
14

A quantity of 85.0 mL of 0.900 M HCl is mixed with 85.0 mL of 0.900 M KOH in a constant-pressure calorimeter that has a heat cap

acity of 325 J/°C. If the initial temperatures of both solutions are the same at 18.24°C, what is the final temperature of the mixed solution?
Chemistry
1 answer:
san4es73 [151]3 years ago
3 0

Answer:

There is a 1:1 neutralization reaction,  so, number of moles:

0.085 l * 0.900 mol/ L = 0.0765 mol HCL

The heat produced:  0.0765 mol * -56.2 kJ/mol = -4.229 kJ (this is the heat of neutralization)

Change in temperature:    The mass to use in the equation Q=cmT.  

4229 J / (4.186 J/gC*170 g) = 6.042 C

Add to the initial temperature:

18.24 + 6.042 = 24.29 C°

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<u>Answer:</u> The percent gallium in gallium bromide is 30.30 %.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of gallium bromide = 0.165 g

Molar mass of titanium gallium bromide = 229.53 g/mol

Putting values in equation 1, we get:

\text{Moles of gallium bromide}=\frac{0.165g}{229.53g/mol}=0.00072mol

  • The chemical equation for the reaction of gallium bromide and silver nitrate follows:

GaBr_2+2AgNO_3\rightarrow 2AgBr(s)+Ga(NO_3)_2

By Stoichiometry of the reaction:

1 moles of gallium bromide produces 1 mole of gallium nitrate

So, 0.00072 moles of gallium bromide will produce = \frac{1}{1}\times 0.00072=0.00072moles of gallium nitrate

  • Now, calculating the mass of gallium nitrate from equation 1, we get:

Molar mass of gallium nitrate = 193.73 g/mol

Moles of gallium nitrate = 0.00072 moles

Putting values in equation 1, we get:

0.00072mol=\frac{\text{Mass of gallium nitrate}}{193.73g/mol}\\\\\text{Mass of gallium nitrate}=0.139g

Calculating the mass of gallium in the reaction, we use unitary method:

In 1 mole of gallium nitrate, 1 mole of gallium atom is present.

In 193.73 grams of gallium nitrate, 69.72 g of gallium atom is present.

So, in 0.139 grams of gallium nitrate, the mass of gallium present will be = \frac{69.72}{193.73}\times 0.139=g

  • To calculate the percentage composition of gallium in gallium bromide, we use the equation:

\%\text{ composition of gallium}=\frac{\text{Mass of gallium}}{\text{Mass of gallium bromide}}\times 100

Mass of gallium bromide = 0.165 g

Mass of gallium = 0.050 g

Putting values in above equation, we get:

\%\text{ composition of gallium}=\frac{0.050g}{0.165g}\times 100=30.30\%

Hence, the percent gallium in gallium bromide is 30.30 %.

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