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3 years ago

1. If 22.5 L of nitrogen at 734 mm Hg are compressed to 702 mm Hg at constant

Chemistry

1 answer:

Alika [10]3 years ago

8 0

Answer:

The answer to your question is V2 = 23.52 l

Explanation:

Data

Volume 1 = V1 = 22.5 l

Pressure 1 = P1 = 734 mmHg

Volume 2 = V2 = ?

Pressure 2 = 702 mmHg

Process

To solve this problem use Boyle's law.

P1V1 = P2V2

-Solve for V2

V2 = P1V1 / P2

-Substitution

V2 = (734 x 22.5) / 702

-Simplification

V2 = 16515 / 702

-Result

V2 = 23.52 l

-Conclusion

If we diminish the pressure, the volume will be higher.

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A 5.000 g mixture contains strontium nitrate and potassium bromide. Excess lead(II) nitrate solution is added to precipitate out

scZoUnD [109]

Answer: The mass percent of potassium bromide in the mixture is 9.996%

Explanation:

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

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

For lead (II) bromide:

Given mass of lead (II) bromide = 0.7822 g

Molar mass of lead (II) bromide = 367 g/mol

Putting values in equation 1, we get:

$\text{Moles of lead (II) bromide}=\frac{0.7822g}{367g/mol}=0.0021mol$

The chemical equation for the reaction of lead (II) nitrate and potassium bromide follows:

$2KBr+Pb(NO_3)_2\rightarrow PbBr_2+2KNO_3$

By Stoichiometry of the reaction:

1 mole of lead (II) bromide is produced from 2 moles of potassium bromide

So, 0.0021 moles of lead (II) bromide will be produced from = $\frac{2}{1}\times 0.0021=0.0042mol$ of potassium bromide

Now, calculating the mass of potassium bromide by using equation 1, we get:

Molar mass of KBr = 119 g/mol

Moles of KBr = 0.0042 moles

Putting values in equation 1, we get:

$0.0042mol=\frac{\text{Mass of KBr}}{119g/mol}\\\\\text{Mass of KBr}=0.4998g$

To calculate the percentage composition of KBr in the mixture, we use the equation:

$\%\text{ composition of KBr}=\frac{\text{Mass of KBr}}{\text{Mass of mixture}}\times 100$

Mass of mixture = 5.000 g

Mass of KBr = 0.4998 g

Putting values in above equation, we get:

$\%\text{ composition of KBr}=\frac{0.4998g}{5.000g}\times 100=9.996\%$

Hence, the percent by mass of KBr in the mixture is 9.996 %

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Answer:

1.395J/g°C

Explanation:

The following were obtained from the question:

Q = 6527J

M = 312g

ΔT = 15°C

C =?

Q = MCΔT

C = Q/MΔT

C = 6527/(312 x 15)

C = 1.395J/g°C

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Answer:

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