If you breathe normally you have about 0.11 moles of air in your lungs and

Chemistry

1 answer:

Firdavs [7]2 years ago

4 0

Answer:

5.89 Liters

Explanation:

Lung capacity=(2.4/0.11) x 0.27

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barxatty [35]

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The freezing point of benzene is 5.5°C. What is the freezing point of a solution of 2.60 g of naphthalene (C10H8) in 675 g of be

Mrac [35]

Answer: The freezing point of solution is 5.35°C

Explanation:

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point elevation constant = 4.90°C/m

$m_{solute}$ = Given mass of solute (naphthalene) = 2.60 g

$M_{solute}$ = Molar mass of solute (naphthalene) = 128.2 g/mol

$W_{solvent}$ = Mass of solvent (benzene) = 675 g

Putting values in above equation, we get:

$5.5-\text{Freezing point of solution}=1\times 4.90^oC/m\times \frac{2.60\times 1000}{128.2g/mol\times 675}\\\\\text{Freezing point of solution}=5.35^oC$