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2 years ago

Calculate the partial pressure of oxygen given the total barometric pressure of 515 mmHg. Round to the nearest hundredth.

Chemistry

1 answer:

Eva8 [605]2 years ago

8 0

The partial pressure of oxygen given the total barometric pressure is : 108.15 mmHg

Given data :

Total barometric pressure = 515 mmHg

Assuming oxygen percentage = 21%

Barometric pressure dry at 37°C

<h3 /><h3>Determine the partial pressure of oxygen </h3>

Applying the relation below

Partial pressure = oxygen percentage * Barometric pressure

= 21% * 515 mmHg

= 108.15 mmHg

Hence we can conclude that the partial pressure of oxygen is 108.15 mmHg.

Learn more about Partial pressure : brainly.com/question/1835226

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Camphor, a white solid with a pleasant odor, is extracted from the roots, branches, and trunk of the camphor tree. Assume you di

katrin [286]

Answer: The molarity of solution is 0.799 M , molality of solution is 1.02 m, mole fraction of camphor is 0.045 and mass percent of camphor in solution is 13.43 %

Explanation:

Calculating the molarity of solution:

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

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Given mass of camphor = 70.0 g

Molar mass of camphor = 152.2 g/mol

Volume of solution = 575 mL

Putting values in above equation, we get:

$\text{Molarity of camphor}=\frac{70\times 1000}{152.2\times 575}\\\\\text{Molarity of camphor}=0.799M$

Calculating the molarity of solution:

To calculate the mass of ethanol, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of ethanol = 0.785 g/mL

Volume of ethanol = 575 mL

Putting values in above equation, we get:

$0.785g/mL=\frac{\text{Mass of ethanol}}{575mL}\\\\\text{Mass of ethanol}=(0.785g/mL\times 575mL)=451.38g$

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

$\text{Molality of solution}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

$m_{solute}$ = Given mass of solute (camphor) = 70 g

$M_{solute}$ = Molar mass of solute (camphor) = 152.2 g/mol

$W_{solvent}$ = Mass of solvent (ethanol) = 451.38 g

Putting values in above equation, we get:

$\text{Molality of camphor}=\frac{70\times 1000}{152.2\times 451.38}\\\\\text{Molality of camphor}=1.02m$

Calculating the mole fraction of camphor:

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

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

For camphor:

Given mass of camphor = 70 g

Molar mass of camphor = 152.2 g/mol

Putting values in equation 1, we get:

$\text{Moles of camphor}=\frac{70g}{152.2g/mol}=0.459mol$

For ethanol:

Given mass of ethanol = 451.38 g

Molar mass of ethanol = 46 g/mol

Putting values in equation 1, we get:

$\text{Moles of ethanol}=\frac{451.38g}{46g/mol}=9.813mol$

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B}$

Moles of camphor = 0.459 moles

Total moles = [0.459 + 9.813] = 10.272 moles

Putting values in above equation, we get:

$\chi_{(camphor)}=\frac{0.459}{10.272}=0.045$ \

Calculating the mass percent of camphor:

To calculate the mass percentage of camphor in solution, we use the equation:

$\text{Mass percent of camphor}=\frac{\text{Mass of camphor}}{\text{Mass of solution}}\times 100$

Mass of camphor = 70 g

Mass of solution = [70 + 451.38] = 521.38 g

Putting values in above equation, we get:

$\text{Mass percent of camphor}=\frac{70g}{521.38g}\times 100=13.43\%$

Hence, the molarity of solution is 0.799 M , molality of solution is 1.02 m, mole fraction of camphor is 0.045 and mass percent of camphor in solution is 13.43 %

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