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2 years ago

6

Which statement best describes a mole?

1 answer:

viva [34]2 years ago

5 0

Answer: A mole is a unit of measurement.

Explanation:

moles= mass/molar mass

for example moles=1000g/342.296 g/mol

moles=2.921 mol

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Now if your lab partner determined that there were 1.552 x 1022 atoms of H in an unknown sample of C2H4Cl2, how many milligrams

STALIN [3.7K]

To determine the mass of the sample in milligrams in this problem, we use the avogadro's number to convert from atoms to moles, relate the moles of element in the sample to the mole present and the molar mass of the sample. We do as follows:

1.552 x 10^22 atoms H ( 1 mol H / 6.022x10^23 atoms H ) ( 1 mol C2H4Cl2 / 4 mol H ) ( 98.96 g C2H4Cl2 / 1 mol C2H4Cl2 ) = 0.625 g C2H4Cl2 = 625 mg <span>C2H4Cl2</span>

3 0

4 years ago

A mouse is placed in a sealed chamber with air at 769.0 torr. This chamber is equipped with enough solid KOH to absorb any CO2 a

svlad2 [7]

<u>Answer:</u> The amount of oxygen gas consumed by mouse is 0.202 grams.

<u>Explanation:</u>

We are given:

Initial pressure of air = 769.0 torr

Final pressure of air = 717.1 torr

Pressure of oxygen = Pressure decreased = Initial pressure - Final pressure = (769.0 - 717.1) torr = 51.9 torr

To calculate the amount of oxygen gas consumed, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 51.9 torr

V = Volume of the gas = 2.20 L

T = Temperature of the gas = 292 K

R = Gas constant = $62.364\text{ L. Torr }mol^{-1}K^{-1}$

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

$51.9torr\times 2.20L=n\times 62.364\text{ L. Torr }mol^{-1}K^{-1}\times 292K\\\\n=\frac{51.9\times 2.20}{62.364\times 292}=0.0063mol$

To calculate the mass from given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of oxygen gas = 0.0063 moles

Molar mass of oxygen gas = 32 g/mol

Putting values in above equation, we get:

$0.0063mol=\frac{\text{Mass of oxygen gas}}{32g/mol}\\\\\text{Mass of oxygen gas}=(0.0063mol\times 32g/mol)=0.202g$

Hence, the amount of oxygen gas consumed by mouse is 0.202 grams.

5 0

3 years ago