Which must be the same when comparing 1 mol of oxygen gas, O2, with 1 mol of carbon monoxide gas, CO?

Chemistry

2 answers:

gayaneshka [121]3 years ago

7 0

Answer: C. The Number of Molecules

Explanation:

I just took the test.

Llana [10]3 years ago

5 0

A mole is a rather unusual unit of measurement. It is defined by a specific number of particles, molecules, atoms, or any other unit.
That number happens to be 6.022×10²³.
When we talk about a mole of water, then, we refer to 6.022×10²³ molecules of water.

Comparing 1 mole of 02 with 1 mole of CO:
This means means that both quantities will have the same number of molecules. The answer is C.

It may be tempting to select D, since it relates to the number of atoms rather than mass or volume, and moles are related to this idea. However, keep in mind that one mole of O2 will actually have twice as much oxygen atoms as one mole of CO, since there are two oxygen atoms per O2 molecule and only 1 oxygen atom per CO molecule.

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How many atoms are in 0.0303 mol K?

Inga [223]

1.661x10 negative 24 squared

8 0

3 years ago

One of the compounds used to increase the octane rating of gasoline is toluene (pictured). Suppose 43.3 mL of toluene (d = 0.867

Thepotemich [5.8K]

<u>Answer:</u>

(A)

Density = Mass / Volume

Mass = Density × Volume

$= 0.867 g/mL \times 43.3mL = 37.5411 g Toluene$

$1C_6 H_5 CH_3 + 9 O_2 > 7 CO_2 + 4 H_2 O$

Mole ratio of toluene : Oxygen is 1 : 9

$37.5411 g \text { Toluene } \times \frac{1 \text {mol} \text {toluene}}{92 g \text { toluene}} \times \frac{9 {mol} O_{2}}{1 \text {mol} \text { toluene }} \times \frac{32 g O_{2}}{1 {mol} O_{2}}=117 g O_{2}(\text {Answer})$

(B)

1 mole of Toluene produces 7 moles of $CO_2$ gas and 4 moles of $H_2 O$ Vapour

So the mole ratio is 1 : 11

$37.5411 g Toluene $\times \frac{1 \text { mol toluene }}{92 g \text { toluene }} \times \frac{11 \mathrm{mol} \text { gas }}{1 \text { mol toluene }} $$\\\\=4.49 \text { mol gaseous products (Answer) } $$

(C)

1mole contains $6.022\times10^{23}$ molecules

$37.5411 g Toluene $\times \frac{1 \text { mol toluene }}{92 g \text { toluene}} \times \frac{4 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{mol} \text { toluene }} \times \frac{6.022 \times 10^{23} \text { molecules } \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}} $\\\\$=9.82 \times 10^{23} \text { molecules } \mathrm{H}_{2} \mathrm{O} \text { (Answer) } $$