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Lina20 [59]
3 years ago
7

HELP PLSSSSS!!

Chemistry
1 answer:
11Alexandr11 [23.1K]3 years ago
7 0

Answer:

68133080.02 g

Explanation:

I believe that the question is to find the mass of air in the room and not the molar mass of air since the molar mass of air was already given in the question as 28.97 g/mol.

Now, if 1 mole of a gas occupies 22.4 L

x moles of air occupies 52,681,428.8 Liters

x = 1 * 52,681,428.8 /22.4

x = 2351849.5 moles of air

Now, number of moles = mass/ molar mass

but molar mass = 28.97 g/mol

2351849.5 = mass/28.97

mass = 2351849.5 * 28.97

mass = 68133080.02 g

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According to the law of conservation of mass. Atoms are neither created nor destroyed in a chemical reaction. What this means is that in a chemical reaction, the number of atoms of each element on the left hand side must be the same as the same as the number of atoms of the same element on the right hand side.

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3 years ago
Write about similarities and the differences of the plant and animal cell
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Explanation:

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5 0
3 years ago
you are given mixtures containin gthe following compounds. Which compound in each pair could be separated by stirring the solid
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Explanation:

We need to see the solubility in water, at similar temperatures, for each compound and see which one is less soluble than the other:

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4 years ago
A filament at which temperature would have the most light absorbed by a piece of yellow glass?
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17) Chlorine and Fluorine react to form gaseous chlorine trifluoride. You start with 1.75 mole of chlorine and 3.68 moles of flu
Mama L [17]

Answer:

a) Cl_2 + 3F_2 \rightarrow 2ClF_3

b) F_2 is the limiting reagent

c) Moles of ClF_3 produced = 2.45 mol

b) Moles of Cl_2 left = 1.75 -1.22 = 0.53

Explanation:

a) Balanced reaction:

Cl_2 + 3F_2 \rightarrow 2ClF_3

b)

No. of mole of Cl_2 = 1.75\ mol

No. of mole of F_2 = 3.68\ mol

Cl_2 + 3F_2 \rightarrow 2ClF_3

As, it is clear from the reaction that,

1 mol of Cl_2 requires 3 moles of F_2

1.75 mol of Cl_2 require = 1.75 × 3 = 5.25 mol of F_2

As, only 3.68 mol of F_2 is present, so F_2 is the limiting reagent.

c)

3 moles of F_2 form 2 moles of ClF_3

3.68 moles of F_2 will form = \frac{2}{3}  \times 3.68 = 2.45\ mol\ of\ ClF_3

d)

Cl_2 is present in excess.

3 moles of F_2 requires 1 mol of Cl_2

3.68 moles of F_2 will require = \frac{1}{3} \times 3.68 = 1.22\ mol\ of\ Cl_2

Cl_2 left = 1.75 -1.22 = 0.53 mol

6 0
4 years ago
Read 2 more answers
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