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Rom4ik [11]
3 years ago
14

The following reaction plays a key role in the destruction of ozone In the atmosphere:

Chemistry
1 answer:
lys-0071 [83]3 years ago
6 0

Answer:

19.91 J/K

Explanation:

The entropy is a measure of the randomness of the system, and it intends to increase in nature, thus for a spontaneous reaction ΔS > 0.

The entropy variation can be found by:

ΔS = ∑n*S° products - ∑n*S° reactants

Where n is the coefficient of the substance. The value of S° (standard molar entropy) can be found at a thermodynamic table.

S°, Cl(g) = 165.20 J/mol.K

S°, O3(g) = 238.93 J/mol.K

S°, O2(g) = 205.138 J/mol.K

So:

ΔS = (1*205.138 + 1*218.9) - (1*165.20 + 1*238.93)

ΔS = 19.91 J/K

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Deffense [45]
The answer for that is 16, math right?
7 0
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What’s the most density thing out of
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Answer:

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Explanation:

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7 0
3 years ago
How many moles of pcl5 can be produced from 28.0 g of p4 (and excess cl2)?
BARSIC [14]
P₄ + 10Cl₂ ---> 4PCl₅
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number of moles of P₄ reacted - 28.0 g / 124 g/mol = 0.22 mol 
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number of PCl₅ moles formed  -0.22 mol x 4 = 0.88 mol 
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5 0
3 years ago
The molar mass of an unknown gas was measured by an effusion experiment. It was found that it took 60 s for the gas to effuse, w
Marizza181 [45]

Answer:

The molar mass of the gas is 44 g/mol

Explanation:

It is possible to solve this problem using Graham's law that says: Rates of effusion are inversely dependent on the square of the mass of each gas. That is:

\frac{r_1}{r_2} =\frac{\sqrt{M_2} }{\sqrt{M_1} }

If rate of effusion of nitrogen is Xdistance / 48s and for the unknown gas is X distance / 60s and mass of nitrogen gas is 28g/mol (N₂):

\frac{X/48s}{X/60s} =\frac{\sqrt{M_2} }{\sqrt{28g/mol} }

6,61 = √M₂

44g/mol = M₂

<em>The molar mass of the gas is 44 g/mol</em>

<em></em>

I hope it helps!

4 0
3 years ago
Which TWO are most likely to form a compound?<br> C<br> Mg<br> S<br> Ar
lbvjy [14]
Answer is Mr and S as MgS ..
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