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Rom4ik [11]
4 years ago
14

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

Chemistry
1 answer:
lys-0071 [83]4 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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What happens to the length of a year for each planet as you get farther away from the sun
Bond [772]

Answer:

they get colder and darker, with less light

Explanation:

3 0
3 years ago
The distance from dalton, ga to Dallas, tx is 1308 km. What is the distance in miles
Oksana_A [137]
Answer:
812.754 miles



Why:
0.621371 miles per km

4 0
3 years ago
What is the mass of carbon dioxide formed when 36g of carbon is burnt?<br> C+O2 -------&gt;C02
Anestetic [448]

Answer:

Mass = 132 g

Explanation:

Given data:

Mass of CO₂ formed = ?

Mass of C burnt = 36 g

Solution:

Chemical equation:

C + O₂    →    CO₂

Number of moles of carbon:

Number of moles = mass/molar mass

Number of moles = 36 g/ 12 g/mol

Number of moles = 3 mol

now we will compare the moles of carbon and carbon dioxide.

                C           :        CO₂

                 1           :         1

                3           :         3

Mass of CO₂:

Mass = number of moles × molar mass

Mass = 3 mol × 44 g/mol

Mass = 132 g

7 0
3 years ago
What is the free energy change if the ratio of the concentrations of the products to the concentrations of the reactants is 22.3
snow_lady [41]

The free energy change(Gibbs free energy-ΔG)=-8.698 kJ/mol

<h3>Further explanation</h3>

Given

Ratio of the concentrations of the products to the concentrations of the reactants is 22.3

Temperature = 37 C = 310 K

ΔG°=-16.7 kJ/mol

Required

the free energy change

Solution

Ratio of the concentration : equilbrium constant = K = 22.3

We can use Gibbs free energy :

ΔG = ΔG°+ RT ln K

R=8.314 .10⁻³ kJ/mol K

\tt \Delta G=-16.7~kJ/mol+8.314.10^{-3}\times 310\times ln~22.3\\\\\Delta G=-8.698~kJ/mol

8 0
3 years ago
NEED HELP ASAFP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!PLEASE!
sp2606 [1]
Here to help! What’s up?
6 0
4 years ago
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