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BlackZzzverrR [31]
3 years ago
10

Calculate ΔSrxn∘ for the reaction

Chemistry
2 answers:
Yanka [14]3 years ago
7 0

The entropy of the reaction can be calculated alike the enthalpy of the reaction which is equal to the difference between the summation of entropies of the products multiplied by their corresponding stoich coeff. and the summation of the entropies of the reactants multiplied by their corresponding stoich coeff. In this case, the answer is -146.8 J / mol K
Effectus [21]3 years ago
6 0

Answer:

ΔS°rxn = -146.8 J/K

Explanation:

Let's consider the following reaction.

2 NO(g) + O₂(g) → 2 NO₂(g)

The standard entropy change of the reaction (ΔS°rxn) is equal to the standard entropies of the products times their stoichiometric coefficients minus the standard entropies of the reactants times their stoichiometric coefficients.

ΔS°rxn = 2 mol × S°(NO₂(g)) - 2 mol × S°(NO(g)) - 1 mol × S°(O₂(g))

ΔS°rxn = 2 mol × 240.0 J/mol.K - 2 mol × 210.8 J/mol.K - 1 mol × 205.2 J/mol.K

ΔS°rxn = -146.8 J/K

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A sample of nitrogen occupies 10.0 liters at 25°C what would be the new volume at 20°C? 7.9 L Ob 9.8 L 10.2 L 10.6 L
PtichkaEL [24]

Answer:

  9.4 liter

Explanation:

1) Data:

V₁ = 10.0 L

T₁ = 25°C = 25 + 273.15 K = 298.15 K

 P₁ = 98.7 Kpa

 T₂ = 20°C = 20 + 273.15 K = 293.15 K

  P₂ = 102.7 KPa

  V₂ = ?

2) Formula:

Used combined law of gases:

  PV / T = constant

  P₁V₁ / T₁ = P₂V₂ / T₂

3) Solution:

Solve the equation for V₂:

  V₂ = P₁V₁ T₂ / (P₂ T₁)

Substitute and compuite:

V₂ = P₁V₁ T₂ / (P₂ T₁)

V₂ = 98.7 KPa × 10.0 L × 293.15 K / (102.7 KPa × 298.15 K)

V₂ =  9.4 liter ← answer

You can learn more about gas law problems reading this other answer on

Explanation:

7 0
2 years ago
Which element has chemical properties that are most<br> similar to potassium, And why?
bekas [8.4K]

Answer:

Brainliest pls

Explanation:

The components potassium and sodium have comparable substance properties since they have a similar number of valence electrons

8 0
2 years ago
2.38g of black copper(iii) oxide is completely reduced to by hydrogen to give copper and water. What are the number of atoms of
OleMash [197]

Explanation:

here is the answer bae. Feel free to ask for more

6 0
3 years ago
The substance water always has a mass ratio of 11% H to 89% O. If 5.00g of a substance containing H and O was decomposed into .2
Gre4nikov [31]

Answer:

                    No the substance is not water.

Explanation:

                   The balance chemical equation for the decomposition of water is as follow;

                                           2 H₂O = 2 H₂ + O₂

Step 1: <u>Calculate moles of H₂O;</u>

               Moles  =  Mass / M.Mass

               Moles  =  5.0 g / 18.01 g/mol

               Moles  =  0.277 moles of H₂O

Step 2: <u>Calculate Moles of O₂ and H₂ produced by 0.277 moles of H₂O:</u>

According to equation,

                        2 moles of H₂O produced  =  1 mole of O₂

So,

                  0.277 moles of H₂O will produce  =  X moles of O₂

Solving for X,

                     X =  0.277 mol × 1 mol / 2 mol

                     X =  0.138 moles of O₂

Also,

According to equation,

                        2 moles of H₂O produced  =  2 mole of H₂

So,

                  0.277 moles of H₂O will produce  =  X moles of H₂

Solving for X,

                     X =  0.277 mol × 2 mol / 2 mol

                     X =  0.227 moles of H₂

Step 3: <u>Calculate Mass of O₂ and H₂ as;</u>

For O₂:

                 Mass  =  Moles × M.Mass

                 Mass  =  0.138 mol × 31.99 g/mol

                 Mass  =  4.44 g of O₂

For H₂:

                 Mass  =  Moles × M.Mass

                 Mass  =  0.227 mol × 2.01 g/mol

                 Mass  =  0.559 g of H₂

Conclusion:

                   From conclusion it is proved that the amount of H₂ produced by decomposition of 5 g of water should be 0.559 g while in statement it is less i.e. 0.290 g.

6 0
3 years ago
How many grams are in 3.14 x 1015 molecules of CO?
Vesnalui [34]

Answer:

Explanation:

Not Many

1 mol of CO has a mass of

C = 12

O = 16

1 mol = 28 grams.

1 mol of molecules = 6.02 * 10^23

x mol of molecules = 3.14 * 10^15        Cross multiply

6.02*10^23 x = 1 * 3.14 * 10^15             Divide by 6.02*10^23

x = 3.14*10^15 / 6.02*10^23

x = 0.000000005 mols

x = 5*10^-9

1 mol of CO has a mass of 28

5*10^-9 mol of CO has a mass of x                        Cross Multiply

x = 5 * 10^-9 * 28

x = 1.46 * 10^-7 grams

Answer: there are 1.46 * 10-7 grams of CO if only 3.14 * 10^15 molecules are in the sample

5 0
3 years ago
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