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4 years ago

What pressure is required to compress 156.0 liters of air at 2.00 atmosphere into a cylinder

Chemistry

1 answer:

kiruha [24]4 years ago

7 0

Answer:

8.67 atm

Explanation:

The data were obtained from the question:

Initial volume (V1) = 156L

Initial pressure (P1) = 2 atm

Final volume (V2) = 36L

Final pressure (P2) =.?

Next, we'll apply the Boyle's law equation in order to obtain the pressure required to compress the air. This is illustrated below:

P1V1 = P2V2

2 x 156 = P2 x 36

Divide both side by 36

P2 = 2 x 156 / 36

P2 = 8.67 atm

Therefore, a pressure of 8.67 atm is required to compress the air into the cylinder.

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The gas is in the standard temperature and pressure condition i.e. at S.T.P

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3 years ago

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4 years ago

The first step in HNO3 production is the catalyzed oxidation of NH3. Without a catalyst, a different reaction predominates:

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Answer:

Kc = 6x10⁻⁶

Explanation:

For the reaction:

4NH₃(g) + 3O₂(g) ⇄ 2N₂(g) + 6H₂O(g)

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Kc =[N₂]² [H₂O]⁶ / [NH₃]⁴ [O₂]³

The equilibrium concentrations of the gases is -Because volume of the container is 1.00L-:

[N₂] = 2X = 1.96x10⁻³; <em>X = 9.8x10⁻⁴</em>

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4 years ago

Consider the decomposition of red mercury(II) oxide under standard state conditions. )H0 T SFE ڮ( H M 0 H (a) Is the decompositi

sertanlavr [38]

The question is incomplete, complete question is :

Consider the decomposition of red mercury(II) oxide under standard state conditions.

$2HgO(s)\rightarrow 2Hg(l)+O_2(g)$

Given :

$\Delta S_{HgO}^o=70.29 J/mol K$

$\Delta S_{Hg}^o=75.9 J/mol K$

$\Delta S_{O_2}^o=205.2 J/mol K$

Enthalpy change of the reaction = ΔH = -90.83 kJ/mol

(a) Is the decomposition spontaneous under standard state conditions?

(b) Above what temperature does the reaction become spontaneous?

Answer:

a) The decomposition is spontaneous under standard state conditions.

b)The reaction will spontaneous above -419.69 Kelvins.

Explanation:

$2HgO(s)\rightarrow 2Hg(l)+O_2(g)$

Given :

$\Delta S_{HgO}^o=70.29 J/mol K$

$\Delta S_{Hg}^o=75.9 J/mol K$

$\Delta S_{O_2}^o=205.2 J/mol K$

Entropy change of the reaction ; ΔS

$\Delta S=[2\times \Delta S_{Hg}^o+1\times \Delta S_{O_2}^o]-[2\times \Delta S_{HgO}^o]$

$=[2\times 75.9 J/mol K+1\times 205.2 J/molK]-[2\times 70.29 J/molK]=216.42 J/mol K$

Enthalpy change of the reaction = ΔH = -90.83 kJ/mol = -90830 J/mol K

1 kJ = 1000 J

At standard condition the value of temperature = T = 298 K

ΔG = ΔH - TΔS

ΔG = -90830 J/mol K - 298 K × 216.42 J/mol K = -155,323.16 J/mol

ΔG < 0 ( spontaneous)

The decomposition is spontaneous under standard state conditions.

b) Above what temperature does the reaction become spontaneous

Let the ΔG = 0

Enthalpy change of the reaction = ΔH = -90.83 kJ/mol = -90830 J/mol K

ΔG = ΔH - TΔS

0 = -90830 J/mol K - T × 216.42 J/mol K

T = -419.69 K

The reaction will spontaneous above -419.69 Kelvins.

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