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Viefleur [7K]
3 years ago
15

CK-12 Boyle and Charles's Laws if Mrs. Pa pe prepares 12.8 L of laughing gas at 100.0 k Pa and -108 °C and then she force s the

gas into an 855 mL cylinder whe re it warms to room temperature at 22 °C, NC what is the pressure of this gas in kilopascals
Chemistry
1 answer:
nirvana33 [79]3 years ago
3 0

Answer:

The answer to your question is   P2 = 2676.6 kPa

Explanation:

Data

Volume 1 = V1 = 12.8 L                        Volume 2 = V2 = 855 ml

Temperature 1 = T1 = -108°C               Temperature 2 = 22°C

Pressure 1 = P1 = 100 kPa                    Pressure 2 = P2 =  ?

Process

- To solve this problem use the Combined gas law.

                     P1V1/T1 = P2V2/T2

-Solve for P2

                     P2 = P1V1T2 / T1V2

- Convert temperature to °K

T1 = -108 + 273 = 165°K

T2 = 22 + 273 = 295°K

- Convert volume 2 to liters

                       1000 ml -------------------- 1 l

                         855 ml --------------------  x

                         x = (855 x 1) / 1000

                         x = 0.855 l

-Substitution

                    P2 = (12.8 x 100 x 295) / (165 x 0.855)

-Simplification

                    P2 = 377600 / 141.075

-Result

                   P2 = 2676.6 kPa

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Soloha48 [4]

The question is incomplete, the complete question is;

1. Given: 2A(g) <-> 2B(g) + C(g). At a particular temperature, K = 1.6x104.

At a higher temperature, K = 1.8x10-5. Placing the equilibrium mixture in an ice bath (thus lowering the temperature) will...

cause [A] to increase

cause [B] to increase

gave no effect

cannot be determined

Answer:

cause [B] to increase

Explanation:

We have to have it behind our minds that;  for an exothermic reaction, when the temperature is increased, the value of the equilibrium constant decreases. On the other hand, for an endothermic reaction, increasing the temperature increases the value of the equilibrium constant.

In this particular case, increasing the temperature decreased the equilibrium constant from K = 1.6x104 to K = 1.8x10-5 meaning that the equilibrium constant decreased with increase in temperature. This means that the reaction is exothermic and a decrease in temperature will favour the forward reaction hence more B is produced.

3 0
3 years ago
How many molecules of water can be produced from 46.72 liters of C8H18? 2\:C_8H_{18}\:+\:25\:O_2\:\:\Longrightarrow\:\:16\:CO_2\
Dmitry [639]

Answer:

1.13×10^25 molecules of water.

Explanation:

Equation of the reaction;

C8H18(g) + 25/2 O2 (g) -------> 8CO2(g) + 9H2O(l)

It is important to first put down the balanced reaction equation. It is not possible to solve any problem on stoichiometric relationship without a balanced reaction equation. Once the equation is obtained, we can now proceed with other steps in the solution of the problem.

From the reaction equation, 1 mole of C8H18 produces 9 moles of water

1 mole of C8H18 occupies 22.4L volume while 1 mole of water contains 6.02×10^23 molecules of water

Hence

22.4 L of C8H18 produces 9(6.02×10^23) molecules of water

46.72 L of C8H18 will produce 46.72 L × 9(6.02×10^23) molecules of water/22.4 L

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5 0
3 years ago
In which of the following choices is the oxidation number incorrect?
8_murik_8 [283]

Cl2(s); oxidation number 1 is the incorrect choices in oxidation number.

Explanation:

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Oxidation number depends on the number of electrons gained or lost by an atom of the element say in compound formation.

If electron is gained oxidation number becomes negative.

If electron is lost then oxidation number is positive.

If the octet rule is fulfilled that valence shell is filled them atomic number gets zero. Since Cl2 is in neutral state the oxidation number is 0.

Oxidation number in general can be made out by checking the valency of the element as oxidation number is also equal to the valency.

4 0
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To know more about ideal gas equation here

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3 0
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