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vlabodo [156]
3 years ago
14

What volume is occupied by 8.7 g of chlorine gas, Cl2, at 23°C and 1.15 atm pressure

Chemistry
1 answer:
Harrizon [31]3 years ago
4 0

Answer:

V = 5.17L

Explanation:

Mass of gas = 8.7g

T = 23°C = (23 + 273.15)K = 296.15K

P = 1.15 atm

V = ?

R = 0.082atm.L / mol.K

From ideal gas equation

PV = nRT

P = pressure of the gas

V = volume of the gas

n = no. Of moles

R = ideal gas constant

T = temperature of the gas

no of moles = mass / molar mass

Molar mass of Chlorine = 35.5g / mol

No. Of moles = 8.7 / 35.5

No. Of moles = 0.245 moles

PV = nRT

V = nRT / P

V = (0.245 * 0.082 * 296.15) / 1.15

V = 5.9496 / 1.15

V = 5.17L

The volume of the gas is 5.17L

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Answer :

The rate law expression for zero order reaction will be:

Rate=k[A]^0

The rate law expression for first order reaction will be:

Rate=k[A]^1

The rate law expression for second order reaction will be:

Rate=k[A]^2

Zero order reaction : There is no affect on the rate law.

First order reaction : The rate law becomes doubled.

Second order reaction : The rate law becomes quadrupled.

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

The given reaction is:

A\rightarrow Products

The rate law expression for zero order reaction will be:

Rate=k[A]^0

The rate law expression for first order reaction will be:

Rate=k[A]^1

The rate law expression for second order reaction will be:

Rate=k[A]^2

Now we have to determine that if doubling of the concentration of A then the rate of reaction will be:

As we know that the zero order reaction does not depend on the concentration of reactant. So, there is no affect on the rate law.

As we know that the first order reaction depend on the concentration of reactant. So, the rate law becomes doubled.

As we know that the second order reaction depend on the concentration of reactant. So, the rate law becomes quadrupled.

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What is 7 2/3=1/5x+2/3-1 1/5x
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4 years ago
Object E has a mass of 4,800 kilograms. Object F has a mass of 600 kilograms. The weight of Object F will be ________ times the
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Answer:The weight of Object F will be \frac{1}{8}times the weight of Object E if both objects are placed on the same planet.

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Mass of the object E,M_E= 4800 kg

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The weight of Object F will be \frac{1}{8}times the weight of Object E if both objects are placed on the same planet.

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