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Anit [1.1K]
3 years ago
11

An increase in temperature results in A) a decrease in the required activation energy while the reaction rate remains constant.

B) an increase in reaction rate due to a decrease in the kinetic energy of the reactants. C) an increase in the rate of reaction because reactant molecules collide with greater energy. D) an increase in both the reaction rate and activation energy due to increased kinetic energy.
Chemistry
1 answer:
N76 [4]3 years ago
7 0

Answer:

C) an increase in rate of reaction because reactant molecules collide with greater energy

Explanation:

Temperature is one of the factors that affect the rate of a reaction. The rate of a reaction increases with an increase in temperature and vice versa. When the temperature of a reaction increases, the kinetic energy of the reactant molecules increases causing them to react at a faster rate.

The reactant molecules respond to an increase in temperature by colliding at a faster rate due to an increased kinetic energy between the reactant molecules.

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A scuba diver with a local Sheriff’s office has been assigned the task of retrieving a sunken vessel in the Ocean. The vessel is
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A balloon containing helium gas expands from 230
Anit [1.1K]

The answer for the following problem is mentioned below.

  • <u><em>Therefore the final  moles of the gas is 14.2 × </em></u>10^{-4}<u><em> moles.</em></u>

Explanation:

Given:

Initial volume (V_{1}) = 230 ml

Final volume (V_{2}) = 860 ml

Initial moles (n_{1}) = 3.8 ×10^{-4} moles

To find:

Final moles (n_{2})

We know;

According to the ideal gas equation;

    P × V = n × R × T

where;

P represents the pressure of the gas

V represents the volume of the gas

n represents the no of the moles of the gas

R represents the universal gas constant

T represents the temperature of the gas

So;

    V ∝ n

\frac{V_{1} }{V_{2} } = \frac{n_{1} }{n_{2} }

where,

(V_{1}) represents the initial volume of the gas

(V_{2}) represents the final volume of the gas

(n_{1}) represents the initial  moles of the gas

(n_{2}) represents the final moles of the gas

Substituting the above values;

   \frac{230}{860} = \frac{3.8 * 10^-4}{n_{2} }

  n_{2} = 14.2 × 10^{-4} moles

<u><em>Therefore the final  moles of the gas is 14.2 × </em></u>10^{-4}<u><em> moles.</em></u>

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3 years ago
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