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spayn [35]
3 years ago
12

The activation energy of a reaction going on its own is 20 kj. If the reaction was treated with a catalyst, which would most lik

ely represent the amount of energy needed to start a reaction?
15 kj
20 kj
25 kj
30kj
Physics
2 answers:
storchak [24]3 years ago
7 0

Answer:

15 kJ

Explanation:

We need to know the following;

what is activation energy?

  • Activation energy is the minimum energy required by reactants for a chemical reaction to take place.

What are catalysts?

  • Catalysts are molecules or substances that speed up the rate of a chemical reaction.
  • In other words, catalysts increase the speed at which products are formed from reactants.

How do catalysts speed up reactions?

  • Catalysts speed the rate of chemical reactions by lowering the activation energy of the reactants.
  • When activation energy is lowered, reactants require less energy for the reaction to occur which easily achieved making the reaction proceed faster.
  • Therefore, in presence of a catalyst, the new activation energy should be lower than the initial value of activation energy.
Arte-miy333 [17]3 years ago
5 0

Answer:

A 15kj

Explanation:

just did the assignment

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If the charge remains the same but the radius of the sphere is doubled, the electric flux coming out of it will be
il63 [147K]

Answer:

Explanation:

We shall apply Gauss's theorem for electric flux to solve the problem . According to this theorem , total electric flux coming out of a charge q can be given by the following relation .

∫ E ds = q / ε

Here q is assumed to be enclosed in a closed surface , E is electric intensity on the surface so

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4 0
3 years ago
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masha68 [24]

Answer:

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2 years ago
1. A large ball was let go on a hill and started rolling down with a constant acceleration of 4.2 m/s². What was the velocity of
pochemuha

Answer:

<em>The velocity after 12s is 50.4m/s</em>.

Explanation:

<em>In acceleration formula make velocity the </em><em>subject.</em>

<em> acceleration(a) = velocity(</em>v)÷time(t)

<h3><em> </em><em>velocity</em><em> </em><em>(</em><em>v)</em><em> </em><em>=</em><em> </em><em>acceleration</em><em>(</em><em>a)</em><em>×</em><em>t</em><em>ime</em><em>(</em><em>t)</em></h3>

<em>V </em><em>=</em><em> </em><em>4</em><em>.</em><em>2</em><em>m</em><em>/</em><em>s²</em><em>×</em><em>1</em><em>2</em><em>s</em>

<em>V </em><em>=</em><em> </em><em>5</em><em>0</em><em>.</em><em>4</em><em>m</em><em>/</em><em>s</em>

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8 0
2 years ago
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