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Kryger [21]
2 years ago
12

The energy required to break existing chemical bonds in reactants is called the ______ energy.

Chemistry
1 answer:
mixer [17]2 years ago
8 0

The energy required to break existing chemical bonds in reactants is called the activation energy.

<h3>What is activation energy?</h3>

Activation energy in chemistry is the energy required to initiate a chemical reaction.

Chemical reactions involve the breaking of chemical bonds in substances called reactants to form new substances called products.

The energy required to break the bond in the existing reactants thus elevating these substances to a state of high activation is known as activation energy.

Therefore, it can be said that energy required to break existing chemical bonds in reactants is called the activation energy.

Learn more about activation energy at: brainly.com/question/11334504

#SPJ1

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Answer:

C.

Explanation:

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You need to prepare .200M solution of hydrochloric acid (HCl). If you took out .830ml of the 12.0M stock solution. How much wate
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Answer:

You must add 48.97 mL of water to make the 0.200 M diluted solution.

Explanation:

In chemistry, dilution is the reduction in concentration of a chemical in a solution. In other words, it is the process of reducing the concentration of solute in solution, simply adding more solvent to the solution.

In a dilution, the quantity or mass of the solute is not changed but only that of the solvent. As only solvent is being added, by not increasing the amount of solute the concentration of the solute decreases.

The expression for the dilution calculations is:

Cinitial* Vinitial = Cfinal* Vfinal

In this case:

  • Cinitial= 12 M
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Replacing:

12 M*0.830 mL= 0.200 M*Vfinal

Solving:

Vfinal=\frac{12 M*0.830 mL}{0.200 M}

Vfinal= 49.8 mL

Since 0.830 mL is the volume you initially have of HCl, the amount of water you must add is:

49.8 mL - 0.830 mL= 48.97 mL

<u><em>You must add 48.97 mL of water to make the 0.200 M diluted solution.</em></u>

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What are four chemical and physical properties of phosphorus?
DedPeter [7]

Answer:

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Explanation:

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