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1 year ago

A nuclear reaction can be controlled using which of the following?

Chemistry

1 answer:

gavmur [86]1 year ago

3 0

Control rods is regarded as a substance which can control nuclear reaction and is denoted as option A.

<h3>What is Nuclear reaction?</h3>

This is a type of reaction which involves the nuclei of an atom being split or joined together and is usually accompanied with a great amount of energy emitted in the form of heat.

The two types of nuclear reaction include:

Nuclear fusion

Nuclear fission

Nuclear fusion involves the collision of two or more nuclei in other form a new atom while nuclear fission involves splitting the atoms to form other products and involves a very high amount of energy and exposure to organisms should be prevented as the radiations can alter DNA.

Control rods are commonly used in nuclear reactors and control the rate of fission in the fuel used. This is therefore the reason why it was chosen as the most appropriate choice.

Read more about Nuclear reaction here brainly.com/question/984564

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$\Delta\text{H}_1+2\Delta\text{H}_2-\Delta\text{H}_3$

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Hess's Law of Constant Heat Summation states that if a chemical equation can be written as the sum of several other chemical equations, the enthalpy change of the first chemical equation is equal to the sum of the enthalpy changes of the other chemical equations. Thus, the reaction that involves the conversion of reactant A to B, for example, has the same enthalpy change even if you convert A to C, before converting it to B. Regardless of how many steps it takes for the reactant to be converted to the product, the enthalpy change of the overall reaction is constant.

With Hess's Law in mind, let's see how A can be converted to 2C +E.

$\bf{\text{A} \rightarrow 2\text{B}}$ (Δ $\text{H}_1$ ) -----(1)

Since we have 2B, multiply the whole of II. by 2:

$\bf{2\text{B} \rightarrow 2\text{C} +2\text{D}}$ (2Δ $\text{H}_2$ ) -----(2)

This step converts all the B intermediates to 2C +2D. This means that the overall reaction at this stage is $\text{A} \rightarrow 2\text{C} +2\text{D}$ .

Reversing III. gives us a negative enthalpy change as such:

$\bf{2\text{D} \rightarrow \text{E}}$ (-Δ $\text{H}_3$ ) -----(3)

This step converts all the D intermediates formed from step (2) to E. This results in the overall equation of $\text{A} \rightarrow 2\text{C} +\text{E}$ , which is also the equation of interest.