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Keith_Richards [23]
3 years ago
7

What is one way that fission and fusion are similar?

Chemistry
2 answers:
nignag [31]3 years ago
8 0
The Correct Answer Is <span>Both Processes Produce Energy From Mass. </span>
Lubov Fominskaja [6]3 years ago
5 0

Explanation:

Fusion reaction is defined as a reaction in which two small nuclei combine together and result in the formation of a large nuclei along with release of huge amount of energy.

For example, ^{2}_{1}H + ^{2}_{1}H \rightarrow ^{4}_{2}He is a nuclear fusion reaction.

On the other hand, a fission reaction is defined as a reaction in which a large nuclei splits into two small nuclei along with release of huge amount of energy.

For example, ^{238}_{92}U \rightarrow ^{234}_{90}Th + ^{4}_{2}He is a nuclear fission reaction.

Therefore, in both these reactions it is common that energy is releasing due to conversion of mass into energy.

Thus, we can conclude that energy is released by converting mass into energy is one way that fission and fusion are similar.

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Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 with aqueous NaI . Include phases. chemical equation: Wha
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<u>Answer:</u> The mass of lead iodide produced is 9.22 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}

Molarity of NaI = 0.200 M

Volume of solution = 0.200 L

Putting values in above equation, we get:

0.200M=\frac{\text{Moles of NaI}}{0.200}\\\\\text{Moles of NaI}=(0.200mol/L\times 0.200L)=0.04moles

The chemical equation for the reaction of NaI and lead chlorate follows:

Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)

By Stoichiometry of the reaction:

2 moles of NaI reacts produces 1 mole of lead iodide

So, 0.04 moles of NaI will react with = \frac{1}{2}\times 0.04=0.02mol of lead iodide

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Molar mass of lead iodide = 461 g/mol

Moles of lead iodide= 0.02 moles

Putting values in above equation, we get:

0.02mol=\frac{\text{Mass of lead iodide}}{461g/mol}\\\\\text{Mass of lead iodide}=(0.02mol\times 461g/mol)=9.22g

Hence, the mass of lead iodide produced is 9.22 grams

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