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Wewaii [24]
3 years ago
12

Which of the following describes nuclear fission?

Chemistry
2 answers:
valentina_108 [34]3 years ago
7 0
Good morning!

<span>It requires extremely high temperatures to begin the reaction.

</span>
While nuclear fusion involves the union of atoms, fission is the breaking of an atomic nucleus into two smaller parts. For this to happen it is necessary that the atom reaches a very high temperature.
Hugs!
kvv77 [185]3 years ago
5 0

Answer: The correct answer is its chain reactions can be initiated by neutrons.

Explanation: There are 2 types of nuclear reactions:

1. Nuclear fission reaction: In these type of reaction, one larger nuclei breaks down into two or more smaller nuclie's. Its chain reaction is initiated when the nucleus is struck with neutrons.

2. Nuclear fusion reactions: In these type of reaction, two small nuclei combine together to form a single large molecule. This type of reaction requires high temperature to occur.

Both these reactions occur due to the change in the nuclei of an atom.

From the above information, the nuclear fission chain reactions re initiated by the neutrons.

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7 0
3 years ago
A reaction of 41.9 g of Na and 30.3 g of Br2 yields 36.4 g of NaBr . What is the percent yield?
Licemer1 [7]

Answer: The percent yield is, 93.4%

Explanation:

First we have to calculate the moles of Na.

\text{Moles of Na}=\frac{\text{Mass of Na}}{\text{Molar mass of Na}}=\frac{41.9g}{23g/mole}=1.82moles

Now we have to calculate the moles of Br_2

{\text{Moles of}Br_2} = \frac{\text{Mass of }Br_2 }{\text{Molar mass of} Br_2} =\frac{30.3g}{160g/mole}=0.189moles

{\text{Moles of } NaBr} = \frac{\text{Mass of } NaBr }{\text{Molar mass of } NaBr} =\frac{36.4g}{103g/mole}=0.353moles

The balanced chemical reaction is,

2Na(s)+Br_2(g)\rightarrow 2NaBr

As, 1 mole of bromine react with = 2 moles of Sodium

So, 0.189 moles of bromine react with = \frac{2}{1}\times 0.189=0.378 moles of Sodium

Thus bromine is the limiting reagent as it limits the formation of product and Na is the excess reagent.

As, 1 mole of bromine give = 2 moles of Sodium bromide

So, 0.189 moles of bromine give = \frac{2}{1}\times 0.189=0.378 moles of Sodium bromide

Now we have to calculate the percent yield of reaction

\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100=\frac{0.353 mol}{0.378}\times 100=93.4\%

Therefore, the percent yield is, 93.4%

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