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3 years ago

URGENT!!!!!!!!Which reaction takes place in a nuclear fission reactor?

Chemistry

2 answers:

ivolga24 [154]3 years ago

6 0

Answer: The correct answer is Option d.

Explanation:

Nuclear fission reaction is defined as the reaction in which 1 heavier nuclei splits into two or more smaller nuclei.

As, nuclear fission reactors require a large amount of energy to split the nucleus, hence it is bombarded with a particle in order to star the reaction.

In a nuclear reaction, a large fissile atomic nucleus like Uranium-235 $(_{92}^{235}\textrm{U})$ or Plutonium-239 $(_{94}^{239}\textrm{Pu})$ when absorbs a neutron, it undergoes nuclear fission reaction.

Hence, in the given options, uranium-235 is absorbing a neutron. Therefore, the correct answer is Option d.

pychu [463]3 years ago

5 0

Answer:

Explanation:

on edge 2020

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A sample of P2Cl5 contains 179 g of phosphorous, how many grams of chlorine are present? Please show step by step thx!

natima [27]

Answer:

About 512 g.

Explanation:

We are given a sample of P₂Cl₅ that contains 179 grams of phosphorus, and we want to determine the grams of chlroine that is present.

Thus, we can convert from grams of phosphorus to moles of phosphorus, moles of phosphorus to moles of chlorine, and moles of chlorine to grams of chlorine.

From the formula, there are two moles of P for every five moles of Cl. The molecular weights of P and Cl are 30.97 g/mol and 35.45 g/mol, respectively. Hence:

$\displaystyle 179\text{ g P} \cdot \frac{1\text{ mol P}}{30.97\text{ g P}} \cdot \frac{5\text{ mol Cl}}{2\text{ mol P}} \cdot \frac{35.45\text{ g Cl}}{1\text{ mol Cl}} = 512\text{ g Cl}$

In conclusion, there is about 512 grams of chlorine present in the sample.

Alternatively, we can mass percentages. The mass percent of phosphorus in P₂Cl₅ is:

$\displaystyle \% \text{P} = \frac{2(30.97)}{5(35.45) + 2(30.97)} = 25.90\%$

Because there are 179 grams of phosphorus, the total amount of sample present is:

$\displaystyle \begin{aligned} 25.90\% \cdot m_T & = 179\text{ g P} \\ \\ m_T & = 691.1 \text{ g}\end{aligned}$

Therefore, the amount of chlorine present is 691.1 g - 179 g, or about 512 g, in agreement with our above answer.

8 0

3 years ago

Why metals are not often used to make clothes

8_murik_8 [283]

Answer:

They would be to heavy and you would be really stiff

Explanation:

3 0

3 years ago

Read 2 more answers

A solution contains 0.0440 M Ca2 and 0.0940 M Ag. If solid Na3PO4 is added to this mixture, which of the phosphate species would

Olenka [21]

Answer:

C. $Ca_3(PO_4)_2$ will precipitate out first

the percentage of $Ca^{2+}$ remaining = 12.86%

Explanation:

Given that:

A solution contains:

$[Ca^{2+}] = 0.0440 \ M$

$[Ag^+] = 0.0940 \ M$

From the list of options , Let find the dissociation of $Ag_3PO_4$

$Ag_3PO_4 \to Ag^{3+} + PO_4^{3-}$

where;

Solubility product constant Ksp of $Ag_3PO_4$ is $8.89 \times 10^{-17}$

Thus;

$Ksp = [Ag^+]^3[PO_4^{3-}]$

replacing the known values in order to determine the unknown ; we have :

$8.89 \times 10 ^{-17} = (0.0940)^3[PO_4^{3-}]$

$\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3} = [PO_4^{3-}]$

$[PO_4^{3-}] =\dfrac{8.89 \times 10 ^{-17}}{(0.0940)^3}$

$[PO_4^{3-}] =1.07 \times 10^{-13}$

The dissociation of $Ca_3(PO_4)_2$

The solubility product constant of $Ca_3(PO_4)_2$ is $2.07 \times 10^{-32}$

The dissociation of $Ca_3(PO_4)_2$ is :

$Ca_3(PO_4)_2 \to 3Ca^{2+} + 2 PO_{4}^{3-}$

Thus;

$Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2$

$2.07 \times 10^{-33} = (0.0440)^3 [PO_4^{3-}]^2$

$\dfrac{2.07 \times 10^{-33} }{(0.0440)^3}= [PO_4^{3-}]^2$

$[PO_4^{3-}]^2 = \dfrac{2.07 \times 10^{-33} }{(0.0440)^3}$

$[PO_4^{3-}]^2 = 2.43 \times 10^{-29}$

$[PO_4^{3-}] = \sqrt{2.43 \times 10^{-29}$

$[PO_4^{3-}] =4.93 \times 10^{-15}$

Thus; the phosphate anion needed for precipitation is smaller i.e $4.93 \times 10^{-15}$ in $Ca_3(PO_4)_2$ than in $Ag_3PO_4$ $1.07 \times 10^{-13}$

Therefore:

$Ca_3(PO_4)_2$ will precipitate out first

To determine the concentration of $[Ca^+]$ when the second cation starts to precipitate ; we have :

$Ksp = [Ca^{2+}]^3 [PO_4^{3-}]^2$

$2.07 \times 10^{-33} = [Ca^{2+}]^3 (1.07 \times 10^{-13})^2$

$[Ca^{2+}]^3 = \dfrac{2.07 \times 10^{-33} }{(1.07 \times 10^{-13})^2}$

$[Ca^{2+}]^3 =1.808 \times 10^{-7}$

$[Ca^{2+}] =\sqrt[3]{1.808 \times 10^{-7}}$

$[Ca^{2+}] =0.00566$

This implies that when the second cation starts to precipitate ; the concentration of $[Ca^{2+}]$ in the solution is 0.00566

Therefore;

the percentage of $Ca^{2+}$ remaining = concentration remaining/initial concentration × 100%

the percentage of $Ca^{2+}$ remaining = 0.00566/0.0440 × 100%

the percentage of $Ca^{2+}$ remaining = 0.1286 × 100%

the percentage of $Ca^{2+}$ remaining = 12.86%

5 0

4 years ago

How can the shape of a molecule determine its polarity.

Natasha_Volkova [10]

Answer:

Letter D

Explanation:

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3 years ago

All of these may be classified as

mixer [17]

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3 years ago

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