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Kitty [74]
1 year ago
7

Someone please explain how this works Which type of reaction occurs in the core of a nuclear reactor in a nuclear power plant?

Chemistry
1 answer:
Paul [167]1 year ago
4 0

An example of a reaction that occurs within the core of a nuclear reactor is the nuclear fission reaction given:

  • ²³⁵₉₂U + ¹₀n ---> ⁹⁰₃₈Sr + ¹⁴³₅₄ + 3 ¹₀n

<h3>What is a nuclear reactor?</h3>

A nuclear reactor is a device which produces electrical energy as a result of the nuclear reactions that take place within it.

In a nuclear reactor, the reaction that takes place within the core is a nuclear fission chain reaction.

In a nuclear fission reaction, the nucleus of larger atoms are split into the nucleus of smaller atoms when fast moving neutrons are used to bombard the nucleus of the large atom. The fission of the nucleus of the large atom results in the formation of atoms of lighter nucleus as well as more protons which then bombard more nucleus of the large atoms resulting in a chain reaction.

The chain reaction occurring within the nuclear reactor core is controlled by the insertion of boron rods which absorbs the excess neutrons produced.

An example of a reaction that occurs within the core of a nuclear reactor is given below:

²³⁵₉₂U + ¹₀n ---> ⁹⁰₃₈Sr + ¹⁴³₅₄ + 3 ¹₀n

Learn more about nuclear fission at: brainly.com/question/913303

#SPJ1

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3 years ago
We want to calculate the concentrations of all species in a 0.58 M Na 2 SO 3 (sodium sulfite) solution. The ionization constants
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Explanation:

Reaction equation is as follows.

      Na_{2}SO_{3}(s) \rightarrow 2Na^{+}(aq) + SO^{2-}_{3}(aq)

Here, 1 mole of Na_{2}SO_{3} produces 2 moles of cations.

[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58

                                  = 1.16 M

[SO^{2-}_{3}] = [Na_{2}SO_{3}] = 0.58 M

The sulphite anion will act as a base and react with H_{2}O to form HSO^{-}_{3} and OH^{-}.

As,     K_{b} = \frac{K_{w}}{K_{a_{2}}}

                       = \frac{10^{-14}}{6.3 \times 10^{-8}}

                       = 1.59 \times 10^{-7}

According to the ICE table for the given reaction,

          SO^{2-}_{3} + H_{2}O \rightleftharpoons HSO^{-}_{3} + OH^{-}

Initial:        0.58             0              0

Change:     -x               +x             +x

Equilibrium: 0.58 - x     x               x

So,

        K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}

 1.59 \times 10^{-7} = \frac{x^{2}}{0.58 - x}

        x^{2} = 1.59 \times 10^{-7} \times (0.58 - x)

                x = 0.0003 M

So,   x = [HSO^{-}_{3}] = [OH^{-}] = 0.0003 M

[SO^{2-}_{3}] = 0.58 - 0.0003

                     = 0.579 M

Now, we will use [HSO^{-}_{3}] = 0.0003 M

The reaction will be as follows.

              HSO^{2-}_{3} + H_{2}O \rightleftharpoons H_{2}SO_{3} + OH^{-}

Initial:   0.0003

Equilibrium:  0.0003 - x        x             x

              K_{b} = \frac{x^{2}}{0.0003 - x}

        K_{b} = \frac{K_{w}}{K_{a_{1}}}

                      = \frac{10^{-14}}{1.4 \times 10^{-2}}

                      = 7.14 \times 10^{-13}

Therefore,  7.14 \times 10^{-13} = \frac{x^{2}}{0.0003 - x}

As,  x <<<< 0.0003. So, we can neglect x.

Therefore,  x^{2} = 7.14 \times 10^{-13} \times 0.0003

                              = 0.00214 \times 10^{-13}

                     x = 0.0146 \times 10^{-6}

x = [OH^{-}] = [H_{2}SO_{3}] = 1.46 \times 10^{-8}

    [H^{+}] = \frac{10^{-14}}{[OH^{-}]}

                = \frac{10^{-14}}{0.0003}

                = 3.33 \times 10^{-11} M

Thus, we can conclude that the concentration of spectator ion is 3.33 \times 10^{-11} M.

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3 years ago
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