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Masteriza [31]
2 years ago
9

Which reaction takes place in a nuclear fission reactor?

Chemistry
1 answer:
Marat540 [252]2 years ago
3 0

The reaction that takes place in a nuclear fission reactor is as follows: 235/92 U + 1/0n 94/36Kr + 139/56 Ba + 3/0n.

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

A nuclear fission reactor is the place where nuclear chain reactions occur that produce energy by fission.

Nuclear fission is the nuclear reaction in which a large nucleus splits into smaller ones with the simultaneous release of energy.

Therefore, the option that involves the splitting of atoms into smaller ones is as follows: 235/92 U + 1/0n 94/36Kr + 139/56 Ba + 3/0n.

Learn more about nuclear fission reactor at: brainly.com/question/10203508

#SPJ1

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The number of protons in an atom is known as its atomic
Fynjy0 [20]
Atomic number should be the answer
4 0
3 years ago
A river with a flow of 50 m3/s discharges into a lake with a volume of 15,000,000 m3. The river has a background pollutant conce
spin [16.1K]

Explanation:

The given data is as follows.

       Volume of lake = 15 \times 10^{6} m^{3} = 15 \times 10^{6} m^{3} \times \frac{10^{3} liter}{1 m^{3}}

        Concentration of lake = 5.6 mg/l

Total amount of pollutant present in lake = 5.6 \times 15 \times 10^{9} mg

                                                                    = 84 \times 10^{9} mg

                                                                    = 84 \times 10^{3} kg

Flow rate of river is 50 m^{3} sec^{-1}

Volume of water in 1 day = 50 \times 10^{3} \times 86400 liter

                                          = 432 \times 10^{7} liter

Concentration of river is calculated as 5.6 mg/l. Total amount of pollutants present in the lake are 2.9792 \times 10^{10} mg or 2.9792 \times 10^{4} kg

Flow rate of sewage = 0.7 m^{3} sec^{-1}

Volume of sewage water in 1 day = 6048 \times 10^{4} liter

Concentration of sewage = 300 mg/L

Total amount of pollutants = 1.8144 \times 10^{10} mg or 1.8144 \times 10^{4}kg

Therefore, total concentration of lake after 1 day = \frac{131936 \times 10^{6}}{1.938 \times 10^{10}}mg/ l

                                        = 6.8078 mg/l

                 k_{D} = 0.2 per day

       L_{o} = 6.8078

Hence, L_{liquid} = L_{o}(1 - e^{-k_{D}t}

             L_{liquid} = 6.8078 (1 - e^{-0.2 \times 1})  

                             = 1.234 mg/l

Hence, the remaining concentration = (6.8078 - 1.234) mg/l

                                                             = 5.6 mg/l

Thus, we can conclude that concentration leaving the lake one day after the pollutant is added is 5.6 mg/l.

5 0
3 years ago
Each astronaut produces 8.8 × 102 g CO2 per day that must be removed from the air on the shuttle. If a typical shuttle mission i
suter [353]

The reaction of removing CO2 using LiOH is the following:

2 LiOH + CO2 -----> Li2CO3 + H2O

By solving the amount of CO2 the LiOH can scrub:

(3.50 × 10^4 g LiOH) (1 mol LiOH/ 24 g LiOH) ( 1 mol CO2 / 2 mol LiOH) ( 44 g CO2 /1 mol CO2) = 32, 083.33 g CO2 it can scrub

<span>Since number of astronaut = 32, 083.33 g / 9 (8.8 × 10^2) = 4 astronaut</span>

7 0
3 years ago
Read 2 more answers
Determine the pH of a 0.048 M hypochlorous acid (HClO) solution. Hypochlorous acid is a weak acid (Ka = 4.0 ✕ 10−8 M).
storchak [24]

pH of 0.048 M HClO is 4.35.

<u>Explanation:</u>

HClO is a weak acid and it is dissociated as,

HClO ⇄ H⁺ + ClO⁻

We can write the equilibrium expression as,

Ka = $\frac{[H^{+}] [ClO^{-}]  }{[HClO]}

Ka = 4.0 × 10⁻⁸ M

4.0 × 10⁻⁸ M = $\frac{x \times x }{0.048}

Now we can find x by rewriting the equation as,

x² =  4.0 × 10⁻⁸ × 0.048

   = 1.92 × 10⁻⁹

Taking sqrt on both sides, we will get,

x = [H⁺] = 4.38 × 10⁻⁵

pH = -log₁₀[H⁺]

     = - log₁₀[ 4.38 × 10⁻⁵]

   = 4.35

8 0
3 years ago
Calculate the molar mass of B(NO3)3 ?​
Stella [2.4K]

The molar mass of B(NO₃)₃ - Boron nitrate : 196.822 g/mol

<h3>Further explanation</h3>

In stochiometry therein includes  

<em>Relative atomic mass (Ar) and relative molecular mass / molar mass (M)  </em>

So the molar mass of a compound is given by the sum of the relative atomic mass of Ar  

M AxBy = (x.Ar A + y. Ar B)  

The molar mass of B(NO₃)₃ - Boron nitrate :

M B(NO₃)₃ = Ar B + 3. Ar N + 9.Ar O

M B(NO₃)₃ = 10.811 + 3. 14,0067 + 9. 15,999

M B(NO₃)₃ = 196.822 g/mol

7 0
2 years ago
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