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san4es73 [151]
2 years ago
12

Part complete When a 235 92U nucleus is bombarded by neutrons (10n) it undergoes a fission reaction, resulting in the formation

of two new nuclei and neutrons. The following equation is an example of one such fission process: 235 92U+10n→AZBa+9436Kr+310n
Enter the isotope symbol for the barium (Ba) nucleus in this reaction.In another process in which 235 92U undergoes neutron bombardment, the following reaction occurs:235 92U+10n --> AZSr+143 54Xe+310nEnter the isotope symbol for the strontium (Sr) nucleus in this reaction.
Chemistry
1 answer:
nikdorinn [45]2 years ago
7 0

<u>Answer:</u> The isotopic symbol of barium is _{56}^{138}\textrm{Ba} and that of strontium is _{38}^{89}\textrm{Sr}

<u>Explanation:</u>

Nuclear fission reactions are defined as the reactions in which a heavier nuclei breaks down in two or more smaller nuclei.

In a nuclear reaction, the total mass and total atomic number remains the same.

  • For the given fission reaction:

^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^A_Z\textrm{Ba}+^{94}_{36}\textrm{Kr}+3^1_0\textrm{n}

  • <u>To calculate A:</u>

Total mass on reactant side = total mass on product side

235 + 1 = A + 94 + 3

A = 139

  • <u>To calculate Z:</u>

Total atomic number on reactant side = total atomic number on product side

92 + 0 = Z + 36 + 0

Z = 56

The isotopic symbol of barium is _{56}^{139}\textrm{Ba}

  • For the given fission reaction:

^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^A_Z\textrm{Sr}+^{143}_{54}\textrm{Xe}+3^1_0\textrm{n}

  • <u>To calculate A:</u>

Total mass on reactant side = total mass on product side

235 + 1 = A + 143 + 3

A = 90

  • <u>To calculate Z:</u>

Total atomic number on reactant side = total atomic number on product side

92 + 0 = Z + 54 + 0

Z = 38

The isotopic symbol of strontium is _{38}^{89}\textrm{Sr}

Hence, the isotopic symbol of barium is _{56}^{138}\textrm{Ba} and that of strontium is _{38}^{89}\textrm{Sr}

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The molecular formula of the given compound is $\mathrm{H}_{2} \mathrm{CO}_{3}$ also known as  Carbonic acid.

<h3>What is empirical formula and molecular formula?</h3>

The simplest whole-number ratio of the various atoms in a compound is represented by an empirical formula.

The precise number of various atom types present in a compound's molecule is indicated by the molecular formula.

Given that,

H = 3.3%

C = 19.3%

O = 77.4%

No. of moles of H = 3.3/1

No. of moles of H = 3.3

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No. of moles of O = 77.4/16

No. of moles of O = 4.83

Therefore, the ratio of the atoms of C, H and O = 3.3 : 1.60 : 4.83

Divide by smallest value which you get =3.3 / 1.60 : 1.60 / 1.60 : 4.83 / 1.60

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So, the empirical formula is $\mathrm{H}_{2} \mathrm{CO}_{2}$

Let the molecular formula is $\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right) \mathrm{n}$

Then, molar mass $=(2 \times 1+1 \times 12+3 \times 16) n\\

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n = 0.96 or n = 1 (rounded off to nearest ones)

So, the molecular formula is $\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right) 1=\mathrm{H}_{2} \mathrm{CO}_{3}$ i.e., the compound is Carbonic acid.

To know more about molecular formula visit:

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