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1 year ago

If U-235 decays into Cs-135 and 4 neutrons, what other nuclide will be produced?

1 answer:

5 0

If U-235 decays into Cs-135 and 4 neutrons, the other nuclide that will be produced is Rb-96 (option D).

<h3>What is radioactive decay?</h3>

A radioactive decay is the process by which an unstable large nuclei emit subatomic particles and disintegrate into one or more smaller nuclei.

According to this question, a radioactive material Uranium- 235 undergoes radioactive decay into Cs- 135 and 4 neutrons (1/0n).

This means that the mass of the products we have is 135 + 4 = 139.

The mass of the nuclide left must be 235 - 139 = 96, hence, the other nuclide that will be produced is Rb-96.

Learn more about radioactive decay at: brainly.com/question/1770619

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Answer: The molecular formula will be $C_2H_4O_2$

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If percentage are given then we are taking total mass is 100 grams.

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Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{40.0g}{12g/mole}=3.33moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{53.3g}{16g/mole}=3.33moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{6.66g}{1g/mole}=6.66moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{3.33}{3.33}=1$

For O = $\frac{3.33}{3.33}=1$

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The ratio of C : O : H = 1: 1: 2

Hence the empirical formula is $COH_2$

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$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{60}{30}=2$

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