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scZoUnD [109]
3 years ago
7

What is the half-life of a 12 g sample of radioisotope that decayed to 6 g in 28

Chemistry
1 answer:
Charra [1.4K]3 years ago
4 0

Answer:

A. 28 years

Explanation:

Applying,

R = R'(2ᵃ/ⁿ).............. Equation 1

Where R = Original sample, R' = Sample left after decay, a = Total time taken to decay, n = half life.

From the question,

Given: R = 12 g, R' = 6 g, a = 28 years.

Substitute into equation 1 and solve for n

12 = 6(2²⁸/ⁿ)

12/6 = 2²⁸/ⁿ

2²⁸/ⁿ = 2

Equation the base,

28/n = 1

n = 28 years.

Hence the half-life is 28 years

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The balanced molecular equation for the given reaction is:

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<h3>FURTHER EXPLANATION</h3>

To check that the equation is balanced count how many atoms are present for each element in the reactant and product side. If they are the same before and after reaction for all elements, then the reaction is deemed balanced.

The atom counting for the equation is shown below:

2CH_3COOH(aq) \ + \ Ba(OH)_2(aq) \rightarrow \ Ba(CH_3COO)_2(aq) + 2H_2O(l).

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C:  (2 x 2) =4                                  C: (2 x 2) = 4

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Since the number of atoms of each element are similar in the reactants and products, the equation is balanced.

To determine the state of the substances, consider their solubility.

The reactants are both aqueous (aq) as indicated in the problem.

The first product, Ba(CH_3COO)_2 is aqueous (aq) because based on the solubility rule, it will dissolve in water. Acetates are generally soluble.

The other product is water which will be liquid (l) since it is the solvent used to dissolve the substances.

<h3>LEARN MORE</h3>
  • Solubility Rules brainly.com/question/12984314
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