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faust18 [17]
4 years ago
8

What is the mass of an original 5.60-gram sample of iron-53 that remains unchanged after 25.53 minutes? 0.35 g 0.70 g 1.40 g 2.8

0 g?
Chemistry
1 answer:
djverab [1.8K]4 years ago
3 0
Answer: 0.70g

The half-life of iron-53 would be 8.51 minutes. So, in 25.53 minutes would be equal to: 25.53 min/ (8.51 minutes/ half-life)= 3 half-life.

Every half-life will reduce the original weight into half. So, the final weight would be:
final weight = original weight * 1/2 ^(time)
final weight = 5.6g * (1/2)^(3 half-life)
final weight = 5.6g * 1/8= 0.7g
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There is some info missing. I think this is the original question.

<em>Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is  4.50  ×  10 ⁻⁴.</em>

<em />

Step 1: Given data

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Acid dissociation constant (Ka): 4.50  ×  10 ⁻⁴

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\alpha = \frac{[A^{-} ]}{[HA]} \times 100\% = \frac{0.0106M}{0.249} \times 100\% = 4.26\%

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