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Stella [2.4K]
3 years ago
8

Which of the following elements matches the criteria below?

Chemistry
1 answer:
Tems11 [23]3 years ago
4 0

Answer:

The answer to your question is below

Explanation:

1. Found in period 2. All the elements in the list are found in period 2.

a. F   This option is correct

b. Be  Beryllium is located in period two.

c. O  also oxygen is found in period 2.

d. C Carbon is found in period 2.

2.- Can gain lose 4 electrons to become its nearest stable noble gas. Only Carbon.

a. F    This option is wrong, F becomes stable when it gains 1 electron.

b. Be  Beryllium becomes stable when it loses 2 electrons.

c. O  Become stable when it gains 2 electrons.

<u>d. C </u><u>Become stable when it gains or loses 4 electrons.</u>

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Exactly how much time must elapse before 16 grams of potassium-42decays, leaving 2 grams of the original isotope?(1) 8 × 12.4 ho
AleksandrR [38]
The answer is <span>(3) 3 × 12.4 hours
</span>
To calculate this, we will use two equations:
(1/2) ^{n} =x
t_{1/2} = \frac{t}{n}
where:
<span>n - number of half-lives
</span>x - remained amount of the sample, in decimals
<span>t_{1/2} - half-life length
</span>t - total time elapsed.

First, we have to calculate x and n. x is <span>remained amount of the sample, so if at the beginning were 16 grams of potassium-42, and now it remained 2 grams, then x is:
2 grams : x % = 16 grams : 100 %
x = 2 grams </span>× 100 percent ÷ 16 grams
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Thus:
<span>(1/2) ^{n} =x
</span>(0.5) ^{n} =0.125
n*log(0.5)=log(0.125)
n= \frac{log(0.5)}{log(0.125)}
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It is known that the half-life of potassium-42 is 12.36 ≈ 12.4 hours.
Thus:
<span>t_{1/2} = 12.4
</span><span>t_{1/2} *n = t
</span>t= 12.4*3

Therefore, it must elapse 3 × 12.4 hours <span>before 16 grams of potassium-42 decays, leaving 2 grams of the original isotope</span>
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