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Neporo4naja [7]
3 years ago
11

There is a 30g of be-11 it has a half-life of about 14 seconds how much will be left in 28 seconds

Chemistry
1 answer:
Lelu [443]3 years ago
6 0

There will be 7.5 g of Be-11 remaining after 28 s.

If 14 s = 1 half-life, 28 s = 2 half-lives.

After the first half-life, ½ of the Be-11 (15 g) will disappear, and 15 g will remain.

After the second half-life, ½ of the 15 g (7.5 g) will disappear, and 7.5 g will remain.

In symbols,

<em>N</em> = <em>N</em>₀(½)^<em>n</em>

where

<em>n</em> = the number of half-lives

<em>N</em>₀ = the original amount

<em>N</em> = the amount remaining after <em>n</em> half-lives

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The answer is strontium dihydrogen phosphate

That is strontium dihydrogen phosphate, is the compound containing the most hydrogen atoms per molecule or formula unit.

The formulas of the following compounds are as follows:

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As it can be seen from the formula of strontium dihydrogen phosphate  Sr(H₂PO4)₂ , that it contains the most hydrogen atoms per molecule or formula unit.


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