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Nina [5.8K]
2 years ago
8

You have 300,000 atoms of a radioactive substance. After 3 half-lives have past, how many atoms remain? Remember that you cannot

have a fraction of an atom, so round the answer to the nearest whole number.
Chemistry
1 answer:
padilas [110]2 years ago
3 0

The number of atoms that remains after 3 half-lives given that it was originally 300000 atoms is 37500 atoms

<h3>Data obtained from the question</h3>
  • Original amount (N₀) = 300000 atoms
  • Number of half-lives (n) = 3
  • Amount remaining (N) =?

<h3>How to determine the amount remaining</h3>

The amount remaining after 3 half-lives can be obtained as illustrated below:

N = N₀ / 2ⁿ

N = 300000 / 2³

N = 300000 / 8

N = 37500 atoms

Learn more about half life:

brainly.com/question/26374513

#SPJ1

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Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho
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Answer:

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Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\

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n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3

c) The leftover is computed as follows:

m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

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