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Lunna [17]
3 years ago
9

Scientists estimate Earth's age to be 4.6 billion years. Samarium has a half-life of approximately 106 billion years. From the p

resent date, approximately how many more years must pass before a sample of samarium that was created at the beginning of Earth's history contains 25 percent of the original parent material? 10 billion 26 billion 100 billion 206 billion
Chemistry
2 answers:
Wewaii [24]3 years ago
8 0

1/4 Ao = Ao * ( 1/2)^(t/106)
t = 212
subtract 4.6 billion years
<span>d)</span>
Alexandra [31]3 years ago
4 0

Answer: Option (D)

Explanation: The following steps are shown how decay takes place-

  1. Samarium (Sm) initially had 100% of its parent material.
  2. Samarium undergoes radioactive decay and becomes half of it i.e 50% of its parent material.
  3. To Decay this 50% of its initial material, it takes 106 billion years.
  4. Again it undergoes radioactive decay and becomes half of it i.e 25%, and during this process it takes another 106 billion years.

So from the above question, the total number of years for Samarium to contain 25% of its parent material will be (106+106)=212 billion years.

it can also be calculated from the radioactive decay formula.

Thus the most approximate option provided here is option (D).

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8 0
2 years ago
Given the following chemical equation, determine how many grams of N2 are produced by 9.24 of H2O2 nd 6.56g of N2H4?
valkas [14]
The balanced chemical equation will b
N2H4 + 2 H2O2-> N2 + 4H2O
 so first of all we have to find the limiting reactant.
so for this 

9.24 g H2O2 ( 1 mol / 34.02 g mass of H2O2 )
 = 0.27 mol H2O2
Now
6.56  g of  N2H4 ( 1mol / 32.06) = 0.20 mol N2H4

so from the above reaction we found 1:2 ratio of the reactants
 so the limiting reactant is hydrogen peroxide. by using this we will find the amount of N2 produced.
so that will be

<span>0.27 mol H2O2 ( 1 mol N2 / 2 mol H2O2 ) ( 14.01 g N2 / 1 mol N2)
 =1.89 g N2
hope it helps</span>
8 0
3 years ago
Read 2 more answers
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