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katovenus [111]
3 years ago
6

The half-life of carbon-14 is 5730 years. How many grams of carbon-14 remain after 28,650 years if the original amount was 28.0

g?
A.5.60 g B.2.80 g C.0.875 g D.14.0 g
Chemistry
1 answer:
alexdok [17]3 years ago
5 0
TLDR: the answer is C. 0.875.

The basis of this problem is testing your fluency with half-life radioactive decay. The half-life of any radioactive substance represents the amount of time required for 1/2 of the amount of the substance to decay. So, if you had a 100 g sample of Carbon-14, after 5,730 years, only 50 g of Carbon-14 would remain; the other 50 g of substance would have decayed into another substance (Nitrogen-14). If another 5,730 years passes, only 25 g would remain, as half of that 50 g radioactive substance would have decayed.

In the problem, 28 g of Carbon-14 decays as 28,650 years passes, which, when divided by 5,730, yields a value of 5. This number represents the number of half lives that passes during this time. So, over the course of this time, the amount of radioactive substance is cut in half five times.

28 g - 14 g (1 HL) - 7 g (2 HL) - 3.5 (3 HL) - 1.75 g (4 HL) - 0.875 g (5 HL)

After 5 half lives, only 0.875 g of the original 28 g remains.
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Answer: The balanced equation for the complete oxidation reaction that occurs when methane (CH4) burns in air is CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O.

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When a substance tends to gain oxygen atom in a chemical reaction and loses hydrogen atom then it is called oxidation reaction.

For example, chemical equation for oxidation of methane is as follows.

CH_{4} + O_{2} \rightarrow CO_{2} + H_{2}O

Number of atoms present on reactant side are as follows.

  • C = 1
  • H = 4
  • O = 2

Number of atoms present on product side are as follows.

  • C = 1
  • H = 2
  • O = 3

To balance this equation, multiply O_{2} by 2 on reactant side. Also, multiply H_{2}O by 2 on product side. Hence, the equation can be rewritten as follows.

CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O

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  • O = 4

Number of atoms present on product side are as follows.

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Since, the atoms present on both reactant and product side are equal. Therefore, this equation is now balanced.

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