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Radioactive phosphorus is used in the study of biochemical reaction mechanisms because phosphorus atoms are components of many biochemical molecules. The location of the phosphorus (and the location of the molecule it is bound in) can be detected from the electrons (beta particles) it produces: ³215P ⟶ ³216S + e⁻ rate = 4.85 × 10⁻2 day⁻1 [³2P]

a) The rate of decay is directly proportional to the square of the concentration of ³215P.

b) The rate of decay is inversely proportional to the concentration of ³216S.

c) The rate of decay is directly proportional to the concentration of ³2P.

d) The rate of decay is independent of time.

User Steffi
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Final answer:

The rate of decay of radioactive phosphorus is directly proportional to the square of the concentration of ³215P. The rate of decay is not affected by the concentration of ³216S. The rate of decay is not independent of time. Option a

Step-by-step explanation:

The rate of decay of radioactive phosphorus is directly proportional to the square of the concentration of ³215P, which is the isotope of phosphorus being studied. This means that as the concentration of ³215P increases, the rate of decay also increases.

The rate of decay is not affected by the concentration of ³216S, which is the product of the radioactive decay of ³215P. Therefore, the rate of decay is not inversely proportional to the concentration of ³216S.

The rate of decay is not directly proportional to the concentration of ³2P either, as ³2P is not mentioned in the given information.

The rate of decay is also not independent of time, as it is stated that the rate of decay is 4.85 × 10⁻² day⁻¹. This means that the rate of decay is dependent on time and will decrease over time. Option a

User Hokwang
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