A patient is given 0.050 mg of technetium-99 m (where m means metastable—an unstable but long-lived state), a radioactive isotope with a half-life of about 6.0 hours.How long until the radioactive isotope decays to 1.3×10−2 mg ?
Answers
The correct answer is 9.6h.
As you know, a radioactive isotope's nuclear half-life tells you exactly how much time must pass in order for an initial sample of this isotope to be halved.
Using the formula , A = Ao.
where , A- final mass after decay
Ao - initial mass
n - the number of half-lives that pass in the given period of time
Now, putting all the values, we get
1.3 × mg = 0.050 mg ×
Take the natural log of both sides of the equation to get,
㏑ = ㏑
㏑ = n. ln
n = 1.6
Since n represents the number of half-lives that pass in a given period of time, you can say that
t= 1.6 × 6 h
t = 9.6h
Hence, it will take 9.6 h until the radioactive isotope decays.
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Final answer:
Using the formula for radioactive decay and the provided half-life of technetium-99m, it can be calculated that it takes approximately 28.5 hours for 0.050 mg of technetium-99m to decay to a quantity of 1.3 x 10^-2 mg.
Explanation:
The decay of a radioactive isotope is an exponential process based on the half-life, which is, in turn, constant for any given isotope. The general formula for the remaining quantity of a radioactive isotope after a given time is given by: N = N0 (0.5) ^(t/t1/2), where (N0) is the initial amount, (N) is the remaining amount, (t) is time, and (t1/2) is the half-life of the isotope. In this case, we are given the initial quantity (N0 = 0.050 mg), the remaining quantity (N = 1.3 x 10^-2 mg), and the half-life (t1/2 = 6.0 hours).
We can solve for time (t) in the equation: N = N0 (0.5) ^(t/t1/2). Plugging in the values, we get 1.3 x 10^-2 = 0.050 x (0.5)^(t/6), and solving for t, we find that it takes approximately 28.5 hours for the technetium-99m to decay to 1.3 x 10^-2 mg.
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Answer:
false
Explanation:
(1) CH3OH (3) H2O
(2) C6H12O6 (4) KOH
Answers
Answer:
The answer is 4) KOH
Explanation:
Electrolytes are substances which ionize when dissolved in water. They are classified as:
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KOH is an electrolyte. Therefore, option (4) is correct.
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B. a nonmetal and a metal.
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Answers
A salt is obtained as a reaction between a base and an acid which is option D.
A salt is obtained as a reaction between a base and an acid. In this reaction, the base donates a hydroxide ion (OH⁻) and the acid donates a hydrogen ion (H⁺). The hydroxide ion and hydrogen ion combine to form water (H₂O), and the remaining ions from the base and acid combine to form the salt. The salt is typically composed of a metal cation from the base and a non-metal anion from the acid. The formation of salt is a neutralization reaction where the acidic and basic properties of the reactants are neutralized, resulting in the formation of a new compound, the salt.
Hence, the correct option is option d.
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Answers
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Answer:
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Explanation: