Answer:
0.0468 g.
Explanation:
- The decay of radioactive elements obeys first-order kinetics.
- For a first-order reaction: k = ln2/(t1/2) = 0.693/(t1/2).
Where, k is the rate constant of the reaction.
t1/2 is the half-life time of the reaction (t1/2 = 1620 years).
∴ k = ln2/(t1/2) = 0.693/(1620 years) = 4.28 x 10⁻⁴ year⁻¹.
- For first-order reaction: <em>kt = lna/(a-x).</em>
where, k is the rate constant of the reaction (k = 4.28 x 10⁻⁴ year⁻¹).
t is the time of the reaction (t = t1/2 x 8 = 1620 years x 8 = 12960 year).
a is the initial concentration (a = 12.0 g).
(a-x) is the remaining concentration.
∴ kt = lna/(a-x)
(4.28 x 10⁻⁴ year⁻¹)(12960 year) = ln(12)/(a-x).
5.54688 = ln(12)/(a-x).
Taking e for the both sides:
256.34 = (12)/(a-x).
<em>∴ (a-x) = 12/256.34 = 0.0468 g.</em>
Answer:
120 gram sample of a radioactive element, how many grams of that element will be left after 3 half-lives have passed? If you have a 300
Explanation:
Hope this helps!
<span>density is defined as mass per volume.. so 94 mL makes no sense.</span>
The given reaction is,
In this reaction, the positive and negative ions of both the reactants rearrange resulting in the formation of new compounds. Such a reaction is referred to as the double replacement or the double displacement reaction. These type of reactions usually take place in aqueous medium and one of the products is a precipitate.
Double displacement reactions are of three types, precipitation reaction, neutralization reaction and reaction involving the formation of a gas. The given reaction is a precipitation reaction as one of the products is a solid.
