This question is missing the part that actually asks the question. The questions that are asked are as follows:
(a) How much of a 1.00 mg sample of americium remains after 4 day? Express your answer using 2 significant figures.
(b) How much of a 1.00 mg sample of iodine remains after 4 days? Express your answer using 3 significant figures.
We can use the equation for a first order rate law to find the amount of material remaining after 4 days:
[A] = [A]₀e^(-kt)
[A]₀ = initial amount
k = rate constant
t = time
[A] = amount of material at time, t.
(a) For americium we begin with 1.00 mg of sample and must convert time to units of years, as our rate constant, k, is in units of yr⁻¹.
4 days x 1 year/365 days = 0.0110
A = (1.00)e^((-1.6x10^-3)(0.0110))
A = 1.0 mg
The decay of americium is so slow that no noticeable change occurs over 4 days.
(b) We can simply plug in the information of iodine-125 and solve for A:
A = (1.00)e^(-0.011 x 4)
A = 0.957 mg
Iodine-125 decays at a much faster rate than americium and after 4 days there will be a significant loss of mass.
It is b i think bc it looks like it would be it and my tudor helped me
The bromide concentration in this solution of calcium bromide dissolved in enough water to give 469.1 mL is 1.75 × 10-⁵M.
<h3>How to calculate concentration?</h3>
The concentration of a solution can be calculated by dividing the number of moles of the substance by its volume.
No of moles of calcium bromide is calculated as follows:
moles = 1.642 ÷ 199.89 = 8.215 × 10-³moles
Molarity = 8.215 × 10-³moles ÷ 469.1mL = 1.75 × 10-⁵M
Therefore, the bromide concentration in this solution of calcium bromide dissolved in enough water to give 469.1 mL is 1.75 × 10-⁵M.
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Answer:
C 5 mol A; 6 mol B
Explanation:
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