PH of solution at 25ºC = 8.3
![[ H_3O^+] = 10 ^{-pH}](https://tex.z-dn.net/?f=%5B%20H_3O%5E%2B%5D%20%3D%2010%20%5E%7B-pH%7D%20)
![{H_3O^+] = 10 ^{-8.3}](https://tex.z-dn.net/?f=%7BH_3O%5E%2B%5D%20%3D%2010%20%5E%7B-8.3%7D)
![[H_3O^+] = 5.011*10^{-9} M](https://tex.z-dn.net/?f=%5BH_3O%5E%2B%5D%20%3D%205.011%2A10%5E%7B-9%7D%20%20M)
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Answer:
read it first
Explanation:
Population change is governed by the balance between birth rates and death rates. If the birth rate stays the same and the death rate decreases, then population numbers will grow. If the birth rate increases and the death rate stays the same, then population will also grow.
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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.
if this is true or false its true hope i helped
