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.
Answer:
The mass of 
in the container is 2.074 gram
Explanation:
Given:
Volume of 
lit
Equilibrium constant 
The reaction in which is produced
⇄ 
Here equal moles of and 
is formed.
From the formula of equilibrium constant,
M
Above value shows,
So in 2 L no. moles of = 
moles.
So mass of 0.122 mole of is = 
g
Therefore, the mass of in the container is 2.074 gram
Answer:
Kb = [OH⁻] . [C₃H₉NH⁺] / [ C₃H₉N ]
Explanation:
The equation for the reaction of trimethylamine when it is dissolved in water is:
C₃H₉N + H₂O ⇄ C₃H₉NH⁺ + OH⁻ Kb
1 mol of trimethylamine catches a proton from the water in order to produce trimethylamonium.
It is a base, because it give OH⁻ to the medium
Expression for Kb (Molar concentration)
Kb = [OH⁻] . [C₃H₉NH⁺] / [ C₃H₉N ]
