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: -
Solubility of a substance depend on the balance of intermolecular forces between the solvent and solute, and the entropy change that accompanies this process.
Temperature and pressure also plays a role in solubility.
A solution having Group 1 cations like lithium, sodium, potassium etc are always soluble.
A solution having NH₄⁺ is soluble.
All salts with anion as nitrates, acetates, chlorates, and perchlorates are soluble in water.
Use blue litmus paper. This is an indicator that can safely determine whether it is a base or an acid by changing color in response to the substance. This color indicates whether it is an acid or a base. Refer to the pH scale to see if the substance is basic or acidic.
