Answer: a) %(C/Co) = (e^(-0.027t)) × 100
b) t1/2 = 25.67years
c) 5.872%
Explanation:
a) Radioactive reactions always follow a first order reaction dynamic
Let the initial concentration of Strontium-90 be Co and the concentration at any time be C
The rate of decay will be given as:
(dC/dt) = -KC (Minus sign because it's a rate of reduction)
The question provides K = 2.7% per year = 0.027/year
(dC/dt) = -0.027C
(dC/C) = -0.027dt
∫ (dC/C) = -0.027 ∫ dt
Solving the two sides as definite integrals by integrating the left hand side from Co to C and the Right hand side from 0 to t.
We obtain
In (C/Co) = -0.027t
(C/Co) = (e^(-0.027t))
In percentage, %(C/Co) = (e^(-0.027t)) × 100
(Solved)
b) Half life of a first order reaction (t1/2) = (In 2)/K
K = 0.027/year
t1/2 = (In 2)/0.027 = 25.67 years
c) percentage that remains after 105years,
%(C/Co) = (e^(-0.027t)) × 100
t = 105
%(C/Co) = (e^(-0.027 × 105)) × 100 = 5.87%
For lead fluoride Ksp = s³
For Aluminum hydroxide Ksp = s⁴
<u>Explanation:</u>
The expression for the solubility product can be written as the product of the concentration of the ions present in the solution.
PbF₂ → Pb²⁺ + 2 F⁻
Then the Ksp can be written as,
Ksp = [ Pb²⁺] [F⁻]²
If the concentration can be considered as s, then the Ksp can be written as,
Ksp = s× s² = s³
For Aluminum hydroxide on dissociation it gives 4 ions as,
Al(OH)₃ → Al³⁺ + 3 OH⁻
Ksp = [Al³⁺] [OH⁻]³
= s × s³ = s⁴
Thus,For lead fluoride Ksp = s³
For Aluminum hydroxide Ksp = s⁴
The answer is 0.2 M.
Molarity is a measure of the concentration of solute in a solution
c = n ÷V<span>
c - concentration of solute,
n - number of moles of solute
V - volume of solution
</span>
We know:
V = 50.00 mL = 0.05 L
c = ?
n = ?
Let's calculate concentration:
c = m/(V * Mr)
The molar mass of <span>(NH4)2SO4 is the sum of atomic masses (Ar) of its elements):
Mr (</span><span>(NH4)2SO4) = 2Ar(N) + 8 Ar(H) + Ar(S) + 4Ar(O)
= 2 * 14 + 8 * 1 + 32 + 4 * 16 =
= 28 + 8 + 32 + 64 =
= 132 g/mol
c = 26.42 g/ (0.05 L * 132g /mol) =26.42 g/ 6.6 L*g/mol = 4 mol/L = 4 M
Now, calculate molarity:
</span>c = n ÷V
n = c * V
n = 4 mol/L * 0.05 L
n = 0.2 mol
NH3 is neutralised by the equation:
HCL + NH3 -> NH4CL
In this equation there is a one to one relationship in terms of the number of moles of each reactant. I.e. To neutralise 1 mole of NH3 we require 1 mole of HCL.
To calculate the concentration of NH3 required, we must first calculate the number of moles of HCL used.
volume HCL = 35.5mL = 0.0355 litres
concentration HCL = 0.23M = 0.23 mole/litre
Note that the term "M" for concentration simply means moles/litre
number moles = concentration x volume
number moles HCL = 0.0355 x 0.23 = 0.008165 moles HCL
based on the equation, we know the number of moles of NH3 must be the same
So,
moles NH3, n = 0.008165
volume NH3, v = 20.0mL = 0.020 litres
n = c x v
c = n / v
c = 0.008165 / 0.020
=0.41
i.e. the concentration of NH3 would be 0.41 moles/litre or 0.41M
This intuitively makes sense because there is less volume of NH3 required to be neturalised, in a one-to-one mole relationship. So the concentration of NH3 would need to be higher than that of HCL.
A cube has a volume of 8.0 cm3 and a mass of 21.6 grams. ... using an electronic analytical balance. ... graduated cylinder ... The density of iron is 7.87 g/mL.
