Answer:
Half-life = 3 minutes
Explanation:
Using the radioactive decay equation we can solve for reaction constant, k. And by using:
K = ln2 / Half-life
We can find half-life of polonium-218
Radioactive decay:
Ln[A] = -kt + ln [A]₀
Where:
[A] could be taken as mass of polonium after t time: 1.0mg
k is Reaction constant, our incognite
t are 12 min
[A]₀ initial amount of polonium-218: 16mg
Ln[A] = -kt + ln [A]₀
Ln[1.0mg] = -k*12min + ln [16mg]
-2.7726 = - k*12min
k = 0.231min⁻¹
Half-life = ln 2 / 0.231min⁻¹
<h3>Half-life = 3 minutes</h3>
The mass number of aluminium hydroxide is 78 thus, the number of moles in 0.745 g is:
no. of moles= mass/ RFM
= 0.745/78
=0.00955moles
Therefore the 0.00955 moles should be in the 35.18 ml
therefore 1000ml of the solution will have:
(0.00955ml×1000ml)/35.18
=0.2715moles
The solution will be 0.27M hydrochloric acid
Answer:
After 2.0 minutes the concentration of N2O is 0.3325 M
Explanation:
Step 1: Data given
rate = k[N2O]
initial concentration of N2O of 0.50 M
k = 3.4 * 10^-3/s
Step 2: The balanced equation
2N2O(g) → 2 N2(g) + O2(g)
Step 3: Calculate the concentration of N2O after 2.0 minutes
We use the rate law to derive a time dependent equation.
-d[N2O]/dt = k[N2O]
ln[N2O] = -kt + ln[N2O]i
⇒ with k = 3.4 *10^-3 /s
⇒ with t = 2.0 minutes = 120s
⇒ with [N2O]i = initial conc of N2O = 0.50 M
ln[N2O] = -(3.4*10^-3/s)*(120s) + ln(0.5)
ln[N2O] = -1.101
e^(ln[N2O]) = e^(-1.1011)
[N2O} = 0.3325 M
After 2.0 minutes the concentration of N2O is 0.3325 M
