Answer:
a) 3.98 x 10^-10
Explanation:
Hello,
In this case, for the given pH, we can compute the concentration of hydronium by using the following formula:
![pH=-log([H^+])](https://tex.z-dn.net/?f=pH%3D-log%28%5BH%5E%2B%5D%29)
Hence, solving for the concentration of hydronium:
![[H^+]=10^{-pH}=10^{-9.40}\\](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-pH%7D%3D10%5E%7B-9.40%7D%5C%5C)
![[H^+]=3.98x10^{-10}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D3.98x10%5E%7B-10%7DM)
Therefore, answer is a) 3.98 x 10^-10
Best regards.
Technically speaking, yes you can. Using a microscope though.
Answer:
See Explanation
Explanation:
Given that;
N/No = (1/2)^t/t1/2
Where;
No = amount of radioactive isotope originally present
N = A mount of radioactive isotope present at time t
t = time taken
t1/2 = half life
N/1000=(1/2)^3/6
N/1000=(1/2)^0.5
N = (1/2)^0.5 * 1000
N= 707 unstable nuclei
Since the value of the initial activity of the radioactive material was not given, the activity of the radioactive material after three months is given by;
Decay constant = 0.693/t1/2 = 0.693/6 months = 0.1155 month^-1
Hence;
A=Aoe^-kt
Where;
A = Activity after a time t
Ao = initial activity
k = decay constant
t = time taken
A = Aoe^-3 *0.1155
A=Aoe^-0.3465
Radium and polonium is the answer to this question. I hope I helped out!
