<span>a) 7.9x10^9
b) 1.5x10^9
c) 3.9x10^4
To determine what percentage of an isotope remains after a given length of time, you can use the formula
p = 2^(-x)
where
p = percentage remaining
x = number of half lives expired.
The number of half lives expired is simply
x = t/h
where
x = number of half lives expired
t = time spent
h = length of half life.
So the overall formula becomes
p = 2^(-t/h)
And since we're starting with 1.1x10^10 atoms, we can simply multiply that by the percentage. So, the answers rounding to 2 significant figures are:
a) 1.1x10^10 * 2^(-5/10.5) = 1.1x10^10 * 0.718873349 = 7.9x10^9
b) 1.1x10^10 * 2^(-30/10.5) = 1.1x10^10 * 0.138011189 = 1.5x10^9
c) 1.1x10^10 * 2^(-190/10.5) = 1.1x10^10 * 3.57101x10^-6 = 3.9x10^4</span>
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Recall; pH + pOH = 14
In this case [OH-] =0.100 m
therefore;
pOH = -LOG[OH-]
= - Log (0.100)
= 1.00
Therefore; the pOH is 1.00
And since, pH +pOH = 14
Then pH = 14-pOH
= 14 -1
= 13
Thus the pH is 13.00
Answer:
1.58×10E18
Explanation:
Since we have the reduction potentials we could make decisions regarding which one will be the anode or cathode. Evidently, bromine having the more positive reduction potential will be the cathode while the iodine will be the anode.
E°cell= 1.07- 0.53= 0.54 V
E°cell= 0.0592/n logK
0.54 = 0.0592/2 logK
logK= 0.54/0.0296
logK= 18.2
K= Antilog (18.2)
K= 1.58×10^18
