Answer:
The amount left after 49.2 years is 3mg.
Explanation:
Given data:
Half life of tritium = 12.3 years
Total mass pf tritium = 48.0 mg
Mass remain after 49.2 years = ?
Solution:
First of all we will calculate the number of half lives.
Number of half lives = T elapsed/ half life
Number of half lives = 49.2 years /12.3 years
Number of half lives = 4
Now we will calculate the amount left after 49.2 years.
At time zero 48.0 mg
At first half life = 48.0mg/2 = 24 mg
At second half life = 24mg/2 = 12 mg
At 3rd half life = 12 mg/2 = 6 mg
At 4th half life = 6mg/2 = 3mg
The amount left after 49.2 years is 3mg.
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Option (i) would have the highest 2nd Ionization Energy.
Option (i) is Sodium.
Can be Written as 2, 8 , 1
For its 1st Ionization energy... It'd be extremely easy to remove that Electron cos its on the outermost shell.
Now After Removing that Electron...
Sodium's Electronic Configuration Reduces to that of Neon Which is 2, 8.
Neon has a very stable Octet.
It would take an ENORMOUS amount of energy to break its Octet stability... that is... Remove 1 electron from its Octet.
So
Option (i) [Sodium] has the highest 2nd Ionization Energy
Answer:
The isotopic mass of 41K is 40.9574 amu
Explanation:
Step 1: Data given
The isotopes are:
39K with an isotopic mass of 38.963707u and natural abundance of 93.2581%
40K with an isotopic mass of 39.963999u
41K wit natural abundance of 6.7302 %
Average atomic mass =39.098 amu
Step 2: Calculate natural abundance of 40 K
100 % - 93.2581 % - 6.7302 %
100 % = 0.0117 %
Step 3: Calculate isotopic mass of 41K
39.098 = 38.963707 * 0.932581 + 39.963999 * 0.000117 + X * 0.067302
39.098 = 36.33681 + 0.0046758 + X * 2.067302
X = 40.9574 amu
The isotopic mass of 41K is 40.9574 amu
Option 1/A (It is the first one)
