Answer:
Electronegativity = 1.87.
Ionic radius = 109 pm.
Atomic radius = -39 pm
First ionization energy = 410 kJ/mol
Explanation:
Hello!
In this case, since electronegativity, ionic radius, atomic radius and first ionization energy are periodic properties that have specific trends, we can summarize it by realizing that oxygen and beryllium belong the same period 2 and differ in group, 6A and 2A respectively.
In such a way, the required comparison is written below:
Electronegativity = 3.44 (oxygen) - 1.57 (beryllium) = 1.87.
Ionic radius = 140 pm (oxygen)- 31 pm (beryllium) = 109.
Atomic radius = 73 pm (oxygen) - 112 pm (beryllium) = -39 pm
First ionization energy = 1310 kJ/mol (oxygen) - 900 kJ/mol (beryllium) = 410 kJ/mol
It means that electronegativity, ionic radius and first ionization energy increases from left to right whereas the atomic radius from right to left.
Best regards!
Answer:
To check if it's properly cleaned.
Explanation:
When you finish using a microscope, you have to take out all your samples you were using and make sure it is completely clean so that no further contamination affects future sample analysis.
Also when you begin to use it, make sure it is clean so that when you analyze your sample(s) are free from other unexpected agents.
Yo sup??
Let the percentage of K-39 be x
then the percentage of K-40 is 100-(x+0.01)
We know that the net weight should be 39.5. Therefore we can say
(39*x+40*(100-(x+0.01))+38*0.01)/100=39.5
(since we are taking it in percent)
39*x+40*(100-(x+0.01))+38*0.01=3950
39x+4000-40x-0.4+0.38=3950
2x=49.98
x=24.99
=25 (approx)
Therefore K-39 is 25% in nature and K-40 is 75% in nature.
Hope this helps.
