<u>Answer:</u>
<em>The situation given here is imaginary such that the life of Rock has to be found using the half-life of the element lokium that has been found inside the rock. </em>
<u>Explanation:</u>
Half-life of any material is the amount of time taken by that particular material to decay. Now the amount of lokium found in rock can show after how many half-lives this amount has been left out.
The time elapsed will be log (L) atoms X half-life.
The answer is D higher potential energy and is unstable
Answer:
Along period electronegativity and ionization energy increases.
Along group electronegativity and ionization energy decreases.
Explanation:
Along period:
As we move from left to right across the periodic table the number of valance electrons in an atom increase. The atomic size tend to decrease in same period of periodic table because the electrons are added with in the same shell. When the electron are added, at the same time protons are also added in the nucleus. The positive charge is going to increase and this charge is greater in effect than the charge of electrons. This effect lead to the greater nuclear attraction. Thus the attraction of the atoms for valance electrons increases. The electrons are pull towards the nucleus and valance shell get closer to the nucleus. As a result of this greater nuclear attraction atomic radius decreases and ionization energy increases because it is very difficult to remove the electron from atom and more energy is required, and electronegativity also increases.
Along group:
As we move from top to bottom in periodic table the atomic sizes increases.The electrons are added in next energy level in every next element. Thus the valance electrons farther away from the nucleus and hold of nucleus becomes weaker, because of weak nuclear attraction atomic radii increases and electronegativity and ionization energy decreases.
Answer:
B) 0.32 %
Explanation:
Given that:
Concentration = 1.8 M
Considering the ICE table for the dissociation of acid as:-
The expression for dissociation constant of acid is:
![K_{a}=\frac {\left [ H^{+} \right ]\left [ {CH_3COO}^- \right ]}{[CH_3COOH]}](https://tex.z-dn.net/?f=K_%7Ba%7D%3D%5Cfrac%20%7B%5Cleft%20%5B%20H%5E%7B%2B%7D%20%5Cright%20%5D%5Cleft%20%5B%20%7BCH_3COO%7D%5E-%20%5Cright%20%5D%7D%7B%5BCH_3COOH%5D%7D)
Solving for x, we get:
<u>x = 0.00568 M</u>
Percentage ionization = 
<u>Option B is correct.</u>
