Arock is thrown horizontally at a speed of 5.0 m/s from the top of a cliff 64.7 m high. the rock hits the ground 18.0 m from the base of the cliff. how would this distance change if the rock was thrown at 10.0 m/s?

Refer to the diagram shown.

Assume g = 9.8 m/s² and ignore air resistance
When the rock is launched from a height of 64.7 m,
u = 5.0 m/s, the horizontal velocity
v = 0,  the initial vertical velocity

If the rock hits the ground 18.0 m  from the base of the cliff, then the time of flight is
t = (18.0 m)/(5.0 m/s) = 3.6 s
The vertical distance traveled is
s = (1/2)*(9.8 m/s²)*(3.6 s)² = 63.504 m

Because this distance is less than 64.7 m, ground level is slightly higher away from the base of the cliff. It is higher by
64.7 - 63.504 =  1.196 m

If the rock is thrown at 10 m/s, the time of flight remains the same because acceleration due to gravity is the same.
Therefore the horizontal distance traveled is
(10.0 m/s)*(3.6 s) = 36.0 m

The distance will be 36.0 m

It won't lose or gain electrons if it's a covalent bond it shares the electrons with another atom or atoms. if it is an ionic bond it will lose the 3 electrons on the outer shell and become a positively charged ion.  credit goes to @sunshinebaby of answers.yahoo.com.