Answer:
The concentration of hydrogen ion at pH is equal to 2 :![= [H^+]=0.01 mol/L](https://tex.z-dn.net/?f=%3D%20%5BH%5E%2B%5D%3D0.01%20mol%2FL)
The concentration of hydrogen ion at pH is equal to 6 : ![[H^+]'=0.000001 mol/L](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%27%3D0.000001%20mol%2FL)
There are 0.009999 more moles of 
ions in a solution at a pH = 2 than in a solution at a pH = 6.
Explanation:
The pH of the solution is the negative logarithm of hydrogen ion concentration in an aqueous solution.
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)
The hydrogen ion concentration at pH is equal to 2 = [H^+]
![2=-\log [H^+]\\](https://tex.z-dn.net/?f=2%3D-%5Clog%20%5BH%5E%2B%5D%5C%5C)
![[H^+]=10^{-2}M= 0.01 M=0.01 mol/L](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-2%7DM%3D%200.01%20M%3D0.01%20mol%2FL)
The hydrogen ion concentration at pH is equal to 6 = [H^+]
![6=-\log [H^+]\\\\](https://tex.z-dn.net/?f=6%3D-%5Clog%20%5BH%5E%2B%5D%5C%5C%5C%5C)
![[H^+]=10^{-6}M= 0.000001 M= 0.000001 mol/L](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-6%7DM%3D%200.000001%20M%3D%200.000001%20mol%2FL)
Concentration of hydrogen ion at pH is equal to 2 =![[H^+]=0.01 mol/L](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.01%20mol%2FL)
Concentration of hydrogen ion at pH is equal to 6 =
The difference between hydrogen ion concentration at pH 2 and pH 6 :
![= [H^+]-[H^+]' = 0.01 mol/L- 0.000001 mol/L = 0.009999 mol/L](https://tex.z-dn.net/?f=%3D%20%5BH%5E%2B%5D-%5BH%5E%2B%5D%27%20%3D%200.01%20mol%2FL-%200.000001%20mol%2FL%20%3D%200.009999%20mol%2FL)
Moles of hydrogen ion in 0.009999 mol/L solution :
There are 0.009999 more moles of ions in a solution at a pH = 2 than in a solution at a pH = 6.
The kinetic energy is
and the height of the building doesn't matter at all.
joules
What is the definition of newton
Assume no air resistance, and g = 9.8 m/s².
Let
x = angle that the initial velocity makes with the horizontal.
u = 30 cos(x), horizontal velocity
v = 30 sin(x), vertical launch velocity
The horizontal distance traveled is 55 m, therefore the time of flight is
t = 55/[30 cos(x)] = 1.8333 sec(x) s
With regard to the vertical velocity, and the time of flight,obtain
[30 sin(x)]*(1.8333 sec(x)) + (1/2)*(-9.8)*(1.8333 sec(x))² = 0
55 tan(x) - 16.469 sec²x = 0
55 tan(x) - 16.469[1 + tan²x] = 0
16.469 tan²x - 55 tan(x) + 16.469 = 0
tan²x - 3.3396 tan(x) + 1 = 0
Solve with the quadratic formula.
tan(x) = 0.5[3.3396 +/- √(7.153)] = 3.007 or 0.3326
Therefore
x = 71.6° or x = 18.4°
The time of flight is
t = 1.8333 sec(x) = 5.8096 s or 1.932 s
The initial vertical velocity is
v = 30 sin(x) = 28.467 m/s or 9.468 m/s
The horizontal velocity is
u = 30 cos(x) = 9.467 m/s or 28.469 m/s
If t = 5.8096 s,
u*t = 9.467*5.8096 = 55 m (Correct)
or
u*t = 28.469*15.8096 = 165.4 m (Incorrect)
Therefore, reject x = 18.4°. The correct solution is
t = 5.8096 s
x = 71.6°
u = 9.467 m/s
v = 28.467 m/s
The height from which the ball was thrown is
h = 28.467*5.8096 - 0.5*9.8*5.8096² = -110.4 m
The ball was thrown from a height of 110.4 m
Answer: h = 110.4 m
