PKa= 4.9 therefore ka= 10^-4.9= 1.259x10^-5
![ka= \frac{[H^+][CH3CH2COO^-]}{[CH3CH2COOH]}](https://tex.z-dn.net/?f=ka%3D%20%5Cfrac%7B%5BH%5E%2B%5D%5BCH3CH2COO%5E-%5D%7D%7B%5BCH3CH2COOH%5D%7D%20)
![[CH3CH2COO^-] ](https://tex.z-dn.net/?f=%5BCH3CH2COO%5E-%5D%0A)
= 0.05
![[CH3CH2COOH]](https://tex.z-dn.net/?f=%5BCH3CH2COOH%5D)
= 0.10
Therefore 1.259x10^-5 =
![\frac{[H^+][0.05]}{[0.1]}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BH%5E%2B%5D%5B0.05%5D%7D%7B%5B0.1%5D%7D%20)
Rearrange the equation to make the concentration of hydrogen the subject.
Therefore
![[H^+] = \frac{(1.259*10^-5)(0.1)}{0.05}](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%20%3D%20%20%5Cfrac%7B%281.259%2A10%5E-5%29%280.1%29%7D%7B0.05%7D%20%20)
Therefore
![[H^+]= 2.513*10^-5](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%202.513%2A10%5E-5)
pH= -log [
] = -log(2.513*10^-5)= 4.59.
Well in this
case, silver
nitrate is reduced:
Ag<span>+ </span><span>+ </span>e<span>− </span>→ Ag(s) ↓
Meanwhile, the aluminum
is oxidized forming a positive ion:
Al(s<span>) → </span>Al<span>3+ </span><span>+ 3</span>e−
To get the
overall reaction, we add the half
equations so that the electrons are eliminated:
Al(s<span>) + 3</span>Ag<span>+ </span><span>→ </span>Al<span>3+ </span><span>+ 3</span>Ag(s)
And similarly:
Al(s<span>) + 3</span>AgNO3(aq<span>) → </span>Al(NO3)3(aq<span>) + 3</span>Ag(s<span>)</span>
Variables we know:
t = 8 seconds
Vi = 0 m/s
g = -9.81
Δy = ?
Vf = ?
Equation we will be using to solve for Vf: Vf = Vi + gt
Steps to solve:
Vf = (0) + (-9.81)(8)
Vf = -78.48 m/s
Hope this helps!! :)
Radium and polonium is the answer to this question. I hope I helped out!
It is the boilimg point now aa
