m = mass of the particle = 2 x 10⁻³⁰ kg
Δv = uncertainty in velocity of the particle = 10² m/s
Δx = uncertainty in position of the particle = ?
h = plank's constant = 6.63 x 10⁻³⁴ J-s
uncertainty principle is given as
m Δx Δv ≥ h/(4π)
inserting the values
(2 x 10⁻³⁰) Δx (10²) ≥ (6.63 x 10⁻³⁴)/(4(3.14))
Δx ≥ 2.64 x 10⁻⁷ m
so minimum possible uncertainty is 2.64 x 10⁻⁷ m
12 pens for writing chemical formulas cost
12*2=24 dollars
12 pads cost
12*3=36 dollars
1)
12*2=24
12*3=36
2)
12*2+12*3=60 dollars
3) 12*(3+5)=60 dollars
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.
I think energy is released
The correct answer is B. the Taliban.