Answer: E
Hope this helped :)
Answer:
The relationship is expressed as follows: ![K_{a} = \frac{[H+][A-]}{[HA]}](https://tex.z-dn.net/?f=K_%7Ba%7D%20%3D%20%5Cfrac%7B%5BH%2B%5D%5BA-%5D%7D%7B%5BHA%5D%7D)
Explanation:
Most acidic substances are weak acids and are therefore only partially ionized in acqeous solution. We cab use the equilibrium constant for the ionization of acid to express the extent to which the weak acid ionizes. If we represent a general weak acid as HA, we can write the equation for its ionization reaction like this:
![K_{a} = \frac{[H+][A-]}{[HA]}](https://tex.z-dn.net/?f=K_%7Ba%7D%20%3D%20%5Cfrac%7B%5BH%2B%5D%5BA-%5D%7D%7B%5BHA%5D%7D)
To calculate the pH of a weak acid, we use the equilibrium concentration of the reacted species and product.
Take for example:
HA → H + A⁻
where A id the conjugate base.
Knowing that x amount of acid reacts, we can solve like this:
HA → H + A⁻
H+ = antilog (pH)
thus, the pH of the acid is equals to H+ (initial) - H+ (equilibrium) ≈ H+ (initial)
212 degrees Fahrenheit is equivalent to 100 degrees Celsius.
Answer:
Average atomic mass of uranium = 237.97 amu
Explanation:
Given data:
Abundance of U-234 = 0.01%
Abundance of U-235 = 0.71%
Abundance of U-238 = 99.28%
Average atomic mass of uranium = ?
Solution:
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass) / 100
Average atomic mass = (234×0.01)+(235×0.71)+(238+99.28) /100
Average atomic mass = 2.34 + 166.85+23628.64 / 100
Average atomic mass = 23797.83 / 100
Average atomic mass = 237.97 amu.