Answer:
400 L
Explanation:
As this problem deals with a dilution process, we can use the following formula:
Where subscript 1 stands for the concentrated concentration and volume, while 2 stands for the diluted values. Meaning that in this case:
We <u>input the data</u>:
And <u>solve for V₂</u>:
Fluorine is the lightest halogen
Answer:
pKa = 4.89.
Explanation:
We can solve this problem by using the <em>Henderson-Hasselbach equation</em>, which states:
pH = pKa + log ![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
In this case [A⁻] is the concentration of sodium benzoate and [HA] is the concentration of benzoic acid.
We <u>input the given data</u>:
4.63 = pKa + log ![\frac{0.16}{0.29}](https://tex.z-dn.net/?f=%5Cfrac%7B0.16%7D%7B0.29%7D)
And <u>solve for pKa</u>:
pKa = 4.89
Answer:
See explanation
Explanation:
-0.7ev = -13.6/n^2
n= 4
4--->1
4--->3, 3---> 2, 2--->1
E4-E1= -13.36(1/4^2 - 1/1^2) = 12.53 eV
E4-E3 = -13.36 (1/4^2 - 1/3^2) = 0.65 ev
E3-E2 = -13.36 (1/3^2 - 1/2^2) = 1.86 eV
E2- E1 = -13.6 ( 1/2^2 - 1/1^2) = 10.02 ev
a)
λ= 6.6 × 10^-34 × 3 × 10^8/12.53 × 1.6 × 10^-19
λ= 19.8 × 10^-26/20.048 ×10^-19
λ= 9.9 × 10^ -8 m
b)
λ= 6.6 × 10^-34 × 3 × 10^8/0.65 × 1.6 × 10^-19
λ= 19.8 × 10^-26/1.04 ×10^-19
λ= 19.0 × 10^ -7 m