Answer:
pH of the final solution = 3.8
Explanation:
Concentration of NaF = 
molar
= 0.3 molar
NaF → Na⁺ + F⁻
HF ⇆ H⁺ + F⁻
- NaF is strong electrolyte so completely ionized but HF weak acid not completely ionized.
- Since F⁻ is common ion here
according to common ion effect dissociation of weak acid decreases.
Ka = ![\frac{[H]^{+}[F]^{-} }{[HF]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BH%5D%5E%7B%2B%7D%5BF%5D%5E%7B-%7D%20%20%7D%7B%5BHF%5D%7D)
⇒ [H⁺] = ![\frac{K_{a} [HF]}{[F]^{-} }](https://tex.z-dn.net/?f=%5Cfrac%7BK_%7Ba%7D%20%5BHF%5D%7D%7B%5BF%5D%5E%7B-%7D%20%7D)
...............(1)
{Ka of HF = 3.5 x 10⁻⁴} & Concentration of HF = 30 x 4 x 10⁻³ = 0.12 molar
from equation 1
⇒ [H⁺] = 
[Concentration of F⁻ ≡ Concentration of NaF]
⇒ [H⁺] = 0.00014
⇒pH = - log 0.00014 = 3.85
Answer: The coordination number of Cu+ is two while the coordination number of O2- is 4
Explanation:
The coordination number of an ion in a unit cell arrangement refers to the number of immediate neighboring ions of opposite charge surrounding a particular ion. In Cu2O, the charge on Cu is +1 and the charge on the oxide ion is -2. Each copper I ion is surrounded by two oxide ions while each oxide ion is surrounded by four copper I ions in each unit cell.
What's the relationship between total and partial pressure? The total pressure is the sum of the parcial pressures!
So for us, it would be:
378= 212+101+x
where x is the parcial pressure of nitrogen.
Now we count:
378= 212+101+x
378=313+x
378-313=x
65=x
So the parcial pressure exerted by nitrogen is 65!
.0002345 I believe this is correct
