When the reaction equation is:
HF ↔ H+ + F-
and when the Ka expression
= concentration of products/concentration of reactions
so, Ka = [H+][F-]/[HF]
when we assume:
[H+] = [F-] = X
and [HF] = 0.35 - X
So, by substitution:
6.8 x 10^-4 = X^2 / (0.35 - X) by solving for X
∴ X = 0.015 M
∴[H+] = X = 0.015
when PH = -㏒[H+]
∴PH = -㏒0.015
= 1.8
Answer:
53.6 g of N₂H₄
Explanation:
The begining is in the reaction:
N₂(g) + 2H₂(g) → N₂H₄(l)
We determine the moles of each reactant:
59.20 g / 28.01 g/mol = 2.11 moles of nitrogen
6.750 g / 2.016 g/mol = 3.35 moles of H₂
1 mol of N₂ react to 2 moles of H₂
Our 2.11 moles of N₂ may react to (2.11 . 2) /1 = 4.22 moles of H₂, but we only have 3.35 moles. The hydrogen is the limiting reactant.
2 moles of H₂ produce at 100 % yield, 1 mol of hydrazine
Then, 3.35 moles, may produce (3.35 . 1)/2 = 1.67 moles of N₂H₄
Let's convert the moles to mass:
1.67 mol . 32.05 g/mol = 53.6 g
Answer:The answer is D.1,3
Explanation:
all the elements in group 18 are Nobel gases or inert gases . all the elements such as neon , helium, argon etc. ,their outermost shell is completely filled . The noble gases have the largest ionization energies, reflecting their chemical inertness
