Initial [ HCO2H] = moles * volume
=0.35 moles * 1 L = 0.35 M
by using ICE table:
HCO2H ↔ H+ + HCO2-
initial 0.35 M 0 0
change - X +X +X
Equ (0.35 - X) X X
∴ Ka = [H+][HCO2-] / [HCO2H]
by substitution:
1.8 x 10^-4 = X^2 / (0.35-X) by solving for X
∴ X = 0.0079 or 7.9 x 10^-3
∴ [H+] = X = 7.9 x 10^-3 M
The solid state where particles are more compact
Answer:
1st page, number a-ENergy is added or absorbed
I cant see the answers on the 2nd page
Explanation:
Answer:
94.2 g/mol
Explanation:
Ideal Gases Law can useful to solve this
P . V = n . R . T
We need to make some conversions
740 Torr . 1 atm/ 760 Torr = 0.974 atm
100°C + 273 = 373K
Let's replace the values
0.974 atm . 1 L = n . 0.082 L.atm/ mol.K . 373K
n will determine the number of moles
(0.974 atm . 1 L) / (0.082 L.atm/ mol.K . 373K)
n = 0.032 moles
This amount is the weigh for 3 g of gas. How many grams does 1 mol weighs?
Molecular weight → g/mol → 3 g/0.032 moles = 94.2 g/mol
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