Ok first, we have to create a balanced equation for the dissolution of nitrous acid.
HNO2 <-> H(+) + NO2(-)
Next, create an ICE table
HNO2 <--> H+ NO2-
[]i 0.139M 0M 0M
Δ[] -x +x +x
[]f 0.139-x x x
Then, using the concentration equation, you get
4.5x10^-4 = [H+][NO2-]/[HNO2]
4.5x10^-4 = x*x / .139 - x
However, because the Ka value for nitrous acid is lower than 10^-3, we can assume the amount it dissociates is negligable,
assume 0.139-x ≈ 0.139
4.5x10^-4 = x^2/0.139
Then, we solve for x by first multiplying both sides by 0.139 and then taking the square root of both sides.
We get the final concentrations of [H+] and [NO2-] to be x, which equals 0.007M.
Then to find percent dissociation, you do final concentration/initial concentration.
0.007M/0.139M = .0503 or
≈5.03% dissociation.
Water molecules are highly packed and are always near each other.
Answer:
Volume = 44.96L
Explanation:
p = 427mmHg = 0.56atm
R = 0.08206
T 6°C = 279k
V = ?
n = 1.10 mol
using ideal gas equation
pv=nRT
0.56 x V = 1.10 x 0.08206 x 279
0.56 x V = 25.18
V = 25.18/0.56
V = 44.96L
Answer:
Explanation:
The monochlorination of n-butane gives two products out of which one is optically active with one chiral carbon atom. Thus, there are 3 products are obtained
