Answer:
Explanation:
Hello!
In this case, since a dilution process implies that the moles of the solute remain the same before and after the addition of diluting water, we can write:
Thus, since we know the volume and concentration of the initial sample, we compute the resulting concentration as shown below:
Best regards!
<span>Van der waal or ideal eqn is given by PV = NRT; P = NRT/ V.
Where N = 1.335 is the number of moles. T = 272K is temperature. V = 4.920L is the volume. And R = 0.08205L. Substiting the values into the eqn; we have,
P = (1.331* 0.08205 * 272)/ 4.920 = 29.7047/ 4.920 = 6.03atm.</span>
Answer:
Hi! I believe this is your answer:
52 kilograms
Hope this helps, sorry if it's wrong!
Explanation:
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.
