It is a graph. It shows observations and then you record your results with any of the graph types.
It’s probably Mercury I’m not sure
Answer:
At the second equivalent point 200 mL of NaOH is required.
Explanation:
at the first equivalent point:
H2A + OH- = HA- + H2O
initial mmoles y*100 y*100 - -
final mmoles 0 0 y*100 y*100
at the second equivalent point:
HA- + OH- = A2- + H2O
initial mmoles y*100 y*100 - -
final mmoles - - y*100 y*100
at the second equivalent point we have that y*100 mmoles of NaOH or 100 mL of NaOH ir required, thus:
at the second equivalent point 200 mL of NaOH is required.
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 0.230M 0M 0M
Δ[] -x +x +x
[]f 0.230-x x x
Then, using the concentration equation, you get
4.5x10^-4 = [H+][NO2-]/[HNO2]
4.5x10^-4 = x*x / .230 - x
However, because the Ka value for nitrous acid is lower than 10^-3, we can assume the amount it dissociates is negligable,
assume 0.230-x ≈ 0.230
4.5x10^-4 = x^2/0.230
Then, we solve for x by first multiplying both sides by 0.230 and then taking the square root of both sides.
We get the final concentrations of [H+] and [NO2-] to be x, which equals 0.01M.
Then to find percent dissociation, you do final concentration/initial concentration.
0.01M/0.230M = .0434 or
≈4.34% dissociation.
Ans: Elements in the same period have the same number of electron shells.
