First, we should get moles acetic acid = molarity * volume
=0.3 M * 0.5 L
= 0.15 mol
then, we should get moles acetate = molarity * volume
= 0.2 M * 0.5L
= 0.1 mol
then, we have to get moles of OH- which added:
moles OH- = molarity * volume
= 1 M * 0.02L
= 0.02 mol
when the reaction equation is:
CH3COOH + OH- → CH3COO- + H2O
moles acetic acid after adding OH- = (0.15-0.02)
= 0.13M
moles acetate after adding OH- = (0.1 + 0.02)
= 0.12 M
Total volume = 0.5 L + 0.02 L= 0.52 L
∴[acetic acid] = moles acetic acid after adding OH- / total volume
= 0.13mol / 0.52L
= 0.25 M
and [acetate ) = 0.12 mol / 0.52L
= 0.23 M
by using H-H equation we can get PH:
PH = Pka + ㏒[salt/acid]
when we have Ka = 1.8 x 10^-5
∴Pka = -㏒Ka
= -㏒ 1.8 x 10^-5
= 4.7
So by substitution:
∴ PH = 4.7 + ㏒[acetate/acetic acid]
= 4.7 + ㏒(0.23/0.25)
= 4.66
I think it's less, more, more. correct me if i'm wrong
Answer: Magnet
Explanation:Using a magnet is the best separating technique to be deployed in this case. The nails are easily picked out by just holding a magnet over the sandbox.
Answer:
about 19 or 20 g
Explanation:
To do this, is neccesary to watch a solubility curve of this compound. This is the only way that you can know how many grams are neccesary to dissolve this compound in 50 mL of water to a given temperature.
Now, if you watched the attached graph, you can see the solubility curve of many compounds in 100 g of water (or 100 mL of water). So, to know how many do you need in 50 mL, it's just the half.
So watching the curve, you can see that at 20 °C, we simply need between 35 g and 40 g. Let's just say we need 38 grams of NH4Cl to be dissolved in 100 mL of water.
So, in 50 mL, it's just the half. So, we only need 19 g or 20 g of NH4Cl at 20 °C, to dissolve this compound in water.
The correct answer is B. if im wrong please correct me if i am wrong
