I'm pretty sure the answer is hypothesis
Volume required for neutralization V will be:
V * 0.2125 M HCl = 25 mL * 0.17 M
V = 20 ml
First part:
When 10 mL is added we can apply Henderson equation to get the result, so:
The pH will be of basic buffer
pOH = pKb + log(salt/base)
or pOH = 4.19 + log (0.2125*10 / 25*0.17 - 10*0.2125 )
pOH = 4.19 and pH = 14 - 4.19 = 9.81
Second part:
When 20 ml is added, there is only salt formed
The pH will be salt of strong acid and weak base
So pH = 7 - 0.5 pKb - 0.5 log C
where C is the concentration of the salt formed so:
pH = 7 - (0.5*4.19) - (0.5 log (25*0.17) / (25+20))
= 5.42
Third part:
When 30 ml of the acid has been added,
The pH will be of the remaining strong acid
pH = - log (0.2125*10 / 25 + 30 )
= 1.326
Answer:
10moles of kcl
Explanation:
2
K
C
l
O3 → 2
K
C
l + 3
O
2
Notice that you have a 2
:
3 mole ratio between potassium chlorate and oxygen gas, which means that, regardless of how many moles of the former react, you'll always produce 2/3 times more moles of the latter.
15 mol of O2 * ((2mol of KCLO3)/(3mol of O2))= 15*2/3=10 Mol