The first energy level
Let me know if this helps !!
(I think you have a mistake in your question as the addition is 30mL, not 100mL)
when PH = - ㏒[H+]
and here we have HClO4 is the strong acid
So PH = - ㏒[HClO4]
moles of HClO4 = 0.1 L *0.18 m = 0.018 M
moles of LiOH = 0.03 L * 0.27 m = 0.0081 M
when the total volume = 0.1L + 0.03L = 0.13 L
∴ [HClO4] = (0.018-0.0081)/0.13 L
= 0.076 M
PH = -㏒ 0.076
= 1.12
Answer:
0.45 M
Explanation:
Now we need to make use of the dilution formula
C1V1= C2V2
C1= initial molarity of the solution which is the unknown
V1 = initial volume of the solution which is 175 ml
C2 =final molarity of the solution= 0.315 M
V2= final volume of solution= 250 ml
From C1V1= C2V2 we have;
C1= C2V2/V1
C1= 0.315 × 250/ 175
C1= 0.45 M
I mole of water has an Avogadro number of molecules.
1 mole = 6.02 * 10^ 23 molecules.
6.02 * 10^ 23 molecules = 1 mole of water
1 molecule = 1/(6.02 * 10^23) mole of water
2.0 * 10^22 molecules would have = (2*10^22) * 1/(6.02*10^23)
= 0.033
2* 10 ^22 molecules of water would have 0.033 moles of water.
