When CaSO4 → Ca2+ + SO4
So when we have Ksp = [Ca2+][SO4]
when Ksp = 4.93 x 10^-5
and [SO4] = 0.02 M
so by substitution we can get [Ca2+]
4.93x10^-5 = [Ca2+] [0.02]
∴ [Ca2+] = 0.0025 mol/L
∴ the moles of calcium chloride = 0.0025 mol / L * 1.5 L
= 0.00167 mol
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Answer:
molarity of diluted solution = 1.25 M
Explanation:
Using,
C1V1 (Stock solution) = C2V2 (dilute solution)
given that
C1 = 2.50M
V1 = 250ML
C2 = ?
V2 = 500ML
2.50 M x 250 mL = C2 x 500 mL
C2 = (2.50 M x 250 mL) / 500 mL
C2 = 1.25 M
Hence, molarity of diluted solution = 1.25 M
Answer:
Explanation:
Hello!
In this case, given this is an acid-base neutralization and we are considering a diprotic acid, we can write the following mole-mole relationship:
It means that the moles of acid can be computed given the volume and concentration of NaOH:
It means that the approximate molar mass of the acid is:
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