This is held together by the charges of anions and cations that are attracted to each other, which is the reason why crystal lattice is being formed beside the fct that it has tightly packed network. The crystal lattice are the crystals which has an arrangement that could be describe with its three dimensional feature.
PV=nRT
n = mole = 2.1 g : 32 g/mole = 0.065
T = 13 + 273 = 286 K
V = nRT/P
V = (0.065 x 0.08205 x 286)/0.1
V = 15.253 L
<span>1,3-cylohexadiene i synthesized starting from cyclohexane in following 4 steps.
1) Free Radical Substitution Rxn: Halogenation of cyclohexane in the presence of UV yield chlorocyclohexane.
2) Elimination Rxn: Dehydrohalogenation of chlorocyclohexane yields cyclohexene.
3) Halogenation of Cyclohexene (
Electrophillic Addition Rxn) gives 1,2-dihalocyclohexane.
4) Elemination Rxn: When dibromocyclohexane is treated with KOH and heated it gives 1,3-cyclohexadiene as shown below,</span>
Phosphoric acid has 3 pKa values (pKa1=2.1, pKa2=6.9, pKa3= 12.4) and after 3 ionization it gives 3 types of ions at different pKa values:
H₃PO₄(aq) + H₂O(l) ⇌ H₃O⁺(aq) + H₂PO₄⁻ (aq) pKₐ₁
<span>
</span>H₂PO₄⁻(aq) + H₂O(l) ⇌ H₃O⁺(aq) + HPO₄²⁻ (aq) pKₐ₂
HPO₄²⁻(aq) + H₂O(l) ⇌ H₃O⁺(aq) + PO₄³⁻ (aq) pKₐ₃
The last equilibrium is associated with the highest pKa value (12.4) of phosphoric acid. There the last OH group will lose its hydrogen and hydrogen phosphate ion (HPO₄²⁻) turns into phosphate ion (PO₄³⁻).
The ph of a saturated solution of Ca(OH)2 is 12.35
CALCULATION:
For the reaction
Ca(OH)2 → Ca2+ + 2OH-
we will use the Ksp expression to solve for the concentration [OH-] and then use the acid base concepts to get the pH:
Ksp = [Ca2+][OH-]^2
The listed Ksp value is 5.5 x 10^-6. Substituting this to the Ksp expression, we have
Ksp = 5.5 x 10^-6 = (s) (2s)^2 = 4s^3
s3 = 5.5x10^-6 / 4
Taking the cube root, we now have
s = cube root of (5.5x10^-6 / 4)s
= 0.01112
We know that the value of [OH-] is actually equal to 2s:
[OH-] = 2s = 2 * 0.01112 = 0.02224 M
We can now calculate for pOH:
pOH = - log [OH-]
= -log(0.02224)
= 1.65
Therefore, the pH is
pH = 14 - pOH
= 14 - 1.65
= 12.35
