CaCO₃ partially dissociates in water as Ca²⁺ and CO₃²⁻. The balanced equation is,
CaCO₃(s) ⇄ Ca²⁺(aq) + CO₃²⁻(aq)
Initial Y - -
Change -X +X +X
Equilibrium Y-X X X
Ksp for the CaCO₃(s) is 3.36 x 10⁻⁹ M²
Ksp = [Ca²⁺(aq)][CO₃²⁻(aq)]
3.36 x 10⁻⁹ M² = X * X
3.36 x 10⁻⁹ M² = X²
X = 5.79 x 10⁻⁵ M
Hence the solubility of CaCO₃(s) = 5.79 x 10⁻⁵ M
= 5.79 x 10⁻⁵ mol/L
Molar mass of CaCO₃ = 100 g mol⁻¹
Hence the solubility of CaCO₃ = 5.79 x 10⁻⁵ mol/L x 100 g mol⁻¹
= 5.79 x 10⁻³ g/L
P1T2 = P2T1
(3.8)(36)=25P2
136.8=25P2
136.8/25=P2
P2=5.472atm
Answer:
b) Phosphorus acid
Explanation:
To distinguish the type of acid of phosphorus with the oxidation state of +3, we need to be familiar with the chemical formula of each of the compounds:
Orthophosphoric acid H₃PO₄
Phosphorus acid H₃PO₃
Metaphosphoric acid HPO₃
Phyrophosphoric acid H₄P₂O₇
Now that we know the formula of the given compounds, the algebraic sum of all the oxidation numbers of all atoms in a neutral compound is zero:
Only phosphorus acid yielded an oxidation state of +3 for phosphorus in the compound.
H₃PO₃:
we know the oxidation state of H = +1
O = -2
The oxidation state of P is unknown. We can express this as an equation:
3(+1) + P + 3(-2) = 0
3 + P -6 = 0
P-3 = 0
P = +3
Answer:
the stabilization of the negative charge in orbitals with higher s character
Explanation:
Acetylide anion is a carbon anion compound or popularly called carbanion. Now Acetylide anion is sp hybridized. However acetylide anion tends to be more acidic as we move from sp³ to sp, hence acidicity increases, which makes sp to have the highest acidity and become the most stable.
So, we can conclude that the acetylide anion is more acidic due to the stabilization of the negative charge in orbitals with higher s character and as the s character increases, acidic nature of acetylide anion also increases.