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
Assuming that you mean 10^-4 M then this would be basic and would have a pH of 10.
pOH = -log[OH].
So pOH = 4
pH=14-pOH
pH = 10
To find moles in this sample, you would divide grams by molar mass of ethyl alcohol
(18.0g)/(46.07g/mol) = 0.391mol C2H6O
Answer: There would have to be three nitrogen atoms in the products. The law of conservation of matter states that the amount of substance before a reaction occurs should be the same as the amount of substance after the reaction.
Explanation: This is the EXACT sample answer from the test, just reword it if you want. ^
Element Atomic Number Valency
Valency of Hydrogen 1 1
Valency of Helium 2 0
Valency of Lithium 3 1
Valency of Beryllium 4 2
Valency of Boron 5 3
Valency of Carbon 6 4
Valency of Nitrogen 7 3
Valency of Oxygen 8 2
Valency of Fluorine 9 1
Valency of Neon 10 0
Valency of Sodium (Na) 11 1
Valency of Magnesium (Mg) 12 2
Valency of Aluminium 13 3
Valency of Silicon 14 4
Valency of Phosphorus 15 3
Valency of Sulphur 16 2
Valency of Chlorine 17 1
Valency of Argon 18 0
Valency of Potassium (K) 19 1
Valency of Calcium 20 2
Valency of Scandium 21 3
Valency of Titanium 22 4
Valency of Vanadium 23 5,4
Valency of Chromium 24 2
Valency of Manganese 25 7, 4, 2
Valency of Iron (Fe) 26 2, 3
Valency of Cobalt 27 3, 2
Valency of Nickel 28 2
Valency of Copper (Cu) 29 2, 1
Valency of Zinc 30 2
