Answer:
The molar mass of the unknown gas is 100.4 g/mol
Explanation:
Step 1: Data given
Molar mass of argon = 39.95 g/mol
After filling with argon the flask gained 3.221 grams
After filling with an unknown gas, the flask gained 8.107 grams
Step 2: Calculate the molar mass of the unknown gas
The gas with the higher molar mass will have the higher density.
Ar - 3.224 g; molar mass = 39.95 g/mol
X = 8.102 g; molar mass = ??
Molar mass of the unknown gas = 8.102g X *(39.95 g/mol / 3.224 g) = 100.4 g/mol
The molar mass of the unknown gas is 100.4 g/mol
Answer:
the solubility of CaCO3 is 0.015g/l 25 °C
is favored at equilibrium
Explanation:
The Ksp of calcium carbonate in water at 25 °C is 2.25 x 10-8. CaCO3(s) <----> Ca2+ (aq) + CO3 2- (aq) What is favored at equilibrium?
solubility is the property of a solute to dissolve in a solvent(liquid, gas ) to form a solution(soution can be saturated ,unsaturated, or supersaturated)
CaCO3(s) <----> Ca2+ (aq) + CO3 2- (aq)
in partial dissociation , we can say
2.25x 10^-8=
let Ca^2+=CO3^-2=S
2.25x10^-8=S*S
S^2=2.25x10^-8
S=0.00015mol/L
Converting that to g/l
the relative molecular mass of CaCO3=100g/mol
0.00015*100g/mol
0.015g/l
the solubility of CaCO3 is 0.015g/l @room temperature
is favored at equilibrium
I don’t know what you mean by classification exactly but it is a redox equation. The reactant side of carbon is losing hydrogen to form carbon dioxide. And oxygen is gaining hydrogen which gives you the water. Redox reactions are also known as combustion reactions.