Answer:
Solubility= 1.08×10-12
Explanation:
Take the cube root of 1.27×10-36
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 only got 50 points (which is not 100). :-)
Look at the graph. At 80 °C, about 38 g of solute is able to dissolve, and that’s for ever 100 g of water. That means that for every 150 grams of water, 57 grams of solute can dissolve (38/2 = 19 + 38 = 57 g) at 80 °C. Since 57 g is greater than 55 g, all for he sodium chloride should dissolve in 150 g of water at 80 °C - you can put all of that into a “mathematical explanation”.
Multiply the mass by the abundance and add each. Make sure to convert the percentage into a decimal. (49.946 * .043) + (51.941 * .838) + (52.941 * .095) + (53.939 * .024) = 51.99 round up using sig figs and the answer is c. 52.00 amu