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
Answer: The net ionic equation will be as follows.

Explanation:
The chemical equation for the given reaction is as follows.

We know that a strong acid or base will dissociate completely into a solvent whereas a weak acid or base dissociates partially into the solvent. Hence, the ionic equation will be as follows.
Now, we will cancel the spectator ions from the above equation. Therefore, the net ionic equation will be as follows.

or,
Answer:
Our planet's rotation produces a force on all bodies moving relative to theEarth. ... The force, called the "Coriolis effect," causes the direction of winds and ocean currents to be deflected.
Explanation:
It’s
1.A
2.C
3.B
hope it’s correct
The variable is what changes during an experiment. Hopefully this helped! :)