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:
маликулярная масса ровна 23
Good electrical conductivity and electronegativities less than 1.7 are the properties and characteristic of Group 2 elements at STP.
<h3>What are the properties of group 2 elements?</h3>
Group 2 elements are metals so they are good conductors of heat and electricity. It has electronegativity values less than 1.7 and very reactive. They form 2+ charge in cationic form and also formed ionic bonds with other negatively charged elements.
So we can conclude that good electrical conductivity and electronegativities less than 1.7 are the properties and characteristic of Group 2 elements at STP.
Learn more about electronegativity here: brainly.com/question/2415812
#SPJ1
The transfer of energy that occurs when a force is applied over a distance is WORK.
Hope this helps!
