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
One correct thging is that there are the same amount of positive and negative atoms
A system that releases heat to the surroundings, an exothermic reaction, has a negative ΔH by convention, because the enthalpy of the products is lower than the enthalpy of the reactants of the system. The enthalpies of these reactions are less than zero, and are therefore exothermic reactions
Answer:
Okay the Answers are top to bottom.
5, 4, 1, 3, 2
Explanation:
Answer:
62.15 m (squared)
Explanation:
We know that the Area of the square is 5.5 x 4= 22
so 90 degree to 30 degree is divide by 3
so u divide 22 by 3 and get 7.3 wich is the semi circle thing
now 7.3 x 5.5 = 40.15
22 + 40.15 = 62.15 m (squared)
Look im not really sure but I think this is the answer
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