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 given statement is true.
Explanation:
Entropy means the measure of randomness present in a substance. That is, an increase in temperature will lead cause more motion in the particles of a substance more will be their kinetic energy.
As a result, there will occur more collisions due to which randomness of molecules will increase. Hence, there will be increase in entropy.
So, when we decrease the temperature then there will be decrease in motion of particles. As a result, lesser number of collisions will take place between them. Hence, degree of randomness will also decrease.
Thus, we can conclude the statement entropy of a system decreases with decrease in temperature, is true.
It is because C4 compromises on water loss and CAM compromises on photorespiration. and Both minimize photorespiration but expend more ATP during carbon fixation.
Answer:
You may need better soil and a more plentiful amount of water coming from another source, and maybe find another way to contain the rain water
Explanation:you may need to draw it on paper sorry bout how it looks
