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:
electrons should be added to one of these spheres to make it electrically neutral.
Explanation:
Total charge on each sphere = +3.0 μC =
In order to neutralize the positive charge equal magnitude of negative charge is to be added.
Total charge electrons, Q= -3.0 μC =
Number of electrons = n
Charge on a single electron, e = 
Q = n × e
electrons should be added to one of these spheres to make it electrically neutral.
Answer:
92gm
Explanation:
Atomic mass of Mg=24g=1 mole of Mg
∴ 24g =1 mole of Mg contain 6.022×10^23 atom
∴ 6gm contains 246.022×1023×6
=4×6.022×10^23 atoms
Now according to question, there are 6.022×1023 atoms of Na
23gm of Na contains 6.022×10^23 atoms
∴6.022×4×10^23 atoms of Na weighs 23×6.022×10^23×4/6.022×10^23⇒92gm
Answer:
AgC2H3O2 is: Silver acetate
