Answer:
Silver chromate, Ag2CrO4 , a sparingly soluble compound, has a solubility product, Ksp , of 1.2 x 10-12
What is the molar solubility of Ag2CrO4 in pure water?
Explanation:
Use the ICE approach: Write the Ksp equilibrium:
Ag2CrO4(s) = 2 Ag+(aq) + CrO4 2- (aq) Ksp = 1.2 x 10 -12
I : 1 0 0
C : (“0”) +2x +x
E : 1 2x x
Note: above, we let x = [CrO4 2- ] and so [Ag+] = 2x )
Ksp = [Ag+] 2 [CrO4 2-]/1 = (2x)2(x) = 4x3 = 1.2 x 10-12
Or, x3 = 3.0 x 10 -13 Take cube root of both sides:
=> x = 6.7 x 10 -5 M = [ CrO4 2- ] => [ Ag+ ] =2x = 1.3 x 10 -4 M
Answer:
anawer is none of the above
It’s D because the other ones don’t aren’t true
Okay thanks for reminding me
Answer:
The relevant equation is:
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂
Explanation:
1 mol of calcium carbonate can react to 2 moles of Hydrochloric acid to produce 1 mol of water, 1 mol of calcium chloride and 1 mol of carbon dioxide.
The formed CO₂ is the reason why you noticed bubbles as the reaction took place