To solve this we use the equation,
M1V1 = M2V2
where M1 is the concentration of the stock
solution, V1 is the volume of the stock solution, M2 is the concentration of
the new solution and V2 is its volume.
65 x V1 = 2 x 200 L
V1 = 6.15 L
Answer:
0.000000540
Explanation:
Step 1: Make an ICE chart for the solution of AgBr
"S" represents the molar solubility of AgBr
AgBr(s) ⇄ Ag⁺(aq) + Br⁻(aq)
I 0 0
C +S +S
E S S
Step 2: Write the expression for the solubility product constant (Ksp)
Ksp = [Ag⁺] [Br⁻] = S × S
Ksp = S² = (0.0007350)² = 0.000000540
Answer:
C) H2S
Explanation:
In chemistry, the dissolution of one substance in another is dependent on the magnitude of intermolecular interaction between the two substances. Hence, if two substances do not interact in one way or the other, then one can not dissolve the other.
Let us consider the fact that NH3 is a polar molecule and it is a general principle that like dissolves like. Hence, only H2S which is also a polar molecule can effectively interact with NH3 due to dipole-dipole interaction between the two molecules.
Also, ammonia reacts with hydrogen sulphide as follows;
2NH3 + H2S → (NH4)2S
Hence H2S is more likely to dissolve in NH3.
Yes because look in the book dh
