Answer:
Check below
Explanation:
Ksp of AgCl = 1.8*10^-10
Ksp = [Ag+] [Cl-]
[Ag+] [Cl-] = 1.8*10^-10
Because [Ag+] = [Cl-] ,
[Ag+]² = 1.8*10^-10
[Ag+] = √((1.8*10^-10)_
[Ag+] = 1.35*10^-5M
That is [Ag+] in the original saturated solution.
NaCl is added to the saturated AgCl solution:
If Cl- = 0.72M , equation becomes
[Ag+] [ 0.72] = 1.8*10^-10
[Ag+] = (1.8*10^-10)/0.72
[Ag+] = 2.5*10^-10M
% Ag+ remaining in solution =(2.5*10^-10) / { (1.35*10^-5) * 100
% Ag+ remaining = 1.85*10^-3% or 0.00185%
Note that an approximation has been made here: [Cl-] is 0.72M + the unprecipitated Cl- from the AgCl. Because this latter is an infinitely small amount compared to 0.72M , it is quite safe to make this approximation without any effect on the answer.
220
A suspension is the mixture with the largest particles.
It would be 1/2beacise I said so
Answer:- (a)
(b)
(c)
(d) 0.663 grams
Solution:- (a) Number of atoms of hydrogen are given and it asks to calculate the number of atoms of carbon. The given molecule has 19 carbons and 28 hydrogens. So, the ratio of atoms of C to H is 19:28.
Let's make multiply the given number of atoms of H by the ratio to get the atoms of C.

= 
(b) One molecule of testosterone contains 28 atoms of H. So, the ratio is 1:28. Let's multiply the given number of atoms of H by this ratio to get the number of molecules of testosterone.

= 
(c) To calculate the moles, we divide the molecules by Avogadro number.
= 
= 
(d) To get the mass of the sample we multiply the moles by the molar mass. Molar mass of testosterone = 19(12.01)+28(1.01)+2(16)
= 228.19+28.28+32
= 288.47 gram per mol
let's multiply the moles of the above part by molar mass to get the mass.

= 0.663 grams