Electrons are only
about 0.054% as massive as neutrons and protons are only 99.86% as massive as
the neutrons. The mass of the Proton is 1.67 x 10^-27 kg and the mass of the electron
is 9.11 x 10^-31 kg. The mass of the electron is so much lighter than the mass
of the proton.
Answer:
The precipitate is CuS.
Sulfide will precipitate at [S2-]= 3.61*10^-35 M
Explanation:
<u>Step 1: </u>Data given
The solution contains 0.036 M Cu2+ and 0.044 M Fe2+
Ksp (CuS) = 1.3 × 10-36
Ksp (FeS) = 6.3 × 10-18
Step 2: Calculate precipitate
CuS → Cu^2+ + S^2- Ksp= 1.3*10^-36
FeS → Fe^2+ + S^2- Ksp= 6.3*10^-18
Calculate the minimum of amount needed to form precipitates:
Q=Ksp
<u>For copper</u> we have: Ksp=[Cu2+]*[S2-]
Ksp (CuS) = 1.3*10^-36 = 0.036M *[S2-]
[S2-]= 3.61*10^-35 M
<u>For Iron</u> we have: Ksp=[Fe2+]*[S2-]
Ksp(FeS) = 6.3*10^-18 = 0.044M*[S2-]
[S2-]= 1.43*10^-16 M
CuS will form precipitates before FeS., because only 3.61*10^-35 M Sulfur Ions are needed for CuS. For FeS we need 1.43*10^-16 M Sulfur Ions which is much larger.
The precipitate is CuS.
Sulfide will precipitate at [S2-]= 3.61*10^-35 M
Answer:
<span>In the addition of hbr to 1-butyne the electrophile in the first step of the mechanism is <u>Hydrogen atom of HBr</u>.
Explanation:
In this reaction first of all HBr approaches the triple bond. A Pi Complex (weak inter-molecular interactions) is formed between the two molecules. And the triple bond attacks the partial positive hydrogen atom creating a negative charge on Bromine along with positive charge on itself (Sigma Complex). In second step the negative Bromide attacks the positive carbon of Butyne.</span>
Five hundred twenty million, three hundred and forty thousand. hope it helps!
