Answer:
See explanation
Explanation:
TLC is a chromatographic method in which the solute is spotted on a plate and the plate is placed in an air tight chamber containing a solvent. The solvent is maintained below the level of the spot. The capillary movement of the solvent through the plate achieves the required separation.
If two spots have Rf values of 0.25 and 0.26 respectively and then the plate was removed from the developing chamber, subsequently, the residual solvent was allowed to evaporate from the plate, and then the plate was returned to the developing chamber.
It will be observed after the second development is complete that the new Rf values will be 0.50 and 0.52 respectively. It will just be as though the second chromatogram picked up from where the first chromatogram stopped.
The correct answer is gonna be C) A lithium cation ion is smaller
A lithium cation has lost its valence electrons, which causes the remaining electrons to be pulled in stronger by the positive charge in the nucleus. As they get closer to the nucleus, the overall size of the atom is decreased.
Answer:
Yes
Explanation:
As long as its a solid, you can count it. It will be hard, but possible.
Answer:
Final temperature of the gas is 576 
.
Explanation:
As the amount of gas and pressure of the gas remains constant therefore in accordance with Charles's law:
where 
and 
are volume of gas at 
and 
temperature (in kelvin scale) respectively.
Here 
, 
and 
So 
849 K = (849-273) = 576
So final temperature of the gas is 576 .
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
