Answer:
See explanation
Explanation:
When the complex ion Co(H2O)6 2+(aq) is placed in solution and chloride ions are added, the following equilobrium is set up;
Co(H2O)62+(aq) + 4 Cl-(aq) <=> CoCl42-(aq) + 6 H2O(g)
Co(H2O)6 2+(aq) solution is pink in colour while CoCl42-(aq) solution is blue in colour.
Since the solubility of CoCl42-(aq) is endothermic, heating the solution will move the equilibrium position towards the right (more CoCl42-(aq) is formed and the solution is blue in colour).
When the solution is cooled, more Co(H2O)62+(aq) is formed and the equilibrium position shifts towards the left and the solution becomes pink in colour.
Option (a) is correct.
A reducing agent is the one which loses electrons to other substance and an oxidizing agent is one which accepts electrons.
Here, In

, Cr has oxidation number 6+ in the L.H.S of the equation, but on R.H.S its oxidation number is 0 i.e. it Cr has gained electrons such that total charge is 0.
And the oxidation state of Al in the left-hand side of equation is 0 and in right-hand side, it is +6.i.e. it has donated its electrons to Cr.
Hence, Cr is the oxidizing agent and Al is the reducing agent.
First equation:
molarity=
no.of moles of solute (Na2S)/volume of solution
0.300/1.75 = 0.171
second equation:
same law
but in volume it is milliliter so you have to convert it by multiplying it with 10^-3
(10 power -3)
to understand it clearly
see this
http://imgur.com/xTS35QM
Answer:
I think is the last one.
Anthracite – Stage Four
Anthracite, the fourth stage in coal formation, is also known as “hard coal” because it is hard and has a high lustre. It appears to have been formed as a result of combined pressure and high temperature. Anthracite burns with a short flame and little smoke.