Answer:
As they absorb energy, they are broken.
Explanation:
Solutes dissolve when they interact with water. This is because, the attraction of solute particles with each other weakens and become replaced by solute-solvent interaction. As the water surrounds the solute, the solute is now said to be hydrated.
Energy is required to break solute-solute bonds. This energy absorbed in breaking solute-solute bonds is compensated for by the energy liberated when solute particles are hydrated.
If energy required to break solute-solute interaction is less than the energy liberated when solute particles are hydrated, the substance will dissolve in water.
Hence, as a substance dissolve in water, energy is absorbed when solute-solute interactions are broken.
Answer:
unsaturated solution
Explanation:
This solution is made by the coffee, which is the solvent and the sugar, which is the solute. The solute dissolves in the solvent.
Sugar starts to precipitate because it cannot dissolve anymore. This means that the solution at the equilibrium point and is saturated. Since more coffee or solvent is added, the solution will now be able to dissolve more sugar. This means that the solution is unsaturated
Answer:
2MnO4^- (aq) + 3C2O4^2- (aq) + 2H2O (l) --> 2MnO2(s) +6CO3^2 -(aq) + 4H^+ (aq)
Explanation:
First, write the half equations for the reduction of MnO4^- and the oxidation of C2O4^2- respectively. Balance it.
Reduction requires H+ ions and e- and gives out water, vice versa for oxidation.
Reduction:
MnO4^- (aq) + 4H^+ (aq) + 3e- ---> MnO2(s) + 2H2O (l)
Oxidation:
C2O4^2- (aq) + 2H2O (l) ---> 2CO3^2 -(aq) + 4H^+ (aq) + 2e-
Balance the no. of electrons on both equations so that electrons can be eliminated. we can do so by multiplying the reduction eq by 2, and oxidation eq by 3.
2MnO4^- (aq) + 8H^+ (aq) + 6e- ---> 2MnO2(s) + 4H2O (l)
3C2O4^2- (aq) + 6H2O (l) ---> 6CO3^2 -(aq) + 12H^+ (aq) + 6e-
Now combine both equations and eliminate repeating H+ and H2O.
2MnO4^- (aq) + 8H^+ (aq) + 3C2O4^2- (aq) + 6H2O (l) --> 2MnO2(s) + 4H2O (l) +6CO3^2 -(aq) + 12H^+ (aq)
turns into:
2MnO4^- (aq) + 3C2O4^2- (aq) + 2H2O (l) --> 2MnO2(s) +6CO3^2 -(aq) + 4H^+ (aq)
Vanadium (V)
Vanadium is the only one in the 4th period here so
