Answer:
A single compound is simultaneously oxidized and reduced.
Explanation:
In chemistry, disproportionation is a simultaneous oxidation and reduction of a single chemical specie.
What this means is that; in a disproportionation reaction, only one compound is both oxidized and reduced. This implies that two products are formed during disproportionation. One is the oxidized product while the other is the reduced product.
Consider the disproportionation of CuCl shown below;
2CuCl -----> CuCl2 + Cu
Here, CuCl2 is the oxidized product while Cu is the reduced product.
The term that best described a 10 gram of KClO3 per 100 grams of water solution at 30 degree Celcius is Saturated. The solubility chart is needed for this work. If the solubility chart is drawn for KClO3, it will be observed that the proportion of KClO3 that is needed to dissolve in 100ml of water to make the solution saturated is 10 grams at 30 degree Celcius.
Answer:
Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)
Explanation:
In writing the cell notation for an electrochemical cell, the anode is written on the left hand side while the cathode is written on the right hand side. The two half cells are separated by two thick lines which represents the salt bridge.
For the cell discussed in the question; the Mn(s)/Mn^2+(aq) is the anode while the Co^2+(aq)/Co(s) half cell is the cathode.
Hence I can write; Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)
Answer:
0.0084
Explanation:
The mole fraction of BaCl₂ (X) is calculated as follows:
X = moles BaCl₂/total moles of solution
Given:
moles of BaCl₂ = 0.400 moles
mass of water = 850.0 g
We have to convert the mass of water to moles, by using the molecular weight of water (Mw):
Mw of water (H₂O) = (2 x 1 g/mol)+ 16 g/mol = 18 g/mol
moles of water = mass of water/Mw of water = 850.0 g/(18 g/mol) = 47.2 mol
The total moles of the solution is given by the addition of the moles of solute (BaCl₂) and the moles of solvent (water):
total moles of solution = moles of BaCl₂ + moles of water = 0.400 + 47.2 mol = 47.6 mol
Finally, we calculate the mole fraction:
X = 0.400 mol/47.6 mol = 0.0084
