Answer:
hope this helps
Explanation:
Water was on earth with the dinosaurs and almost at the very start.
Water starts anywhere from clouds to ponds
I choose ponds, water evaporates from the sun and moves its way through the air being collected into clouds in the atmosphere, clouds are evaporated water molecules, then the water combines to form rain wich then end up on a hill and travel back to a water source or goes into the ground and makes groundwater and if there is enough rain overtime will form a pond.
Balanced Half reactions are:
At anode 2
==> Cl₂+ 
+ H₂O ==> 
+ 2
+
At Cathode: 2 + ==> H₂
Since the question states that you are using an aqueous solution of MnCl₂, so ions will have present are, H₂O, , 
and
Now at Anode reaction will occur as given:
2 ==> Cl₂+ + H₂O ==> + 2 + (will occur)
At Cathode:
2 + ==> H₂ (will occur)
At Cathode:
+ ==> Mn (This reaction will not occur)
The deposition of solid Mn will not occur because in aqueous solution, will be reduced before .
The reduction potentials for is zero whereas reduction potential for is - 1.18V.
The reduction potential of a species is its tendency to gain electrons and get reduced. It is measured in millivolts or volts. Larger positive values of reduction potential are indicative of a greater tendency to get reduced.
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Answer:
The rapid movement of excess charge from one place to another is an <em>electric discharge.</em>
Explanation:
A material in which electrons <em>CANNOT</em> move easily from place to place is an insulator. A material in which electrons <em>CAN </em>move easily from place to place is a conductor.
First, we should get moles acetic acid = molarity * volume
=0.3 M * 0.5 L
= 0.15 mol
then, we should get moles acetate = molarity * volume
= 0.2 M * 0.5L
= 0.1 mol
then, we have to get moles of OH- which added:
moles OH- = molarity * volume
= 1 M * 0.02L
= 0.02 mol
when the reaction equation is:
CH3COOH + OH- → CH3COO- + H2O
moles acetic acid after adding OH- = (0.15-0.02)
= 0.13M
moles acetate after adding OH- = (0.1 + 0.02)
= 0.12 M
Total volume = 0.5 L + 0.02 L= 0.52 L
∴[acetic acid] = moles acetic acid after adding OH- / total volume
= 0.13mol / 0.52L
= 0.25 M
and [acetate ) = 0.12 mol / 0.52L
= 0.23 M
by using H-H equation we can get PH:
PH = Pka + ㏒[salt/acid]
when we have Ka = 1.8 x 10^-5
∴Pka = -㏒Ka
= -㏒ 1.8 x 10^-5
= 4.7
So by substitution:
∴ PH = 4.7 + ㏒[acetate/acetic acid]
= 4.7 + ㏒(0.23/0.25)
= 4.66
