The reaction is as follow,
Ca(OH)₂ + Al₂(SO₄)₃ → CaSO₄ + Al(OH)₃
Ca are balance on both sides,
There are 2 Al at left side and one at right so, multiply Al(OH)₃ by 2 to balance,
So,
Ca(OH)₂ + Al₂(SO₄)₃ → CaSO₄ + 2 Al(OH)₃
Now, Ca and Al are balanced, now balance SO₄, which is 3 at left hand side and one at right hand side, so multiply CaSO₄ on right side by 3, so,
Ca(OH)₂ + Al₂(SO₄)₃ → 3 CaSO₄ + 2 Al(OH)₃
Again Ca got imbalance, so multiply Ca(OH)₂ by 3 to balance Ca, So,
3 Ca(OH)₂ + Al₂(SO₄)₃ → 3 CaSO₄ + 2 Al(OH)₃
The Equation is balance now with respect to every element.
Result:
The Ratio is 3 : 1 : 3 : 2, so, Option-G is correct.
False
because the smaller bowl of water has less volume to cool off with the ice than the large fish tank
<span>Your answer would be ....A.the moon’s shadow covers all of Earth during a solar eclipse.</span>
Pfsst
bombal
chow
ddoggone
ernst
floxxit
goldy
nuutye
apstrom
logon
byyou
zapper
yazzer
highho
magnificon
oz
fratt
jeptum
quackzil
doadeer
rhaap
terriblum
sississ
urrp
vulcania
wobble
xtalt
pie
anatom
eldorado (I think these are the answers)
Answer: The potential of the following electrochemical cell is 1.08 V.
Explanation:
=-0.74V[/tex]
=0.34V[/tex]
The element with negative reduction potential will lose electrons undergo oxidation and thus act as anode.The element with positive reduction potential will gain electrons undergo reduction and thus acts as cathode.
Here Cr undergoes oxidation by loss of electrons, thus act as anode. copper undergoes reduction by gain of electrons and thus act as cathode.
Where both 
are standard reduction potentials, when concentration is 1M.
![E^0=E^0_{[Cu^{2+}/Ni]}- E^0_{[Cr^{3+}/Cr]}](https://tex.z-dn.net/?f=E%5E0%3DE%5E0_%7B%5BCu%5E%7B2%2B%7D%2FNi%5D%7D-%20E%5E0_%7B%5BCr%5E%7B3%2B%7D%2FCr%5D%7D)
Thus the potential of the following electrochemical cell is 1.08 V.
