Answer:
I HOPE THE ABOVE INFORMATION WILL HELP YOU A LOT.
HAVE A NICE DAY.
V= 50. L n=45 mol T= 200°C = 473k P=?
CP)X 50.L)= (45 mol)(0.0821 light_kimol)(473k)
P = 30am or 4000 kPa
Answer:
- C₃H₈ (g) + 5O₂(g) → 3CO₂ (g) + 4H₂O (l)
(option D. with the proviso that the subscripts of propane's chemical formula must be corrected)
Explanation:
<em>Propane</em> is the saturated hydrocarbon, alkane, with chemical formula C₃H₈ or CH₃CH₂CH₃.
The complete combustion of the hydrocarbons yield carbon dioxide (CO₂) and water (H₂O).
The chemical equation that represents this combustion is:
- C₃H₈ (g) + O₂(g) → CO₂ (g) + H₂O (l) (skeleton equation: unbalanced)
Once you balance it, you get:
- C₃H₈ (g) + 5O₂(g) → 3CO₂ (g) + 4H₂O (l)
Left side Right side
C 3 3
H 8 4×2 = 8
O 5×2 = 10 3×2 + 4 = 10
That equation corresponds to the option D. of the list, with the proviso that the subscripts of propane's chemical formula must be corrected
One major characteristic they have in common is they are metals.
Answer: 0.18 V
Explanation:-
Here Cd undergoes oxidation by loss of electrons, thus act as anode. nickel undergoes reduction by gain of electrons and thus act as cathode.
=-0.40V[/tex]
=-0.24V[/tex]
Here Cd undergoes oxidation by loss of electrons, thus act as anode. nickel undergoes reduction by gain of electrons and thus act as cathode.
Where both 
are standard reduction potentials.
![E^0=E^0_{[Ni^{2+}/Ni]}- E^0_{[Cd^{2+}/Cd]}](https://tex.z-dn.net/?f=E%5E0%3DE%5E0_%7B%5BNi%5E%7B2%2B%7D%2FNi%5D%7D-%20E%5E0_%7B%5BCd%5E%7B2%2B%7D%2FCd%5D%7D)
Using Nernst equation :
![E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cd^{2+}]}{[Ni^{2+]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.0592%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BCd%5E%7B2%2B%7D%5D%7D%7B%5BNi%5E%7B2%2B%5D%7D)
where,
n = number of electrons in oxidation-reduction reaction = 2
= standard electrode potential = 0.16 V
![E_{cell}=0.16-\frac{0.0592}{2}\log \frac{[0.10]}{[0.5]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3D0.16-%5Cfrac%7B0.0592%7D%7B2%7D%5Clog%20%5Cfrac%7B%5B0.10%5D%7D%7B%5B0.5%5D%7D)
Thus the potential of the following electrochemical cell is 0.18 V.
