35.9g x 1 mol/ 2.016g x 22.4 L/ 1 mol= 398.89 L
The answer is Latitude (B)
Answer:
-2, -1, 0, 1, 2
Explanation:
There are four types of quantum numbers;
1) Principal quantum number (n)
2) Azimuthal quantum number (l)
3) magnetic quantum number (ml)
4) Spin quantum number (s)
The azimuthal quantum number (l) describes the orbital angular momentum and shape of an orbital while the magnetic quantum number shows the projections of the orbital angular momentum along a specified axis. This implies that the magnetic quantum number shows the orientation of various orbitals along the Cartesian axes. The values of the magnetic quantum number ranges from -l to + l
For l= 2, the possible values of the magnetic quantum number are; -2, -1, 0, 1, 2
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The word is "pollution"
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Answer : The correct option is, (b) +0.799 V
Solution :
The values of standard reduction electrode potential of the cell are:
![E^0_{[H^{+}/H_2]}=+0.00V](https://tex.z-dn.net/?f=E%5E0_%7B%5BH%5E%7B%2B%7D%2FH_2%5D%7D%3D%2B0.00V)
![E^0_{[Ag^{+}/Ag]}=+0.799V](https://tex.z-dn.net/?f=E%5E0_%7B%5BAg%5E%7B%2B%7D%2FAg%5D%7D%3D%2B0.799V)
From the cell representation we conclude that, the hydrogen (H) undergoes oxidation by loss of electrons and thus act as anode. Silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.
The half reaction will be:
Reaction at anode (oxidation) :
Reaction at cathode (reduction) :
The balanced cell reaction will be,

Now we have to calculate the standard electrode potential of the cell.

![E^o_{cell}=E^o_{[Ag^{+}/Ag]}-E^o_{[H^{+}/H_2]}](https://tex.z-dn.net/?f=E%5Eo_%7Bcell%7D%3DE%5Eo_%7B%5BAg%5E%7B%2B%7D%2FAg%5D%7D-E%5Eo_%7B%5BH%5E%7B%2B%7D%2FH_2%5D%7D)

Therefore, the standard cell potential will be +0.799 V