Answer:
pH = 8.314
Explanation:
equil: S S 3S
∴ Ksp = [ Y+ ] * [ OH- ]³ = 6.0 E-24
⇒ 6.0 E-24 = ( S )*( 3S )³
⇒ 6.0 E-24 = 27S∧4
⇒ 2.22 E-25 = S∧4
⇒ ( 2.22 E-25 )∧(1/4) = S
⇒ S = 6.866 E-7 M
⇒ [ OH- ] = 3*S =2.06 E-6 M
⇒ pOH = - Log [ OH- ]
⇒ pOH = - Log ( 2.06 E-6 )
⇒ pOH = 5.686
∴ pH = 14 - pOH
⇒ pH = 8.314
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
Answer:
1.818x10²³ particles
Explanation:
Add the two atomic masses given to get 74.55 g.
Divide the number of grams given by the molar mass of KCl:
22.5g ÷ 74.55 = 0.302 mol
Multiply the moles by 6.022×10²³ particles
Cells have lots of things to do. Some cells make the whole organism, so that one cell must do everything that organism needs to do to live. Other cells perform specific functions, so they must be designed to do that specific activity.
There you go :) just reword it and try make it into your own words like just switch some sentences around
