Answer:
Volume and number of moles of gas have direct relation.
When the amount of gas increases its volume will increase and when the amount of gas decreases its volume will decrease.
Explanation:
According to Avogadro Law,
Equal volume of all the gases at same temperature and pressure have equal number of molecules.
This law state that volume and number of moles of gas have direct relation.
When the amount of gas increases its volume will increase and when the amount of gas decreases its volume will decrease.
Mathematical relation:
V ∝ n
V/n = K
K is proportionality constant.
When number of moles change from n₁ to n₂ and volume from V₁ to V₂
expression will be,
V₁/n₁ = K , V₂/n₂ = K
V₁/n₁ = V₂/n₂
So, we know the reduction potentials of the electrodes :
Electrode A = -0.21 V
Electrode B = -0.15 V
We want our cell to be spontaneous - this means that the voltage has to be positive. In order to do so, we need to turn the more negative reduction half reaction to be an oxidation half reaction.
-0.21 is more negative than -0.15, so electrode A will be the oxidation half reaction.
Anode is where the oxidation half reaction takes place, so electrode A is the correct answer . A fun way to remember this is An(ode)Ox(idation) = AnOx and Red(uction)Cat(hode) = RedCat
The E° cell will be 0.21 -0.15 V = 0.6 V
Answer:
S AgBr = 2.82 E-3 mol/L
Explanation:
- AgBr ↔ Ag+ + Br- .....(1)
∴ Ksp = [ Ag+ ] * [ Br- ] = 5.0 E-13
- Ag+ + 2NH3 ↔ [ Ag(NH3)2 ]+ ..........(2)
∴ Kf = 1.6 E7 = α[Ag(NH3)2]+ / ( αAg+ )*( αNH3 )²
∴ <em>C </em> NH3(sln) = 1.00 M
from (1) + (2):
- AgBr(s) + 2NH3(aq) ↔ Ag(NH3)2+(aq) + Br-(aq)
1 M 0 0
1 - 2x x x
∴ K = Ksp*Kf = ( 5.0 E-13 )*( 1.6 E7 ) = 8.0 E-6
⇒ K = ( [ Br- ] * [ Ag(NH3)2+] ) / [ NH3 ]² = 8.0 E-6
⇒ K = (( x )*( x )) / ( 1 - 2x ) = 8.0 E-6
⇒ x² = 8.0 E-6*( 1- 2x )
⇒ x² + 1.6 E-5x - 8.0 E-6 = 0
⇒ x = 2.82 E-3 M
⇒ [ Br- ] = [ AgBr ] = 2.82 E-3 M
The answer is C. It bans the exploitation of mineral resources
Answer:+3 oxidation state
Explanation: