Answer:
27.0
Explanation:
Because Mass can neither be created nor be destroyed hence total mass of sample of iodine and tube remain equal as it is sealed.
The equilibrium constant of a reaction is defined as:
"The ratio between equilibrium concentrations of products powered to their reaction quotient and equilibrium concentration of reactants powered to thier reaction quotient".
The reaction quotient, Q, has the same algebraic expressions but use the actual concentrations of reactants.
To solve this question we need this additional information:
<em>For this reaction, K = 6.0x10⁻² and the initial concentrations of the reactants are:</em>
<em>[N₂] = 4.0M; [NH₃] = 1.0x10⁻⁴M and [H₂] = 1.0x10⁻²M</em>
<em />
Thus, for the reaction:
N₂ + 3H₂ ⇄ 2NH₃
The equilibrium constant, K, of this reaction, is defined as:
![K = 6.0x10^{-2} = \frac{[NH_3]^2}{[N_2][H_2]^3}](https://tex.z-dn.net/?f=K%20%3D%206.0x10%5E%7B-2%7D%20%3D%20%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5BH_2%5D%5E3%7D)
And Q, is:
![Q = \frac{[NH_3]^2}{[N_2][H_2]^3}](https://tex.z-dn.net/?f=Q%20%3D%20%5Cfrac%7B%5BNH_3%5D%5E2%7D%7B%5BN_2%5D%5BH_2%5D%5E3%7D)
Where actual concentrations are:
[NH₃] = 1.0x10⁻⁴M
[N₂] = 4.0M
[H₂] = 2.5x10⁻¹M
Replacing:
![Q = \frac{[1.0x10^{-4}]^2}{[4.0][2.5x10^{-1}]^3}](https://tex.z-dn.net/?f=Q%20%3D%20%5Cfrac%7B%5B1.0x10%5E%7B-4%7D%5D%5E2%7D%7B%5B4.0%5D%5B2.5x10%5E%7B-1%7D%5D%5E3%7D)
<h3>Q = 1.6x10⁻⁷</h3>
As Q < K,
<h3>The chemical system will shift to the right in order to produce more NH₃</h3>
Learn more about chemical equililbrium in:
brainly.com/question/24301138
H h I ointment o on NB j. Non b kkkk NB nis o ok kno kno
Answer:
C is the element thats has been oxidized.
Explanation:
MnO₄⁻ (aq) + H₂C₂O₄ (aq) → Mn²⁺ (aq) + CO₂(g)
This is a reaction where the manganese from the permanganate, it's reduced to Mn²⁺.
In the oxalic acid, this are the oxidation states:
H: +1
C: +3
O: -2
In the product side, in CO₂ the oxidation states are:
C: +4
O: -2
Carbon from the oxalate has increased the oxidation state, so it has been oxidized.
Answer:
Explanation:
Fe⁺² (aq) + 2e⁻ = Fe (s) ; E⁰ = - .44 V
Fe⁺³ (aq) + e⁻ = ® Fe²⁺ (aq) ; E⁰ = + .77 V
Reduction potential of second reaction is more , so it will take place , ie Fe⁺³ will be reduced and Fe will be oxidised .
So reaction in the combined cell will be
2Fe⁺³ + Fe = 3Fe⁺²
cell potential = .77 - ( - .44 )
= 1.21 V .