Answer:
E° = 1.24 V
Explanation:
Let's consider the following galvanic cell: Fe(s) | Fe²⁺(aq) || Ag⁺(aq) | Ag(s)
According to this notation, Fe is in the anode (where oxidation occurs) and Ag is in the cathode (where reduction occurs). The corresponding half-reactions are:
Anode: Fe(s) ⇒ Fe²⁺(aq) + 2 e⁻
Cathode: Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)
The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an
E° = 0.80 V - (-0.44 V) = 1.24 V
Rate law for the given 2nd order reaction is:
Rate = k[a]2
Given data:
rate constant k = 0.150 m-1s-1
initial concentration, [a] = 0.250 M
reaction time, t = 5.00 min = 5.00 min * 60 s/s = 300 s
To determine:
Concentration at time t = 300 s i.e. ![[a]_{t}](https://tex.z-dn.net/?f=%5Ba%5D_%7Bt%7D)
Calculations:
The second order rate equation is:
![1/[a]_{t} = kt +1/[a]](https://tex.z-dn.net/?f=1%2F%5Ba%5D_%7Bt%7D%20%3D%20kt%20%2B1%2F%5Ba%5D)
substituting for k,t and [a] we get:
1/[a]t = 0.150 M-1s-1 * 300 s + 1/[0.250]M
1/[a]t = 49 M-1
[a]t = 1/49 M-1 = 0.0204 M
Hence the concentration of 'a' after t = 5min is 0.020 M
Answer:
The correct answer is b) 2
Explanation:
When is dissolved in water, silver acetate (AgCH₃COO) is dissociated into ions according to the following equilibrium:
AgCH₃COO ⇄ Ag⁺ + CH₃COO⁻
Where Ag⁺ is a silver cation and CH₃COO⁻ is the acetate anion (an organic anion). As we can see, from one single formula unit are obtained 2 ions (1 cation and 1 anion).
Therefore, the correct option is b) - 2
The new volume of a 250 Ml sample of gas at 300k and 1atm if heated to 350 k at 1 atm is 291.67 Ml
<u>calculation</u>
This is solved using the Charles law formula since the pressure is constant.
that is V1/T1 = V2/T2 where,
V1 =250 ml
T1=300 K
V2=?
T2= 350 k
by making V2 the subject of the formula by multiplying both side by T2
V2= T2V1/T1
V2= (350 K x 250 ml) / 300K =291.67 Ml
