Answer:
a. 113 min
Explanation:
Considering the equilibrium:-
2N₂O₅ ⇔ 4NO₂ + O₂
At t = 0 125 kPa
At t = teq 125 - 2x 4x x
Thus, total pressure = 125 - 2x + 4x + x = 125 - 3x
125 - 3x = 176 kPa
x = 17 kPa
Remaining pressure of N₂O₅ = 125 - 2*17 kPa = 91 kPa
Using integrated rate law for first order kinetics as:
Where,
is the concentration at time t
is the initial concentration
Given that:
The rate constant, k =
min⁻¹
Initial concentration
= 125 kPa
Final concentration
= 91 kPa
Time = ?
Applying in the above equation, we get that:-

Answer:
0.40 L
Explanation:
Calculation of the moles of
as:-
Mass = 51.24 g
Molar mass of
= 171.34 g/mol
The formula for the calculation of moles is shown below:
Thus,

Volume = 1.20 L
The expression for the molarity is:


Thus,
Considering
Given that:
So,
<u>The volume of 0.24925M stock solution added = 0.40 L
</u>
Answer:
3 e⁻ transfer has occurred.
Explanation
This is a redox reaction.
- Oxidation (loss of electrons or increase in the oxidation state of entity)
- Reduction (gain of electrons or decrease in the oxidation state of the entity)
- An element undergoes oxidation or reduction in order to achieve a stable configuration. It can be an octet or duplet configuration. An octet configuration is that of outer shell configuration of noble gas.
- [Ne]= (1s²) (2s² 2p⁶)
A combination of both the reactions( Half-reactions) leads to a redox reaction.
Let us look at initial configurations of Al and Cl
[Al]= 1s² 2s² 2p⁶ 3s² 3p¹
[Cl]= 1s² 2s² 2p⁶ 3s² 3p⁵
Hence, Al can lose 3 electrons to achieve octet config.
and, Cl can gain 1e to achieve nearest noble gas config. [Ar]
This reaction can be rewritten, by clearly mentioning the oxidation states of all the entities involved.
Al⁰ + Cl⁰ → (Al⁺³)(Cl⁻)₃
Here, Aluminum is undergoing an oxidation(i.e loss of electrons) from: 0→(+3)
Chlorine undergoes a reduction half reaction (i.e gain of electrons) from: 0→(-1). There are 3 such chlorine atoms, hence 3 e⁻ transfer has occurred.