Answer:
The answer to your question is Volume = 11.4 L
Explanation:
Data
Volume 1 = V1 = 6 L
Pressure 1 = P1 = 1 atm
Temperature 1 = T1 = 22°C
Volume 2 = V2 = ?
Pressure 2 = 0.45 atm
Temperature 2 = -21°C
Process
1.- Convert temperature (°C) to °K
T1 = 273 + 22 = 295°K
T2 = 273 + (-21) = 252°K
2.- Use the combined gas law to solve this problem
P1V1 / T1 = P2V2 / T2
-Solve for V2
V2 = P1V1T2 / T1P2
-Substitution
V2 = (6)(1)(252) / (295)(0.45)
- Simplification
V2 = 1512 / 132.75
- Result
V2 = 11.38 L
Answer:

Explanation:
It often helps to write the heat as if it were a reactant or a product in the thermochemical equation.
Then you can consider it to be 11018 "moles" of "kJ"
We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.
M_r: 32.00
2C₈H₁₈ + 25O₂ ⟶ 16CO₂ + 8H₂O + 11 018 kJ
n/mol: 7280
1. Moles of O₂
The molar ratio is 25 mol O₂:11 018 kJ

2. Mass of O₂

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.