Molecular geometry about the left carbon atom in CH₃CO₂CH₃ is tetrahedral.
The geometry around left carbon that is CH₃ is tetrahedral.
As the hybridization around left carbon is sp³ that shows its geometry should be tetrahedral and as there are 4 ligands around carbon and there is no lone pair present so the geometry is tetrahedral. So, the molecular geometry about the left carbon atom in CH₃CO₂CH₃ is tetrahedral.
Answer:
Explanation:
Oxidation:
Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.
Reduction:
Reduction involve the gain of electron and oxidation number is decreased.
Oxidizing agents:
Oxidizing agents oxidize the other elements and itself gets reduced.
Reducing agents:
Reducing agents reduced the other element are it self gets oxidized.
Consider the following reaction:
2AgCl + Zn → 2Ag + ZnCl₂
In this reaction oxidation state of Zn on left side is 0 while on right side +2 so it gets oxidized and oxidation state of Ag on left side is +1 and on right side 0 so it get reduced.
4NH₃ + 3O₂ → 2N₂ + 6H₂O
In this reaction oxidation state of nitrogen on left side is -3 while on right side 0 so it gets oxidized and oxidation state of oxygen on left side is 0 and on right side -2 so it get reduced.
Fe₂O₃ + 2Al → Al₂O₃ + 2Fe
In this reaction oxidation state of iron on left side is +3 while on right side 0 so it gets reduced and oxidation state of Al on left side is 0 and on right side +3 so it get oxidized.
Answer: It loses electrons to another element.
Explanation:- Oxidation is the process in which an element loses electrons and there is an increase in the oxidation state. On losing electrons it combines with a electronegative element such as oxygen, sulphur or nitrogen etc.
Reduction is the process in which an element gains electrons and there is a decrease in the oxidation state.
Answer : The correct option is, (D) 3600 kJ
Explanation :
Mass of octane = 75 g
Molar mass of octane = 114.23 g/mole
Enthalpy of combustion = -5500 kJ/mol
First we have to calculate the moles of octane.
Now we have to calculate the heat released in the reaction.
As, 1 mole of octane released heat = -5500 kJ
So, 0.656 mole of octane released heat = 0.656 × (-5500 kJ)
= -3608 kJ
≈ -3600 kJ
Therefore, the heat released in the reaction is 3600 kJ
