The reaction of sodium hypochlorite (NaOCl, bleach) with iodide ion (sodium iodide) leads to the formation of electrophile. The I⁺ ion formed in this reaction is a strong electrophile that reacts quickly in an electrophilic aromatic substitution reaction.
NaOCl + NaI = I⁺
Answer:
-800 kJ/mol
Explanation:
To solve the problem, we have to express the enthalpy of combustion (ΔHc) in kJ per mole (kJ/mol).
First, we have to calculate the moles of methane (CH₄) there are in 2.50 g of substance. For this, we divide the mass into the molecular weight Mw) of CH₄:
Mw(CH₄) = 12 g/mol C + (1 g/mol H x 4) = 16 g/mol
moles CH₄ = mass CH₄/Mw(CH₄)= 2.50 g/(16 g/mol) = 0.15625 mol CH₄
Now, we divide the heat released into the moles of CH₄ to obtain the enthalpy per mole of CH₄:
ΔHc = heat/mol CH₄ = 125 kJ/(0.15625 mol) = 800 kJ/mol
Therefore, the enthalpy of combustion of methane is -800 kJ/mol (the minus sign indicated that the heat is released).
H3C -- C -- CH2-- CH2-- CH2-- CH2-- CH2--CH3
|
CH2 -- CH3
Observe the longest chain. Then start counting from the branching side.
1 2 3 4 5 6 7 8
H3C -- C -- CH2-- CH2-- CH2-- CH2-- CH2--CH3
|
CH2 -- CH3
So there are 8 carbons in the long chain means Octane.
Now, there is a ethyl group at 2nd carbon.
So, it is 2-ethylocatane
2-ethylocatane