The empirical formula : C₆H₂Cl₂O
The molecular formula : C₁₂H₄Cl₄O₂
<h3>Further explanation</h3>
Given
The percentage composition
Required
The empirical formula and the molecular formula
Solution
The mol ratio of the components :
C : H : Cl : O
=44.76/12 : 1.25/1 : 44.05/35.5 : 9.94/16
=3.73 : 1.25 : 1.241 : 0.621 divide by 1.241
= 3 : 1 : : 1 : 0.5 x 2
= 6 : 2 : 2 : 1
The empirical formula : C₆H₂Cl₂O
(Empirical formula)n=molecular formula
(C₆H₂Cl₂O)n=321.97
(160.986)n=321.97
n=2
(C₆H₂Cl₂O)₂=C₁₂H₄Cl₄O₂
The molecular formula : C₁₂H₄Cl₄O₂
Answer:(d) has a high melting point because it has a giant covalent structure ... (i) Calculate the maximum mass of propyl ethanoate that can be made from 7.20 g of ethanoic ... (c) Anhydrous aluminium chloride contains 20.2% by mass of aluminium. (i) Show that the empirical formula for anhydrous aluminium chloride is A
Explanation:
Answer : The activation energy for the reaction is, 43.4 KJ
Explanation :
According to the Arrhenius equation,
or,
![\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7BK_2%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%20R%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= rate constant at 
= rate constant at 
= 
= activation energy for the reaction = ?
R = gas constant = 8.314 J/mole.K
= initial temperature = 
= final temperature = 
Now put all the given values in this formula, we get:
![\log (\frac{3K_1}{K_1})=\frac{Ea}{2.303\times 8.314J/mole.K}[\frac{1}{291K}-\frac{1}{310K}]](https://tex.z-dn.net/?f=%5Clog%20%28%5Cfrac%7B3K_1%7D%7BK_1%7D%29%3D%5Cfrac%7BEa%7D%7B2.303%5Ctimes%208.314J%2Fmole.K%7D%5B%5Cfrac%7B1%7D%7B291K%7D-%5Cfrac%7B1%7D%7B310K%7D%5D)
Therefore, the activation energy for the reaction is, 43.4 KJ
