The temperature at which a liquid becomes a solid is the:
1 answer:
You might be interested in
Answer:
using bond energies = - 37 kJ
using enthalpies of formation = - 45.7 kJ
Explanation:
From the reaction;
CH2═CH2(g) + H2O(g) -------> CH3CH2OH(g)
From the above reaction; there are 4 (C-H) bonds , 1 ( C=C) and 2 (H-O) of ethene which forms 5(C-H) bonds, 1 (C-C) and 1 (C-O) and 1(H-O) bonds.
Using Bond Energies; the heat of the reaction can be written as:
∑ energy of old bond breaking + ∑ energies of the new bond formation.
∑
+ ∑
∑
+ ∑
-37 kJ
To calculate the heats of reaction by using enthalpies of formation; we have:
∑
- ∑
∑
- ∑
(-235.1 kJ) - [(+52.47 kJ) + (-241.826 kJ)]
-45.7 kJ
Answer: " 13.5 g Al " ;
→ that is: "13.5 grams of aluminum."
<u>____________________________</u>
Explanation:
<u>____________________________</u>
<u>Note</u>: What is missing from the question is the "balanced chemical equation" for the "chemical reaction" that contains:
The reactants: "aluminum (Al) " ; and "chlorine (Cl) " ; and:
The product: "aluminum choloride (AlCl₃) " .
____________________________
The "balanced chemical equation" is:
____________________________
2 Al + 3 Cl₂ → 2 AlCl₃ ;
_____________________________
<u>Note</u>: The molecular weight of "aluminum (Al)" is: " 26.98 g /mol " .
____________________________
So: We call solve using a technique known as: "dimensional analysis" :
____________________________
0.500 mol AlCl₃ *
____________________________
<u>Note</u>: The units of "mol AlCl₃" cancel out to "1' ; and:
The units of "mol Al" cancel out to "1" ; and we are left with:
____________________________
" g Al ["grams of aluminum"] ;
____________________________
<u>Note</u>: We can "cancel out the "2's" ; since "2/2 = 1 " ; and we have:
→ (0.500 * 26.98) g Al ;
= 13.49 g Al ;
→ Round to 3 (Three) significant figures;
→ Since: "0.500" has 3 (Three) significant figures:
____________________________
= 13.5 g Al ; that is: "13.5 grams of aluminum."
____________________________
Hope this is helpful!
Best wishes to you in your academic pursuits—and within the "Brainly" community!
____________________________