Answer:
the heat of formation of isopropyl alcohol is -317.82 kJ/mol
Explanation:
The heat of combustion of isopropyl alcohol is given as follows;
C₃H₇OH (l) +(9/2)O₂ → 3CO₂(g) + 4H₂O (g)
The heat of combustion of CO₂ and H₂O are given as follows
C (s) + O₂ (g) → CO₂(g) = −393.50 kJ
H₂ (g) + 1/2·O₂(g) → H₂O (l) = −285.83 kJ
Therefore we have
3CO₂(g) + 4H₂O (g) → C₃H₇OH (l) +(9/2)O₂ which we can write as
3C (s) + 3O₂ (g) → 3CO₂(g) = −393.50 kJ × 3 =
4H₂ (g) + 2·O₂(g) → 4H₂O (l) = −285.83 kJ × 4
3CO₂(g) + 4H₂O (g) → C₃H₇OH (l) +(9/2)O₂ = +2006 kJ/mol
-1180.5 - 1143.32 +2006 = -317.82 kJ/mol
Therefore, the heat of formation of isopropyl alcohol = -317.82 kJ/mol.
The answer is -60.57 = -60.6 KJ.
CaC2(s) + 2 H2O(l) ---> Ca(OH)2(s) +C2H2(g) H= -127.2 KJ
Hf C2H2 = 226.77
Hf Ca(OH)2 = -986.2
<span>Hf H2O = -285.83
Now,
</span><span>add them up. 226.77 - 986.2 + (2*285.83) = -187.77
</span><span>Add back the total enthalpy that is given in the question
-187.77+127.2 = -60.57 </span>
Ammonium sulfide
68,154 g/mol
41.10 percent nitrogen, 11.83 percent hydrogen, and 47.04 percent sulfur.
Carbonyl compounds are those compounds in which carbon aton is doubly bonded to oxygen atom.
The hybridization of carbon and oxygen in carbonyl compounds is
Sp². Folowing are examples of carbonyl compounds,
1. Ketones
2. Aldehydes
3. Carboxylic Acids
4. Acyl Halides
5. Amides
6. Esters
7. Anhydrides e.t.c
Result: In given the structure the carbonyl groups are highlighted RED, as shown below.
Answer:
molarity = moles of solute/liters of solution
0.25 M NaOH = 0.25 mol NaOH/L
mass = moles (n) × molar mass (M)
n(NaOH) = 0.25 mol
M(NaOH) = 39.997 g/mol
m(NaOH) = 0.25 mol × 39.997 g/mol = 9.999 g