Answer:It’s C on edge 2020
(Combustion of car engines producing pollutants in the air)
Explanation:
I got it right :))
Answer:
C₃H₄O₄
Explanation:
In order to get the empirical formula of a compound, we have to follow a series of steps.
Step 1: Divide the percent by mass of each element by its atomic mass.
C: 34.6/12.01 = 2.88
H: 3.9/1.01 = 3.86
O: 61.5/16.00 = 3.84
Step 2: Divide all the numbers by the smallest one, i.e., 2.88
C: 2.88/2.88 = 1
H: 3.86/2.88 ≈ 1.34
O: 3.84/2.88 ≈ 1.33
Step 3: Multiply all the numbers by a number that makes all of them integer
C: 1 × 3 = 3
H: 1.34 × 3 = 4
O: 1.33 × 3 = 4
The empirical formula is C₃H₄O₄.
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.
Answer:
Pneumonoultramicroscopicsilicovolcanoconiosis
Explanation:
