Answer:
Molecular formula => C₁₂H₁₂O₃
Explanation:
From the question given above, the following data were obtained:
Molar mass of compound = 205gmol¯¹
Mass of Carbon (C) = 3.758 g
Mass of Hydrogen (H) = 0.316 g
Mass of Oxygen (O) = 1.251 g
Molecular formula =?
We'll begin by determining the empirical formula of the compound. This can be obtained as follow:
C = 3.758 g
H = 0.316 g
O = 1.251 g
Divide by their molar masses
C = 3.758 /12 = 0.313
H = 0.316 /1 = 0.316
O = 1.251 /16 = 0.078
Divide by the smallest.
C = 0.313 / 0.078 = 4
H = 0.316 / 0.078 = 4
O = 0.078 / 0.078 = 1
Thus, the empirical formula of the compound is C₄H₄O
Finally, we shall determine the molecular formula of the compound. This can be obtained as follow:
Molar mass of compound = 205gmol¯¹
Empirical formula => C₄H₄O
Molecular formula =>?
[C₄H₄O]ₙ = 205
[(12×4) + (4×1) +16]n = 205
[48 + 4 +16]n = 205
68n = 205
Divide both side by n
n = 205 / 68
n = 3
Molecular formula => [C₄H₄O]ₙ
Molecular formula => [C₄H₄O]₃
Molecular formula => C₁₂H₁₂O₃
The relation between vapour pressure , enthalpy of vapourisation and temperature is
ln (88/ 39) = DeltaH / 8.314 (1 / 318 - 1 / 298)
0.814 = DeltaH / 8.314 (2.11 X 10^-4 )
DeltaH = -32.07 kJ
Answer:
B
Explanation:
bonding is a process of two different atoms sharing electrons for stability and these electrons are attracted by one atom losing it's electrons to another
hi im breanna
Answer:
The mole is simply a very large number that is used by chemists as a unit of measurement.
Explanation:
The mole is simply a very large number,
6.022
×
10
23
, that has a special property. If I have
6.022
×
10
23
hydrogen atoms, I have a mass of 1 gram of hydrogen atoms . If I have
6.022
×
10
23
H
2
molecules, I have a mass of 2 gram of hydrogen molecules. If I have
6.022
×
10
23
C
atoms, I have (approximately!) 12 grams.
The mole is thus the link between the micro world of atoms and molecules, and the macro world of grams and litres, the which we can easily measure by mass or volume. The masses for a mole of each element are given on the periodic table as the atomic weight. So, if have 12 g of
C
, I know, fairly precisely, how many atoms of carbon I have. Given this quantity, I know how many molecules of
O
2
are required to react with the
C
, which I could measure by mass or by volume.
