How many liters of water vapor can be produced if 8.9 liters of methane gas (CH4) are combusted, if all measurements are taken a
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Answer:
The molecular formula =
Explanation:
% of C = 11.79
Molar mass of C = 12.0107 g/mol
<u>% moles of C = 11.79 / 12.0107 = 0.9816</u>
% of Cl = 69.57
Molar mass of Cl = 35.453 g/mol
<u>% moles of Cl = 69.57 / 35.453 = 1.9623</u>
Given that the gaseous chlorofluorocarbon only contains chlorine, flourine and carbon. So,
% of F = 100% - % of C - % of C = 100 - 11.79 - 69.57 = 18.64
Molar mass of F = 18.998 g/mol
<u>% moles of F = 18.64 / 18.998 = 0.9812</u>
Taking the simplest ratio for C, Cl and F as:
0.9816 : 1.9623 : 0.9812
= 1 : 2 : 1
The empirical formula is =
Also, Given that:
Pressure = 21.3 mm Hg
Also, P (mm Hg) = P (atm) / 760
Pressure = 21.3 / 760 = 0.02803 atm
Temperature = 25 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (25 + 273.15) K = 298.15 K
Volume = 458 mL = 0.458 L (1 mL = 0.001 L)
Using ideal gas equation as:
PV=nRT
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Applying the equation as:
0.02803 atm × 0.458 L = n × 0.0821 L.atm/K.mol × 298.15 K
⇒n = 0.00052445 moles
Given that :
Amount = 0.107 g
Molar mass = ?
The formula for the calculation of moles is shown below:
Thus,
Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.
Thus,
Molecular mass = n × Empirical mass
Where, n is any positive number from 1, 2, 3...
Mass from the Empirical formula = 1×12.0107 + 2×35.453 + 1×18.998 = 101.9147 g/mol
Molar mass = 204.0233 g/mol
So,
Molecular mass = n × Empirical mass
204.0233 = n × 101.9147
⇒ n = 2
<u>The molecular formula = </u>
Alkenes on reacting with ozone results in the formation of ozonide which undergo reductive cleavage in presence of dimethyl sulfide to form carbonyl compounds (aldehyde or ketone). Whereas in presence of hydrogen peroxide it undergoes oxidative cleavage to form carboxylic acids or ketones.
Since, A alkene yields 4-heptanone only on treatment with ozone and DMS thus, it implies that both the chains on the side of the double-bond are similar the product is 4-heptanone that means the double bond is present between the chains at the 4th carbon. Therefore the structure of compound A is 4,5-dipropyloct-4-ene.
The reaction is as shown in the image.
The reaction of A with m-CPBA (meta-perchlorobenzoic acid) followed by aqueous acid is shown in the image.
m-CPBA (meta-perchlorobenzoic acid) is a peracid and forms epoxides on reacting with alkenes.
