What is the atomic number of an atom?
2 answers:
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Answer:
The formula of hydrocarbon =
Explanation:
% of C = 85.7
Molar mass of C = 12.0107 g/mol
% moles of C = 85.7 / 12.0107 = 7.14
% of H = 14.3
Molar mass of H = 1.00784 g/mol
% moles of H = 14.3 / 1.00784 = 14.19
Taking the simplest ratio for C and H as:
7.14 : 14.19 = 1 : 2
The empirical formula is =
Also, Given that:
Pressure = 508 Torr
Temperature = 17 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T = (17 + 273.15) K = 290.15 K
Volume = 1.500 L
Using ideal gas equation as:
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 = 62.3637 L.torr/K.mol
Applying the equation as:
508 Torr × 1.500 L = n × 62.3637 L.torr/K.mol × 290.15 K
⇒n = 0.0421 moles
Given that :
Amount = 1.77 g
Molar mass = ?
The formula for the calculation of moles is shown below:
Thus,
Molar mass of the hydrocarbon = 42.04 g/mol
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 + 2×1= 14 g/mol
Molar mass = 42.04 g/mol
So,
Molecular mass = n × Empirical mass
42.04 = n × 14
⇒ n = 3
<u>The formula of hydrocarbon = </u>
Answer:
None of these
Explanation:
Friedel–Craft reaction is a reaction involves the attachment of substituents to the benzene ring.
Mechanism of the reaction of methylbenzene with 1-chlorodecane in the presence of ether and aluminum chloride :
Step -1 : Generation of stable carbocation.
Aluminium chloride acts as Lewis acid which removes the chloride ion from the alkyl halide forming carbocation. The primary carbocation thus formed gets rearranged to secondary primary carbocation which is more stable due to hyperconjugation.
Step-2: Attack of the ring to the carbocation
The pi electrons of the ring behave as a nucleophile and attacks the carbocation. Since, the group attached on the benzene is methyl (+R effect) , the attack is from the ortho and the para positions. Para product is more stable due to less steric hinderance.
The product formed is shown in mechanism does not mention in any of the options.
So, None of these is the answer
