29.81 g/mol
48.81 g/mol
67.81 g/mol
The molar mass of BF3 is 67.81 g/mol
Explanation
that is;
atomic mass of B = 10.81 g/mol
atomic mass of F = 19 g/mol
67.81 g/mol is the molar mass of BF . Therefore, the correct option is option D among all the given options.
The ratio among the mass with the quantity of substance (measured within moles) in any sample of a compound of chemicals is known as the molar mass (M) in chemistry. The molar mass of a material is a bulk attribute rather than a molecular one. The compound's molecular weight is an average over numerous samples, which frequently have different masses because of isotopes. A terrestrial average or a function of the relative proportion of the isotopes of the component atoms on Earth, the molar mass is most frequently calculated using the standard atomic weights.
atomic mass of B = 10.81 g/mol
atomic mass of F = 19 g/mol
atomic massof F in BF= 3 x19 = 57 g/mol
total molar mass of BF = 10.81 g/mol + 57 g/mol = 67.81 g/mol
Therefore, the correct option is option D.
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40
20Ca
The symbol for an isotope of Calcium-42 corresponding to the isotope notation is
Isotopes are atoms of the same element that have the same number of protons (atomic number) but different numbers of neutrons (mass number).
In case of Calcium-42,
Therefore, the chemical symbol for the isotope Calcium-42 is
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Answer : The number of moles of is, 0.1 moles
Solution : Given,
Mass of = 6.2 g
Molar mass of = 62 g/mole
Now we have to calculate the moles of .
Therefore, the number of moles of present in 6.2 g is, 0.1 moles
Answer:
Empirical Formula = C₃H₆O₁
Solution:
Data Given:
Mass of Ethyl Butyrate = 3.61 mg = 0.00361 g
Mass of CO₂ = 8.22 mg = 0.00822 g
Mass of H₂O = 3.35 mg = 0.00335 g
Step 1: Calculate %age of Elements as;
%C = (mass of CO₂ ÷ Mass of sample) × (12 ÷ 44) × 100
%C = (0.00822 ÷ 0.00361) × (12 ÷ 44) × 100
%C = (2.277) × (12 ÷ 44) × 100
%C = 2.277 × 0.2727 × 100
%C = 62.09 %
%H = (mass of H₂O ÷ Mass of sample) × (2.02 ÷ 18.02) × 100
%H = (0.00335 ÷ 0.00361) × (2.02 ÷ 18.02) × 100
%H = (0.9279) × (2.02 ÷ 18.02) × 100
%H = 0.9279 × 0.1120 × 100
%H = 10.39 %
%O = 100% - (%C + %H)
%O = 100% - (62.09% + 10.39%)
%O = 100% - 72.48%
%O = 27.52 %
Step 2: Calculate Moles of each Element;
Moles of C = %C ÷ At.Mass of C
Moles of C = 62.09 ÷ 12.01
Moles of C = 5.169 mol
Moles of H = %H ÷ At.Mass of H
Moles of H = 10.39 ÷ 1.01
Moles of H = 10.287 mol
Moles of O = %O ÷ At.Mass of O
Moles of O = 27.52 ÷ 16.0
Moles of O = 1.720 mol
Step 3: Find out mole ratio and simplify it;
C H O
5.169 10.287 1.720
5.169/1.720 10.287/1.720 1.720/1.720
3.00 5.98 1
3 ≈ 6 1
Result:
Empirical Formula = C₃H₆O₁