Answer:
when 3-phenylpropanoic, PhCH₂CH₂COOH , is treated with NaOH, H₂0 it produce sodium salt of 3 - phenylpropanoic acid with water as a major product.
Explanation:
Carboxylic acid when treated with a base like NaOH or KOH gives carboxylate ion because H⁺ from acid reacts with OH⁻ from base gives H₂0 and salt of sodium carboxylate.
The mechanism of the above reaction given below
Answer: The empirical formula and the molecular formula of the organic compound is and
respectively.
Explanation:
The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:
where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.
We are given:
Mass of = 21.71 g
Mass of = 5.926 g
Molar mass of carbon dioxide = 44 g/mol
Molar mass of water = 18 g/mol
For calculating the mass of carbon:
In 44g of carbon dioxide, 12 g of carbon is contained.
So, in 21.71 g of carbon dioxide, = of carbon will be contained.
For calculating the mass of hydrogen:
In 18g of water, 2 g of hydrogen is contained.
So, in 5.926 g of water, = of hydrogen will be contained.
Mass of oxygen in the compound = (17.11) - (5.921+0.658) = 10.53 g
Mass of C = 5.921 g
Mass of H = 0.658 g
Mass of O = 10.53 g
Step 1 : convert given masses into moles.
Moles of C =
Moles of H=
Mass of O=
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C =
For H =
For O=
The ratio of C : H: O = 1: 1: 1
Hence the empirical formula is .
empirical mass of CHO = 12(1) + 1(1) + 1 (16) = 29
Molecular mass = 104.1 g/mol
Thus molecular formula =
Answer:
pharynx > larynx>trachea> bronchi>alveoli (so B)
Explanation:
nasal cavities (or oral cavity) > pharynx > trachea > primary bronchi (right & left) > secondary bronchi > tertiary bronchi > bronchioles > alveoli (site of gas exchange)
