Answer:
Empirical formula is C₃H₃O
Explanation:
Given data:
Mass of ethyl butyrate = 1.95 mg
Mass of CO₂ = 4.42 mg
Mass of water = 1.81 mg
Empirical formula = ?
Solution:
Percentage of C = 4.42/1.95 × 12/44×100
= 2.27× 0.273 ×100
= 62
percentage of H = 1.81/1.95 × 2/18 × 100
= 0.93 × 0.11 ×100
= 10.23
percentage of oxygen = 100 - (72.23)
= 27.77
number of gram atoms of C = 62/12 =5.17
number of gram atoms of H = 10.23 / 2 =5.115
number of gram atoms of O = 27.77 / 16 =1.74
Atomic ratio:
C : H : O
5.17/1.74 : 5.115/1.74 : 1.74/1.74
3 : 3 : 1
Empirical formula is C₃H₃O
(a) The molecular shapes of cis and Trans forms of N2F2 are-
b) There is EN difference between N atom and F atom. Hence the molecule will have a bond dipole.
In Cis form two fluorine atoms are on the same side of N N bond. Hence, bond moment in N2F2 cis form) do not cancel each other. Thus the cis form is polar.
In the trans form two fluorine atoms are on the opposite side of N-N bond. Bond moment of NF cancels each other.
Thus: trans N2F2 is non polar molecule.
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