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
Answer:
macro-evolution
Explanation:
This happens when evolution occurs because of separation of a population.
The answer is 1296.75 g
Molar mass of <span>Fe2(SO4)3 is the sum of atomic masses of its elements:
Mr(</span>Fe2(SO4)3) = 2Ar(Fe) + 3Ar(S) + 12Ar(O)
Ar(Fe) = 55.84 g/mol
Ar(S) = 32.06 g/mol
Ar(O) = 16 g/mol
Mr(Fe2(SO4)3) = 2 * 55.84 + 3 * 32.06 + 12 * 16
= 111.68 + 96.18 + 192
= 399.9 g/mol
Thus there are 399.9 g in 1 mol and x grams in 3.25 mol:
399 g : 1 mol = x : 3.25 mol
x = 399 g * 3.25 mol : 1 mol
x = 1296.75 g