Answer:
Molecular formula = C₁₂H₁₂O₄
Empirical formula is C₃H₃O.
Explanation:
Given data:
Mass of C = 91.63 g
Mass of H = 7.69 g
Mass pf O = 40.81 g
Molar mass of compound = 220 g/mol
Empirical formula = ?
Molecular formula = ?
Solution:
Number of gram atoms of H = 7.69 / 1.01 = 7.61
Number of gram atoms of O = 40.81 / 16 = 2.55
Number of gram atoms of C = 91.63 / 12 = 7.64
Atomic ratio:
C : H : O
7.64/2.55 : 7.61 /2.55 : 2.55/2.55
3 : 3 : 1
C : H : O = 3 : 3 : 1
Empirical formula is C₃H₃O.
Molecular formula:
Molecular formula = n (empirical formula)
n = molar mass of compound / empirical formula mass
Empirical formula mass = 3×12+ 3×1.01 +16 = 55.03
n = 220 / 55.03
n = 4
Molecular formula = 4 (empirical formula)
Molecular formula = 4 (C₃H₃O)
Molecular formula = C₁₂H₁₂O₄
the oxidation states of the elements before and after the reaction is;
Pb oxidation state changes from 0 to +2
SO₄²⁻ ion there's no change in the oxidation state during the reaction
Au oxidation state changes from +3 to 0
reduction reactions are when there's a decrease in the oxidation state of the species
oxidation reactions are when theres an increase in the oxidation state of the species
the element where there's a decrease in oxidation state is Au.
Therefore Au gets reduced.
answer is B) Au
2H2O=2H2+O2
37.4g H2O(1 mol/18.02)=2.07547 mol H2O
PV=nRT
(1.30)(V)=(2.07547)(.0821)(297)
Vwater=38.92898L
38.92898L (1 mol O2/2 mol H2O)=19.46449L O2 gas
Answer: Option (d) is the correct answer.
Explanation:
When two or more different substances are mixed together then it results in the formation of a mixture.
Mixture are of two types, that is, homogeneous mixture and heterogeneous mixture.
In homogeneous mixture, the constituent particles are distributed evenly throughout the mixture.
Whereas in heterogeneous mixture, the constituent particles are non-uniformly distributed.
Thus, we can conclude that mixtures are classified based on the distribution of particles in them.
