For Hydrocarbon combustion:
CₓHₐ + O₂ → xCO₂ + a/2 H₂O
Moles of CO₂ = 16.2 / 44 = 0.37
Moles of Carbon = 0.37
Moles of H₂O = 4.976 / 18 = 0.28
Moles of Hydrogen = 0.28 x 2 = 0.56
Molar ratio = C : H = 1 : 1.5
= 2 : 3
C₂H₃
Mass of empirical unit = 12 x 2 + 1 x 3
= 27
Mr = 54.09
Number of empirical units repeated: 54.09 / 27
= 2
Molecular formula = C₄H₆
The electron configuration that corresponds to the valence electrons of an element for which there is an especially large jump between the second and third ionization energies is ns^2.
The valence electron configuration of an atom refers to the arrangement of electrons on the outermost shell of the atom.
Recall that a large jump in ionization energy occurs when electrons are removed from inner shells of the atom.
If we study our options closely, we will discover that option A has only two electrons in the valence shell (ns^2).
This means that the third ionization energy involves removing electrons from an inner shell which leads to large jump.
Learn more: brainly.com/question/14283892
Answer:
1.2 M
Explanation:
If you use the dilution equation (M1V1=M2V2), you end up with (50)(12)=(500)(M2), and when you solve for M2 you get 1.2 M.
