Isotopes of any given factor all incorporate the equal variety of protons, so they have the identical atomic wide variety (for example, the atomic wide variety of helium is usually 2). Isotopes of a given factor include exceptional numbers of neutrons, therefore, special isotopes have special mass numbers.
Answer is: <span>the empirical formula of the hydrocarbon is CH</span>₂.<span>
Chemical reaction: C</span>ₓHₐ + O₂ → xC + a/2H₂O.<span>
m(CO</span>₂) = 33.01 g.
n(CO₂) = m(CO₂) ÷ M(CO₂).
n(CO₂) = 33.01 g ÷ 44.01 g/mol.
n(CO₂) = n(C) = 0.75 mol.
m(H₂O) = 13.52 g.
n(H₂O) = 13.52 g ÷ 18 g/mol.
n(H₂O) = 0.75 mol.
n(H) = 2 · n(H₂O) = 1.5 mol.
n(C) : n(H) = 0.75 mol : 1.5 mol /0.75 mol.
n(C) : n(H) = 1 : 2.
(2) Mg, Ca, and Ba have the most similar chemical properties, and we know this because of their coexistence in the group alkali earth metals (group 2) on the periodic table.
Answer:
See explanation below
Explanation:
In both cases the central atoms, C in CHCl₃ and O in H₂O, are sp³ hybridized .
Since they are sp³ hybridized we predict an angle between the H-C-Cl and H-O-H of 109.5 º ( tetrahedral ), but two of the sp³ orbitals in water are occupied by lone pairs.
These lone pairs do excercise more repulsion ( need more room ) than the bonds oxygen is making with hydrogen.
As a consequence of this repulsion the angles H-O-H are less than the predicted 109.5º in tetrahedra. ( Actually is 104.5 º)
