Alkenes must undergo addition because they have easily broken tt bonds.
Markonikov's rule states in the addition of HX to an unsymmetrical alkene, the H atom bonds to the less substituted carbon atom.
alkenes are unsaturated hydrocarbons because they have fewer than the maximum number of hydrogen atoms per carbon.
Alkyl halides have good leaving groups and therefore readily undergo substitution and elimination reactions.
In hydroboration, the boron atom bonds to the substituted carbon.
Hydroxides, amines and alcoxides undergo substitution and elimination, but can do so only when the heteroatom is made into a good leaving group.
Answer:
0.4694 moles of CrCl₃
Explanation:
The balanced equation is:
Cr₂O₃(s) + 3CCl₄(l) → 2CrCl₃(s) + 3COCl₂(aq)
The stoichiometry of the equation is how much moles of the substances must react to form the products, and it's represented by the coefficients of the balanced equation. So, 1 mol of Cr₂O₃ must react with 3 moles of CCl₄ to form 2 moles of CrCl₃ and 3 moles of COCl₂.
The stoichiometry calculus must be on a moles basis. The compounds of interest are Cr₂O₃ and CrCl₃. The molar masses of the elements are:
MCr = 52 g/mol
MCl = 35.5 g/mol
MO = 16 g/mol
So, the molar mass of the Cr₂O₃ is = 2x52 + 3x35.5 = 210.5 g/mol.
The number of moles is the mass divided by the molar mass, so:
n = 49.4/210.5 = 0.2347 mol of Cr₂O₃.
For the stoichiometry:
1 mol of Cr₂O₃ ------------------- 2 moles of CrCl₃
0.2347 mol of Cr₂O₃----------- x
By a simple direct three rule:
x = 0.4694 moles of CrCl₃
Answer:
Potassium
1s2 2s2 2p6 3s2 3p6 4s1
Explanation:
The atom having only one electron its outermost shell must belong to an element in group one of the periodic table.
Having noted that, we proceed to find out what element in group one that has the atom just described in the question.
That atom must belong to an element in the fourth period. The only group 1 element in the fourth period is potassium.
The electron configuration of potassium is;
1s2 2s2 2p6 3s2 3p6 4s1
