A free-radical substitution reaction is likely to be responsible for the observations. The reaction mechanism of a reaction like this can be grouped into three phases:
- Initiation; the "light" on the mixture deliver sufficient amount of energy such that the halogen molecules undergo homologous fission. It typically takes ultraviolet radiation to initiate fissions of the bonds.
- Propagation; free radicals react with molecules to produce new free radicals and molecules.
- Termination; two free radicals combine and form covalent bonds to produce stable molecules. Note that it is possible for two carbon-containing free-radicals to combine, leading to the production of trace amounts of long carbon chains in the product.
Initiation
where the big black dot indicates unpaired electrons attached to the atom.
Propagation
Termination
Answer:
1. The electronic configuration of X is: 1s2 2s2 sp6 3s2
2. The configuration of the anion of Y (i.e Y^2-) is 1s2 2s2 2p6
3. The formula of the compound form by X and Y is given as: XY
Explanation:
For X to loss two electrons, it means X is a group 2 element. X can be any element in group 2. The electronic configuration of X is:
1s2 2s2 sp6 3s2
To get the electronic configuration of the anion of element Y, let us find the configuration of element Y. This is done as follows:
Y receives two electrons from X to complete its octet. Therefore Y is a group 6 element. The electronic configuration of Y is given below
1s2 2s2 2p4
The configuration of the anion of Y (i.e Y^2-) is 1s2 2s2 2p6
The formula of the compound form by X and Y is given below :
X^2+ + Y^2- —> XY
Their valency will cancel out thus forming XY
Answer:
86.2 or 431/5
Explanation:
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Answer:
1, 3, 2
Explanation:
N2 + H2 → NH3
I usually find that the best way to systematically balance an equation by inspection is to start with the most complicated-looking formula and then balance atoms in the order:
- All atoms other than O and H
- O
- H
(a) The most complicated formula is NH3.
(b) Balance N.
We have 1 H in NH3, but 2 N on the left. We need 2 N on the right. Put a 1 in front of N2 and a 2 in front of NH3.
1N2 + H2 → 2NH3
(c) Balance H.
We have fixed 6 H on the right, so we need 6 H on the left. Put a 3 in front of H2.
1N2 + 3H2 → 2NH3
The equation is now balanced, and the coefficients are 1, 3, 2.
