Answer:
Explan ionization energy, atomic radius, and electron affinityation:
12
Neon is the 10th element and has 10 protons. The mass of an element is the sum of the protons and neutrons.
22-10=12
So, the compound is HCN.
H can only form one bond, while carbon can form 4 bonds and N can form three
bons.
That drives you to conclude that H will bond the carbon and the carbon will form a triple bond
with N:
H - C (triple bond) N
Now, you must count the valence electrons to determine how many and where valence electrons are not forming bonds.
H has one valence electron which is in the H - C bond.
C has four valence electrons which are one in the H - C bond and three in the CN bonds.
N
has five valence electrons and it has three of them in the CN bonds,
therefore you have to add two electrons (points or other marks) to the N
atom.
This is the resultant Lewis structrue:
**
H - C (triple bond) N (draw the two ** close to the N symbol)
You have to draw three hyphens to represent the triple bond and include
the two ** over the N atom to represent the two electrons that are not
forming a bond.
The second shell is left at 7, it should be filled to 8 to go to the next shell.
Answer: see below
<u>Explanation:</u>
When multiplying or dividing, use the smallest number of significant figures (sf) of the digits provided.
37(a) 24 × 3.26 = 78.24 <em>2 sf × 3 sf </em>--> round to 2 sf's
≈ 78
(b) 120 × 0.10 = 12 <em>2 sf × 2 sf </em>--> round to 2 sf's
= 12
(c) 1.23 × 2.0 = 2.46 <em>3 sf × 2 sf </em>--> round to 2 sf's
= 2.5
(d) 53 × 1.53 = 81.09 <em>2 sf × 3 sf </em>--> round to 2 sf's
= 81
38(a) 4.84 ÷ 2.4 = 2.0167 <em>3 sf × 2 sf </em>--> round to 2 sf's
= 2.0
(b) 60.2 ÷ 20.1 = 2.995 <em>3 sf × 3 sf </em>--> round to 3 sf's
= 3.00
(c) 102.4 ÷ 51.2 = 2 <em>4 sf × 3 sf </em>--> round to 3 sf's
= 2.00
(d) 168 ÷ 58 = 2.896 <em>3 sf × 2 sf </em>--> round to 2 sf's
= 2.9