Chemistry, 21.09.2019 09:00 funnybugy16

Complete these structures by adding electron dots as needed.

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This is hard to show but here is how you would determine these. NOTE each dot is an electron.
Question 1)
F-H
1) determine the valance electrons for each. F has 7 and H has 1
2) one electron from both F and H form the bond "-" which means that you still have 6 electrons to place around F and none to place around H. Place the 6 in sets of 2 around the F
..
F-H
¨
Question 2)

2) H-O-H
H has 1 valence electron minus 1 used in the bond to O = 0 electrons to place
H has 1 valence electron minus 1 used in the bond to O = 0 electrons to place
O has 6 valence electrons minus 2 used in the bonds to the H's = 4 electrons to place

H-O-H: place two dots above and below the oxygen

Question 3)
3) O=NH : NOTE: a double bond requires O and N to share two of their electrons each
O has 6 valence electrons minus 2 used in the bonds to N = 4 electrons to place
N has 5 valence electrons minus 3 used in the bonds to O and H = 2 electrons to place
H has 1 valence electron minus 1 used in the bond to N = 0 electrons to place
place the 2 dots on top and bottom of oxygen.
place 2 above the N

Answer from: Quest

a rain shadow is a dry region of land on the side that is protected against the prevailing winds of a mountain range. the prevailing winds are the winds that occur most of the time in a particular place of the earth

prevailing winds carry air toward the mountain range. as the air rises up over a mountain range, the air cools, water vapor condenses, and clouds form. on this side of the mountains, called the windward side, precipitation falls in the form of rain or snow. this dry air produces a rain shadow.

Answer from: Quest

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o

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Complete these structures by adding electron dots as needed....