2
C
4
H
10
(
g
) + 13
O
2
(
g
) = 8
C
O
2
(
g
) + 10
H
2
O
(
g
)
This is hard to show but here is how you would determine these. NOTE each dot is an electron.
<span>Question 1) </span>
<span>F-H </span>
<span>1) determine the valance electrons for each. F has 7 and H has 1 </span>
<span>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 </span>
<span>.. </span>
<span>F-H </span>
<span>¨ </span>
<span>Question 2) </span>
<span>2) H-O-H </span>
<span>H has 1 valence electron minus 1 used in the bond to O = 0 electrons to place </span>
<span>H has 1 valence electron minus 1 used in the bond to O = 0 electrons to place </span>
<span>O has 6 valence electrons minus 2 used in the bonds to the H's = 4 electrons to place </span>
<span>H-O-H: place two dots above and below the oxygen </span>
<span>Question 3) </span>
<span>3) O=N----H : NOTE: a double bond requires O and N to share two of their electrons each </span>
<span>O has 6 valence electrons minus 2 used in the bonds to N = 4 electrons to place </span>
<span>N has 5 valence electrons minus 3 used in the bonds to O and H = 2 electrons to place </span>
<span>H has 1 valence electron minus 1 used in the bond to N = 0 electrons to place </span>
<span>place the 2 dots on top and bottom of oxygen. </span>
<span>place 2 above the N </span>
Try adding spaces next time! That's iodine. Check all of the numbers to make sure all of the orbitals are filled, then find the ones which aren't. In this one, only the 5p5 subshell isn't full. 5p5 is the fifth row on the right side, count across the nonmetals and metalloids until the fifth one (a halogen). That's iodine, and that's your answer!
0.14
Explanation:
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