Transcript Atom Drawing Worksheet ANSWER KEY DRB

Daniel R. Barnes
Init:10/2/09
PT
. . . draw an atom when given a chemical symbol
with a mass number and an electric charge.
“SWBAT = “Students will be able to”
H
He
Li Be
Mg
B C N O F Ne
S
Ar
neutral, so #
Li
3
=#
3
3
Li
7S
-3
4
SHOW ME
THE END
PRODUCT!
NOW!
6.94
Li+
Li
PT
neutral, so #
Li
3
=#
3
3
Li
7S
-3
4
6.94
Li+
Li
PT
1 missing
Li
+
3
, so 3 – 1 =
2
3
Li
7S
-3
4
SHOW ME
THE END
PRODUCT!
NOW!
6.94
Li+
Li
PT
1 missing
Li
+
3
, so 3 – 1 =
2
3
Li
7S
-3
4
6.94
Li+
Li
PT
neutral, so #
C
6
=#
6
6
C
12S
-6
6
12.01
SHOW ME
THE END
PRODUCT!
NOW!
14C
C
PT
neutral, so #
C
6
=#
6
6
C
12S
-6
6
12.01
14C
C
PT
neutral, so #
14
C
6
=#
6
6
C
14S
-6
8
12.01
SHOW ME
THE END
PRODUCT!
NOW!
14C
C
PT
neutral, so #
14
C
6
=#
6
6
C
14S
-6
8
12.01
14C
C
PT
neutral, so #
O
8
=#
8
8
O
16S
-8
8
16.00
18O2-
18O
SHOW ME
THE END
PRODUCT!
NOW!
O2O
PT
neutral, so #
O
8
=#
8
8
O
16S
-8
8
16.00
18O2-
18O
O2O
PT
2 extra
2-
O
8
, so 8 + 2 = 10
8
O
16S
-8
8
16.00
18O2-
18O
SHOW ME
THE END
PRODUCT!
NOW!
O2O
PT
2 extra
2-
O
8
, so 8 + 2 = 10
8
O
16S
-8
8
16.00
18O2-
18O
O2O
PT
neutral, so #
18
O
8
=#
8
8
O
18S
-8
10
16.00
18O2-
18O
SHOW ME
THE END
PRODUCT!
NOW!
O2O
PT
neutral, so #
18
O
8
=#
8
8
O
18S
-8
10
16.00
18O2-
18O
O2O
PT
2 extra
2-
18
O
8
, so 8 + 2 = 10
8
O
18S
-8
10
16.00
18O2-
18O
SHOW ME
THE END
PRODUCT!
NOW!
O2O
PT
2 extra
18
2-
O
8
, so 8 + 2 = 10
8
O
18S
-8
10
16.00
18O2-
18O
O2O
PT
neutral, so #
He
2
=#
2
2
He
4S
-2
2
SHOW ME
THE END
PRODUCT!
NOW!
4.00
He2+
He
PT
neutral, so #
He
2
=#
2
2
He
4S
-2
2
4.00
He2+
He
PT
neutral, so #
H
1
=#
1
1
H
1S
-1
0
1.01
H33H-
H+
3H
SHOW ME
THE END
PRODUCT!
NOW!
2H
H
PT
neutral, so #
H
1
=#
1
1
H
1S
-1
0
1.01
H33H-
H+
3H
2H
H
PT
neutral, so #
2
H
1
=#
1
1
H
2S
-1
1
1.01
H33H-
H+
3H
SHOW ME
THE END
PRODUCT!
NOW!
2H
H
PT
neutral, so #
2
H
1
=#
1
1
H
2S
-1
1
1.01
H33H-
H+
3H
2H
H
PT
neutral, so #
3
H
1
=#
1
1
H
3S
-1
2
1.01
H33H-
H+
3H
SHOW ME
THE END
PRODUCT!
NOW!
2H
H
PT
neutral, so #
3
H
1
=#
1
1
H
3S
-1
2
1.01
H33H-
H+
3H
2H
H
PT
2 missing
2+
Be
4
, so 4 – 2 =
2
4
Be
9S
-4
5
SHOW ME
THE END
PRODUCT!
NOW!
9.01
Be2+
Be
PT
2 missing
2+
Be
4
, so 4 – 2 =
2
4
Be
9S
-4
5
9.01
Be2+
Be
PT
neutral, so #
Ne
10
=#
10
10
Ne
20S
-10
10
20.18
SHOW ME
THE END
PRODUCT!
NOW!
Ne
PT
neutral, so #
Ne
10
=#
10
10
Ne
20S
-10
10
20.18
Ne
PT
1 missing electron, so 1 – 1 = 0
+
H
1
1
H
1S
-1
0
1.01
H33H-
“H+”
is just another way of saying “proton”, isn’t it?
H+
3H
SHOW ME
THE END
PRODUCT!
NOW!
2H
H
PT
1 missing electron, so 1 – 1 = 0
+
H
1
1
H
1S
-1
0
1.01
H33H-
“H+”
is just another way of saying “proton”, isn’t it?
H+
3H
2H
H
PT
1 extra
3
H
1
, so 1 + 1 = 2
1
H
3S
-1
2
1.01
H33H-
H+
3H
SHOW ME
THE END
PRODUCT!
NOW!
2H
H
PT
1 extra
3
H
1
, so 1 + 1 = 2
1
H
3S
-1
2
1.01
H33H-
H+
3H
2H
H
PT
neutral, so #
Be
4
=#
4
4
Be
9S
-4
5
SHOW ME
THE END
PRODUCT!
NOW!
9.01
Be2+
Be
PT
neutral, so #
Be
4
=#
4
4
Be
9S
-4
5
9.01
Be2+
Be
PT
3 extra electrons, so
3-
H
1
1+3= 4
1
H
1S
-1
0
1.01
H33H-
H+
3H
SHOW ME
THE END
PRODUCT!
NOW!
2H
H
PT
3 extra electrons, so
3-
H
1
1+3= 4
1
H
1S
-1
0
1.01
H33H-
H+
3H
2H
H
PT
2 missing electrons, so 2 - 2 = 0
2+
He
2
2
He
4S
-2
2
SHOW ME
THE END
PRODUCT!
NOW!
4.00
He2+, which is basically just a naked
helium nucleus, is also referred to as
an “alpha particle” when it is shot out
of a decaying, radioactive atom.
He2+
He
PT
2 missing electrons, so 2 - 2 = 0
2+
He
2
2
He
4S
-2
2
4.00
He2+, which is basically just a naked
helium nucleus, is also referred to as
an “alpha particle” when it is shot out
of a decaying, radioactive atom.
He2+
He
PT
neutral, so #
B
11S
-5
6
5
=#
5
5
B
10.81
SHOW ME
THE END
PRODUCT!
NOW!
6B3+
B
PT
neutral, so #
B
11S
-5
6
5
=#
5
5
B
10.81
6B3+
B
PT
3 missing
6
B
3+
5
, so 5 – 3 =
2
5
B
6S
-5
1
10.81
SHOW ME
THE END
PRODUCT!
NOW!
6B3+
B
PT
3 missing
6
B
3+
5
, so 5 – 3 =
2
5
B
6S
-5
1
10.81
6B3+
B
PT
, so 7 – 13 = -6
13 missing
13+
N
7
7
N
14S
-7
7
14.01
N3SHOW ME
THE END
PRODUCT!
NOW!
N13+
N
PT
, so 7 – 13 = -6
13 missing
13+
N
7
7
N
14S
-7
7
14.01
N3N13+
THIS ATOM IS
NOT
POSSIBLE!
N
PT
3 extra
N
3-
7
, so 7 + 3 = 10
7
N
14S
-7
7
14.01
N3SHOW ME
THE END
PRODUCT!
NOW!
N13+
N
PT
3 extra
N
3-
7
, so 7 + 3 = 10
7
N
14S
-7
7
14.01
N3N13+
N
PT
neutral, so #
F
9
=#
9
9
F
19S
-9
10
19.00
SHOW ME
THE END
PRODUCT!
NOW!
FF
PT
neutral, so #
F
9
=#
9
9
F
19S
-9
10
19.00
FF
PT
1 extra
F
-
9
, so 9 + 1 = 10
9
F
19S
-9
10
19.00
SHOW ME
THE END
PRODUCT!
NOW!
FF
PT
1 extra
F
-
9
, so 9 + 1 = 10
9
F
19S
-9
10
19.00
FF
PT
neutral, so #
Mg
12
=#
12
12
Mg
24S
-12
12
24.31
SHOW ME
THE END
PRODUCT!
NOW!
Mg
PT
neutral, so #
Mg
12
=#
12
12
Mg
24S
-12
12
24.31
Mg
PT
2 extra
30
2-
S
30S
-16
14
16
, so 16 + 2 = 18
16
S
32.07
30S2-
SHOW ME
THE END
PRODUCT!
NOW!
S2S
PT
2 extra
30
2-
S
30S
-16
14
16
, so 16 + 2 = 18
16
S
32.07
30S2-
S2S
PT
neutral, so #
Ar
40S
-18
22
18
=#
18
18
Ar
39.95
SHOW ME
THE END
PRODUCT!
NOW!
Ar
PT
neutral, so #
Ar
40S
-18
22
18
=#
18
18
Ar
39.95
Ar
PT
From here on, the slides do not correspond to the
order of the atoms on the worksheet.
I’m working on it, okay?
neutral, so #
S
32S
-16
16
16
=#
16
16
S
32.07
30S2-
SHOW ME
THE END
PRODUCT!
NOW!
S2S
PT
neutral, so #
S
32S
-16
16
16
=#
16
16
S
32.07
30S2-
S2S
PT
2 extra
2-
S
32S
-16
16
16
, so 16 + 2 = 18
16
S
32.07
30S2-
SHOW ME
THE END
PRODUCT!
NOW!
S2S
PT
2 extra
2-
S
32S
-16
16
16
, so 16 + 2 = 18
16
S
32.07
30S2-
S2S
PT
neutral, so #
N
7
=#
7
7
N
14S
-7
7
14.01
N3SHOW ME
THE END
PRODUCT!
NOW!
N13+
N
PT
neutral, so #
N
7
=#
7
7
N
14S
-7
7
14.01
N3N13+
N
PT
3 missing
B
3+
5
B
11S
-5
6
10.81
SHOW ME
THE END
PRODUCT!
NOW!
, so 5 – 3 =
5
2
3 missing
B
3+
5
B
11S
-5
6
10.81
, so 5 – 3 =
5
2
1 extra electron, so
-
H
1
H
1S
-1
0
1.01
1
1+1= 2
3 extra
13
C
3-
6
C
13S
-6
7
12.01
, so 6 + 3 = 9
6
Sorry.
I haven’t done N, N3-, or O yet.
I need to finish grades for the 5 week progress report.
I don’t care if we don’t know how to use PowerSchool yet. I’m still
sticking to the deadline, gull durnit.
Yeah, yeah, I know. I haven’t done any of the atoms after O2either. So sue me. Go ahead! Steal an old lady’s Polident &
Ensure money!