chem 6a fall 2009 practice test with key

chem 6a fall 2009 practice test with key -...

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Unformatted text preview: Chemistry
6A:
Midterm
Review
 “Pop
Quiz”
 
 
 2)

Write
the
electron
configuration
for
Ag+
(with
and
without
the
abbreviation).
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 1s22s22p63s23p64s23d104p65s24d8




or
 [Kr]
5s24d8
 
 
 
 1)

How
many
electrons
can
have
the
following
quantum
numbers:
(
3,
1,
‐1)
 
 
 
 
 2
electrons
(with
+
½
and
–
½
spin
quantum
numbers)
 3)

Draw
the
orbital
diagram
for
ground
state
oxygen.
 
 
 
 
 
 








 












2p
 

 2s



 
 
 1s
 4)

How
many
nitrogen
atoms
are
there
in
5.36g
of
Ba(NO3)2?
 
 
 
 
 Molar
Mass
of
Ba(NO3)2
:

1
(Ba)
+
2
(N)
+
6
(O)

 
 (137.3
g/mol)
+
2
(14.01
g/mol)
+
6
(16.00
g/mol)=
261.32
g/mol
 
 
 Moles
of
Ba(NO3)2
:

5.36
g
Ba(NO3)2
÷
(261.32
g/mol)
=
0.0205
mol
Ba(NO3)2

 
 
 (0.0205
mol
Ba(NO3)2
)(2
moles
N)(6.02214
x
1023
mol‐1)
=
2.47
x
1022
atoms
N
 5)

Arrange
the
following
(Sr,
Mg,
Mg2+)
in
order
of
increasing:







 
 a)
Radius:
 
 
 b)
Ionization
Energy:
 
 
 
 2+ 2+ 
 
 Mg 
< Mg

<

Sr
 
 Sr

< Mg

< Mg 
 
 
 6)

Why
is
the
2nd
ionization
energy
so
much
higher
than
the
1st?
 
 After
you
pull
off
the
first
electron,
the
remaining
electrons
feel
a
stronger
pull
from
the
positive
nucleus
 (larger
Zeff).


 7)
You
decide
to
make
some
lemonade.

You
like
things
sweet
so
you
add
10g
of
sugar
(sucrose,
C12H22O11)
to
6mL
 lemon‐water.

After
a
large
sip,
you
realize
that
you
might
go
into
a
sugar
coma
if
you
drink
this,
so
you
decide
to
 dilute
it.

You
take
1.5mL
of
that
solution
and
add
it
to
20mL
of
lemon
water.

This
time
it
is
just
right!

What
is
 the
final
concentration
of
sucrose?

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Molar
Mass
of
sucrose:

C12H22O11
:

12(12.01
g/mol)
+
22(1.008
g/mol)
+

11(16.00
g/mol)=
342.30
g/mol
 Moles
Sucrose:

10
g
÷
342.30
g/mol
=
0.0292
moles
 ciVi
=
cfVf
 Initial
Concentration:

0.0292
moles
/
(0.006
L)
=
4.87
M
 
 Initial
Volume
(volume
taken
from
initial
concentration)
=
1.5
mL
=
0.0015
L
 
 Final
Volume
=
0.0015
L
+
0.020
L
=
0.0215
L
 
 
 
 
 
 (4.87
M)(0.0015
L)
=
cf
(0.0215
L)
 

 Final
Concentration:
cf
=
0.30
M
 ...
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This note was uploaded on 01/26/2010 for the course CHEM6A 6A taught by Professor Stuartchristina during the Fall '09 term at UCSD.

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