M3 Problem Set
pdf
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Portage Learning *
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Course
121
Subject
Chemistry
Date
Jan 9, 2024
Type
Pages
12
Uploaded by ERProfParrotPerson911
M3:
Problem
Set
Due
No
due
date
Points
5
Questions
13
Time
Limit
None
Instructions
You
may
find
the
following
resources
helpful
while
working
through
these
problem
sets:
|
—
e
[
——
]
—————————————
]
———————————
)
S——
]
e
————
e
||
S
]
|
Equation
Table
=_(https://previous.nursingabc.com/upload/images/Help_file_picture/
CHEM121%20-
%20Equation%20Bank%5B20719%5D.pdf)
How
to
Draw
Lewis
Structures
(https://portagelearning.instructure.com/courses/1952/files/637411/download
?wrap=1)_
(https://portagelearning.instructure.com/courses/1952/files/637411/download?download_frd=1)
Table
of
Water
Vapor
Pressures
&
—_—f_—_—_—_—_e—_——--
-
—
——
s
re—_——eeeeeee
e
e
-
——_—,_—_—_—_—,———
e
—1
Attempt
History
Attempt
Time
Score
LATEST
Attempt
1
2,498
minutes
S5outof
5
Score
for
this
quiz:
5
out
of
5
Submitted
Dec
9
at
10:28am
This
attempt
took
2,498
minutes.
Question
1
0/0
pts
Show
your
work
and
be
sure
to
give
answer
with
the
correct
units.
A
gas
sample
has
an
original
volume
of
680
ml
when
collected
at
720
mm
and
28°C.
What
will
be
the
volume
of
the
gas
sample
if
the
pressure
increases
to
820
mm
and
the
temperature
increases
to
55°C?
Your
Answer:
Pi
x
Vi
=
Pf
x
Vf
Ti
Tf
680
ml/1000
=
0.680
liters
=
Vi
720
mm/760
=
0.947
atm
=
Pi
820
mm/760
=
1.08
atm
=
Pf
28
C
+
273
=
301
K
=
Ti
55
C
+
273
=
328
K
=
Tf
Put
in
the
data:
(0.947)
x
(0.680)
=
(1.08)
x
Vf
(301)
(328)
Solve
for
Vf:
0.002139402
=
0.003292683
x
Vf Vf
=
0.650
liters
P,
x
V,
P,xV,
T
T,
680
mI/1000
=
0.680
liters
=V,
720
mm/760
=
0.947
atm
=
P;
28°C
+
273
=
301K
=T,
820
mm/760
=
1.08
atm
=
P;
55°C
+
273
=
328K
=T
Put
in
the
data:
(0.947)
x
(0.680)
=
(1.08)
x
V;
(301)
(328)
Solve
for
V.
0.002139402
=
0.003292683
x
Vs:
Vs
=
0.002139402
=
0.650
liter
0.003292683
Question
2
0/0
pts
Show
your
work
and
be
sure
to
give
answer
with
the
correct
units.
A
sample
of
CO,
gas
which
weighs
1.62
grams
has
a
volume
of
1.02
liters
when
collected
at
20°C.
What
would
be
the
pressure
of
the
gas
sample?
Your
Answer:
CO2
C
12.011
O
15.999
x
2
44.009
1.62
/
44.009
=
0.0368
mol
=
n
V=
1.02LR=0.0821T=20+
273
=293
K?
x
1.02
=0.0368
x
0.0821
x
293
Px1.02=0.8850.885/1.02
=P
=
0.868
atm
PxV=nxRxT
CO>
molecular
weight
=
44
1.62
grams/44
=
0.03682
mole
=
n
R
=0.0821
1.02
liters
=V
20°C
+
273
=293K=T
P
x
(1.02)
=
(0.03682)
x
(0.0821)
x
(293)
P
x
(1.02)
=
0.8857
P
=0.8857
=
0.868
atm
1.02
Question
3
0/0
pts
Show
your
work
and
be
sure
to
give
answer
with
the
correct
units.
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The
combustion
of
benzene
(CgHg)
takes
place
by
the
following
reaction
equation.
2
CeHg
(9)
+
15
02
(g)
—
12
CO2
(g)
+
6
H20
(9)
What
is
the
volume
of
CO,
gas
formed
by
the
combustion
of
18.5
grams
of
CgHg
at
30°C
and
1.50
atm?
Your
Answer:
(MW
=78)
(MW
=
32)
(MW
=44)
(MW
=18)
2
C6H6
(g)
+
15
02
(g)
—
12
CO2
(g)
+
6
H20
(g)
18.5
grams/78
=
0.2372
mol
C6H6
0.2372
mol
C6H6
X
(12
mol
CO2/2
mol
C6HB)
=
1.4232
mol
CO2
PV=nRT
V
=nRT/P
V
=
(1.4232)(0.0821)(303)
/
1.50
V
=
23.6
liters
(MW
=78)
(MW=32)
(MW=44)
(MW
=18)2
CgHg(g)
+
15
O5
(g)
=
12
CO5(s)
+
6
H,O
(I)
18.5
grams
23.6
liters
|
1
0.2372
mol
—»
-
—
12/2
x
0.2372
mol
V
=
nRT
/P
=
(1.4232)(0.0821)(303)
/
1.50
=
23.6
liters
Question
4
0/0
pts
Show
your
work
and
be
sure
to
give
answer
with
the
correct
units.
A
1.00
liter
container
holds
a
mixture
of
0.52
mg
of
He
and
2.05
mg
of
Ne
at
25°C.
Determine
the
partial
pressures
of
He
and
Ne
in
the
flask.
What
is
the
total
pressure?
Your
Answer:
n(He)
=
g(He)
/
[MW(He)]
=
(0.52
mg
x
1
g/
1000
mg)
/
4.002
=
0.000129935
mol
n(Ne)
=
g(Ne)
/
[MW(Ne)]
=
(2.05
mgx
1
g/
1000
mg)
/
20.18
=
0.000101586
mol
PT=n
R
T/V
=
(0.000129935
+
0.000101586)
(0.0821)
(298K)
/
1.00
=
0.005664
atm
PT
=
0.005664
atm
x
760
mm/1
atm
=
4.305
mm
X(He)
=
0.000129935
/
(0.000129935
+
0.000101586)
=
0.5612
X(Ne)
=
0.000101586
/
(0.000129935
+
0.000101586)
=
0.4388
P(He)
=
X(He)
(4.305
mm)
=
0.5612
(4.305
mm)
=
2.416
mm
P(Ne)
=
X(Ne)
(4.305
mm)
=
0.4388
(4.305
mm)
=
1.889
mm
n(He)
=
g(He)
/
[MW(He)]
=
(0.52
mg
x
1
g/
1000
mqg)
/
4.002
=
©
0.000129935
mol
n(Ne)
=
g(Ne)
/
[MW(Ne)]
=
(2.05
mgx
1
g/
1000
mg)/
20.18
=
~
0.000101586
mol
Pr=nRT/V=(0.000129935
+
0.000101586)
(0.0821)
(298K)
/
1.00
=
0.005664
atm
Pt
=
0.005664
atm
x
760
mm/1
atm
=
4.305
mm
X(He)
=
0.000129935
/
(0.000129935
+
0.000101586)
=
0.5612
X(Ne)
=
0.000101586
/
(0.000129935
+
0.000101586)
=
0.4388
P(He)
=
X(He)
(4.305
mm)
=
0.5612
(4.305
mm)
=
2.416
mm
P(Ne)
=
X(Ne)
(4.305
mm)
=
0.4388
(4.305
mm)
=
1.889
mm
Question
5
0/0
pts
Show
your
work
and
be
sure
to
give
answer
with
the
correct
units.
A
sample
of
hydrogen
(H»>)
gas
is
collected
over
water
at
35°C
and
745
mm.
The
volume
of
the
gas
collected
is
72.0
ml.
How
many
moles
of
H»
gas
has
been
collected?
How
many
grams
of
H,
gas
has
been
collected?
(Click
here
(https://portagelearning.instructure.com/courses/1952/files/637992?wrap=1)
J
N—
(https://portagelearning.instructure.com/courses/1952/files/637992/download
download_frd=1)
for
Water
Vapor
Pressure
Table).
Your
Answer:
PH2
=725
-
P
H20
(from
table)
=
725
-42.2
=682.8
x
1
atm
/
760
mm
=
0.898
atm
H2
from
Ideal
Gas
Law:
n
H2
=
PV
/
RT
=
(0.898
atm)
(72.0
ml
X
1
liter
/
1000
ml)
/
(0.0821)
(3080K)
n
H2
=
0.002557
moles
grams
H2
=
moles
x
MW
=
0.002557
moles
x
2.016
grams
/
1
mole
=
0.00515
grams
P(Hz)
=
745
-
PHo0)
(from
table)
=
745
-
42.2
=
702.8
x
1
atm
/
760
mm
=
0.925
atm
H,
from
ldeal
Gas
Law:
n(H5)
=
PV
/
RT
=
(0.925
atm)
(72.0
ml
x
1
liter
/
1000
ml)
/
(0.0821)
(308K)
n(H2)
=
0.002634
moles
grams(H»>)
=
moles
x
MW
=
0.002634
moles
x
2.016
grams
/
1
mole
=
0.0053
grams
Question
6
0/0
pts
Show
your
work
and
be
sure
to
give
answer
with
the
correct
units.
The
rate
of
effusion
of
nitrogen
gas
(N»)
is
1.253
times
faster
than
that
of
an
unknown
gas.
What
is
the
molecular
weight
of
the
unknown
gas®?
Your
Answer:
1.253)"2
=
MW(unknown)
/
28.014
MW(unknown)=
28.014
x
(1.253)2
28.014
x
1.570009
=
43.99
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(rn2)
/
runknown)2
=
MW
nknown
/
MWp2
(1.253/1)?
=
MW
nknown
/
28.02
MW
.nknown
=
(1.253)?
x
28.02
=
43.99
Question
7
0/0
pts
Write
electron
(subshell)
configurations
for
the
following
elements:
19K
35Br
Your
Answer:
19K
=
1s22s522p63s23p64s1
35Br
=
1522s22p63s23p64s23d104p5
19K
=
15?2
252
2p®
3s?
3p°
45
35Br
=
12
2s?
2p®
3s?
3p®
452
3d'°
4p°
Question
8
0/0
pts
The
following
document
will
be
helpful
in
understanding
how
to
insert
arrows
as
part
of
your
answers.
CHEM
121
-
Module
3.7
-
How
to
Insert
Arrows
(https://portagelearning.instructure.com/courses/1952/files/637394/download
wrap=1)_
{,
(https://portagelearning.instructure.com/courses/1952/files/637394/download
download_frd=1)
Write
orbital
diagrams
for
the
following
atoms:
15P
2gNIi
(«
G
>
Your
Answer:
15P
=
1s22s22p63s23p3
1|
11
TITITL
T4-
11T
1
1s2s.
2p
3s.
3p
28Ni
=
1522522p63s23p64s23d8
1|
1|
TITITL
T
NI
UL
LT
T
1s
2s
2p
3s
3p
4s
3d
15P
=
182
252
2p®
352
3p3
a5
W
o
MO
21
o
s
O
I
W
o
.
1s
25
2p
3
3p
2sNi
=
152
252
2p®
352
3p°®
452
37°
5
e
4
A
o
6
3
A
N
s
e
s
o
o
G
A
s
1s
23
2p
3s
3p
4s
3d
Question
9
0/0
pts
Write
the
values
for
the
four
quantum
numbers
for
a
4d®
electron.
Your
Answer:
NNty
Ttn=41=2Mi=0Ms
=-1/2
A
4d3
electron
is
in
the
4th
shell
so
n
=
4.
Its
"d"
orbital
has
2
nodal
planes
so
|1
=2.
The
electron
is
in
the 3rd
of
the
orbitals
of
the
d
subshell
so
m;
=
0.
And
since
this
is
the
2nd
electron
to
occupy
this
"d"
orbital,
mg
=
-1/2.
In
summary,
the
values
for
the
four
quantum numbers
for
the
4d®
electron
are:
n
=4,
|
=2,
m=0and
mg
=-1/2.
Question
10
0/0pts
Write
the
values
for
the
four
quantum
numbers
for
the
last
electron
to
fill
2gNi.
Your preview ends here
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- Unlimited textbook solutions
- 24/7 expert homework help
Your
Answer:
28Ni
=
1822522p63s23p64s23d8
1]
1.
TITITL-
TL-
NI
1s
2s
2p
3s
3p
4s
3d
n=31=2Mi=0Ms=-1/2
The
last
electron
to
fill
ogNi
is
a
3d®
electron
which
is
in
the
3rd
shell
so
n
=
3.
Its
"d"
orbital
has
2
nodal
planes
so
|
=2.
The
electron
is
in
the
3rd
of
the
orbitals
of
the
d
subshell
som;
=
0.
And
since
this
is
the
2nd
electron
to
occupy
this
"d"
orbital,
mg
=
-1/2.
In
summary,
the
values
for
the
four
quantum
numbers
for
the
3d®
electron
are:n=3,1=2,
m;=0
and
mg
=
-1/2.
Question
11
0/0
pts
Choose
the
atom
that
matches
the
identified
atomic
property
and
explain
your
answer
based
on
the
periodic
trend
and
its
position
within
a
group
or
period
in
the
periodic
table
&
(http://www.nursingabc.com/upload/images/Help_file_
picture/Periodic_Table.
1.
Highest
lonization
Energy:
Li,
Be,
B
2.
Lowest
Electronegativity:
Li,
Be,
B
3.
Largest
Atomic
Size:
Li,
Be,
B
«
B
Your
Answer:
1.
Boron
(B)
has
the
highest
ionization
energy.
lonization
energy
iIncreases
across
a
period
from
left
to
right.
2.
Lithium
(Li)
has
the
lowest
electronegativity.
Since
Lithium
is
the
leftmost
element
among
the
three
in
the
same
period,
it
has
the
lowest
electronegativity.
3.
Lithium
(Li)
has
the
largest
atomic
size.
Lithium
is
the
leftmost
among
these
elements,
it
has
the
largest
atomic
size
in
this
group.
1.
B
has
the
highest
ionization
energy
since
it
is
farthest
to
right
of
these
elements
and
ionization
energy
increases
as
you
go
from
left
to
right
in
a
period.
2.
Li
has
the
lowest
electronegativity
since
it
is
farthest
to
left
of
these
elements
and
electronegativity
increases
as
you
go
from
left
to
right
in
a
period.
3.
Li
has
the
largest
atomic
size
since
it
is
farthest
to
left
of
these
elements
and
atomic
size
decreases
as
you
go
from
left
to
right
in
a
period.
Question
12
0/0pts
Create
an
orbital
diagram
and
Lewis
dot
structure
for
4¢S.
Your
Answer:
16S
=
1522522p63s23p4
1.
1L
TITITL
TL
14
11
1s2s
2p
3s
3p
S
16S
=
152
2522pd
3s23p4
s
$0tL
MLTLTL
H
oMTt
.
S
)
1s
2s
2p
35
3p
o0
Question
13
515
pts
As
a
reminder,
the
questions
in
this
review
quiz
are
a
requirement
of
the
course
and
the
best
way
to
prepare
for
the
module
exam.
Did
you
complete
all
questions
in
their
entirety
and
show
your
work?
Your
Answer:
Yes
Quiz
Score:
5
out
of
5
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Your answer is wrong. In addition to checking your math, check that you used the right data and DID NOT round any intermediate calculations.
A certain drug has a half-life in the body of 3.5 h. Suppose a patient takes one 300. mg pill at 7:00PM and another identical pill 45 min later. Calculate the
amount of drug left in his body at 10:00PM.
Be sure your answer has a unit symbol, if necessary, and round it to 3 significant digits.
559. mg
x10
X
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Answer question 7 please
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Plot your values of In(Ksp) VS. 1/Tand find the slope and y-intercept of the best fit line. Use the equation for the best fit
line and the following equation
In(K) = - AH°
RT
AS°
R
to calculate AH and AS for dissolving Borax
What is the slope of your best fit line in the plot?
What is AH (kJ/mol)?
What is the y-intercept of your best fit line in the plot?
What is AS (J/mol)?
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Show complete solution.
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Choose the factor below that is likely to have led to the greatest number of mass extinctions.
Earthquakes
UV radiation
Temperature change
Asteroids
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The wow expert Hand written solution is not allowed.
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Hydrochloric acid can be prepared by the following reaction:
2NACI(s) + H2SO4(aq)2HCI(g) + Na2SO4(s)
What mass of HCI can be prepared from 2.00 mol H SO4 and 150. g NaCl?
Multiple Choice
2.57 g
167 g
93.6 g
11
asus COLLECTION
prt sc
delete
112/A
insert
f7
f8
f11
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Which one is it?
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Part A
Which route would be better: one that produces a small amount of hazardous waste or one that produces a large amount of nonhazardous waste? Which green chemistry metric accounts for this dilemma?
Match the items in the left column to the appropriate blanks in the sentences on the right.
Reset Help
minimize
The best route is
because it should
the E-factor and
maximize
the effective mass yield.
the one that produces a small
The
accounts for this dilemma as it looks at the amount of
relative to
amount of hazardous waste
the amount of
E-factor
product made
the one that produces a large
amount
nonhazardous
waste
materials used
difficult to choose
hazardous reagents used
waste produced
Submit
Previous Answers Request Answer
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ssignment/takeCovalent Activity.do?locator=assignment-take
d...
Communication s...
1.00 x 10-8 +1.71 x 10-
-4
[References]
Use the References to access important values if needed for this question.
It is often necessary to do calculations using scientific notation when working
chemistry problems. For practice, perform each of the following calculations.
2.37 x 104 +2.12 × 105
9.60 × 104
(1.00
QUT_Teamwork_...
28
7.00 × 10-5
0-8) (1.71 x 10-4)
x 10
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Experiment 4: Preparation and Properties of Hydrogen and Oxygen Gases
Toriana Mieer -Jones
Name:
Section:
Lab Instructor:
L/26
Date:
Pre-Lab Exercise 4
1. In this experiment, simple chemical reactions are used to generate small quantities of hydrogen and
oxygen gases. These methods would be too expensive and too complicated if large quantities of these
gases were needed, however. Use your textbook, a chemical encyclopedia, or an online source to look
up the major industrial methods of preparing very large quantities of these gases. Summarize your
findings here.
ono od te
led
bateono to n
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What is the major ingredient of Tums tablets used in today's experiment ?
- Mg(OH)2
- NaHCO3
- KHCO3
- CaCO3
- Ca(OH)3
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Can you identify the acceptable exposure limits for affected populations in occupational and nonoccupational environments?
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I need the correct answer in 3-4 lines fast please
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Search for Life Transfer Task 1 - Poisoned Shopworkers
Back in 1920, one of the most sought-after jobs on the East Coast was a dial painter at a watch factory. These workers were all female, and they absolutely loved their jobs. These girls were instructed to paint the number on a watch with paint laced with an element called radium. Years later, many of these workers developed many medical problems with their bones. Some girls lost their jaw bones, and others’ bones broke when they walked across a room. Several doctors thought that phosphorus poisoned the girls’ bones. Their companies insisted that radium was not damaging their bones in order to avoid a lawsuit. Answer the questions below to answer the big questions:
Was it the radium or the phosphorus that poisoned the girls’ bones?
Were the factories responsible for the girls’ sicknesses?
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If the intensity of radiation is 255 microcuries (μCi) at a distance of 1.74 m, what is the intensity at a distance of 2.87 m?
Intensity = _____ μCi
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I'm having a really hard time with this problem. Please help!!
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