Lab 11
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Course
171
Subject
Chemistry
Date
Dec 6, 2023
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15
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ENTHALPY
OF
FORMATION
OF
AMMONIUM
SALTS
BACKGROUND
INFORMATION
There
are
a
very
large
number
of
chemical
reactions
that
have
never
been
investigated.
There
may
be
good
reasons
for
not
performing
certain
chemical
reactions.
For
example,
some
reactions
may
pose
great
risks
to
the
environment
or
to
living
organisms.
Other
reactions
may
be
too
costly
in
terms
of
time
or
money.
Studying
the
chemical
reactivity
of
diamonds
would
certainly
be an
expensive
undertaking.
Fortunately,
there
are
methods
that
allow
us
to
study
certain
aspects
of
chemical
reactions
without
ever
having
to
perform
the
reactions.
Hess’s
Law
(also
known
as
the
Law
of
Constant
Heat
Summation)
is
an
example
of
a
method
that
allows
chemists
to
study
the
thermodynamics
of
a
chemical
reaction
without
having
to
perform
the
reaction
itself.
Germain
Henri
Hess
discovered
this
fundamental
law
of
nature
in
1840
while
studying
the
heat
evolved
when
acids
and
bases
are
mixed
with
water.
Hess’s
Law
states
that the
enthalpy
change
(AH)
of
individual
steps
in
a
process
can
be
added
or
subtracted
to
determine
the
overall
enthalpy
change
of
the
process.
Hess’s
Law
is
a
direct
consequence
of
the
First
Law
of
Thermodynamics,
which
requires
that the
enthalpy
change
of
a
reaction
be
the
same
regardless
of
the
reaction
pathway
that
leads
from
reactants
to
products.
Thus,
the
net
amount
of
heat
liberated
or
absorbed
in
a
chemical
process
is
the
same,
regardless
of
whether
the
process
is
performed
in
one
step
or
multiple
steps.
Because
of
the
additive
property
of
AH,
we
can
use
a
standard
set
of
reactions
from
which
we
can
gener-
ate
any
other reaction.
All
we
have
to
do
is
add
the
enthalpies
of
these
standard
reactions
and
we
have
the
enthalpy
of
our
reaction
of
interest.
The
standard
reactions
we
use
are
called
formation
reactions
and
the
enthalpies
are
enthalpies
of
formation
symbolized
by
AHf.
A
formation
reaction
has
one
single
product:
one
mole
of
the
compound.
The
reactants
are
elements
in
their
standard
states.
Thus,
on
paper,
we
can
decompose
any
compound
to
its
elements
(AH
=
—AHy
and
then
take
those
elements
and
rearrange
them
into
the
product
compounds
(AH
=
AH?).
This
leads
to
the
equation
1.
AH°=
Y
AH?-
Y,
AH}
(1)
Product
Reactants
This
process
for the
reaction
3C,H,
—
Cg¢Hg
is
shown
in
figure
1.
In
today’s
experiment,
you
will
use
Hess’s
Law
to
determine
the
enthalpy
change
(AH)
of
a
chemical
reaction
that
cannot
be
easily
measured
by
experimental
methods.
This
reaction
is
represented
by
equa-
tion
(2).
From
General
Chemistry,
2nd
Edition
by
Department
of
Chemistry,
University
of
North
Carolina
at
Wilmington.
Copyright
©
2004
by
University
of
North
Carolina
at
Wilmington.
Reprinted
by
permission
of
Kendall/Hunt
Publishing
Company.
115
116
LAB
11:
ENTHALPY
OF
FORMATION
OF
AMMONIUM
SALTS
6C
+
3H,
3C,H,
Enthalpy
CeHe
FIGURE
1.
Enthalpy
of
conversion
of
-
acetylene
to
benzene.
%Nz(g)
+
2H,(g)
+
%Clz(g)
—
NH,CI(s)
or
Nx(g)
+
2H,(g)
+
%
O,(g)
-
NH4,NOs(s)
AH,
(2)
You
will
determine
the
value
of
AH
for
this
reaction
by
measuring
the
enthalpy
changes
of
two
related
chemical
reactions.
These
reactions
are
represented
by
equations
(3)
and
(4)
below.
NH;(aq)
+
HCl(aq)
-
NH,Cl(aq)
or
NHs(aq)
+
HNO;(aq)
-
NH;NOs(aq)
AH,
(3)
NH,Cl(aq)
—
222
5
NH,CI(s)
or
NH4;NO;(aq)
—:°
y
NH,NO;
(s)
AH;
(4)
Equation
(2)
can
be
obtained
by
adding
equation
(3)
to
equation
(4)
along
with
the
enthalpies
of
forma-
tion
of the
aqueous
and
ammonia
(shown
in
equations
(5)
and
(6),
below).
|
1
3
Hy(g)
+
-
Cla(g) —
205
HCl(aq)
Ha(g)
+
>
N28)
+
5
0:(8)
—
%0
s
HNOs(ag)
AH;
(5)
1
3
7
Na(g)
+
5
Ha(g)
—=2
5
NHs(aq)
AH,
(6)
We
simply
add
the
enthalpies
for
(3),
(5)
and
(6)
together
with
the
reverse
of
(4)
AH})
=
AH3
+
AH4
+
AHS
<+
AH6
(7)
The
enthalpies
for
(5)
and
(6)
are
readily
available
from
the
literature.
Equation
(1)
can
be
obtained
by
adding
equation
(2)
to
the
reverse
of
equation
(3).
If
we
reverse
equa-
tion
(3),
we
must
also
change
the
sign
of
the
enthalpy
change
for
the
reaction.
According
to
Hess’s
Law,
if
we
add
the
equations
for
two
reactions
together,
we
can
also
add
their
enthalpy
changes
together.
Thus,
the
enthalpy
change
for
reaction
(1)
is
simply
the
sum
of
the
enthalpy
change
for
reaction
(2)
and
the
reverse
of
reaction
(3).
AH,
=
AH,
+
(-AH;)
(8)
LAB
11:
ENTHALPY
OF
FORMATION
OF
AMMONIUM
SALTS
117
To
determine
the
enthalpy
changes
(AH,
and
AH3)
of
reactions
(2)
and
(3),
you
must
measure
the
heat
(q)
released
by
each
reaction.
You
will
do
this
using
the
same
techniques
and
concepts
you
used
in
the
experiment
on
calorimetry.
(You
might
want
to
review
the
Introduction
to
Calorimetry.)
Recall
that
the
heat
exchanged
in
a
process
can
be
calculated
from
the
following
equation
g=cxmxAT
9)
where
m
is
the
mass
of
the
substance
in
grams,
the
specific
heat,
¢,
of
the
substance
is
expressed
in
units
of
J/ge°C,
and
AT
is
the
temperature
change
the
substance
undergoes
(Tfipa
—
Tiniial)-
Some
heat
will
be
lost
through
the
walls
and
lid
of
the
calorimeter.
A
correction
term
C,
=
15
J/K
will
be
used
to
estimate
this
loss.
Our
corrected
relationship
between
heat
and
temperature
becomes:
g
=
(¢
x
m
+
15)AT.
The
specific
heat
capacities
can
be
found
in
Table
2.
Gwater
=
(418
J/g.oc)(mwater)(ATwater)
(10)
The
heat
gained
by
the
water
is
equal
in
magnitude
by
opposite
in
sign
to
the
heat
lost
by
the
reaction.
Therefore,
we
can
calculate
ge,cion
from
the
following
relationship.
Yreaction
=
—Ywater
(1 1)
Finally,
the
enthalpy
change
of
each
reaction,
AH,.,.ion,
can
be
determined
by
dividing
the
heat
lost
by
the
reaction
(Greaciion)
by
the
number
of
moles
of
product
formed
(either
NH,Cl(aq)
or
NH;NOs(aq)).
For
reaction
(2),
this
calculation
takes
the
following
form.
AH,
=
4
(12)
moles
NH,Cl
or
NH,NO;
PROCEDURE
Caurion:
Wear
departmentally
approved
eye
protection.
If
you
are
found
without
the
eye
protection,
penalty
will
be
imposed
without
warning.
Gloves
and
proper
attire
must
be
worn
during
this
experiment.
This
experiment
will
be
performed
in
pairs.
The
write-up,
calculations,
graphing,
etc.,
must
be
done
totally
independently.
A
series
of
two
experiments
will
be
performed.
Assume
the
calorimeter
constant,
Cec,
to
be
157
K.
Part
A.
Determination
of
the
enthalpy
of
reaction
of
aqueous
ammonia
with
a
strong
acid
(HCI
or
HNO;)
assigned
by
the
instructor.
Part
B.
Determination
of
the
enthalpy
of
solution
of
the
ammonium
salt
of
the
acid
used
in
Part
A
(NH4C1
or
NH4NO3).
In
addition
to
reagents,
the
equipment
needed
is:
two
6-0z
pressed
Polystyrene
cups
(and
one
cup
lid),
a
150-mL
beaker,
two
100-mL
graduated
cylinders,
and
one
thermometer.
Keep
the
cups
together,
as
though
they
were
glued
together.
You
need
2
cups
for
the
additional
insulation.
118
LAB
11:
ENTHALPY
OF
FORMATION
OF
AMMONIUM
SALTS
Solution
temperature
as
a
function
of
time
will
be
recorded
on
the
data
sheet.
In
addition
to
recording
the
data,
a
temperature-time
plot
on
graph
paper
should
be
prepared;
this
can
be
done
while
the
data
are
being
collected.
The
temperature
should
be
plotted
on
the
y-axis
and
the
time
in
minutes
on
the
x-axis.
The
T
observed
will
be
between
7
and
14°
for
Part
A
and
between
—1
and
—6°
for
Part
B.
The
scale
units
on
the
y-axis
should
be
selected
so
that
all
points
can
be
plotted
on
scale;
the
scale
units
on
the
x-axis
should
cover
the
range
0—20
minutes.
The
procedure
is
essentially
the
same
for
both
parts:
1.
Rinse
one
Polystyrene
cup
and
beaker
with
distilled
water
and
wipe
dry
with
a
clean
piece
of
absorbent
paper.
Nest
it
in
the
other
cup
(see
figure
2).
Clean
a
beaker
with
distilled
water
and
wipe
dry
with
absorbent
paper.
Place
the
thermometer
through
the
lid
of
the
cup
with
the
rubber
ring
over
the
hole
set
so
that the
thermometer
will
not
touch
the
bottom
of
the
cup
when
the
lid
is
placed
on
the
cup.
Do
not
clamp
the
thermometer
to
the
ring
stand.
They
are
fragile
and
will
likely
break
if
clamped.
2.
Using
graduated
cylinders,
measure
solution
1
(see
Table
1)
into
the
cup and
(except
in
Part
B)
mea-
sure
solution
2
into
the
beaker.
Record
the
concentrations
of
the
solutions
used
on
the
Data
Sheet.
Note:
Calculate
the
concentration
of
the
ammonium
salt—NH,C]
or
NH,NO;—formed
in
the
reaction
of
Part
A.
Determine
the
mass
of
salt
in
100
mL
of
solution;
weigh
out
that
amount
of
salt
and
use
in
place
of
solution
2
in
Part
B.
3.
Check
the
temperature
of
solution
1.
Assume
the
temperature
of the
two
solutions
before
mixing
is
essentially
the
same.
Record
the
temperature
of
solution
1
at
1-minute
intervals
for
5
minutes.
4.
After
5
minutes
of
temperature
readings,
pour
solution
2
into
the
calorimeter
cup.
Immediately
swirl
the
cup
gently
to
mix
the
solution
well.
Record
the
time
of
mixing.
(Never
use
a
thermometer
as
a
stirring
rod!)
Continue
to
record
the
temperature
at
1-minute
intervals
for
10
minutes.
Do
one
determination
of
the
temperature
change
for
each
part
of
the
experiment.
Lid——>»
"
\
/4—
Inner
Cup
Thermometer
<«—
Outer
Cup
FIGURE
2.
Nested
polystyrene
cup.
LAB
11:
ENTHALPY
OF
FORMATION
OF
AMMONIUM
SALTS
119
TABLE
1
REAGENTS
TO
BE
USED
Part
A
B
Solution
1
50.0
mL
acid
(2.00M
HCI
or
2.00M
HNO;)
100.0
mL
distilled
water
Solution
2
50.0
mL
2.05M
NH;
See
Note,
p.
114
CALCULATIONS
Study
carefully
the
calculations
for
NH
Br
at
the
end
of
this
section.
1.
Find
the
temperature
change
AT
from
the
temperature-time
plots
by
extrapolating
the
plotted
data
to
the
time
of
mixing,
reading
the
initial
and
final
temperatures,
and
taking
the
temperature
change
for
each
determination.
See
figure
3
for
an
example.
Use
good
graphing
technique.
Data should
take
up
about
2/3rd
of
the
paper.
The
extrapolation
should
not
run
off
the
top
or
the
bottom
of
the
page.
Graph
should
have
a
title
and
axes
should
be
labeled.
Origin
does
not
need
to
be
(0,
0).
2.
Use
the
calorimeter
constant,
C,
of
15
J/K
for
the
calculation.
3.
Use
the
data
from
Part
A
and
the
calorimeter
constant
to
calculate
the
enthalpy
of
neutralization
of
NH;
and
the
assigned
acid.
4.
Use
the
data
from
Part
B
to
compute
the
enthalpy
of
solution
of
the
ammonium
salt.
5.
Use
the
results
for
the
enthalpy
of
neutralization
and
the
enthalpy
of
dissolution,
along
with
the
liter-
ature
values
of
AHy
for
NH;(aq)
and
the
assigned
acid
(given
in
Table
3)
to
compute
the
enthalpy
of
formation
of
the
ammonium
salt.
AT
=104K
25
T
(°C)
-
FIGURE
3.
Plot
of
temperature-time
0
5
10
15
data
showing
determination
of
AT.
Time
(min)
120
LAB
11:
ENTHALPY
OF
FORMATION
OF
AMMONIUM
SALTS
Typical
Calculations
for
NH,Br
Part
A
AT=119K,
d=1018g/mL
C=3911J/gK,
MM.
=97.8
g/mol
Mass
of
solution
=
100
mL
x
1.018
g/mL
=
101.8
g
Cs (heat
capacity
of the
solution)
=
101.8
g
x
3.91
J/g
K
=398
J/K
AH,;
=
—(Cs
+
C))AT
=—-(398
J/K
+
15.0
J/K)11.9
K
=-4915J
=-4915kJ
Students
mixed
50
mL
of
2.00M
acid
with
50.00
mL
of
2.05M
base,
1.e.,
0.100
mol
of the
reactants
were
used.
Therefore,
the
measured
AH,
is
for
0.100
mol
of
product
formed.
molar
AH,
=
—-4.915
kJ/
0.100
mol
=
—49.2
kJ/mol
Part
B
TABLE
2
MOLECULAR
WEIGHTS,
HEAT
CAPACITIES,
AND
DENSITIES
OF
SOLUTIONS
Molecular
wt.
Concentration
Sp.
heat
capacity”
density”
Solute
g
mol™
mol
L™
Jg'
K
gmL™
NH,Cl
53.49
1.00
3.93
1.013
NH,NO;
80.04
1.00
3.90
1.029
H,O
18.02
—
4.18
0.997
At
25°C
TABLE
3
ENTHALPIES
OF
FORMATION
AT
25
°C®
Reaction
AH?
kJ
mole™
INL(2)
+
2Hy(2)
+
—=2
5
NH;(ag,
2.0M)
—80.7
JHa(g)
+
3Cly(g)
—=2
5
HCl(aq,
2.0M)
~164.7
JHy(2)
+
3Nx(2)
+
50,(g)
—22
5
HNO;
(aq,
2.0M)
~206.0
Ha(g)
+305(g)
=
H0()
-285.8
*
“Selected
Values
of
Chemical
Thermodynamic
Properties,”
Circular
500,
National
Bureau
of
Standards,
Washington,
D.C.,
1961
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A marked rise in the overall temperature of the planet
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20. Dissolving the salt potassium nitrate in water is an endothermic reaction. What would be
observed if potassium nitrate was dissolved in water?
a) The water would get very hot.
b) The water would get colder.
c) The salt would float to the top of the water.
d) Bubbles of hydrogen gas would be produced.
arrow_forward
Chemical reactions occur everywhere in the world around us, this are the most important types of
events in the universe. It is through chemical reactions that plants grow, produce fruit, and become
compost for new plants. How are they related to our daily living? Why are they so essential for our
everyday lives? E
arrow_forward
Part IV
Cabbage Juice Indicator
Chop cabbage into small pieces until you have about 2 cups of chopped
cabbage. Boil for at least ten minutes to allow time for the color to leach out of
the cabbage. Allow it to cool for 15 min
• Filter out the plant material to obtain a red-purple-bluish colored liquid.
1. Add 1-2 drops of cabbage juice into a small bowl
2. Add 3-5 drops of vinegar into the bowl with the cabbage juice
3. Observe
Is this a chemical change or physical change? Explain?
Store the excess cabbage juice in your refrigerator, you will need them for future
experiments.
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What is Environmental chemistry?
What is Stoichiometry?
What are the types of chemical reactions?
Include sources and cite references.
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When potassium metal is placed in water, a large amount of energy is released as potassium hydroxide and hydrogen gas are produced in the reaction 2K(s) + 2H2O(l) → 2KOH(aq) + H2(g). Your lab partner says this is a redox reaction and a combustion reaction. Do you agree? Defend your answer by explaining whether or not it meets the requirements of each type of reaction.
Source StylesFormat
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Q29
Santa Monica Big Blue Bus operates a fleet of CNG buses, which run on compressed
natural gas (CNG), natural gas that has been compressed into a much smaller
volume in the fuel tank. A newly designed CNG vehicle carries about several
thousand cubic feet of compressed natural gas which contain 72 kg (kilograms) of
carbon. If all of the natural gas is burned, how many kilograms of carbon dioxide will
be emitted into the atmosphere? Assume no residue remains and the carbon dioxide
is the only product in the combustion.
If you are presenting this question, please explain the following:
The steps in the calculation process
O Please write out the steps involved in the conversion process
including the units
O
Please explain how to operate on the units
Please explain how to manipulate the powers of ten by hand
(e.g. product rule and the quotient rule)
How to apply the appropriate rule to the calculation:
O addition-subtraction rule ("the decimal places rule"), OR
multiplication-division…
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Catalysts are substances that increase the rate of reaction but can be
recovered unchanged at the end of the reaction. Catalysts can be
classified as either homogeneous (same state as reactants) or
heterogeneous (different state than reactants).
Part A
Platinum is used to catalyze the hydrogenation of ethylene:
H₂(g) + CH₂CH₂(g)
Chlorofluorocarbons (CFCs) catalyze the conversion of ozone (03) to oxygen gas (0₂):
CFC(g)
203(g) 30₂(g)
Magnesium catalyzes the disproportionation of hydrogen peroxide to produce water and oxygen:
Mg(s)
2H₂O₂(aq) 2H₂O(1) + O₂(g)
What type of catalysts are platinum, CFCs, and magnesium under these conditions?
Drag the appropriate items to their respective bins.
▸ View Available Hint(s)
platinum
Homogeneous catalysts
Submit
CFCs
magnesium.
Pt(s)
Reset Help
Heterogeneous catalysts
CH3CH3(g)
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(please type answer no write by hand)
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Consider the following scenario:
Carbon and oxygen combine to form carbon dioxide.
A chemical equation you might write for this reaction is C(s) + O2(g) → CO2(g)-
If you were watching this reaction occur, how would you know when it was finished?
Hint: There is evidence in the equation!
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In experiment #7, you observed the formation of carbon dioxide by reacting baking soda with vinegar. Why was it not possible to relight the candle?
A. Carbon dioxide is heavier than air so it stays in the beaker preventing the candle from relighting.
B. Carbon dioxide is lighter than air so it escaped from the beaker and the candle should have relit.
C. Carbon dioxide is heavier than air so it stays in the beaker and the candle should have relit.
D. Carbon dioxide is lighter than air so it escaped from the beaker preventing the candle from relighting.
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The reaction between nitrogen and oxygen is given below:
N2(g) +2 O2(g)
+ 2 NO2(g)
We therefore know that which of the following reactions can also occur?
2 N20(g) → 2 N2(g) + O2(g)
N20(g) → 2 NH3(g) + 2 O2(g)
O2 N2(g) + 02(g) → 2 N20(g)
None of the Above
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Which of the following best describes the formation of coal and natural gas?
A
Coal and natural gas form from decaying marine organisms. The material is buried deep
under the surface of Earth and exposed to extreme heat and pressure over millions of
years.
Natural gas forms from decaying plant material, and coal forms from decaying marine
organisms. The material is buried deep under the surface of Earth and exposed to ex-
treme heat and pressure over millions of years.
Coal and natural gas form from decaying plant material. The material is buried deep un-
der the surface of Earth and exposed to extreme heat and pressure over millions of
years.
Coal forms from decaying plant material, and natural gas forms from decaying marine
organisms. The material is buried deep under the surface of Earth and exposed to ex-
treme heat and pressure over millions of years.
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The reaction between ammonia and nitrogen monoxide is given below:4 NH3(g) + 6 NO(g) 5 N2(g) + 6 H2O(l) We therefore know that which of the following reactions can also occur?
4 N2(g) + 3 H2O(g) 2 NH3(g) + 3 N2O(g)
2 NH3(g) + 3 N2O(g) 4 N2(g) + 3 H2O(g)
2 N2O(g) 2 N2(g) + O2(g)
None of the Above
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Suppose a pair of chemical compounds A and B can react in two different ways:
A+B → C
Reaction 1 gives product C.
A+B → D
Reaction 2 gives product D.
The following facts are known about the two reactions:
●
Reaction 1 is endothermic and Reaction 2 is exothermic.
• If a reaction vessel is charged ("filled") with A and B, then at first C is produced faster than D.
Use these facts to sketch a qualitative reaction energy diagram for both reactions.
Note: because these sketches are only qualitative, the energies don't have to be exact. They only have to have the right relationship to each other. For exam
if one energy is less than another, that fact should be clear in your sketch.
Reaction 1
Reaction 2
energy
energy
reaction coordinate
A + B
A + B
reaction coordinate
D
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Reaction 1
FeO(s) + CO(g) → Fe(1) + CO2(g)
AG" > 0
Reaction 2:
C(s) + CO2(g) → 2 CO(g)
AGzn <0
Overall reaction:
FeO(s) + C(s) → Fe(l) + CO(g)
AG n < 0
Subm
The chemical equations above represent the main reactions that occur during the production of Fe(l) under certain conditions. The overall reaction couples reactions 1 and 2, resulting in a
< 0?
thermodynamically favorable process. Which of the following best explains whether or not a particle diagram could represent how the coupling of reaction 1 and reaction 2 results in AG
A particle diagram that represents the increase in the volume of gaseous product particles would be a good representation of how the coupling of reactions 1 and 2 results in a
thermodynamically favorable process.
A particle diagram that represents the decrease in the average kinetic energy of the particles would be a good representation of how the coupling of reactions 1 and 2 results in a
thermodynamically favorable process.
A particle diagram cannot represent…
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7. Applying the principles of green chemistry requires some consideration of
the way in which they can be applied. Figure 1 represents a generic industrial
process. Using the principles of green chemistry and circled letters, describę
how four of the principles could be applied to this process to make it a green
operation. E kaEA
product
raw
materials
energy
Source
waste
гeactor
Figure 1
st
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Define the following terms
Centrifuge
Supernatant liquid
Precipitation
Decant
Washing precipitate
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unuunRINO
Consider these reactions, where M represents a generic metal.
2 M(s) + 6HCl(aq) →→2 MC1₂ (aq) + 3 H₂(g)
HCl(g) →→→ HCl(aq)
-
H₂(g) + Cl₂(g) →→→ 2 HCl(g)
MC1₂ (s) → MC1₂ (aq)
Use the given information to determine the enthalpy of the reaction
2 M(s) + 3 Cl₂(g) → 2 MC1₂ (s)
esc
1.
2.
3.
4.
AH =
!
x10
TOOLS
k
2
→
#3
C
4
AH₁ = -651.0 kJ
AH₂ = -74.8 kJ
AH3 = -1845.0 kJ
AH4 = -152.0 kJ
G Search or type URL
do LC
MacBook Pro
%
< (C
6
&
I
7
11
kJ
☆
*
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This section in a Safety Data Sheet
contains information on the
possibility of the material to
undergo hazardous reactions, and
its hazardous decomposition
products (if any), among others. *
Transport Information
Stability and Reactivity
Disposal Considerations
Ecological Information
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Why is that the surrounding temperature of ice cubes is cold and has condensation? What process (reaction) takes place? Is it endothermic or exothermic ?
In the photo given, describe which of the experiments are exothermic or endothermic. Explain on how each reaction should be defined.
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The reaction between nitrogen and oxygen is given below:2 N2(g) + O2(g) 2 N2O(g)We therefore know that which of the following reactions can also occur?
2 N2O(g) 2 N2(g) + O2(g)
N2(g) + 2 O2(g) 2 NO2(g)
2 NO2(g) N2(g) + 2 O2(g)
None of the Above
The reaction between ammonia and nitrogen monoxide is given below:4 NH3(g) + 6 NO(g) 5 N2(g) + 6 H2O(l)We therefore know that which of the following reactions can also occur?
4 N2(g) + 3 H2O(g) 2 NH3(g) + 3 N2O(g)
2 NH3(g) + 3 N2O(g) 4 N2(g) + 3 H2O(g)
5 N2(g) + 6 H2O(l) 4 NH3(g) + 6 NO(g)
None of the Above
arrow_forward
Section 1: Nuclear Energy
What types of reactions occur in a nuclear
power plant? Is the process endothermic or
exothermic?
What is your opinion about the use of this
energy?
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1. A 2.15-gram cashew nut is burned. The heat released raises the temperature of a 100.0-gram sample of water from 18.2°C to 31.5°C. The mass of the nut after the experiment is 1.78 grams. Determine the calorie content of the nut in Calories/gram. Assume that the water is only able to absorb 25% of the heat released by the burning nut. Given 1.00 Calorie=4.18 kJ.
This is another question
2. A large paraffin candle has a mass of 96.83 gram. A metal cup with 100.0 mL of water at 16.2°C absorbs the heat from the burning candle and increases its temperature to 35.7°C. Once the burning is ceased, the temperature of the water was 35.7°C and the paraffin had a mass of 96.14 gram. Determine the heat of combustion of paraffin in kJ/gram. GIVEN: density of water = 1.0 g/mL.
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