Chlorofluoromethanes (CFMs) are carbon compounds of chlorine and fluorine and are also known as Freons. Examples are Freon-11 (CFCl 3 ) and Freon-12 ( CF 2 Cl 2 ) , which were used as aerosol propellants. Freons have also been used in refrigeration and air-conditioning systems. In 1995 Mario Molina, F. Sherwood Rowland, and Paul Crutzen were awarded the Nobel Prize mainly for demonstrating how these and other CFMs contribute to the “ozone hole” that develops at the end of the Antarctic winter. In other pares of the world, reactions such as those shown below occur in the upper atmosphere where ozone protects the earths inhabitants from harmful ultraviolet radiation. In the stratosphere CFMs absorb high-energy radiation from the sun and split off chlorine atoms that hasten the decomposition of ozone, O 3 . Possible reactions are O 3 ( g ) + Cl ( g ) → O 2 ( g ) + ClO ( g ) Δ H ° = − 126 kJ ClO ( g ) + O ( g ) → Cl ( g ) + O 2 ( g ) Δ H ° = − 268 kJ O 3 ( g ) + O ( g ) → 2 O 2 ( g ) Δ H ° = ? The O atoms in the second equation come from the breaking apart of O 2 molecules caused by ultraviolet radiation from the sun. Use the first two equations to calculate the value of Δ H ° (in kilojoules) for the last equation, the net reaction for the removal of O 3 from the atmosphere.
Chlorofluoromethanes (CFMs) are carbon compounds of chlorine and fluorine and are also known as Freons. Examples are Freon-11 (CFCl 3 ) and Freon-12 ( CF 2 Cl 2 ) , which were used as aerosol propellants. Freons have also been used in refrigeration and air-conditioning systems. In 1995 Mario Molina, F. Sherwood Rowland, and Paul Crutzen were awarded the Nobel Prize mainly for demonstrating how these and other CFMs contribute to the “ozone hole” that develops at the end of the Antarctic winter. In other pares of the world, reactions such as those shown below occur in the upper atmosphere where ozone protects the earths inhabitants from harmful ultraviolet radiation. In the stratosphere CFMs absorb high-energy radiation from the sun and split off chlorine atoms that hasten the decomposition of ozone, O 3 . Possible reactions are O 3 ( g ) + Cl ( g ) → O 2 ( g ) + ClO ( g ) Δ H ° = − 126 kJ ClO ( g ) + O ( g ) → Cl ( g ) + O 2 ( g ) Δ H ° = − 268 kJ O 3 ( g ) + O ( g ) → 2 O 2 ( g ) Δ H ° = ? The O atoms in the second equation come from the breaking apart of O 2 molecules caused by ultraviolet radiation from the sun. Use the first two equations to calculate the value of Δ H ° (in kilojoules) for the last equation, the net reaction for the removal of O 3 from the atmosphere.
Chlorofluoromethanes (CFMs) are carbon compounds of chlorine and fluorine and are also known as Freons. Examples are Freon-11
(CFCl
3
)
and Freon-12
(
CF
2
Cl
2
)
, which were used as aerosol propellants. Freons have also been used in refrigeration and air-conditioning systems. In 1995 Mario Molina, F. Sherwood Rowland, and Paul Crutzen were awarded the Nobel Prize mainly for demonstrating how these and other CFMs contribute to the “ozone hole” that develops at the end of the Antarctic winter. In other pares of the world, reactions such as those shown below occur in the upper atmosphere where ozone protects the earths inhabitants from harmful ultraviolet radiation. In the stratosphere CFMs absorb high-energy radiation from the sun and split off chlorine atoms that hasten the decomposition of ozone,
O
3
. Possible reactions are
O
3
(
g
)
+
Cl
(
g
)
→
O
2
(
g
)
+
ClO
(
g
)
Δ
H
°
=
−
126
kJ
ClO
(
g
)
+
O
(
g
)
→
Cl
(
g
)
+
O
2
(
g
)
Δ
H
°
=
−
268
kJ
O
3
(
g
)
+
O
(
g
)
→
2
O
2
(
g
)
Δ
H
°
=
?
The O atoms in the second equation come from the breaking apart of
O
2
molecules caused by ultraviolet radiation from the sun. Use the first two equations to calculate the value of
Δ
H
°
(in kilojoules) for the last equation, the net reaction for the removal of
O
3
from the atmosphere.
Experiment #8 Electrical conductivity & Electrolytes.
Conductivity of solutions
FLINN Scientific
Scale RED LED
Green LED
LED Conductivity
0
OFF
OFF
1
Dim
OFF
2
medium
OFF
3
Bright
Dim
Low or Nowe
Low
Medium
High
4
Very Bright
Medium
nd
very high
AA
Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ Δ
SE=Strong Electrolyte, FE = Fair Electrolyte
CWE = Weak Electrolyte, NE= Noni Electrolyte,
#Solutions
1 0.1 M NaCl
2/1x 102 M NaCl,
3/1X103 M Nall
Can
Prediction
M
Observed Conductivity
Very bright red
Bright red
Dim red
you help me understand how I'm supposed
to find the predictions of the following solutions?
I know this is an Ionic compound and that
the more ions in a solution means it is able
to carry a charge, right?
AAAA
D
(SE EX 2) Prblsm 4-7: Can you please explain problems 4-7 and color code if needed for me. (step by step) detail explanations
(SE EX 2) Problems 8-11, can you please explain them to me in detail and color-code anything if necessary?
Applications and Investigations in Earth Science (9th Edition)
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