Step1 Moles of H2 = 35.02 =17.5 ;Moles of O2 = 2.pdf
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Step1 Moles of H2 = 35.0/2 =17.5 ;Moles of O2 = 276/32 =8.625 Step2 H2 is in excess Solution Step1 Moles of H2 = 35.0/2 =17.5 ;Moles of O2 = 276/32 =8.625 Step2 H2 is in excess.
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the sea of electrons shared between metal atoms .pdf
the sea of electrons shared between metal atoms
Solution
the sea of electrons shared between metal atoms.
isotopes .pdf
This very short document does not contain enough substantive information to generate a multi-sentence summary. It appears to be discussing isotopes but provides no additional context or details.
HOMO and LUMO are acronyms for highest occupied m.pdf
HOMO and LUMO are acronyms for highest occupied molecular orbital and lowest
unoccupied molecular orbital, respectively. The energy difference between the HOMO and
LUMO is termed the HOMO-LUMO gap. HOMO and LUMO are sometimes referred to as
frontier orbitals.[1] Roughly, the HOMO level is to organic semiconductors what the valence
band is to inorganic semiconductors and quantum dots. The same analogy exists between the
LUMO level and the conduction band. The atomic orbitals and molecular orbitals pictured above
require complicated calculations and assumptions that far exceed the equation shown for ?nlm.
Spartan calculates these orbitals as HOMOs or LUMOs. The HOMO (highest occupied
molecular orbital) is the valence orbital that received the last valence electron(s). The orbital
directly beneath the HOMO in energy is labeled HOMO(-1). The LUMO (lowest unoccupied
molecular orbital) is the empty orbital just above the HOMO. The orbital just above the LUMO
is labeled LUMO(+1). The \"1\" can be changed to 2, 3, 4, ... to create a complete set of all MOs
in the molecule. The HOMO and LUMO images (surfaces) created by Spartan can be
correlated/represented with the lines on a molecular orbital (MO) diagram. In this experiment,
you will be asked to create the MO diagrams complete with Spartan orbital surfaces for H2 + ,
H2, and H2 - . Spartan also can generate electron density surfaces. Electron density is the square
of the wave function (|?| 2 , or |?| 2 ). Squaring the wave function results in the loss of phase
information (i 2 = - 1). The lack of phase information is indicated in Spartan by the creation of
gray only density surfaces (Note: the \"electron\" is omitted in Spartan). The surface created
represents the relative size and shape of the molecule, atom, or ion being investigated.
Solution
HOMO and LUMO are acronyms for highest occupied molecular orbital and lowest
unoccupied molecular orbital, respectively. The energy difference between the HOMO and
LUMO is termed the HOMO-LUMO gap. HOMO and LUMO are sometimes referred to as
frontier orbitals.[1] Roughly, the HOMO level is to organic semiconductors what the valence
band is to inorganic semiconductors and quantum dots. The same analogy exists between the
LUMO level and the conduction band. The atomic orbitals and molecular orbitals pictured above
require complicated calculations and assumptions that far exceed the equation shown for ?nlm.
Spartan calculates these orbitals as HOMOs or LUMOs. The HOMO (highest occupied
molecular orbital) is the valence orbital that received the last valence electron(s). The orbital
directly beneath the HOMO in energy is labeled HOMO(-1). The LUMO (lowest unoccupied
molecular orbital) is the empty orbital just above the HOMO. The orbital just above the LUMO
is labeled LUMO(+1). The \"1\" can be changed to 2, 3, 4, ... to create a complete set of all MOs
in the molecule. The HOMO and LUMO images (surfaces) created by Spartan can be
correlated/represented with the.
YearCash flow0-1000000115000021500003150000415.pdf
Year
Cash flow
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-1000000
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150000
53
150000
54
150000
55
150000
56
150000
57
150000
58
150000
59
150000
60
150000
IRR
15%
Computation of inflation rate between 1939 and 1957 i.e. for 18 years
= (Ending value/ Beginning value)^(1/n) -1
= (1,000,000/80,805.12)^(1/18) -1
=12.375^(1/18) -1
= 1.15 -1
= 0.15 = 15%
The rate of return during the period of 60 years is 15% and the inflation rate between 1939 and
1957 is 15%.
I have computed IRR using excel formula. Please post as comments for any clarifications.
Year
Cash flow
0
-1000000
1
150000
2
150000
3
150000
4
150000
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150000
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150000
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150000
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150000
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150000
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150000
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150000
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150000
52
150000
53
150000
54
150000
55
150000
56
150000
57
150000
58
150000
59
150000
60
150000
IRR
15%
Solution
Year
Cash flow
0
-1000000
1
150000
2
150000
3
150000
4
150000
5
150000
6
150000
7
150000
8
150000
9
150000
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150000
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150000
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150000
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150000
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150000
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150000
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150000
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150000
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150000
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150000
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150000
44
150000
45
150000
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150000
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150000
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150000
49
150000
50
150000
51
150000
52
150000
53
150000
54
150000
55
150000
56
150000
57
150000
58
150000
59
150000
60
150000
IRR
15%
Computation of inflation rate between 1939 and 1957 i.e. for 18 years
= (Ending value/ Beginning value)^(1/n) -1
= (1,000,000/80,805.12)^(1/18) -1
=12.375^(1/18) -1
= 1.15 -1
= 0.15 = 15%
The rate of return during the period of 60 years is 15% and the inflation rate between 1939 and
1957 is 15%.
I have computed IRR using excel formula. Please post as comments for any clarifications.
Year
Cash flow
0
-1000000
1
150000
2
150000
3
150000
4
150000
5
150000
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150000
7
150000
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150000
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33
.
Toxic shock syndrome (TSS) is a toxin-mediated acute life-threatenin.pdf
Toxic shock syndrome (TSS) is a toxin-mediated acute life-threatening illness, usually
precipitated by infection with either Staphylococcus aureus or group A Streptococcus (GAS),
also called Streptococcus pyogenes. It is characterized by high fever, rash, hypotension,
multiorgan failure (involving at least 3 or more organ systems), and desquamation, typically of
the palms and soles, 1-2 weeks after the onset of acute illness. The clinical syndrome can also
include severe myalgia, vomiting, diarrhea, headache, and non focal neurologic abnormalities.
Toxic shock syndrome toxin 1(TSST-1), a prototype superantigen secreted by a Staphylococcus
aureus bacterium strain in susceptible hosts, acts on the vascular system by causing
inflammation, fever, and shock.This bacterium strain that produces the TSST-1\'s can be found
in any area of the body. TSST-1 is a bacterial exotoxin found in patients who have developed
toxic shock syndrome (TSS), which can be found in menstruating women or any man or child for
that matter.
GAS is an aerobic gram-positive organism that forms chains and is an important cause of soft
tissue infections. Diabetes, alcoholism, varicella infections, and surgical procedures all increase
the risk of severe GAS infections and hence may potentially increase the risk of GAS TSS.
Severe, invasive GAS infections can cause necrotizing fasciitis and spontaneous gangrenous
myositis. An increasing number of severe GAS infections associated with shock and organ
failure have been reported. These infections are termed streptococcal TSS.
Toxic shock syndrome toxin (TSST) is a superantigen with a size of 22 kDa produced by 5 to
25% of Staphylococcus aureus isolates. It causes toxic shock syndrome (TSS) by stimulating the
release of large amounts of interleukin-1, interleukin-2 and tumour necrosis factor. In general,
the toxin is not produced by bacteria growing in the blood; rather, it is produced at the local site
of an infection, and then enters the blood stream.
Pathogensis:
M proteins: Filamentous proteins on the cell membrane with antiphagocytic properties.
More than 80 M types.
Solution
Toxic shock syndrome (TSS) is a toxin-mediated acute life-threatening illness, usually
precipitated by infection with either Staphylococcus aureus or group A Streptococcus (GAS),
also called Streptococcus pyogenes. It is characterized by high fever, rash, hypotension,
multiorgan failure (involving at least 3 or more organ systems), and desquamation, typically of
the palms and soles, 1-2 weeks after the onset of acute illness. The clinical syndrome can also
include severe myalgia, vomiting, diarrhea, headache, and non focal neurologic abnormalities.
Toxic shock syndrome toxin 1(TSST-1), a prototype superantigen secreted by a Staphylococcus
aureus bacterium strain in susceptible hosts, acts on the vascular system by causing
inflammation, fever, and shock.This bacterium strain that produces the TSST-1\'s can be found
in any area of the body. TSST-1 is a bacter.
The trophozites (active form) are present only in the host and fresh.pdf
The trophozites (active form) are present only in the host and fresh loose faeces. E. coli is found
in the faeces of the carrier. In unsanitary conditions (like open defecation), the cysts are released
and they may be persistent for a few months in the environment. When a person drinks untreated
water contaminated with E. coli, the cysts enter his digestive system. once inside the digestive
system, the cysts release the trophozite stage within the digestive tract which may lead to further
infection. the trophozites are generally present in the large intestine. Sometimes, they invade the
intestinal mucosa, enter the blood stream and pass on to the different parts of the body causing a
number of outcomes, the most common of which is liver abcess.
Solution
The trophozites (active form) are present only in the host and fresh loose faeces. E. coli is found
in the faeces of the carrier. In unsanitary conditions (like open defecation), the cysts are released
and they may be persistent for a few months in the environment. When a person drinks untreated
water contaminated with E. coli, the cysts enter his digestive system. once inside the digestive
system, the cysts release the trophozite stage within the digestive tract which may lead to further
infection. the trophozites are generally present in the large intestine. Sometimes, they invade the
intestinal mucosa, enter the blood stream and pass on to the different parts of the body causing a
number of outcomes, the most common of which is liver abcess..
The set Z15ZSolutionThe set Z15Z.pdf
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Superior colliculusoptic tectum is a midbrain structure. The term s.pdf
Superior colliculus/optic tectum is a midbrain structure. The term superior colliculus is used
when discussing mammals and optic tectum when discussing either specific non-mammalian
species or vertebrates in general. It is a layered structure, with a number of layers that varies by
species. The layers can be grouped into the superficial layers and the deeper layers. Neurons in
the superficial layers receive direct input from the retina and respond almost exclusively to visual
stimuli. The deeper layers contain a population of motor-related neurons, capable of activating
eye movements as well as other responses. The superior colliculus mainly has its role in directing
eye movements. Visual input from the retina, or an input from the cerebral cortex, creates a
\"bump\" of activity in the tectal map, which, if strong enough, induces a saccadic eye movement
(A quick and simultaneous movement of both eyes). In primates however, the superior colliculus
is also involved in generating spatially directed head turns, arm-reaching movements, and shifts
in attention that do not involve any overt movements. In other species, the tectum is involved in
a wide range of responses, including whole-body turns etc.
Solution
Superior colliculus/optic tectum is a midbrain structure. The term superior colliculus is used
when discussing mammals and optic tectum when discussing either specific non-mammalian
species or vertebrates in general. It is a layered structure, with a number of layers that varies by
species. The layers can be grouped into the superficial layers and the deeper layers. Neurons in
the superficial layers receive direct input from the retina and respond almost exclusively to visual
stimuli. The deeper layers contain a population of motor-related neurons, capable of activating
eye movements as well as other responses. The superior colliculus mainly has its role in directing
eye movements. Visual input from the retina, or an input from the cerebral cortex, creates a
\"bump\" of activity in the tectal map, which, if strong enough, induces a saccadic eye movement
(A quick and simultaneous movement of both eyes). In primates however, the superior colliculus
is also involved in generating spatially directed head turns, arm-reaching movements, and shifts
in attention that do not involve any overt movements. In other species, the tectum is involved in
a wide range of responses, including whole-body turns etc..
Recommended
the sea of electrons shared between metal atoms .pdf
the sea of electrons shared between metal atoms
Solution
the sea of electrons shared between metal atoms.
isotopes .pdf
This very short document does not contain enough substantive information to generate a multi-sentence summary. It appears to be discussing isotopes but provides no additional context or details.
HOMO and LUMO are acronyms for highest occupied m.pdf
HOMO and LUMO are acronyms for highest occupied molecular orbital and lowest
unoccupied molecular orbital, respectively. The energy difference between the HOMO and
LUMO is termed the HOMO-LUMO gap. HOMO and LUMO are sometimes referred to as
frontier orbitals.[1] Roughly, the HOMO level is to organic semiconductors what the valence
band is to inorganic semiconductors and quantum dots. The same analogy exists between the
LUMO level and the conduction band. The atomic orbitals and molecular orbitals pictured above
require complicated calculations and assumptions that far exceed the equation shown for ?nlm.
Spartan calculates these orbitals as HOMOs or LUMOs. The HOMO (highest occupied
molecular orbital) is the valence orbital that received the last valence electron(s). The orbital
directly beneath the HOMO in energy is labeled HOMO(-1). The LUMO (lowest unoccupied
molecular orbital) is the empty orbital just above the HOMO. The orbital just above the LUMO
is labeled LUMO(+1). The \"1\" can be changed to 2, 3, 4, ... to create a complete set of all MOs
in the molecule. The HOMO and LUMO images (surfaces) created by Spartan can be
correlated/represented with the lines on a molecular orbital (MO) diagram. In this experiment,
you will be asked to create the MO diagrams complete with Spartan orbital surfaces for H2 + ,
H2, and H2 - . Spartan also can generate electron density surfaces. Electron density is the square
of the wave function (|?| 2 , or |?| 2 ). Squaring the wave function results in the loss of phase
information (i 2 = - 1). The lack of phase information is indicated in Spartan by the creation of
gray only density surfaces (Note: the \"electron\" is omitted in Spartan). The surface created
represents the relative size and shape of the molecule, atom, or ion being investigated.
Solution
HOMO and LUMO are acronyms for highest occupied molecular orbital and lowest
unoccupied molecular orbital, respectively. The energy difference between the HOMO and
LUMO is termed the HOMO-LUMO gap. HOMO and LUMO are sometimes referred to as
frontier orbitals.[1] Roughly, the HOMO level is to organic semiconductors what the valence
band is to inorganic semiconductors and quantum dots. The same analogy exists between the
LUMO level and the conduction band. The atomic orbitals and molecular orbitals pictured above
require complicated calculations and assumptions that far exceed the equation shown for ?nlm.
Spartan calculates these orbitals as HOMOs or LUMOs. The HOMO (highest occupied
molecular orbital) is the valence orbital that received the last valence electron(s). The orbital
directly beneath the HOMO in energy is labeled HOMO(-1). The LUMO (lowest unoccupied
molecular orbital) is the empty orbital just above the HOMO. The orbital just above the LUMO
is labeled LUMO(+1). The \"1\" can be changed to 2, 3, 4, ... to create a complete set of all MOs
in the molecule. The HOMO and LUMO images (surfaces) created by Spartan can be
correlated/represented with the.
YearCash flow0-1000000115000021500003150000415.pdf
Year
Cash flow
0
-1000000
1
150000
2
150000
3
150000
4
150000
5
150000
6
150000
7
150000
8
150000
9
150000
10
150000
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150000
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150000
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150000
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150000
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150000
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150000
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150000
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150000
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150000
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150000
45
150000
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150000
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150000
48
150000
49
150000
50
150000
51
150000
52
150000
53
150000
54
150000
55
150000
56
150000
57
150000
58
150000
59
150000
60
150000
IRR
15%
Computation of inflation rate between 1939 and 1957 i.e. for 18 years
= (Ending value/ Beginning value)^(1/n) -1
= (1,000,000/80,805.12)^(1/18) -1
=12.375^(1/18) -1
= 1.15 -1
= 0.15 = 15%
The rate of return during the period of 60 years is 15% and the inflation rate between 1939 and
1957 is 15%.
I have computed IRR using excel formula. Please post as comments for any clarifications.
Year
Cash flow
0
-1000000
1
150000
2
150000
3
150000
4
150000
5
150000
6
150000
7
150000
8
150000
9
150000
10
150000
11
150000
12
150000
13
150000
14
150000
15
150000
16
150000
17
150000
18
150000
19
150000
20
150000
21
150000
22
150000
23
150000
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150000
25
150000
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150000
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150000
28
150000
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150000
30
150000
31
150000
32
150000
33
150000
34
150000
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150000
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150000
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150000
38
150000
39
150000
40
150000
41
150000
42
150000
43
150000
44
150000
45
150000
46
150000
47
150000
48
150000
49
150000
50
150000
51
150000
52
150000
53
150000
54
150000
55
150000
56
150000
57
150000
58
150000
59
150000
60
150000
IRR
15%
Solution
Year
Cash flow
0
-1000000
1
150000
2
150000
3
150000
4
150000
5
150000
6
150000
7
150000
8
150000
9
150000
10
150000
11
150000
12
150000
13
150000
14
150000
15
150000
16
150000
17
150000
18
150000
19
150000
20
150000
21
150000
22
150000
23
150000
24
150000
25
150000
26
150000
27
150000
28
150000
29
150000
30
150000
31
150000
32
150000
33
150000
34
150000
35
150000
36
150000
37
150000
38
150000
39
150000
40
150000
41
150000
42
150000
43
150000
44
150000
45
150000
46
150000
47
150000
48
150000
49
150000
50
150000
51
150000
52
150000
53
150000
54
150000
55
150000
56
150000
57
150000
58
150000
59
150000
60
150000
IRR
15%
Computation of inflation rate between 1939 and 1957 i.e. for 18 years
= (Ending value/ Beginning value)^(1/n) -1
= (1,000,000/80,805.12)^(1/18) -1
=12.375^(1/18) -1
= 1.15 -1
= 0.15 = 15%
The rate of return during the period of 60 years is 15% and the inflation rate between 1939 and
1957 is 15%.
I have computed IRR using excel formula. Please post as comments for any clarifications.
Year
Cash flow
0
-1000000
1
150000
2
150000
3
150000
4
150000
5
150000
6
150000
7
150000
8
150000
9
150000
10
150000
11
150000
12
150000
13
150000
14
150000
15
150000
16
150000
17
150000
18
150000
19
150000
20
150000
21
150000
22
150000
23
150000
24
150000
25
150000
26
150000
27
150000
28
150000
29
150000
30
150000
31
150000
32
150000
33
.
Toxic shock syndrome (TSS) is a toxin-mediated acute life-threatenin.pdf
Toxic shock syndrome (TSS) is a toxin-mediated acute life-threatening illness, usually
precipitated by infection with either Staphylococcus aureus or group A Streptococcus (GAS),
also called Streptococcus pyogenes. It is characterized by high fever, rash, hypotension,
multiorgan failure (involving at least 3 or more organ systems), and desquamation, typically of
the palms and soles, 1-2 weeks after the onset of acute illness. The clinical syndrome can also
include severe myalgia, vomiting, diarrhea, headache, and non focal neurologic abnormalities.
Toxic shock syndrome toxin 1(TSST-1), a prototype superantigen secreted by a Staphylococcus
aureus bacterium strain in susceptible hosts, acts on the vascular system by causing
inflammation, fever, and shock.This bacterium strain that produces the TSST-1\'s can be found
in any area of the body. TSST-1 is a bacterial exotoxin found in patients who have developed
toxic shock syndrome (TSS), which can be found in menstruating women or any man or child for
that matter.
GAS is an aerobic gram-positive organism that forms chains and is an important cause of soft
tissue infections. Diabetes, alcoholism, varicella infections, and surgical procedures all increase
the risk of severe GAS infections and hence may potentially increase the risk of GAS TSS.
Severe, invasive GAS infections can cause necrotizing fasciitis and spontaneous gangrenous
myositis. An increasing number of severe GAS infections associated with shock and organ
failure have been reported. These infections are termed streptococcal TSS.
Toxic shock syndrome toxin (TSST) is a superantigen with a size of 22 kDa produced by 5 to
25% of Staphylococcus aureus isolates. It causes toxic shock syndrome (TSS) by stimulating the
release of large amounts of interleukin-1, interleukin-2 and tumour necrosis factor. In general,
the toxin is not produced by bacteria growing in the blood; rather, it is produced at the local site
of an infection, and then enters the blood stream.
Pathogensis:
M proteins: Filamentous proteins on the cell membrane with antiphagocytic properties.
More than 80 M types.
Solution
Toxic shock syndrome (TSS) is a toxin-mediated acute life-threatening illness, usually
precipitated by infection with either Staphylococcus aureus or group A Streptococcus (GAS),
also called Streptococcus pyogenes. It is characterized by high fever, rash, hypotension,
multiorgan failure (involving at least 3 or more organ systems), and desquamation, typically of
the palms and soles, 1-2 weeks after the onset of acute illness. The clinical syndrome can also
include severe myalgia, vomiting, diarrhea, headache, and non focal neurologic abnormalities.
Toxic shock syndrome toxin 1(TSST-1), a prototype superantigen secreted by a Staphylococcus
aureus bacterium strain in susceptible hosts, acts on the vascular system by causing
inflammation, fever, and shock.This bacterium strain that produces the TSST-1\'s can be found
in any area of the body. TSST-1 is a bacter.
The trophozites (active form) are present only in the host and fresh.pdf
The trophozites (active form) are present only in the host and fresh loose faeces. E. coli is found
in the faeces of the carrier. In unsanitary conditions (like open defecation), the cysts are released
and they may be persistent for a few months in the environment. When a person drinks untreated
water contaminated with E. coli, the cysts enter his digestive system. once inside the digestive
system, the cysts release the trophozite stage within the digestive tract which may lead to further
infection. the trophozites are generally present in the large intestine. Sometimes, they invade the
intestinal mucosa, enter the blood stream and pass on to the different parts of the body causing a
number of outcomes, the most common of which is liver abcess.
Solution
The trophozites (active form) are present only in the host and fresh loose faeces. E. coli is found
in the faeces of the carrier. In unsanitary conditions (like open defecation), the cysts are released
and they may be persistent for a few months in the environment. When a person drinks untreated
water contaminated with E. coli, the cysts enter his digestive system. once inside the digestive
system, the cysts release the trophozite stage within the digestive tract which may lead to further
infection. the trophozites are generally present in the large intestine. Sometimes, they invade the
intestinal mucosa, enter the blood stream and pass on to the different parts of the body causing a
number of outcomes, the most common of which is liver abcess..
The set Z15ZSolutionThe set Z15Z.pdf
The document discusses the set Z/15Z. This set contains the equivalence classes of the integers modulo 15. It represents the integers from 0 to 14 where addition and multiplication are performed modulo 15. This set has 15 elements.
Superior colliculusoptic tectum is a midbrain structure. The term s.pdf
Superior colliculus/optic tectum is a midbrain structure. The term superior colliculus is used
when discussing mammals and optic tectum when discussing either specific non-mammalian
species or vertebrates in general. It is a layered structure, with a number of layers that varies by
species. The layers can be grouped into the superficial layers and the deeper layers. Neurons in
the superficial layers receive direct input from the retina and respond almost exclusively to visual
stimuli. The deeper layers contain a population of motor-related neurons, capable of activating
eye movements as well as other responses. The superior colliculus mainly has its role in directing
eye movements. Visual input from the retina, or an input from the cerebral cortex, creates a
\"bump\" of activity in the tectal map, which, if strong enough, induces a saccadic eye movement
(A quick and simultaneous movement of both eyes). In primates however, the superior colliculus
is also involved in generating spatially directed head turns, arm-reaching movements, and shifts
in attention that do not involve any overt movements. In other species, the tectum is involved in
a wide range of responses, including whole-body turns etc.
Solution
Superior colliculus/optic tectum is a midbrain structure. The term superior colliculus is used
when discussing mammals and optic tectum when discussing either specific non-mammalian
species or vertebrates in general. It is a layered structure, with a number of layers that varies by
species. The layers can be grouped into the superficial layers and the deeper layers. Neurons in
the superficial layers receive direct input from the retina and respond almost exclusively to visual
stimuli. The deeper layers contain a population of motor-related neurons, capable of activating
eye movements as well as other responses. The superior colliculus mainly has its role in directing
eye movements. Visual input from the retina, or an input from the cerebral cortex, creates a
\"bump\" of activity in the tectal map, which, if strong enough, induces a saccadic eye movement
(A quick and simultaneous movement of both eyes). In primates however, the superior colliculus
is also involved in generating spatially directed head turns, arm-reaching movements, and shifts
in attention that do not involve any overt movements. In other species, the tectum is involved in
a wide range of responses, including whole-body turns etc..
ROE = 14.3 Payout Ratio = 42Sustainable Growth Rate = ROE (1.pdf
ROE = 14.3%
Payout Ratio = 42%
Sustainable Growth Rate = ROE * (1 - Payout Ratio)
Sustainable Growth Rate = 14.30% * (1 - 0.42)
Sustainable Growth Rate = 14.30% * 0.58
Sustainable Growth Rate = 8.29%
Solution
ROE = 14.3%
Payout Ratio = 42%
Sustainable Growth Rate = ROE * (1 - Payout Ratio)
Sustainable Growth Rate = 14.30% * (1 - 0.42)
Sustainable Growth Rate = 14.30% * 0.58
Sustainable Growth Rate = 8.29%.
Sample code for Discrete Cosine Transform for an imageSolution.pdf
Sample code for Discrete Cosine Transform for an image:
Solution
Sample code for Discrete Cosine Transform for an image:.
Present PbNot present HgSolutionPresent PbNot present .pdf
The document reports on the results of testing a solution for the presence of lead and mercury. It states that lead was present in the solution but mercury was not detected in the solution. The conclusion is repeated that the solution contained lead but did not contain mercury.
dear pls put prob 14 and 20... i cant solve witho.pdf
dear pls put prob 14 and 20... i cant solve without tht data
Solution
dear pls put prob 14 and 20... i cant solve without tht data.
D) is the answer. note BF3 is a strong Lewis aci.pdf
D) is the answer. note: BF3 is a strong Lewis acid (electron pair acceptor) that
reacts with a Lewis base (electron pair donor). CH3NH2 has a lone pair of electrons at N atom
and thus acts as a Lewis base to react with BF3.
Solution
D) is the answer. note: BF3 is a strong Lewis acid (electron pair acceptor) that
reacts with a Lewis base (electron pair donor). CH3NH2 has a lone pair of electrons at N atom
and thus acts as a Lewis base to react with BF3..
One major similarity with soils dominated by calcium and sodium ions.pdf
One major similarity with soils dominated by calcium and sodium ions are that they will be
saline soils without much moveable waer within them and plants would have to apply more
energy to exude salts and take up pure water.
The main difference is sodium ions will disperse into separate particles rather than sticking
together. In terms of affecting water flow and hydrology in soils, the most important impact of
high Na+ is that aggregates will be weaker or nonexistent if they are held together by clay.
Without aggregation pore structure of soil is highly affected and it is seen that water drainage
doesnt take place as fast as that in calcium rich soils. Calcium ions have a tendency to
‘flocculate’ (clump together) soil colloids (fine clay and organic matter particles), thus,
increasing aggregation and macroporosity.In turn, soil porosity, structural stability and water
movement may actually be improved in saline soils.
Solution
One major similarity with soils dominated by calcium and sodium ions are that they will be
saline soils without much moveable waer within them and plants would have to apply more
energy to exude salts and take up pure water.
The main difference is sodium ions will disperse into separate particles rather than sticking
together. In terms of affecting water flow and hydrology in soils, the most important impact of
high Na+ is that aggregates will be weaker or nonexistent if they are held together by clay.
Without aggregation pore structure of soil is highly affected and it is seen that water drainage
doesnt take place as fast as that in calcium rich soils. Calcium ions have a tendency to
‘flocculate’ (clump together) soil colloids (fine clay and organic matter particles), thus,
increasing aggregation and macroporosity.In turn, soil porosity, structural stability and water
movement may actually be improved in saline soils..
Interval.SolutionInterval..pdf
This very short document does not provide enough context or information to generate a meaningful 3 sentence summary. It contains only the word "Interval" repeated twice without any other details.
C) II and IV .pdf
This document provides the solution to a multi-part question, indicating that options II and IV are correct. No other context or details are given about the problem or answer choices.
Hi!Since Lithium has 1 single valence electron (it is located in G.pdf
Hi!
Since Lithium has 1 single valence electron (it is located in Group/Column 1 on the periodic
table) and Fluorine is lacking 1 single valence electron (it is located in Group 7A), the most
likely ionic compound to form here would be: LiF, Lithium Fluoride
I hope this helps! Please rate if it does :)
Solution
Hi!
Since Lithium has 1 single valence electron (it is located in Group/Column 1 on the periodic
table) and Fluorine is lacking 1 single valence electron (it is located in Group 7A), the most
likely ionic compound to form here would be: LiF, Lithium Fluoride
I hope this helps! Please rate if it does :).
I think that 0.434SolutionI think that 0.434.pdf
This very short document does not contain enough contextual information to generate a meaningful 3 sentence summary. The document consists of two identical sentences that state "I think that 0.434" but does not provide any other details about the topic, perspective or intended meaning.
Determine decimal fraction of C-14 remaining 43.3 58.2 = 0.74.pdf
Determine decimal fraction of C-14 remaining:
43.3 / 58.2 = 0.744
Determine how many half-lives have elapsed
(1/2)n = 0.744
n log (1/2) = log (0.744)
n = 0.42
Determine length of time elapsed:
5715 yr X 0.42 = 2438.16
Solution
Determine decimal fraction of C-14 remaining:
43.3 / 58.2 = 0.744
Determine how many half-lives have elapsed
(1/2)n = 0.744
n log (1/2) = log (0.744)
n = 0.42
Determine length of time elapsed:
5715 yr X 0.42 = 2438.16.
c) All colonies sub-cultured from blood culture broth media must be .pdf
c) All colonies sub-cultured from blood culture broth media must be identified. Multiple bottles
improves the likelihood that diphtheroids and coagulase negative Staphylococcus will be treated.
During sepsis, multiple organsims may be present in the infection. Time separation and culture
of blood in different environement allows different types of bacteria to be easily separated and
identified.
Solution
c) All colonies sub-cultured from blood culture broth media must be identified. Multiple bottles
improves the likelihood that diphtheroids and coagulase negative Staphylococcus will be treated.
During sepsis, multiple organsims may be present in the infection. Time separation and culture
of blood in different environement allows different types of bacteria to be easily separated and
identified..
AnswerFour phenotypes are possible in a flock of ducks that contain.pdf
Answer:Four phenotypes are possible in a flock of ducks that contains all the possible
combinations of these three alleles.As in the ABO blood system 4 phenotypes are possible in this
case.GenotypePhenotypeHH,Hi(H)II,Ii(I)HI(HI)ii(i)
Solution
Answer:Four phenotypes are possible in a flock of ducks that contains all the possible
combinations of these three alleles.As in the ABO blood system 4 phenotypes are possible in this
case.GenotypePhenotypeHH,Hi(H)II,Ii(I)HI(HI)ii(i).
Answer1. b. AgrobacteriumThe ability of Agrobacterium to transf.pdf
Answer:
1. b. Agrobacterium
The ability of Agrobacterium to transfer genes to plants and fungi is used in biotechnology, in
particular, genetic engineering for plant improvement. A modified Ti or Ri plasmid can be used.
2. c. Artificial active
Artificially acquired active immunity can be induced by a vaccine, a substance that contains the
antigen.
3. c. Attenuated
An attenuated vaccine is a vaccine created by reducing the virulence of a pathogen, but still
keeping it viable (or \"live\"). Attenuation takes an infectious agent and alters it so that it
becomes harmless or less virulent.
4. a. Sign
Symptom is nothing but the abnormal phenotype expressed in living organisms. Symdrome
means the all abnormalities associated with the disease.
5. b. The eye
Normal flora can be found in many sites of the human bodyincluding the skin (especially the
moist areas, such as the groin and between the toes), respiratory tract (particularly the nose),
urinary tract, and the digestive tract (primarily the mouth and the colon).
Solution
Answer:
1. b. Agrobacterium
The ability of Agrobacterium to transfer genes to plants and fungi is used in biotechnology, in
particular, genetic engineering for plant improvement. A modified Ti or Ri plasmid can be used.
2. c. Artificial active
Artificially acquired active immunity can be induced by a vaccine, a substance that contains the
antigen.
3. c. Attenuated
An attenuated vaccine is a vaccine created by reducing the virulence of a pathogen, but still
keeping it viable (or \"live\"). Attenuation takes an infectious agent and alters it so that it
becomes harmless or less virulent.
4. a. Sign
Symptom is nothing but the abnormal phenotype expressed in living organisms. Symdrome
means the all abnormalities associated with the disease.
5. b. The eye
Normal flora can be found in many sites of the human bodyincluding the skin (especially the
moist areas, such as the groin and between the toes), respiratory tract (particularly the nose),
urinary tract, and the digestive tract (primarily the mouth and the colon)..
H2SO4 is stronger than HSO4- since HSO4 already has one negative cha.pdf
H2SO4 is stronger than HSO4- since HSO4 already has one negative charge - gaining a second
is unfavorable compared to H2SO4 which would gain only a -1 charge after deprotonation.
HNO3 is stronger than HNO2 since the greater number of O atoms exerts a stronger electron-
withdrawing effect on the O-H electrons. Additionally, the NO3- ion has more resonance
structures and thus greater resonance stability than the NO2- ion. H2S is a stronger acid than
H2O since hydrogen forms weaker bonds to elements lower in the periodic table (larger
elements). This is why HI is stronger than HBr, for example. H2S is a stronger acid than PH3
since S is more electronnegative (further to the right in the periodic table) than P. Thus, S prefers
to have a negative charge compared to P. For part 2: We know that Kw = [H3O*][OH-] = 10^-
14 M. In pure water, these ions are at equal concentrations, so [H+] = [OH-] = 10^-7 M. Now,
we have 1.00 mL = 0.001 L. So, we have 10-7 mol/L * 0.001 L = 10^-10 moles of each ion.
Now, 1 mol = 6.02 * 10^23 ions, so we have: 6.02*10^23 * 10^-10 = 6.02 * 10^13 [H3O*] ions
and 6.02 * 10^13 [OH-] ions.
Solution
H2SO4 is stronger than HSO4- since HSO4 already has one negative charge - gaining a second
is unfavorable compared to H2SO4 which would gain only a -1 charge after deprotonation.
HNO3 is stronger than HNO2 since the greater number of O atoms exerts a stronger electron-
withdrawing effect on the O-H electrons. Additionally, the NO3- ion has more resonance
structures and thus greater resonance stability than the NO2- ion. H2S is a stronger acid than
H2O since hydrogen forms weaker bonds to elements lower in the periodic table (larger
elements). This is why HI is stronger than HBr, for example. H2S is a stronger acid than PH3
since S is more electronnegative (further to the right in the periodic table) than P. Thus, S prefers
to have a negative charge compared to P. For part 2: We know that Kw = [H3O*][OH-] = 10^-
14 M. In pure water, these ions are at equal concentrations, so [H+] = [OH-] = 10^-7 M. Now,
we have 1.00 mL = 0.001 L. So, we have 10-7 mol/L * 0.001 L = 10^-10 moles of each ion.
Now, 1 mol = 6.02 * 10^23 ions, so we have: 6.02*10^23 * 10^-10 = 6.02 * 10^13 [H3O*] ions
and 6.02 * 10^13 [OH-] ions..
ANS#includestdio.h#includeconio.h int main() { int.pdf
ANS:
#include
#include
int main()
{
int x,y=31;
printf(\"input %d\",l_bit);
binar_ar(y);
for (int i = 7; i >= 0; i--)
printf(\"%d/t%d\",get_bit(y,i),l_bit);
x=1;
printf(\"%d th bit %d\",x,l_bit);
y = set_bit(y, x,true);
for (int i =15; i >= 0; i--)
printf(\"%d/t%d\",get_bit(y,i),l_bit);
x = 1;
printf(\"%d th bit %d\",x,l_bit);
y = set_bit(y, x, false);
for (int i=31; i >= 0; i--)
printf(\"%d/t%d\",get_bit(y,i),l_bit);
}
void binar_ar(int n)
{
for(int i = 31; i > 0; i = i/2) {
if(n & i)
printf(\"1\");
else
printf(\"0\");
}
printf(\"%d\",l_bit);
}
bool get_bit(int n, int x)
{
return ( (n & (1 << x) ) > 0);
}
int set_bit(int n, int x, bool b)
{
if(b)
return (n | (1 << x)) ;
else {
int ma = ~(1 << x);
return n & ma;
}
}
Solution
ANS:
#include
#include
int main()
{
int x,y=31;
printf(\"input %d\",l_bit);
binar_ar(y);
for (int i = 7; i >= 0; i--)
printf(\"%d/t%d\",get_bit(y,i),l_bit);
x=1;
printf(\"%d th bit %d\",x,l_bit);
y = set_bit(y, x,true);
for (int i =15; i >= 0; i--)
printf(\"%d/t%d\",get_bit(y,i),l_bit);
x = 1;
printf(\"%d th bit %d\",x,l_bit);
y = set_bit(y, x, false);
for (int i=31; i >= 0; i--)
printf(\"%d/t%d\",get_bit(y,i),l_bit);
}
void binar_ar(int n)
{
for(int i = 31; i > 0; i = i/2) {
if(n & i)
printf(\"1\");
else
printf(\"0\");
}
printf(\"%d\",l_bit);
}
bool get_bit(int n, int x)
{
return ( (n & (1 << x) ) > 0);
}
int set_bit(int n, int x, bool b)
{
if(b)
return (n | (1 << x)) ;
else {
int ma = ~(1 << x);
return n & ma;
}
}.
a. An interval or ratio scale.Solutiona. An interval or ratio .pdf
This 1 sentence document discusses that the scale being referred to is either an interval or ratio scale.
A) Germ cell specification in Drosophila occurs by maternally contri.pdf
A) Germ cell specification in Drosophila occurs by maternally contributed determinants. In
Drosophilla, germ cells can easily be identified very early in embryogenesis, when their
differentiation as germ cells is assured by the localisation of maternally inherited determinants
before, or immediately following, fertilisation.
During oogenesis in Drosophila melanogaster, RNAs and proteins are synthesised by the nurse
cells. These products are transported through cytoplasmic bridges to the oocyte where They
become localised to the posterior of the ooplasm both by molecular anchoring at the posterior of
the oocyte, and by posterior-specific translational and transcriptional regulation. This posterior
ooplasm is the germ plasm, or germ line determinant. During early embryogenesis, cells which
inherit the germ plasm become the primordial germ cells.
One of the important gene that contributes to germ cell specification in Drosophila is Nanos.
b) Neural crest cell specification is due to the cell signalling and depends on BMP signalling in
the prospective epidermis and Wnt signalling from the underlying mesoderm.
C) Anterior -posterior specification in Drosophila is initiated by maternal effect genes that
produce messenger RNA that are placed in different regions of egg . These mRNA encode
transcriptional and translational regulatory protiens that diffuse through the syncytial blastoderm
and activates or repress the expression of certain zygotic genes.
One important maternal effect gene in this respect is bicoid gene.
Solution
A) Germ cell specification in Drosophila occurs by maternally contributed determinants. In
Drosophilla, germ cells can easily be identified very early in embryogenesis, when their
differentiation as germ cells is assured by the localisation of maternally inherited determinants
before, or immediately following, fertilisation.
During oogenesis in Drosophila melanogaster, RNAs and proteins are synthesised by the nurse
cells. These products are transported through cytoplasmic bridges to the oocyte where They
become localised to the posterior of the ooplasm both by molecular anchoring at the posterior of
the oocyte, and by posterior-specific translational and transcriptional regulation. This posterior
ooplasm is the germ plasm, or germ line determinant. During early embryogenesis, cells which
inherit the germ plasm become the primordial germ cells.
One of the important gene that contributes to germ cell specification in Drosophila is Nanos.
b) Neural crest cell specification is due to the cell signalling and depends on BMP signalling in
the prospective epidermis and Wnt signalling from the underlying mesoderm.
C) Anterior -posterior specification in Drosophila is initiated by maternal effect genes that
produce messenger RNA that are placed in different regions of egg . These mRNA encode
transcriptional and translational regulatory protiens that diffuse through the syncytial blastoderm
and activates or repress the expression of certa.
97.5 mL of water is added to 125 mL of 3.00 M HCl.pdf
97.5 mL of water is added to 125 mL of 3.00 M HCl solution Molartiy = (3.00
M)*(0.125 L)/(0.125 L + 0.097 L) = 1.69 M
Solution
97.5 mL of water is added to 125 mL of 3.00 M HCl solution Molartiy = (3.00
M)*(0.125 L)/(0.125 L + 0.097 L) = 1.69 M.
Wound healing PPT
This document provides an overview of wound healing, its functions, stages, mechanisms, factors affecting it, and complications.
A wound is a break in the integrity of the skin or tissues, which may be associated with disruption of the structure and function.
Healing is the body’s response to injury in an attempt to restore normal structure and functions.
Healing can occur in two ways: Regeneration and Repair
There are 4 phases of wound healing: hemostasis, inflammation, proliferation, and remodeling. This document also describes the mechanism of wound healing. Factors that affect healing include infection, uncontrolled diabetes, poor nutrition, age, anemia, the presence of foreign bodies, etc.
Complications of wound healing like infection, hyperpigmentation of scar, contractures, and keloid formation.
How to Build a Module in Odoo 17 Using the Scaffold Method
Odoo provides an option for creating a module by using a single line command. By using this command the user can make a whole structure of a module. It is very easy for a beginner to make a module. There is no need to make each file manually. This slide will show how to create a module using the scaffold method.
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ROE = 14.3 Payout Ratio = 42Sustainable Growth Rate = ROE (1.pdf
ROE = 14.3%
Payout Ratio = 42%
Sustainable Growth Rate = ROE * (1 - Payout Ratio)
Sustainable Growth Rate = 14.30% * (1 - 0.42)
Sustainable Growth Rate = 14.30% * 0.58
Sustainable Growth Rate = 8.29%
Solution
ROE = 14.3%
Payout Ratio = 42%
Sustainable Growth Rate = ROE * (1 - Payout Ratio)
Sustainable Growth Rate = 14.30% * (1 - 0.42)
Sustainable Growth Rate = 14.30% * 0.58
Sustainable Growth Rate = 8.29%.
Sample code for Discrete Cosine Transform for an imageSolution.pdf
Sample code for Discrete Cosine Transform for an image:
Solution
Sample code for Discrete Cosine Transform for an image:.
Present PbNot present HgSolutionPresent PbNot present .pdf
The document reports on the results of testing a solution for the presence of lead and mercury. It states that lead was present in the solution but mercury was not detected in the solution. The conclusion is repeated that the solution contained lead but did not contain mercury.
dear pls put prob 14 and 20... i cant solve witho.pdf
dear pls put prob 14 and 20... i cant solve without tht data
Solution
dear pls put prob 14 and 20... i cant solve without tht data.
D) is the answer. note BF3 is a strong Lewis aci.pdf
D) is the answer. note: BF3 is a strong Lewis acid (electron pair acceptor) that
reacts with a Lewis base (electron pair donor). CH3NH2 has a lone pair of electrons at N atom
and thus acts as a Lewis base to react with BF3.
Solution
D) is the answer. note: BF3 is a strong Lewis acid (electron pair acceptor) that
reacts with a Lewis base (electron pair donor). CH3NH2 has a lone pair of electrons at N atom
and thus acts as a Lewis base to react with BF3..
One major similarity with soils dominated by calcium and sodium ions.pdf
One major similarity with soils dominated by calcium and sodium ions are that they will be
saline soils without much moveable waer within them and plants would have to apply more
energy to exude salts and take up pure water.
The main difference is sodium ions will disperse into separate particles rather than sticking
together. In terms of affecting water flow and hydrology in soils, the most important impact of
high Na+ is that aggregates will be weaker or nonexistent if they are held together by clay.
Without aggregation pore structure of soil is highly affected and it is seen that water drainage
doesnt take place as fast as that in calcium rich soils. Calcium ions have a tendency to
‘flocculate’ (clump together) soil colloids (fine clay and organic matter particles), thus,
increasing aggregation and macroporosity.In turn, soil porosity, structural stability and water
movement may actually be improved in saline soils.
Solution
One major similarity with soils dominated by calcium and sodium ions are that they will be
saline soils without much moveable waer within them and plants would have to apply more
energy to exude salts and take up pure water.
The main difference is sodium ions will disperse into separate particles rather than sticking
together. In terms of affecting water flow and hydrology in soils, the most important impact of
high Na+ is that aggregates will be weaker or nonexistent if they are held together by clay.
Without aggregation pore structure of soil is highly affected and it is seen that water drainage
doesnt take place as fast as that in calcium rich soils. Calcium ions have a tendency to
‘flocculate’ (clump together) soil colloids (fine clay and organic matter particles), thus,
increasing aggregation and macroporosity.In turn, soil porosity, structural stability and water
movement may actually be improved in saline soils..
Interval.SolutionInterval..pdf
This very short document does not provide enough context or information to generate a meaningful 3 sentence summary. It contains only the word "Interval" repeated twice without any other details.
C) II and IV .pdf
This document provides the solution to a multi-part question, indicating that options II and IV are correct. No other context or details are given about the problem or answer choices.
Hi!Since Lithium has 1 single valence electron (it is located in G.pdf
Hi!
Since Lithium has 1 single valence electron (it is located in Group/Column 1 on the periodic
table) and Fluorine is lacking 1 single valence electron (it is located in Group 7A), the most
likely ionic compound to form here would be: LiF, Lithium Fluoride
I hope this helps! Please rate if it does :)
Solution
Hi!
Since Lithium has 1 single valence electron (it is located in Group/Column 1 on the periodic
table) and Fluorine is lacking 1 single valence electron (it is located in Group 7A), the most
likely ionic compound to form here would be: LiF, Lithium Fluoride
I hope this helps! Please rate if it does :).
I think that 0.434SolutionI think that 0.434.pdf
This very short document does not contain enough contextual information to generate a meaningful 3 sentence summary. The document consists of two identical sentences that state "I think that 0.434" but does not provide any other details about the topic, perspective or intended meaning.
Determine decimal fraction of C-14 remaining 43.3 58.2 = 0.74.pdf
Determine decimal fraction of C-14 remaining:
43.3 / 58.2 = 0.744
Determine how many half-lives have elapsed
(1/2)n = 0.744
n log (1/2) = log (0.744)
n = 0.42
Determine length of time elapsed:
5715 yr X 0.42 = 2438.16
Solution
Determine decimal fraction of C-14 remaining:
43.3 / 58.2 = 0.744
Determine how many half-lives have elapsed
(1/2)n = 0.744
n log (1/2) = log (0.744)
n = 0.42
Determine length of time elapsed:
5715 yr X 0.42 = 2438.16.
c) All colonies sub-cultured from blood culture broth media must be .pdf
c) All colonies sub-cultured from blood culture broth media must be identified. Multiple bottles
improves the likelihood that diphtheroids and coagulase negative Staphylococcus will be treated.
During sepsis, multiple organsims may be present in the infection. Time separation and culture
of blood in different environement allows different types of bacteria to be easily separated and
identified.
Solution
c) All colonies sub-cultured from blood culture broth media must be identified. Multiple bottles
improves the likelihood that diphtheroids and coagulase negative Staphylococcus will be treated.
During sepsis, multiple organsims may be present in the infection. Time separation and culture
of blood in different environement allows different types of bacteria to be easily separated and
identified..
AnswerFour phenotypes are possible in a flock of ducks that contain.pdf
Answer:Four phenotypes are possible in a flock of ducks that contains all the possible
combinations of these three alleles.As in the ABO blood system 4 phenotypes are possible in this
case.GenotypePhenotypeHH,Hi(H)II,Ii(I)HI(HI)ii(i)
Solution
Answer:Four phenotypes are possible in a flock of ducks that contains all the possible
combinations of these three alleles.As in the ABO blood system 4 phenotypes are possible in this
case.GenotypePhenotypeHH,Hi(H)II,Ii(I)HI(HI)ii(i).
Answer1. b. AgrobacteriumThe ability of Agrobacterium to transf.pdf
Answer:
1. b. Agrobacterium
The ability of Agrobacterium to transfer genes to plants and fungi is used in biotechnology, in
particular, genetic engineering for plant improvement. A modified Ti or Ri plasmid can be used.
2. c. Artificial active
Artificially acquired active immunity can be induced by a vaccine, a substance that contains the
antigen.
3. c. Attenuated
An attenuated vaccine is a vaccine created by reducing the virulence of a pathogen, but still
keeping it viable (or \"live\"). Attenuation takes an infectious agent and alters it so that it
becomes harmless or less virulent.
4. a. Sign
Symptom is nothing but the abnormal phenotype expressed in living organisms. Symdrome
means the all abnormalities associated with the disease.
5. b. The eye
Normal flora can be found in many sites of the human bodyincluding the skin (especially the
moist areas, such as the groin and between the toes), respiratory tract (particularly the nose),
urinary tract, and the digestive tract (primarily the mouth and the colon).
Solution
Answer:
1. b. Agrobacterium
The ability of Agrobacterium to transfer genes to plants and fungi is used in biotechnology, in
particular, genetic engineering for plant improvement. A modified Ti or Ri plasmid can be used.
2. c. Artificial active
Artificially acquired active immunity can be induced by a vaccine, a substance that contains the
antigen.
3. c. Attenuated
An attenuated vaccine is a vaccine created by reducing the virulence of a pathogen, but still
keeping it viable (or \"live\"). Attenuation takes an infectious agent and alters it so that it
becomes harmless or less virulent.
4. a. Sign
Symptom is nothing but the abnormal phenotype expressed in living organisms. Symdrome
means the all abnormalities associated with the disease.
5. b. The eye
Normal flora can be found in many sites of the human bodyincluding the skin (especially the
moist areas, such as the groin and between the toes), respiratory tract (particularly the nose),
urinary tract, and the digestive tract (primarily the mouth and the colon)..
H2SO4 is stronger than HSO4- since HSO4 already has one negative cha.pdf
H2SO4 is stronger than HSO4- since HSO4 already has one negative charge - gaining a second
is unfavorable compared to H2SO4 which would gain only a -1 charge after deprotonation.
HNO3 is stronger than HNO2 since the greater number of O atoms exerts a stronger electron-
withdrawing effect on the O-H electrons. Additionally, the NO3- ion has more resonance
structures and thus greater resonance stability than the NO2- ion. H2S is a stronger acid than
H2O since hydrogen forms weaker bonds to elements lower in the periodic table (larger
elements). This is why HI is stronger than HBr, for example. H2S is a stronger acid than PH3
since S is more electronnegative (further to the right in the periodic table) than P. Thus, S prefers
to have a negative charge compared to P. For part 2: We know that Kw = [H3O*][OH-] = 10^-
14 M. In pure water, these ions are at equal concentrations, so [H+] = [OH-] = 10^-7 M. Now,
we have 1.00 mL = 0.001 L. So, we have 10-7 mol/L * 0.001 L = 10^-10 moles of each ion.
Now, 1 mol = 6.02 * 10^23 ions, so we have: 6.02*10^23 * 10^-10 = 6.02 * 10^13 [H3O*] ions
and 6.02 * 10^13 [OH-] ions.
Solution
H2SO4 is stronger than HSO4- since HSO4 already has one negative charge - gaining a second
is unfavorable compared to H2SO4 which would gain only a -1 charge after deprotonation.
HNO3 is stronger than HNO2 since the greater number of O atoms exerts a stronger electron-
withdrawing effect on the O-H electrons. Additionally, the NO3- ion has more resonance
structures and thus greater resonance stability than the NO2- ion. H2S is a stronger acid than
H2O since hydrogen forms weaker bonds to elements lower in the periodic table (larger
elements). This is why HI is stronger than HBr, for example. H2S is a stronger acid than PH3
since S is more electronnegative (further to the right in the periodic table) than P. Thus, S prefers
to have a negative charge compared to P. For part 2: We know that Kw = [H3O*][OH-] = 10^-
14 M. In pure water, these ions are at equal concentrations, so [H+] = [OH-] = 10^-7 M. Now,
we have 1.00 mL = 0.001 L. So, we have 10-7 mol/L * 0.001 L = 10^-10 moles of each ion.
Now, 1 mol = 6.02 * 10^23 ions, so we have: 6.02*10^23 * 10^-10 = 6.02 * 10^13 [H3O*] ions
and 6.02 * 10^13 [OH-] ions..
ANS#includestdio.h#includeconio.h int main() { int.pdf
ANS:
#include
#include
int main()
{
int x,y=31;
printf(\"input %d\",l_bit);
binar_ar(y);
for (int i = 7; i >= 0; i--)
printf(\"%d/t%d\",get_bit(y,i),l_bit);
x=1;
printf(\"%d th bit %d\",x,l_bit);
y = set_bit(y, x,true);
for (int i =15; i >= 0; i--)
printf(\"%d/t%d\",get_bit(y,i),l_bit);
x = 1;
printf(\"%d th bit %d\",x,l_bit);
y = set_bit(y, x, false);
for (int i=31; i >= 0; i--)
printf(\"%d/t%d\",get_bit(y,i),l_bit);
}
void binar_ar(int n)
{
for(int i = 31; i > 0; i = i/2) {
if(n & i)
printf(\"1\");
else
printf(\"0\");
}
printf(\"%d\",l_bit);
}
bool get_bit(int n, int x)
{
return ( (n & (1 << x) ) > 0);
}
int set_bit(int n, int x, bool b)
{
if(b)
return (n | (1 << x)) ;
else {
int ma = ~(1 << x);
return n & ma;
}
}
Solution
ANS:
#include
#include
int main()
{
int x,y=31;
printf(\"input %d\",l_bit);
binar_ar(y);
for (int i = 7; i >= 0; i--)
printf(\"%d/t%d\",get_bit(y,i),l_bit);
x=1;
printf(\"%d th bit %d\",x,l_bit);
y = set_bit(y, x,true);
for (int i =15; i >= 0; i--)
printf(\"%d/t%d\",get_bit(y,i),l_bit);
x = 1;
printf(\"%d th bit %d\",x,l_bit);
y = set_bit(y, x, false);
for (int i=31; i >= 0; i--)
printf(\"%d/t%d\",get_bit(y,i),l_bit);
}
void binar_ar(int n)
{
for(int i = 31; i > 0; i = i/2) {
if(n & i)
printf(\"1\");
else
printf(\"0\");
}
printf(\"%d\",l_bit);
}
bool get_bit(int n, int x)
{
return ( (n & (1 << x) ) > 0);
}
int set_bit(int n, int x, bool b)
{
if(b)
return (n | (1 << x)) ;
else {
int ma = ~(1 << x);
return n & ma;
}
}.
a. An interval or ratio scale.Solutiona. An interval or ratio .pdf
This 1 sentence document discusses that the scale being referred to is either an interval or ratio scale.
A) Germ cell specification in Drosophila occurs by maternally contri.pdf
A) Germ cell specification in Drosophila occurs by maternally contributed determinants. In
Drosophilla, germ cells can easily be identified very early in embryogenesis, when their
differentiation as germ cells is assured by the localisation of maternally inherited determinants
before, or immediately following, fertilisation.
During oogenesis in Drosophila melanogaster, RNAs and proteins are synthesised by the nurse
cells. These products are transported through cytoplasmic bridges to the oocyte where They
become localised to the posterior of the ooplasm both by molecular anchoring at the posterior of
the oocyte, and by posterior-specific translational and transcriptional regulation. This posterior
ooplasm is the germ plasm, or germ line determinant. During early embryogenesis, cells which
inherit the germ plasm become the primordial germ cells.
One of the important gene that contributes to germ cell specification in Drosophila is Nanos.
b) Neural crest cell specification is due to the cell signalling and depends on BMP signalling in
the prospective epidermis and Wnt signalling from the underlying mesoderm.
C) Anterior -posterior specification in Drosophila is initiated by maternal effect genes that
produce messenger RNA that are placed in different regions of egg . These mRNA encode
transcriptional and translational regulatory protiens that diffuse through the syncytial blastoderm
and activates or repress the expression of certain zygotic genes.
One important maternal effect gene in this respect is bicoid gene.
Solution
A) Germ cell specification in Drosophila occurs by maternally contributed determinants. In
Drosophilla, germ cells can easily be identified very early in embryogenesis, when their
differentiation as germ cells is assured by the localisation of maternally inherited determinants
before, or immediately following, fertilisation.
During oogenesis in Drosophila melanogaster, RNAs and proteins are synthesised by the nurse
cells. These products are transported through cytoplasmic bridges to the oocyte where They
become localised to the posterior of the ooplasm both by molecular anchoring at the posterior of
the oocyte, and by posterior-specific translational and transcriptional regulation. This posterior
ooplasm is the germ plasm, or germ line determinant. During early embryogenesis, cells which
inherit the germ plasm become the primordial germ cells.
One of the important gene that contributes to germ cell specification in Drosophila is Nanos.
b) Neural crest cell specification is due to the cell signalling and depends on BMP signalling in
the prospective epidermis and Wnt signalling from the underlying mesoderm.
C) Anterior -posterior specification in Drosophila is initiated by maternal effect genes that
produce messenger RNA that are placed in different regions of egg . These mRNA encode
transcriptional and translational regulatory protiens that diffuse through the syncytial blastoderm
and activates or repress the expression of certa.
97.5 mL of water is added to 125 mL of 3.00 M HCl.pdf
97.5 mL of water is added to 125 mL of 3.00 M HCl solution Molartiy = (3.00
M)*(0.125 L)/(0.125 L + 0.097 L) = 1.69 M
Solution
97.5 mL of water is added to 125 mL of 3.00 M HCl solution Molartiy = (3.00
M)*(0.125 L)/(0.125 L + 0.097 L) = 1.69 M.
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