Submit Search
Upload
isotopes .pdf
•
0 likes
•
3 views
A
angelcolluctions
Follow
isotopes Solution isotopes.
Read less
Read more
Education
Report
Share
Report
Share
1 of 1
Download now
Download to read offline
Recommended
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.
Step1 Moles of H2 = 35.02 =17.5 ;Moles of O2 = 2.pdf
Step1 Moles of H2 = 35.02 =17.5 ;Moles of O2 = 2.pdf
angelcolluctions
the sea of electrons shared between metal atoms Solution the sea of electrons shared between metal atoms.
the sea of electrons shared between metal atoms .pdf
the sea of electrons shared between metal atoms .pdf
angelcolluctions
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.
HOMO and LUMO are acronyms for highest occupied m.pdf
HOMO and LUMO are acronyms for highest occupied m.pdf
angelcolluctions
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 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% 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 .
YearCash flow0-1000000115000021500003150000415.pdf
YearCash flow0-1000000115000021500003150000415.pdf
angelcolluctions
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.
Toxic shock syndrome (TSS) is a toxin-mediated acute life-threatenin.pdf
Toxic shock syndrome (TSS) is a toxin-mediated acute life-threatenin.pdf
angelcolluctions
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 trophozites (active form) are present only in the host and fresh.pdf
The trophozites (active form) are present only in the host and fresh.pdf
angelcolluctions
The set Z/15Z Solution The set Z/15Z.
The set Z15ZSolutionThe set Z15Z.pdf
The set Z15ZSolutionThe set Z15Z.pdf
angelcolluctions
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..
Superior colliculusoptic tectum is a midbrain structure. The term s.pdf
Superior colliculusoptic tectum is a midbrain structure. The term s.pdf
angelcolluctions
Recommended
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.
Step1 Moles of H2 = 35.02 =17.5 ;Moles of O2 = 2.pdf
Step1 Moles of H2 = 35.02 =17.5 ;Moles of O2 = 2.pdf
angelcolluctions
the sea of electrons shared between metal atoms Solution the sea of electrons shared between metal atoms.
the sea of electrons shared between metal atoms .pdf
the sea of electrons shared between metal atoms .pdf
angelcolluctions
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.
HOMO and LUMO are acronyms for highest occupied m.pdf
HOMO and LUMO are acronyms for highest occupied m.pdf
angelcolluctions
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 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% 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 .
YearCash flow0-1000000115000021500003150000415.pdf
YearCash flow0-1000000115000021500003150000415.pdf
angelcolluctions
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.
Toxic shock syndrome (TSS) is a toxin-mediated acute life-threatenin.pdf
Toxic shock syndrome (TSS) is a toxin-mediated acute life-threatenin.pdf
angelcolluctions
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 trophozites (active form) are present only in the host and fresh.pdf
The trophozites (active form) are present only in the host and fresh.pdf
angelcolluctions
The set Z/15Z Solution The set Z/15Z.
The set Z15ZSolutionThe set Z15Z.pdf
The set Z15ZSolutionThe set Z15Z.pdf
angelcolluctions
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..
Superior colliculusoptic tectum is a midbrain structure. The term s.pdf
Superior colliculusoptic tectum is a midbrain structure. The term s.pdf
angelcolluctions
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%.
ROE = 14.3 Payout Ratio = 42Sustainable Growth Rate = ROE (1.pdf
ROE = 14.3 Payout Ratio = 42Sustainable Growth Rate = ROE (1.pdf
angelcolluctions
Sample code for Discrete Cosine Transform for an image: Solution Sample code for Discrete Cosine Transform for an image:.
Sample code for Discrete Cosine Transform for an imageSolution.pdf
Sample code for Discrete Cosine Transform for an imageSolution.pdf
angelcolluctions
Present: Pb Not present: Hg Solution Present: Pb Not present: Hg.
Present PbNot present HgSolutionPresent PbNot present .pdf
Present PbNot present HgSolutionPresent PbNot present .pdf
angelcolluctions
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.
dear pls put prob 14 and 20... i cant solve witho.pdf
dear pls put prob 14 and 20... i cant solve witho.pdf
angelcolluctions
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..
D) is the answer. note BF3 is a strong Lewis aci.pdf
D) is the answer. note BF3 is a strong Lewis aci.pdf
angelcolluctions
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..
One major similarity with soils dominated by calcium and sodium ions.pdf
One major similarity with soils dominated by calcium and sodium ions.pdf
angelcolluctions
Interval. Solution Interval..
Interval.SolutionInterval..pdf
Interval.SolutionInterval..pdf
angelcolluctions
C) II and IV Solution C) II and IV.
C) II and IV .pdf
C) II and IV .pdf
angelcolluctions
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 :).
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 G.pdf
angelcolluctions
I think that 0.434 Solution I think that 0.434.
I think that 0.434SolutionI think that 0.434.pdf
I think that 0.434SolutionI think that 0.434.pdf
angelcolluctions
B)diffusion a)effusion Solution B)diffusion a)effusion.
B)diffusion a)effusion .pdf
B)diffusion a)effusion .pdf
angelcolluctions
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.
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 = 0.74.pdf
angelcolluctions
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..
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 .pdf
angelcolluctions
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).
AnswerFour phenotypes are possible in a flock of ducks that contain.pdf
AnswerFour phenotypes are possible in a flock of ducks that contain.pdf
angelcolluctions
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)..
Answer1. b. AgrobacteriumThe ability of Agrobacterium to transf.pdf
Answer1. b. AgrobacteriumThe ability of Agrobacterium to transf.pdf
angelcolluctions
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..
H2SO4 is stronger than HSO4- since HSO4 already has one negative cha.pdf
H2SO4 is stronger than HSO4- since HSO4 already has one negative cha.pdf
angelcolluctions
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; } }.
ANS#includestdio.h#includeconio.h int main() { int.pdf
ANS#includestdio.h#includeconio.h int main() { int.pdf
angelcolluctions
a. An interval or ratio scale. Solution a. An interval or ratio scale..
a. An interval or ratio scale.Solutiona. An interval or ratio .pdf
a. An interval or ratio scale.Solutiona. An interval or ratio .pdf
angelcolluctions
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.
A) Germ cell specification in Drosophila occurs by maternally contri.pdf
A) Germ cell specification in Drosophila occurs by maternally contri.pdf
angelcolluctions
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.
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.pdf
angelcolluctions
Odoo 17 Studio App introduces enhancements to its approval flow, empowering businesses to manage approvals effectively and boost productivity. This slide dives into these improvements and highlighting their benefits.
Improved Approval Flow in Odoo 17 Studio App
Improved Approval Flow in Odoo 17 Studio App
Celine George
Photos from our Spring Gala 2024. Thank you for making our 2024 Spring Gala unforgettable! Whether you attended or supported us with a donation, we extend our heartfelt thanks for celebrating School-Community Partnerships alongside ExpandED Schools. Special appreciation to our honorees, Rachel Skaistis and Cravath, Swaine, & Moore LLP, for their unwavering support in advancing educational equity. Your contributions have profoundly impacted countless young New Yorkers, and we’re proud to have you as part of our community. Our gratitude also extends to our Board of Directors, sponsors, city partners, community-based organizations, and the inspiring young people who joined us. Your support fuels our mission, and we’re endlessly thankful for your dedication. While we celebrate our gala’s success, our work continues. As you view the evening’s highlights, consider how you can further empower young New Yorkers and educators with your donation, no matter the size. Donate today: https://www.expandedschools.org/support-our-work/donate/
Spring gala 2024 photo slideshow - Celebrating School-Community Partnerships
Spring gala 2024 photo slideshow - Celebrating School-Community Partnerships
expandedwebsite
More Related Content
More from angelcolluctions
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%.
ROE = 14.3 Payout Ratio = 42Sustainable Growth Rate = ROE (1.pdf
ROE = 14.3 Payout Ratio = 42Sustainable Growth Rate = ROE (1.pdf
angelcolluctions
Sample code for Discrete Cosine Transform for an image: Solution Sample code for Discrete Cosine Transform for an image:.
Sample code for Discrete Cosine Transform for an imageSolution.pdf
Sample code for Discrete Cosine Transform for an imageSolution.pdf
angelcolluctions
Present: Pb Not present: Hg Solution Present: Pb Not present: Hg.
Present PbNot present HgSolutionPresent PbNot present .pdf
Present PbNot present HgSolutionPresent PbNot present .pdf
angelcolluctions
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.
dear pls put prob 14 and 20... i cant solve witho.pdf
dear pls put prob 14 and 20... i cant solve witho.pdf
angelcolluctions
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..
D) is the answer. note BF3 is a strong Lewis aci.pdf
D) is the answer. note BF3 is a strong Lewis aci.pdf
angelcolluctions
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..
One major similarity with soils dominated by calcium and sodium ions.pdf
One major similarity with soils dominated by calcium and sodium ions.pdf
angelcolluctions
Interval. Solution Interval..
Interval.SolutionInterval..pdf
Interval.SolutionInterval..pdf
angelcolluctions
C) II and IV Solution C) II and IV.
C) II and IV .pdf
C) II and IV .pdf
angelcolluctions
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 :).
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 G.pdf
angelcolluctions
I think that 0.434 Solution I think that 0.434.
I think that 0.434SolutionI think that 0.434.pdf
I think that 0.434SolutionI think that 0.434.pdf
angelcolluctions
B)diffusion a)effusion Solution B)diffusion a)effusion.
B)diffusion a)effusion .pdf
B)diffusion a)effusion .pdf
angelcolluctions
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.
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 = 0.74.pdf
angelcolluctions
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..
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 .pdf
angelcolluctions
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).
AnswerFour phenotypes are possible in a flock of ducks that contain.pdf
AnswerFour phenotypes are possible in a flock of ducks that contain.pdf
angelcolluctions
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)..
Answer1. b. AgrobacteriumThe ability of Agrobacterium to transf.pdf
Answer1. b. AgrobacteriumThe ability of Agrobacterium to transf.pdf
angelcolluctions
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..
H2SO4 is stronger than HSO4- since HSO4 already has one negative cha.pdf
H2SO4 is stronger than HSO4- since HSO4 already has one negative cha.pdf
angelcolluctions
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; } }.
ANS#includestdio.h#includeconio.h int main() { int.pdf
ANS#includestdio.h#includeconio.h int main() { int.pdf
angelcolluctions
a. An interval or ratio scale. Solution a. An interval or ratio scale..
a. An interval or ratio scale.Solutiona. An interval or ratio .pdf
a. An interval or ratio scale.Solutiona. An interval or ratio .pdf
angelcolluctions
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.
A) Germ cell specification in Drosophila occurs by maternally contri.pdf
A) Germ cell specification in Drosophila occurs by maternally contri.pdf
angelcolluctions
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.
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.pdf
angelcolluctions
More from angelcolluctions
(20)
ROE = 14.3 Payout Ratio = 42Sustainable Growth Rate = ROE (1.pdf
ROE = 14.3 Payout Ratio = 42Sustainable Growth Rate = ROE (1.pdf
Sample code for Discrete Cosine Transform for an imageSolution.pdf
Sample code for Discrete Cosine Transform for an imageSolution.pdf
Present PbNot present HgSolutionPresent PbNot present .pdf
Present PbNot present HgSolutionPresent PbNot present .pdf
dear pls put prob 14 and 20... i cant solve witho.pdf
dear pls put prob 14 and 20... i cant solve witho.pdf
D) is the answer. note BF3 is a strong Lewis aci.pdf
D) is the answer. note BF3 is a strong Lewis aci.pdf
One major similarity with soils dominated by calcium and sodium ions.pdf
One major similarity with soils dominated by calcium and sodium ions.pdf
Interval.SolutionInterval..pdf
Interval.SolutionInterval..pdf
C) II and IV .pdf
C) II and IV .pdf
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 G.pdf
I think that 0.434SolutionI think that 0.434.pdf
I think that 0.434SolutionI think that 0.434.pdf
B)diffusion a)effusion .pdf
B)diffusion a)effusion .pdf
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 = 0.74.pdf
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 .pdf
AnswerFour phenotypes are possible in a flock of ducks that contain.pdf
AnswerFour phenotypes are possible in a flock of ducks that contain.pdf
Answer1. b. AgrobacteriumThe ability of Agrobacterium to transf.pdf
Answer1. b. AgrobacteriumThe ability of Agrobacterium to transf.pdf
H2SO4 is stronger than HSO4- since HSO4 already has one negative cha.pdf
H2SO4 is stronger than HSO4- since HSO4 already has one negative cha.pdf
ANS#includestdio.h#includeconio.h int main() { int.pdf
ANS#includestdio.h#includeconio.h int main() { int.pdf
a. An interval or ratio scale.Solutiona. An interval or ratio .pdf
a. An interval or ratio scale.Solutiona. An interval or ratio .pdf
A) Germ cell specification in Drosophila occurs by maternally contri.pdf
A) Germ cell specification in Drosophila occurs by maternally contri.pdf
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.pdf
Recently uploaded
Odoo 17 Studio App introduces enhancements to its approval flow, empowering businesses to manage approvals effectively and boost productivity. This slide dives into these improvements and highlighting their benefits.
Improved Approval Flow in Odoo 17 Studio App
Improved Approval Flow in Odoo 17 Studio App
Celine George
Photos from our Spring Gala 2024. Thank you for making our 2024 Spring Gala unforgettable! Whether you attended or supported us with a donation, we extend our heartfelt thanks for celebrating School-Community Partnerships alongside ExpandED Schools. Special appreciation to our honorees, Rachel Skaistis and Cravath, Swaine, & Moore LLP, for their unwavering support in advancing educational equity. Your contributions have profoundly impacted countless young New Yorkers, and we’re proud to have you as part of our community. Our gratitude also extends to our Board of Directors, sponsors, city partners, community-based organizations, and the inspiring young people who joined us. Your support fuels our mission, and we’re endlessly thankful for your dedication. While we celebrate our gala’s success, our work continues. As you view the evening’s highlights, consider how you can further empower young New Yorkers and educators with your donation, no matter the size. Donate today: https://www.expandedschools.org/support-our-work/donate/
Spring gala 2024 photo slideshow - Celebrating School-Community Partnerships
Spring gala 2024 photo slideshow - Celebrating School-Community Partnerships
expandedwebsite
multiple sclerosis (undergraduate)
demyelinated disorder: multiple sclerosis.pptx
demyelinated disorder: multiple sclerosis.pptx
Mohamed Rizk Khodair
This presentation was provided by William Mattingly of the Smithsonian Institution, during the seventh segment of the NISO training series "AI & Prompt Design." Session 7: Open Source Language Models, was held on May 16, 2024.
Mattingly "AI and Prompt Design: LLMs with Text Classification and Open Source"
Mattingly "AI and Prompt Design: LLMs with Text Classification and Open Source"
National Information Standards Organization (NISO)
Word Stress rules
Word Stress rules esl .pptx
Word Stress rules esl .pptx
Nicholas Montgomery
https://app.box.com/s/cbgl8f0rgcll2fzdqp83sjxx8nom8188
TỔNG HỢP HƠN 100 ĐỀ THI THỬ TỐT NGHIỆP THPT VẬT LÝ 2024 - TỪ CÁC TRƯỜNG, TRƯ...
TỔNG HỢP HƠN 100 ĐỀ THI THỬ TỐT NGHIỆP THPT VẬT LÝ 2024 - TỪ CÁC TRƯỜNG, TRƯ...
Nguyen Thanh Tu Collection
https://app.box.com/s/71kthbth9ww0fyjrppmh1p2gasinqj5z
ĐỀ THAM KHẢO KÌ THI TUYỂN SINH VÀO LỚP 10 MÔN TIẾNG ANH FORM 50 CÂU TRẮC NGHI...
ĐỀ THAM KHẢO KÌ THI TUYỂN SINH VÀO LỚP 10 MÔN TIẾNG ANH FORM 50 CÂU TRẮC NGHI...
Nguyen Thanh Tu Collection
Chapter 7 Pharmacosy Traditional System of Medicine & Ayurvedic Preparations (1).pdf notes
Chapter 7 Pharmacosy Traditional System of Medicine & Ayurvedic Preparations ...
Chapter 7 Pharmacosy Traditional System of Medicine & Ayurvedic Preparations ...
Sumit Tiwari
The Graduate Outcomes survey exists to improve the experience of future students.
Graduate Outcomes Presentation Slides - English (v3).pptx
Graduate Outcomes Presentation Slides - English (v3).pptx
neillewis46
Poster to be presented at Stochastic Numerics and Statistical Learning: Theory and Applications Workshop 2024, Kaust, Saudi Arabia, https://cemse.kaust.edu.sa/stochnum/events/event/snsl-workshop-2024. In this work we have considered a setting that mimics the Henry problem \cite{Simpson2003,Simpson04_Henry}, modeling seawater intrusion into a 2D coastal aquifer. The pure water recharge from the ``land side'' resists the salinisation of the aquifer due to the influx of saline water through the ``sea side'', thereby achieving some equilibrium in the salt concentration. In our setting, following \cite{GRILLO2010}, we consider a fracture on the sea side that significantly increases the permeability of the porous medium. The flow and transport essentially depend on the geological parameters of the porous medium, including the fracture. We investigated the effects of various uncertainties on saltwater intrusion. We assumed uncertainties in the fracture width, the porosity of the bulk medium, its permeability and the pure water recharge from the land side. The porosity and permeability were modeled by random fields, the recharge by a random but periodic intensity and the thickness by a random variable. We calculated the mean and variance of the salt mass fraction, which is also uncertain. The main question we investigated in this work was how well the MLMC method can be used to compute statistics of different QoIs. We found that the answer depends on the choice of the QoI. First, not every QoI requires a hierarchy of meshes and MLMC. Second, MLMC requires stable convergence rates for $\EXP{g_{\ell} - g_{\ell-1}}$ and $\Var{g_{\ell} - g_{\ell-1}}$. These rates should be independent of $\ell$. If these convergence rates vary for different $\ell$, then it will be hard to estimate $L$ and $m_{\ell}$, and MLMC will either not work or be suboptimal. We were not able to get stable convergence rates for all levels $\ell=1,\ldots,5$ when the QoI was an integral as in \eqref{eq:integral_box}. We found that for $\ell=1,\ldots 4$ and $\ell=5$ the rate $\alpha$ was different. Further investigation is needed to find the reason for this. Another difficulty is the dependence on time, i.e. the number of levels $L$ and the number of sums $m_{\ell}$ depend on $t$. At the beginning the variability is small, then it increases, and after the process of mixing salt and fresh water has stopped, the variance decreases again. The number of random samples required at each level was estimated by calculating the decay of the variances and the computational cost for each level. These estimates depend on the minimisation function in the MLMC algorithm. To achieve the efficiency of the MLMC approach presented in this work, it is essential that the complexity of the numerical solution of each random realisation is proportional to the number of grid vertices on the grid levels.
Poster_density_driven_with_fracture_MLMC.pdf
Poster_density_driven_with_fracture_MLMC.pdf
Alexander Litvinenko
STL algorithm SWAP MAX MAX_ELEMENT SORT COUNT COUNT_IF
Stl Algorithms in C++ jjjjjjjjjjjjjjjjjj
Stl Algorithms in C++ jjjjjjjjjjjjjjjjjj
Mohammed Sikander
Telehealth.org's slide deck for PSYPACT- Practicing Over State Lines LIVE event.
PSYPACT- Practicing Over State Lines May 2024.pptx
PSYPACT- Practicing Over State Lines May 2024.pptx
Marlene Maheu
Sales margin plays a crucial role in the corporate world as a compass that directs enterprises towards profitability. It is the discrepancy between a good or service's selling price and its production or acquisition costs. This margin shows a company's capacity to produce income in addition to how well it manages expenses.
How to Analyse Profit of a Sales Order in Odoo 17
How to Analyse Profit of a Sales Order in Odoo 17
Celine George
These are philosophical chalenges faced by person in his life
philosophy and it's principles based on the life
philosophy and it's principles based on the life
NitinDeodare
For the Closest Location removal strategy, products are picked based on the alphanumeric order of storage location titles. The goal of this strategy is to save the warehouse worker from taking a long journey to a farther shelf when the product is also available at a closer location.
How to Manage Closest Location in Odoo 17 Inventory
How to Manage Closest Location in Odoo 17 Inventory
Celine George
Mastering JNKVV Entrance Exam: Previous Year Question Paper Analysis for MSc Ag Genetics and Plant Breeding Welcome to our comprehensive guide to cracking the JNKVV Entrance Examination for MSc in Agricultural Genetics and Plant Breeding. we'll dive deep into the analysis of previous year question papers, providing you with valuable insights, tips, and strategies to excel in this competitive exam. The Jawaharlal Nehru Krishi Vishwa Vidyalaya (JNKVV) is one of India's premier agricultural universities, and gaining admission to its MSc program in Genetics and Plant Breeding is a significant achievement. This video aims to help you prepare effectively by analyzing past question papers, highlighting important topics, and suggesting the best study strategies. Understanding the Exam Pattern Before we delve into the questions, it's crucial to understand the structure of the JNKVV entrance exam. The exam typically comprises multiple-choice questions (MCQs) that test your knowledge across various domains of genetics and plant breeding. The key areas usually covered include: Fundamentals of Genetics: - Mendelian inheritance - Chromosome structure and function - Genetic variation and mutation Plant Breeding Techniques - Hybridization methods - Selection techniques - Breeding for disease resistance Molecular Genetics - DNA replication, transcription, and translation - Molecular markers in plant breeding - Genomics and genetic engineering 4. Quantitative Genetics: - Heritability and genetic advance - Statistical methods in plant breeding - QTL mapping #### Detailed Question Paper Analysis Now, let's analyze a previous year’s question paper to understand the types of questions asked and the best ways to approach them. Section 1: Fundamentals of Genetics Example Question: Which of the following is NOT a part of Mendel's Laws? - A) Law of Segregation - B) Law of Independent Assortment - C) Law of Dominance - D) Law of Uniformity Analysis: This question tests your basic understanding of Mendelian genetics. To tackle such questions, ensure you are well-versed with the core principles of genetics. Revising basic concepts from standard textbooks like "Principles of Genetics" by Gardner et al. can be highly beneficial. Section 2: Plant Breeding Techniques Example Question: In plant breeding, the method where two plants with desirable traits are crossed is known as: - A) Backcrossing - B) Hybridization - C) Mutation breeding - D) Polyploidy Analysis: Questions in this section often require applied knowledge of breeding techniques. Focus on understanding the different methods of plant breeding, their applications, and examples. Books like "Plant Breeding: Principles and Methods" by B.D. Singh can provide comprehensive coverage of these topics.
MSc Ag Genetics & Plant Breeding: Insights from Previous Year JNKVV Entrance ...
MSc Ag Genetics & Plant Breeding: Insights from Previous Year JNKVV Entrance ...
Krashi Coaching
This is the question set of the IPL Online Quiz organised by Pragya, UEMK Official Quiz Club on 10th May, 2024.
IPL Online Quiz by Pragya; Question Set.
IPL Online Quiz by Pragya; Question Set.
Pragya - UEM Kolkata Quiz Club
diagnosting testing help to determine / monitoring the patient condition and find out disease, and evaluate progress of patient.
diagnosting testing bsc 2nd sem.pptx....
diagnosting testing bsc 2nd sem.pptx....
Ritu480198
ANTI PARKINSONIAN DRUGS
ANTI PARKISON DRUGS.pptx
ANTI PARKISON DRUGS.pptx
PoojaSen20
OER Benefits and Challenges by Sbabel
Benefits and Challenges of OER by Shweta Babel.pptx
Benefits and Challenges of OER by Shweta Babel.pptx
sbabel
Recently uploaded
(20)
Improved Approval Flow in Odoo 17 Studio App
Improved Approval Flow in Odoo 17 Studio App
Spring gala 2024 photo slideshow - Celebrating School-Community Partnerships
Spring gala 2024 photo slideshow - Celebrating School-Community Partnerships
demyelinated disorder: multiple sclerosis.pptx
demyelinated disorder: multiple sclerosis.pptx
Mattingly "AI and Prompt Design: LLMs with Text Classification and Open Source"
Mattingly "AI and Prompt Design: LLMs with Text Classification and Open Source"
Word Stress rules esl .pptx
Word Stress rules esl .pptx
TỔNG HỢP HƠN 100 ĐỀ THI THỬ TỐT NGHIỆP THPT VẬT LÝ 2024 - TỪ CÁC TRƯỜNG, TRƯ...
TỔNG HỢP HƠN 100 ĐỀ THI THỬ TỐT NGHIỆP THPT VẬT LÝ 2024 - TỪ CÁC TRƯỜNG, TRƯ...
ĐỀ THAM KHẢO KÌ THI TUYỂN SINH VÀO LỚP 10 MÔN TIẾNG ANH FORM 50 CÂU TRẮC NGHI...
ĐỀ THAM KHẢO KÌ THI TUYỂN SINH VÀO LỚP 10 MÔN TIẾNG ANH FORM 50 CÂU TRẮC NGHI...
Chapter 7 Pharmacosy Traditional System of Medicine & Ayurvedic Preparations ...
Chapter 7 Pharmacosy Traditional System of Medicine & Ayurvedic Preparations ...
Graduate Outcomes Presentation Slides - English (v3).pptx
Graduate Outcomes Presentation Slides - English (v3).pptx
Poster_density_driven_with_fracture_MLMC.pdf
Poster_density_driven_with_fracture_MLMC.pdf
Stl Algorithms in C++ jjjjjjjjjjjjjjjjjj
Stl Algorithms in C++ jjjjjjjjjjjjjjjjjj
PSYPACT- Practicing Over State Lines May 2024.pptx
PSYPACT- Practicing Over State Lines May 2024.pptx
How to Analyse Profit of a Sales Order in Odoo 17
How to Analyse Profit of a Sales Order in Odoo 17
philosophy and it's principles based on the life
philosophy and it's principles based on the life
How to Manage Closest Location in Odoo 17 Inventory
How to Manage Closest Location in Odoo 17 Inventory
MSc Ag Genetics & Plant Breeding: Insights from Previous Year JNKVV Entrance ...
MSc Ag Genetics & Plant Breeding: Insights from Previous Year JNKVV Entrance ...
IPL Online Quiz by Pragya; Question Set.
IPL Online Quiz by Pragya; Question Set.
diagnosting testing bsc 2nd sem.pptx....
diagnosting testing bsc 2nd sem.pptx....
ANTI PARKISON DRUGS.pptx
ANTI PARKISON DRUGS.pptx
Benefits and Challenges of OER by Shweta Babel.pptx
Benefits and Challenges of OER by Shweta Babel.pptx
isotopes .pdf
1.
isotopes Solution isotopes
Download now