b.HCN Only two atoms bind to C. And there are no.pdf
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b.HCN Only two atoms bind to C. And there are no lone pairs left on the C atom to bend the molecule, so it is linear or straight. Solution b.HCN Only two atoms bind to C. And there are no lone pairs left on the C atom to bend the molecule, so it is linear or straight..
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Given the evidence available at the time (early 1800s) nothing was.pdf
Given the evidence available at the time (early 1800\'s) nothing was wrong with Dalton\'s
theory. It accounted for the variations among elements, their relative masses, the observed
conservation of mass, and the observations made by earlier scientists about the proportions of
elements found in compounds. Dalton did conjecture that atoms somehow stuck together to make
compounds, but he couldn\'t explain how. It was only much later that electrons were discovered,
showing that atoms weren\'t indestructible solid objects, and later still that it was shown that
atoms had a specific internal structure, the exact nature of which is still under investigation.
None of these discoveries invalidate Dalton\'s atomic theory -- they expand and enrich it.
Dalton\'s theory did exactly what good scientific theories do -- explain and relate existing
evidence, and provide a foundation for further inquiry.
Solution
Given the evidence available at the time (early 1800\'s) nothing was wrong with Dalton\'s
theory. It accounted for the variations among elements, their relative masses, the observed
conservation of mass, and the observations made by earlier scientists about the proportions of
elements found in compounds. Dalton did conjecture that atoms somehow stuck together to make
compounds, but he couldn\'t explain how. It was only much later that electrons were discovered,
showing that atoms weren\'t indestructible solid objects, and later still that it was shown that
atoms had a specific internal structure, the exact nature of which is still under investigation.
None of these discoveries invalidate Dalton\'s atomic theory -- they expand and enrich it.
Dalton\'s theory did exactly what good scientific theories do -- explain and relate existing
evidence, and provide a foundation for further inquiry..
1. Enzymatic digestion of proteins in to amino acids by pepsin.3. .pdf
1. Enzymatic digestion of proteins in to amino acids by pepsin.
3. Secretion of bile.
4. Enzymatic digestion of triglycerides to fatty acids and glycerol by gastric lipase.
8. Secretion of HCl.
Solution
1. Enzymatic digestion of proteins in to amino acids by pepsin.
3. Secretion of bile.
4. Enzymatic digestion of triglycerides to fatty acids and glycerol by gastric lipase.
8. Secretion of HCl..
1. Assume that those DNA molecules are same in length as well as in .pdf
1. Assume that those DNA molecules are same in length as well as in other properities such as
supercoiling...etc, in gel after electrophoresis, we can observe one thick band.
2. As DNA is -vely charged molecules due to P- ions, if the electrodes connects in wrong
direction, it won\'t enter in to gel. Those molecules sometimes may migrate to buffer chambers.
Solution
1. Assume that those DNA molecules are same in length as well as in other properities such as
supercoiling...etc, in gel after electrophoresis, we can observe one thick band.
2. As DNA is -vely charged molecules due to P- ions, if the electrodes connects in wrong
direction, it won\'t enter in to gel. Those molecules sometimes may migrate to buffer chambers..
Try to save the file using .vbs extension rather than .wsfType thi.pdf
Try to save the file using .vbs extension rather than .wsf
Type this in notepad and save the file as SHutdownReminder.vbs and add this file into local
group policy.
Solution
Try to save the file using .vbs extension rather than .wsf
Type this in notepad and save the file as SHutdownReminder.vbs and add this file into local
group policy..
There are two types of respirationAerobic respiration takes plac.pdf
There are two types of respiration:
Aerobic respiration: takes place in presence of Oxygen to produce ATP. The electron transport
chain helps to convert NADH and FADH2 in to ATP. Most of the ATP produced by cellular
respiration are made with the help of Oxidative phosphorylation.
Anaerobic respiration where electron acceptors are other than oxygen .In anaerobes less
oxidizing agents are used such as nitrates where substrate level phosphorylation is used to
produce ATP.
Based on the above descriptions, the correct answers are:
a. Aerobic respiration: Substrate level phosphorylation
b. Anaerobic respiration with nitrate used as the terminal electron acceptor: Substrate level
phosphorylation
c. Embden Meyerhoff pathway for glucose catabolism: Oxidative phosphorylation
d. Anaerobic respiration with iron used as the terminal electon acceptor: Substrate level
phosphorylation
Solution
There are two types of respiration:
Aerobic respiration: takes place in presence of Oxygen to produce ATP. The electron transport
chain helps to convert NADH and FADH2 in to ATP. Most of the ATP produced by cellular
respiration are made with the help of Oxidative phosphorylation.
Anaerobic respiration where electron acceptors are other than oxygen .In anaerobes less
oxidizing agents are used such as nitrates where substrate level phosphorylation is used to
produce ATP.
Based on the above descriptions, the correct answers are:
a. Aerobic respiration: Substrate level phosphorylation
b. Anaerobic respiration with nitrate used as the terminal electron acceptor: Substrate level
phosphorylation
c. Embden Meyerhoff pathway for glucose catabolism: Oxidative phosphorylation
d. Anaerobic respiration with iron used as the terminal electon acceptor: Substrate level
phosphorylation.
The triangle DEF -- TRIANGLE HIG is a glide reflection.Reflection.pdf
The triangle DEF --> TRIANGLE HIG is a glide reflection.
Reflection line:
Each point of a reflected image is the same distance from the line of reflection as the
corresponding point of the original figure. In other words, the line of reflection lies directly in the
middle between the figure and its image -- it is the perpendicular bisector of the segment joining
any point to its image.
The reflection line = IG
Translation rule:
The translation rule is shift.
Shifts
A shift is a rigid translation in that it does not change the shape or size of the graph of the
function. All that a shift will do is change the location of the graph. A vertical shift
adds/subtracts a constant to/from every y-coordinate while leaving the x-coordinate unchanged.
A horizontal shift adds/subtracts a constant to/from every x-coordinate while leaving the y-
coordinate unchanged. Vertical and horizontal shifts can be combined into one expression.
Shifts are added/subtracted to the x or f(x) components. If the constant is grouped with the x,
then it is a horizontal shift, otherwise it is a vertical shift.
center and angle of rotation:
Glide translation rule:and reflection line:
When a translation (a slide or glide) and a reflection are performed one after the other, a
transformation called a glide reflection is produced. In a glide reflection, the line of reflection is
parallel to the direction of the translation. It does not matter whether you glide first and then
reflect, or reflect first and then glide. This transformation is commutative.
When two or more transformations are combined to form a new transformation, the result is
called a composition of transformations. Since translations and reflections are both isometries, a
glide reflection is also an isometry.
Solution
The triangle DEF --> TRIANGLE HIG is a glide reflection.
Reflection line:
Each point of a reflected image is the same distance from the line of reflection as the
corresponding point of the original figure. In other words, the line of reflection lies directly in the
middle between the figure and its image -- it is the perpendicular bisector of the segment joining
any point to its image.
The reflection line = IG
Translation rule:
The translation rule is shift.
Shifts
A shift is a rigid translation in that it does not change the shape or size of the graph of the
function. All that a shift will do is change the location of the graph. A vertical shift
adds/subtracts a constant to/from every y-coordinate while leaving the x-coordinate unchanged.
A horizontal shift adds/subtracts a constant to/from every x-coordinate while leaving the y-
coordinate unchanged. Vertical and horizontal shifts can be combined into one expression.
Shifts are added/subtracted to the x or f(x) components. If the constant is grouped with the x,
then it is a horizontal shift, otherwise it is a vertical shift.
center and angle of rotation:
Glide translation rule:and reflection line:
When a translation (a slide or glide) and a reflectio.
The process of making a chip begins with sand. Sand contains 25 sil.pdf
The process of making a chip begins with sand. Sand contains 25% silicon by mass. Sand
especially Quartz contains high percentage of Silicon dioxide which is the basic ingredient of
semiconductor manufacturing.
Silicon is purified in multiple stages to obtain the quality called Electronic Grade Silicon in the
form of Ingots. Ingots are big mono crystals. Mono crystal Silicon Ingots have highly pure
silicon of purity 99.99%.
The Ingots are cut into individual silicon discs or slices called wafers. These wafers are polished
untile mirror smooth surfaces are obtained. A photo resis layer is applied by rotating the wafers.
Wafers are rotated for an even finish. The photo resist finish is exposed to ultra violet light which
triggers a chemical reaction which transforms the photo resist to a soluble . UV light exposure is
for creating various circuit patterns on the wafer.
Solvents are used to clean out the Photo resist to reveal the patterns. made using UV light.
Unwanted material is removed by etching.
Ion implantation is done by bombarding the various chemical impurities called Ions. Ions are
implanted by shooting them onto the surface with a very high speed of over 3 x 10^5 km/h. The
wafers are put into a copper sulphate solution . As a result the copper ions are deposited onto the
transistor through a process called electroplating. The copper ions settle as a thin layer of copper
and the excess material is polished off.
Multiple metal layers are created to interconnect. the architecture and design is dependent on the
functionality of the transistor or part in which the chip has to work. Functionality tests are done
to evaluate the response of every single chip. Wafers are cut into pieces. Furthermore , the
defaulty pieces are discarded off. Groups of pieces are arranged to cater for the respective
functionality. Hence , the procedure for sand to chip
Solution
The process of making a chip begins with sand. Sand contains 25% silicon by mass. Sand
especially Quartz contains high percentage of Silicon dioxide which is the basic ingredient of
semiconductor manufacturing.
Silicon is purified in multiple stages to obtain the quality called Electronic Grade Silicon in the
form of Ingots. Ingots are big mono crystals. Mono crystal Silicon Ingots have highly pure
silicon of purity 99.99%.
The Ingots are cut into individual silicon discs or slices called wafers. These wafers are polished
untile mirror smooth surfaces are obtained. A photo resis layer is applied by rotating the wafers.
Wafers are rotated for an even finish. The photo resist finish is exposed to ultra violet light which
triggers a chemical reaction which transforms the photo resist to a soluble . UV light exposure is
for creating various circuit patterns on the wafer.
Solvents are used to clean out the Photo resist to reveal the patterns. made using UV light.
Unwanted material is removed by etching.
Ion implantation is done by bombarding the various chemical impurities called Ions. Ions ar.
The Permian Period was the ultimate period of the Paleozoic Era whic.pdf
The Permian Period was the ultimate period of the Paleozoic Era which lasted from 299 million
to 251 million years ago.It preceded the Triassic Period. The evolutionary innovations of seed
and pollen are as follows.
The giant swamp forests began to dry out totally. The mossy plants that depended on spores for
reproduction were being replaced by the the gymnosperms which were the first seed-bearing
plants. They are vascular plants, that transport water internally. They have exposed seeds that
develop on the scales of cones and are fertilized when pollen sifts down and lands directly on the
seed. Today’s conifers are gymnosperms, as are the short palm like cycads.
Solution
The Permian Period was the ultimate period of the Paleozoic Era which lasted from 299 million
to 251 million years ago.It preceded the Triassic Period. The evolutionary innovations of seed
and pollen are as follows.
The giant swamp forests began to dry out totally. The mossy plants that depended on spores for
reproduction were being replaced by the the gymnosperms which were the first seed-bearing
plants. They are vascular plants, that transport water internally. They have exposed seeds that
develop on the scales of cones and are fertilized when pollen sifts down and lands directly on the
seed. Today’s conifers are gymnosperms, as are the short palm like cycads..
Recommended
Given the evidence available at the time (early 1800s) nothing was.pdf
Given the evidence available at the time (early 1800\'s) nothing was wrong with Dalton\'s
theory. It accounted for the variations among elements, their relative masses, the observed
conservation of mass, and the observations made by earlier scientists about the proportions of
elements found in compounds. Dalton did conjecture that atoms somehow stuck together to make
compounds, but he couldn\'t explain how. It was only much later that electrons were discovered,
showing that atoms weren\'t indestructible solid objects, and later still that it was shown that
atoms had a specific internal structure, the exact nature of which is still under investigation.
None of these discoveries invalidate Dalton\'s atomic theory -- they expand and enrich it.
Dalton\'s theory did exactly what good scientific theories do -- explain and relate existing
evidence, and provide a foundation for further inquiry.
Solution
Given the evidence available at the time (early 1800\'s) nothing was wrong with Dalton\'s
theory. It accounted for the variations among elements, their relative masses, the observed
conservation of mass, and the observations made by earlier scientists about the proportions of
elements found in compounds. Dalton did conjecture that atoms somehow stuck together to make
compounds, but he couldn\'t explain how. It was only much later that electrons were discovered,
showing that atoms weren\'t indestructible solid objects, and later still that it was shown that
atoms had a specific internal structure, the exact nature of which is still under investigation.
None of these discoveries invalidate Dalton\'s atomic theory -- they expand and enrich it.
Dalton\'s theory did exactly what good scientific theories do -- explain and relate existing
evidence, and provide a foundation for further inquiry..
1. Enzymatic digestion of proteins in to amino acids by pepsin.3. .pdf
1. Enzymatic digestion of proteins in to amino acids by pepsin.
3. Secretion of bile.
4. Enzymatic digestion of triglycerides to fatty acids and glycerol by gastric lipase.
8. Secretion of HCl.
Solution
1. Enzymatic digestion of proteins in to amino acids by pepsin.
3. Secretion of bile.
4. Enzymatic digestion of triglycerides to fatty acids and glycerol by gastric lipase.
8. Secretion of HCl..
1. Assume that those DNA molecules are same in length as well as in .pdf
1. Assume that those DNA molecules are same in length as well as in other properities such as
supercoiling...etc, in gel after electrophoresis, we can observe one thick band.
2. As DNA is -vely charged molecules due to P- ions, if the electrodes connects in wrong
direction, it won\'t enter in to gel. Those molecules sometimes may migrate to buffer chambers.
Solution
1. Assume that those DNA molecules are same in length as well as in other properities such as
supercoiling...etc, in gel after electrophoresis, we can observe one thick band.
2. As DNA is -vely charged molecules due to P- ions, if the electrodes connects in wrong
direction, it won\'t enter in to gel. Those molecules sometimes may migrate to buffer chambers..
Try to save the file using .vbs extension rather than .wsfType thi.pdf
Try to save the file using .vbs extension rather than .wsf
Type this in notepad and save the file as SHutdownReminder.vbs and add this file into local
group policy.
Solution
Try to save the file using .vbs extension rather than .wsf
Type this in notepad and save the file as SHutdownReminder.vbs and add this file into local
group policy..
There are two types of respirationAerobic respiration takes plac.pdf
There are two types of respiration:
Aerobic respiration: takes place in presence of Oxygen to produce ATP. The electron transport
chain helps to convert NADH and FADH2 in to ATP. Most of the ATP produced by cellular
respiration are made with the help of Oxidative phosphorylation.
Anaerobic respiration where electron acceptors are other than oxygen .In anaerobes less
oxidizing agents are used such as nitrates where substrate level phosphorylation is used to
produce ATP.
Based on the above descriptions, the correct answers are:
a. Aerobic respiration: Substrate level phosphorylation
b. Anaerobic respiration with nitrate used as the terminal electron acceptor: Substrate level
phosphorylation
c. Embden Meyerhoff pathway for glucose catabolism: Oxidative phosphorylation
d. Anaerobic respiration with iron used as the terminal electon acceptor: Substrate level
phosphorylation
Solution
There are two types of respiration:
Aerobic respiration: takes place in presence of Oxygen to produce ATP. The electron transport
chain helps to convert NADH and FADH2 in to ATP. Most of the ATP produced by cellular
respiration are made with the help of Oxidative phosphorylation.
Anaerobic respiration where electron acceptors are other than oxygen .In anaerobes less
oxidizing agents are used such as nitrates where substrate level phosphorylation is used to
produce ATP.
Based on the above descriptions, the correct answers are:
a. Aerobic respiration: Substrate level phosphorylation
b. Anaerobic respiration with nitrate used as the terminal electron acceptor: Substrate level
phosphorylation
c. Embden Meyerhoff pathway for glucose catabolism: Oxidative phosphorylation
d. Anaerobic respiration with iron used as the terminal electon acceptor: Substrate level
phosphorylation.
The triangle DEF -- TRIANGLE HIG is a glide reflection.Reflection.pdf
The triangle DEF --> TRIANGLE HIG is a glide reflection.
Reflection line:
Each point of a reflected image is the same distance from the line of reflection as the
corresponding point of the original figure. In other words, the line of reflection lies directly in the
middle between the figure and its image -- it is the perpendicular bisector of the segment joining
any point to its image.
The reflection line = IG
Translation rule:
The translation rule is shift.
Shifts
A shift is a rigid translation in that it does not change the shape or size of the graph of the
function. All that a shift will do is change the location of the graph. A vertical shift
adds/subtracts a constant to/from every y-coordinate while leaving the x-coordinate unchanged.
A horizontal shift adds/subtracts a constant to/from every x-coordinate while leaving the y-
coordinate unchanged. Vertical and horizontal shifts can be combined into one expression.
Shifts are added/subtracted to the x or f(x) components. If the constant is grouped with the x,
then it is a horizontal shift, otherwise it is a vertical shift.
center and angle of rotation:
Glide translation rule:and reflection line:
When a translation (a slide or glide) and a reflection are performed one after the other, a
transformation called a glide reflection is produced. In a glide reflection, the line of reflection is
parallel to the direction of the translation. It does not matter whether you glide first and then
reflect, or reflect first and then glide. This transformation is commutative.
When two or more transformations are combined to form a new transformation, the result is
called a composition of transformations. Since translations and reflections are both isometries, a
glide reflection is also an isometry.
Solution
The triangle DEF --> TRIANGLE HIG is a glide reflection.
Reflection line:
Each point of a reflected image is the same distance from the line of reflection as the
corresponding point of the original figure. In other words, the line of reflection lies directly in the
middle between the figure and its image -- it is the perpendicular bisector of the segment joining
any point to its image.
The reflection line = IG
Translation rule:
The translation rule is shift.
Shifts
A shift is a rigid translation in that it does not change the shape or size of the graph of the
function. All that a shift will do is change the location of the graph. A vertical shift
adds/subtracts a constant to/from every y-coordinate while leaving the x-coordinate unchanged.
A horizontal shift adds/subtracts a constant to/from every x-coordinate while leaving the y-
coordinate unchanged. Vertical and horizontal shifts can be combined into one expression.
Shifts are added/subtracted to the x or f(x) components. If the constant is grouped with the x,
then it is a horizontal shift, otherwise it is a vertical shift.
center and angle of rotation:
Glide translation rule:and reflection line:
When a translation (a slide or glide) and a reflectio.
The process of making a chip begins with sand. Sand contains 25 sil.pdf
The process of making a chip begins with sand. Sand contains 25% silicon by mass. Sand
especially Quartz contains high percentage of Silicon dioxide which is the basic ingredient of
semiconductor manufacturing.
Silicon is purified in multiple stages to obtain the quality called Electronic Grade Silicon in the
form of Ingots. Ingots are big mono crystals. Mono crystal Silicon Ingots have highly pure
silicon of purity 99.99%.
The Ingots are cut into individual silicon discs or slices called wafers. These wafers are polished
untile mirror smooth surfaces are obtained. A photo resis layer is applied by rotating the wafers.
Wafers are rotated for an even finish. The photo resist finish is exposed to ultra violet light which
triggers a chemical reaction which transforms the photo resist to a soluble . UV light exposure is
for creating various circuit patterns on the wafer.
Solvents are used to clean out the Photo resist to reveal the patterns. made using UV light.
Unwanted material is removed by etching.
Ion implantation is done by bombarding the various chemical impurities called Ions. Ions are
implanted by shooting them onto the surface with a very high speed of over 3 x 10^5 km/h. The
wafers are put into a copper sulphate solution . As a result the copper ions are deposited onto the
transistor through a process called electroplating. The copper ions settle as a thin layer of copper
and the excess material is polished off.
Multiple metal layers are created to interconnect. the architecture and design is dependent on the
functionality of the transistor or part in which the chip has to work. Functionality tests are done
to evaluate the response of every single chip. Wafers are cut into pieces. Furthermore , the
defaulty pieces are discarded off. Groups of pieces are arranged to cater for the respective
functionality. Hence , the procedure for sand to chip
Solution
The process of making a chip begins with sand. Sand contains 25% silicon by mass. Sand
especially Quartz contains high percentage of Silicon dioxide which is the basic ingredient of
semiconductor manufacturing.
Silicon is purified in multiple stages to obtain the quality called Electronic Grade Silicon in the
form of Ingots. Ingots are big mono crystals. Mono crystal Silicon Ingots have highly pure
silicon of purity 99.99%.
The Ingots are cut into individual silicon discs or slices called wafers. These wafers are polished
untile mirror smooth surfaces are obtained. A photo resis layer is applied by rotating the wafers.
Wafers are rotated for an even finish. The photo resist finish is exposed to ultra violet light which
triggers a chemical reaction which transforms the photo resist to a soluble . UV light exposure is
for creating various circuit patterns on the wafer.
Solvents are used to clean out the Photo resist to reveal the patterns. made using UV light.
Unwanted material is removed by etching.
Ion implantation is done by bombarding the various chemical impurities called Ions. Ions ar.
The Permian Period was the ultimate period of the Paleozoic Era whic.pdf
The Permian Period was the ultimate period of the Paleozoic Era which lasted from 299 million
to 251 million years ago.It preceded the Triassic Period. The evolutionary innovations of seed
and pollen are as follows.
The giant swamp forests began to dry out totally. The mossy plants that depended on spores for
reproduction were being replaced by the the gymnosperms which were the first seed-bearing
plants. They are vascular plants, that transport water internally. They have exposed seeds that
develop on the scales of cones and are fertilized when pollen sifts down and lands directly on the
seed. Today’s conifers are gymnosperms, as are the short palm like cycads.
Solution
The Permian Period was the ultimate period of the Paleozoic Era which lasted from 299 million
to 251 million years ago.It preceded the Triassic Period. The evolutionary innovations of seed
and pollen are as follows.
The giant swamp forests began to dry out totally. The mossy plants that depended on spores for
reproduction were being replaced by the the gymnosperms which were the first seed-bearing
plants. They are vascular plants, that transport water internally. They have exposed seeds that
develop on the scales of cones and are fertilized when pollen sifts down and lands directly on the
seed. Today’s conifers are gymnosperms, as are the short palm like cycads..
RNs role in kidney transplant recipient in the immediate post-operat.pdf
RNs role in kidney transplant recipient in the immediate post-operative period;
RN\'s role in organ transplantation to assess, plan, implement, practice evidence based approach
to evaluate care interventions in the care of the transplant patient . To analyse the biological,
psychological and sociological effects of transplantation on the patient.
Post-transplant nursing care for the patient begins in the post-anesthesia care unit.
The first 24 hours after transplantation represent a critical period, marked by hemodynamic and
respiratory instability, and there is a great risk of developing complications, mainly of graft
rejection. The nurse treating the patient in the early post-transplant period needs specialized
knowledge to reduce the problems, prevent or anticipate and intervene immediately to maximize
the result of long-term graft and provide quality care throughout the hospitalization period.
The behavior of diuresis is the most important element in the monitoring of renal
functions, since it causes therapeutic behaviors such as hydration, medication and even surgery
for urologic complications.
The urinary catheter placement provides accurate measurement of urine output and determines
the presence of hemorrhage and blood clot, acute graft rejection, and vascular thrombosis, and is
a predictor of the development of transplantation.
The urinary catheter is removed after proper healing of the anastomosis of the ureter into the
bladder, when continued monitoring of urine volume is required. During the hospitalization of
the patient, daily blood tests must be done, with the most relevant results including levels of
serum creatinine, polymerase chain reaction (PCR), white cell count, and levels of
immunosuppressive drugs. These parameters determine that the kidney function is effective and
show the first signs of a possible rejection or infection, and whether the drug levels are within the
therapeutic range required for the maintenance of immunosuppression or induction, maintenance
and treatment of rejection.
The nurse should moniter the patient\'s hemodynamic status and fluid volume to avoid post-
transplant complications while maintaining central venous pressure at 10 mmHg and systolic
blood pressure above 120 mmHg.
Urine output is replaced on an hourly milliliter-for-milliliter basis and should be recorded
hourly.
RNs role in kidney transplant donor in the immediate post-operative period
Donors face the possibility of post-operative complications such as bleeding, wound infection,
fever, etc. Most of the post-operative complications are generally short-term and can be
addressed with quality medical care.
Closely monitor the renal function to assess for impairment and monitor the hematocrit to asses
for bleeding .
Donors experience more pain than recipient. Donor who had an open surgical approach may
experience more pain than when a laproscopic approach used. Donors who had open approach
will be usually discharged from hospital in 4 or 5 day.
Corporate form of business is where an entity is owned by a small .pdf
Corporate form of business is where an entity is owned by a small or large group of people
,called owners /shareholders ,each of whose liability is limited by shares--subject to laws of the
state in which it operates. It has access to finance through issue of shares in the public It is a
separate legal entity ,distinct from its shareholders.So, unlike ,sole-proprietorships&
partnerships, the shareholders are not personally liable,if the business goes into liquidation.
But,the primary disadvantage of this type of business form is double taxation of earnings &
dividends of the company. ie. Any distribution of profits to shareholders ,is only from after-
tax(corporate tax rate) income of the company. But they are also taxed at the personal level,on
receipt of dividends,at the hands of the shareholders.So, the shareholder is subject to double
taxation . So, ANSWER: b) double taxation
Solution
Corporate form of business is where an entity is owned by a small or large group of people
,called owners /shareholders ,each of whose liability is limited by shares--subject to laws of the
state in which it operates. It has access to finance through issue of shares in the public It is a
separate legal entity ,distinct from its shareholders.So, unlike ,sole-proprietorships&
partnerships, the shareholders are not personally liable,if the business goes into liquidation.
But,the primary disadvantage of this type of business form is double taxation of earnings &
dividends of the company. ie. Any distribution of profits to shareholders ,is only from after-
tax(corporate tax rate) income of the company. But they are also taxed at the personal level,on
receipt of dividends,at the hands of the shareholders.So, the shareholder is subject to double
taxation . So, ANSWER: b) double taxation.
Well first thing to remember is that C,N,S, and P.pdf
Well first thing to remember is that C,N,S, and P are more often the central atom
because they are more likely to bond with more atoms than O,F,H, and Cl. If it wants to bond
with more atoms naturally, it will be at the center more often. (If a C always wants 4 atoms
attached to it, it will usually get 4 atoms to attach to it, making it the center). Well choice a. can
be thrown out because its saying that C,N,S, and P cannot hybridize atomic orbitals (which is
already wrong) to allow as many bonds as allowed (meaning that because they can\'t hybridize,
they cant hold as many bonds). This is an argument against why they are not central atoms.
Choice d. can be thrown out too as well because O,F,H, and Cl are less metalic than the others,
yet they do not take as many electrons (because the more electrons an atom shares, the more
bonds it makes). If this was true, O,F,H, and Cl would act more as central atoms. This is also a
negative argument. Left is B and C. C states that more metallic nonmetals have more desire for
electrons, meaning C,N,S, and P want more electrons. This seems true, but then why does B
usually only become BF3, meaning they do not want more electrons. B is true because it says
that O,F,H, and Cl are less likely to share electrons with more than 2 atoms. It doesn\'t say why,
just that it is less likely to. This is true because the sharing electrons with an atom = forming a
bond with an atom. So if they don\'t want to share electrons with more than 2, it means they
don\'t want to naturally form bonds with more than 2 atoms, which is absolutely true. Compared
to C,N,S, and P who are more likely to share electrons with more than 2 atoms and bonding with
more atoms meaning they are more often bonded to several atoms aka being the center.
Solution
Well first thing to remember is that C,N,S, and P are more often the central atom
because they are more likely to bond with more atoms than O,F,H, and Cl. If it wants to bond
with more atoms naturally, it will be at the center more often. (If a C always wants 4 atoms
attached to it, it will usually get 4 atoms to attach to it, making it the center). Well choice a. can
be thrown out because its saying that C,N,S, and P cannot hybridize atomic orbitals (which is
already wrong) to allow as many bonds as allowed (meaning that because they can\'t hybridize,
they cant hold as many bonds). This is an argument against why they are not central atoms.
Choice d. can be thrown out too as well because O,F,H, and Cl are less metalic than the others,
yet they do not take as many electrons (because the more electrons an atom shares, the more
bonds it makes). If this was true, O,F,H, and Cl would act more as central atoms. This is also a
negative argument. Left is B and C. C states that more metallic nonmetals have more desire for
electrons, meaning C,N,S, and P want more electrons. This seems true, but then why does B
usually only become BF3, meaning they do not want more electrons. B is true because it s.
They both have dipole-dipole interactions, interm.pdf
They both have dipole-dipole interactions, intermolecular hydrogen bonding, and
vander wals force. However, water has two lone pairs that can do hydrogen bonding whereas
ammonia has only 1 lone pair.
Solution
They both have dipole-dipole interactions, intermolecular hydrogen bonding, and
vander wals force. However, water has two lone pairs that can do hydrogen bonding whereas
ammonia has only 1 lone pair..
The violin is a string instrument, usually with f.pdf
The violin is a string instrument, usually with four strings tuned in perfect fifths. It
is the smallest, highest-pitched member of the violin family of string instruments, which also
includes the viola and cello. The violin is sometimes informally called a fiddle, regardless of
the type of music played on it. The word violin comes from the Medieval Latin word vitula,
meaning stringed instrument;this word is also believed to be the source of the Germanic
\"fiddle\".The violin, while it has ancient origins, acquired most of its modern characteristics in
16th-century Italy, with some further modifications occurring in the 18th and 19th centuries.
Violinists and collectors particularly prize the instruments made by the Gasparo da Sal
Solution
The violin is a string instrument, usually with four strings tuned in perfect fifths. It
is the smallest, highest-pitched member of the violin family of string instruments, which also
includes the viola and cello. The violin is sometimes informally called a fiddle, regardless of
the type of music played on it. The word violin comes from the Medieval Latin word vitula,
meaning stringed instrument;this word is also believed to be the source of the Germanic
\"fiddle\".The violin, while it has ancient origins, acquired most of its modern characteristics in
16th-century Italy, with some further modifications occurring in the 18th and 19th centuries.
Violinists and collectors particularly prize the instruments made by the Gasparo da Sal.
the element is phosphorus (P) Solution .pdf
Phosphorus is the element being discussed. The document states that the element is phosphorus and repeats that the element is phosphorus.
Step1 False ; An elemental sample consists of one.pdf
An elemental sample consists of only one kind of atom. The statement "An elemental sample consists of one kind of atoms" is false because an elemental sample contains only one type of atom, not multiple kinds of atoms. The solution confirms that the first step is false.
solution or homogeneous mixture Solution .pdf
A solution is a homogeneous mixture that is made up of two or more substances. The substances that make up a solution are evenly distributed throughout and cannot be easily separated. A solution forms a single phase rather than being made up of separate components.
PV ^ n = constant using pv = N R T we get P ^ .pdf
PV ^ n = constant using pv = N* R * T we get P ^ 1-n * T ^ n = contant T1= 150
T1= 293 P2= 600 T2 = 393 k P1/P2 = (T2/T1 ) ^ n/1-n 1/4 = ( 1.34)^ n/1-n taking log on both
side log( 1/4) = n/1-n * ( log (1.34) n/1-n =-4.736 n= 1.267
Solution
PV ^ n = constant using pv = N* R * T we get P ^ 1-n * T ^ n = contant T1= 150
T1= 293 P2= 600 T2 = 393 k P1/P2 = (T2/T1 ) ^ n/1-n 1/4 = ( 1.34)^ n/1-n taking log on both
side log( 1/4) = n/1-n * ( log (1.34) n/1-n =-4.736 n= 1.267.
methyl-m-nitrobenzoate is more polar than methyl .pdf
methyl-m-nitrobenzoate is more polar than methyl benzoate. There is a large
separation of charge in the nitro group (between the nitrogen and oxygens). This leads to a
higher level of polarity, and the compound containing the nitro is more polar. Hope this helps!
Solution
methyl-m-nitrobenzoate is more polar than methyl benzoate. There is a large
separation of charge in the nitro group (between the nitrogen and oxygens). This leads to a
higher level of polarity, and the compound containing the nitro is more polar. Hope this helps!.
L(f(t)) = 6s^4 - 2s^3 .pdf
This document provides the Laplace transform of a function f(t) which is equal to 6/s^4 - 2/s^3. The Laplace transform of the function f(t) is given as L(f(t)) = 6/s^4 - 2/s^3.
It was a 19th century geologist who first recogni.pdf
It was a 19th century geologist who first recognized periodicity in the physical
properties of the elements. Alexandre Beguyer de Chancourtois (1820-1886), professor of
geology at the School of Mines in Paris, published in 1862 a list of all the known elements. The
list was constructed as a helical graph wrapped around a cylinder--elements with similar
properties occupied positions on the same vertical line of cylinder (the list also included some
ions and compounds). Using geological terms and published without the diagram, de
Chancourtois ideas were completely ignored until the work of Mendeleev. it is used to
distinguish elements by their properties
Solution
It was a 19th century geologist who first recognized periodicity in the physical
properties of the elements. Alexandre Beguyer de Chancourtois (1820-1886), professor of
geology at the School of Mines in Paris, published in 1862 a list of all the known elements. The
list was constructed as a helical graph wrapped around a cylinder--elements with similar
properties occupied positions on the same vertical line of cylinder (the list also included some
ions and compounds). Using geological terms and published without the diagram, de
Chancourtois ideas were completely ignored until the work of Mendeleev. it is used to
distinguish elements by their properties.
here the solution is 0.015 M HCl so we have HCl -.pdf
here the solution is 0.015 M HCl so we have HCl --> H+ + Cl- now, [H+] = 0.0015
M so we have pH = -log[H+] = - log [ 0.0015] = 2.82 and [H3O +]
= [H2O][H+] = 0.0015 * 1.4 x 10-14 = 2.1 x 10-17 M
Solution
here the solution is 0.015 M HCl so we have HCl --> H+ + Cl- now, [H+] = 0.0015
M so we have pH = -log[H+] = - log [ 0.0015] = 2.82 and [H3O +]
= [H2O][H+] = 0.0015 * 1.4 x 10-14 = 2.1 x 10-17 M.
H2SO4 + 2 NaOH = Na2SO4 + 2 H2O moles of NaOH = .pdf
H2SO4 + 2 NaOH = Na2SO4 + 2 H2O moles of NaOH = 2.6 * .45 = 1.17 So
moles of Na2So4 formed = 1.17/2 weight = 1.17/2 * 142 = 83.07 g
Solution
H2SO4 + 2 NaOH = Na2SO4 + 2 H2O moles of NaOH = 2.6 * .45 = 1.17 So
moles of Na2So4 formed = 1.17/2 weight = 1.17/2 * 142 = 83.07 g.
D) H2SO4, heat followed by H2, Pt .pdf
This procedure involves heating sulfuric acid followed by adding hydrogen gas in the presence of a platinum catalyst. The overall reaction is producing something, but the summary does not provide enough context to identify the reactants or products.
CCl4 4. molecular CaBr2 3. ionic RbF 3. ionic .pdf
CCl4: 4. molecular CaBr2: 3. ionic RbF: 3. ionic Kr:1. nonbonding atomic Ni: 2.
metallic atomic
Solution
CCl4: 4. molecular CaBr2: 3. ionic RbF: 3. ionic Kr:1. nonbonding atomic Ni: 2.
metallic atomic.
NO3-(aq) + 2H+(aq) + e- --- NO2(g) + H2O(l) Ered = +0.80 VCu(s) -.pdf
NO3-(aq) + 2H+(aq) + e- ---> NO2(g) + H2O(l) Ered = +0.80 V
Cu(s) ---> Cu2+(aq) + 2e- Eox = -0.34 v
Ecell = Eox + Ered = 0.80 - 0.34 = + 0.46 V
Solution
NO3-(aq) + 2H+(aq) + e- ---> NO2(g) + H2O(l) Ered = +0.80 V
Cu(s) ---> Cu2+(aq) + 2e- Eox = -0.34 v
Ecell = Eox + Ered = 0.80 - 0.34 = + 0.46 V.
internet protocol stack contains 5 layersi)application layerii)t.pdf
internet protocol stack contains 5 layers
i)application layer
ii)transport layer
iii)network layer
iv)link layer
v)physical layer
a) Route selection is done in network layer
b)Provide user services such as e-mail and file transfer is done in application layer
c) Mechanical and electrical interface is done in physical layer
Solution
internet protocol stack contains 5 layers
i)application layer
ii)transport layer
iii)network layer
iv)link layer
v)physical layer
a) Route selection is done in network layer
b)Provide user services such as e-mail and file transfer is done in application layer
c) Mechanical and electrical interface is done in physical layer.
in water, HC2H3O2 does not ionize since it is a weak acid CsOH, ho.pdf
in water, HC2H3O2 does not ionize since it is a weak acid
CsOH, however, dissociates.
Molecular: HC2H3O2(aq) + CsOH(aq) --> CsC2H3O2(aq) + H2O(l)
Ionic: HC2H3O2(aq) + Cs+(aq) + OH-(aq) --> Cs+(aq) + C2H3O2-(aq) + H2O(l)
Net Ionic: HC2H3O2(aq) + OH-(aq) --> C2H3O2-(aq) + H2O(l)
Molecular fomula is where all compound are written as you would see them in reagent bottles.
Ionic equation is where all compounds that dissociate in water are written as ions if in aqueous
environment
Net ionic equation is same as ionic equation, but species identical on both sides of equation
(spectator ions) are not written.
Hope it helps :D
Solution
in water, HC2H3O2 does not ionize since it is a weak acid
CsOH, however, dissociates.
Molecular: HC2H3O2(aq) + CsOH(aq) --> CsC2H3O2(aq) + H2O(l)
Ionic: HC2H3O2(aq) + Cs+(aq) + OH-(aq) --> Cs+(aq) + C2H3O2-(aq) + H2O(l)
Net Ionic: HC2H3O2(aq) + OH-(aq) --> C2H3O2-(aq) + H2O(l)
Molecular fomula is where all compound are written as you would see them in reagent bottles.
Ionic equation is where all compounds that dissociate in water are written as ions if in aqueous
environment
Net ionic equation is same as ionic equation, but species identical on both sides of equation
(spectator ions) are not written.
Hope it helps :D.
In general, ligands are viewed as donating electrons andelectrostati.pdf
In general, ligands are viewed as donating electrons andelectrostatic molecules to the central
atom. Bonding is oftendescribed using the formalisms of molecular orbital theory. Ingeneral,
electron pairs occupy the HOMO (Highest Occupied MolecularOrbital) of the ligands.
Ligands and metal ions can be ordered in many ways; one rankingsystem focuses on ligand
\'hardness\' (see also hard/soft acid/basetheory). Metal ions preferentially bind certain ligands.
Ingeneral, \'hard\' metal ions prefer weak field ligands, whereas\'soft\' metal ions prefer strong
field ligands. According to themolecular orbital theory, the HOMO of the ligand should have
anenergy that overlaps with the LUMO (Lowest Unoccupied MolecularOrbital) of the metal
preferential. Metal ions bound tostrong-field ligands follow the Aufbau principle, whereas
complexesbound to weak-field ligands follow Hund\'s rule.
Binding of the metal with the ligands results in a set ofmolecular orbitals, where the metal can be
identified with a newHOMO and LUMO (the orbitals defining the properties and reactivityof the
resulting complex) and a certain ordering of the 5d-orbitals (which may be filled, or partially
filled withelectrons). In an octahedral environment, the 5 otherwisedegenerate d-orbitals split in
sets of 2 and 3 orbitals (for a morein depth explanation, see crystal field theory).
The energy difference between these 2 sets of d-orbitals iscalled the splitting parameter, o. The
magnitudeof o is determined by the field-strength of theligand: strong field ligands, by
definition, increaseo more than weak field ligands. Ligands can now besorted according to the
magnitude of o (see thetable below). This ordering of ligands is almost invariable for allmetal
ions and is called spectrochemical series.
For complexes with a tetrahedral surrounding, the d-orbitalsagain split into two sets, but this time
in reverse order:
The energy difference between these 2 sets of d-orbitals is nowcalled t. The magnitude of t
issmaller than for o, because in a tetrahedralcomplex only 4 ligands influence the d-orbitals,
whereas in anoctahedral complex the d-orbitals are influenced by 6 ligands. Whenthe
coordination number is neither octahedral nor tetrahedral, thesplitting becomes correspondingly
more complex. For the purposes ofranking ligands, however, the properties of the
octahedralcomplexes and the resulting o has been of primaryinterest.
Solution
In general, ligands are viewed as donating electrons andelectrostatic molecules to the central
atom. Bonding is oftendescribed using the formalisms of molecular orbital theory. Ingeneral,
electron pairs occupy the HOMO (Highest Occupied MolecularOrbital) of the ligands.
Ligands and metal ions can be ordered in many ways; one rankingsystem focuses on ligand
\'hardness\' (see also hard/soft acid/basetheory). Metal ions preferentially bind certain ligands.
Ingeneral, \'hard\' metal ions prefer weak field ligands, whereas\'soft\' metal ions prefer strong
field ligands. .
The History of Stoke Newington Street Names
Presented at the Stoke Newington Literary Festival on 9th June 2024
www.StokeNewingtonHistory.com
LAND USE LAND COVER AND NDVI OF MIRZAPUR DISTRICT, UP
This Dissertation explores the particular circumstances of Mirzapur, a region located in the
core of India. Mirzapur, with its varied terrains and abundant biodiversity, offers an optimal
environment for investigating the changes in vegetation cover dynamics. Our study utilizes
advanced technologies such as GIS (Geographic Information Systems) and Remote sensing to
analyze the transformations that have taken place over the course of a decade.
The complex relationship between human activities and the environment has been the focus
of extensive research and worry. As the global community grapples with swift urbanization,
population expansion, and economic progress, the effects on natural ecosystems are becoming
more evident. A crucial element of this impact is the alteration of vegetation cover, which plays a
significant role in maintaining the ecological equilibrium of our planet.Land serves as the foundation for all human activities and provides the necessary materials for
these activities. As the most crucial natural resource, its utilization by humans results in different
'Land uses,' which are determined by both human activities and the physical characteristics of the
land.
The utilization of land is impacted by human needs and environmental factors. In countries
like India, rapid population growth and the emphasis on extensive resource exploitation can lead
to significant land degradation, adversely affecting the region's land cover.
Therefore, human intervention has significantly influenced land use patterns over many
centuries, evolving its structure over time and space. In the present era, these changes have
accelerated due to factors such as agriculture and urbanization. Information regarding land use and
cover is essential for various planning and management tasks related to the Earth's surface,
providing crucial environmental data for scientific, resource management, policy purposes, and
diverse human activities.
Accurate understanding of land use and cover is imperative for the development planning
of any area. Consequently, a wide range of professionals, including earth system scientists, land
and water managers, and urban planners, are interested in obtaining data on land use and cover
changes, conversion trends, and other related patterns. The spatial dimensions of land use and
cover support policymakers and scientists in making well-informed decisions, as alterations in
these patterns indicate shifts in economic and social conditions. Monitoring such changes with the
help of Advanced technologies like Remote Sensing and Geographic Information Systems is
crucial for coordinated efforts across different administrative levels. Advanced technologies like
Remote Sensing and Geographic Information Systems
9
Changes in vegetation cover refer to variations in the distribution, composition, and overall
structure of plant communities across different temporal and spatial scales. These changes can
occur natural.
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RNs role in kidney transplant recipient in the immediate post-operat.pdf
RNs role in kidney transplant recipient in the immediate post-operative period;
RN\'s role in organ transplantation to assess, plan, implement, practice evidence based approach
to evaluate care interventions in the care of the transplant patient . To analyse the biological,
psychological and sociological effects of transplantation on the patient.
Post-transplant nursing care for the patient begins in the post-anesthesia care unit.
The first 24 hours after transplantation represent a critical period, marked by hemodynamic and
respiratory instability, and there is a great risk of developing complications, mainly of graft
rejection. The nurse treating the patient in the early post-transplant period needs specialized
knowledge to reduce the problems, prevent or anticipate and intervene immediately to maximize
the result of long-term graft and provide quality care throughout the hospitalization period.
The behavior of diuresis is the most important element in the monitoring of renal
functions, since it causes therapeutic behaviors such as hydration, medication and even surgery
for urologic complications.
The urinary catheter placement provides accurate measurement of urine output and determines
the presence of hemorrhage and blood clot, acute graft rejection, and vascular thrombosis, and is
a predictor of the development of transplantation.
The urinary catheter is removed after proper healing of the anastomosis of the ureter into the
bladder, when continued monitoring of urine volume is required. During the hospitalization of
the patient, daily blood tests must be done, with the most relevant results including levels of
serum creatinine, polymerase chain reaction (PCR), white cell count, and levels of
immunosuppressive drugs. These parameters determine that the kidney function is effective and
show the first signs of a possible rejection or infection, and whether the drug levels are within the
therapeutic range required for the maintenance of immunosuppression or induction, maintenance
and treatment of rejection.
The nurse should moniter the patient\'s hemodynamic status and fluid volume to avoid post-
transplant complications while maintaining central venous pressure at 10 mmHg and systolic
blood pressure above 120 mmHg.
Urine output is replaced on an hourly milliliter-for-milliliter basis and should be recorded
hourly.
RNs role in kidney transplant donor in the immediate post-operative period
Donors face the possibility of post-operative complications such as bleeding, wound infection,
fever, etc. Most of the post-operative complications are generally short-term and can be
addressed with quality medical care.
Closely monitor the renal function to assess for impairment and monitor the hematocrit to asses
for bleeding .
Donors experience more pain than recipient. Donor who had an open surgical approach may
experience more pain than when a laproscopic approach used. Donors who had open approach
will be usually discharged from hospital in 4 or 5 day.
Corporate form of business is where an entity is owned by a small .pdf
Corporate form of business is where an entity is owned by a small or large group of people
,called owners /shareholders ,each of whose liability is limited by shares--subject to laws of the
state in which it operates. It has access to finance through issue of shares in the public It is a
separate legal entity ,distinct from its shareholders.So, unlike ,sole-proprietorships&
partnerships, the shareholders are not personally liable,if the business goes into liquidation.
But,the primary disadvantage of this type of business form is double taxation of earnings &
dividends of the company. ie. Any distribution of profits to shareholders ,is only from after-
tax(corporate tax rate) income of the company. But they are also taxed at the personal level,on
receipt of dividends,at the hands of the shareholders.So, the shareholder is subject to double
taxation . So, ANSWER: b) double taxation
Solution
Corporate form of business is where an entity is owned by a small or large group of people
,called owners /shareholders ,each of whose liability is limited by shares--subject to laws of the
state in which it operates. It has access to finance through issue of shares in the public It is a
separate legal entity ,distinct from its shareholders.So, unlike ,sole-proprietorships&
partnerships, the shareholders are not personally liable,if the business goes into liquidation.
But,the primary disadvantage of this type of business form is double taxation of earnings &
dividends of the company. ie. Any distribution of profits to shareholders ,is only from after-
tax(corporate tax rate) income of the company. But they are also taxed at the personal level,on
receipt of dividends,at the hands of the shareholders.So, the shareholder is subject to double
taxation . So, ANSWER: b) double taxation.
Well first thing to remember is that C,N,S, and P.pdf
Well first thing to remember is that C,N,S, and P are more often the central atom
because they are more likely to bond with more atoms than O,F,H, and Cl. If it wants to bond
with more atoms naturally, it will be at the center more often. (If a C always wants 4 atoms
attached to it, it will usually get 4 atoms to attach to it, making it the center). Well choice a. can
be thrown out because its saying that C,N,S, and P cannot hybridize atomic orbitals (which is
already wrong) to allow as many bonds as allowed (meaning that because they can\'t hybridize,
they cant hold as many bonds). This is an argument against why they are not central atoms.
Choice d. can be thrown out too as well because O,F,H, and Cl are less metalic than the others,
yet they do not take as many electrons (because the more electrons an atom shares, the more
bonds it makes). If this was true, O,F,H, and Cl would act more as central atoms. This is also a
negative argument. Left is B and C. C states that more metallic nonmetals have more desire for
electrons, meaning C,N,S, and P want more electrons. This seems true, but then why does B
usually only become BF3, meaning they do not want more electrons. B is true because it says
that O,F,H, and Cl are less likely to share electrons with more than 2 atoms. It doesn\'t say why,
just that it is less likely to. This is true because the sharing electrons with an atom = forming a
bond with an atom. So if they don\'t want to share electrons with more than 2, it means they
don\'t want to naturally form bonds with more than 2 atoms, which is absolutely true. Compared
to C,N,S, and P who are more likely to share electrons with more than 2 atoms and bonding with
more atoms meaning they are more often bonded to several atoms aka being the center.
Solution
Well first thing to remember is that C,N,S, and P are more often the central atom
because they are more likely to bond with more atoms than O,F,H, and Cl. If it wants to bond
with more atoms naturally, it will be at the center more often. (If a C always wants 4 atoms
attached to it, it will usually get 4 atoms to attach to it, making it the center). Well choice a. can
be thrown out because its saying that C,N,S, and P cannot hybridize atomic orbitals (which is
already wrong) to allow as many bonds as allowed (meaning that because they can\'t hybridize,
they cant hold as many bonds). This is an argument against why they are not central atoms.
Choice d. can be thrown out too as well because O,F,H, and Cl are less metalic than the others,
yet they do not take as many electrons (because the more electrons an atom shares, the more
bonds it makes). If this was true, O,F,H, and Cl would act more as central atoms. This is also a
negative argument. Left is B and C. C states that more metallic nonmetals have more desire for
electrons, meaning C,N,S, and P want more electrons. This seems true, but then why does B
usually only become BF3, meaning they do not want more electrons. B is true because it s.
They both have dipole-dipole interactions, interm.pdf
They both have dipole-dipole interactions, intermolecular hydrogen bonding, and
vander wals force. However, water has two lone pairs that can do hydrogen bonding whereas
ammonia has only 1 lone pair.
Solution
They both have dipole-dipole interactions, intermolecular hydrogen bonding, and
vander wals force. However, water has two lone pairs that can do hydrogen bonding whereas
ammonia has only 1 lone pair..
The violin is a string instrument, usually with f.pdf
The violin is a string instrument, usually with four strings tuned in perfect fifths. It
is the smallest, highest-pitched member of the violin family of string instruments, which also
includes the viola and cello. The violin is sometimes informally called a fiddle, regardless of
the type of music played on it. The word violin comes from the Medieval Latin word vitula,
meaning stringed instrument;this word is also believed to be the source of the Germanic
\"fiddle\".The violin, while it has ancient origins, acquired most of its modern characteristics in
16th-century Italy, with some further modifications occurring in the 18th and 19th centuries.
Violinists and collectors particularly prize the instruments made by the Gasparo da Sal
Solution
The violin is a string instrument, usually with four strings tuned in perfect fifths. It
is the smallest, highest-pitched member of the violin family of string instruments, which also
includes the viola and cello. The violin is sometimes informally called a fiddle, regardless of
the type of music played on it. The word violin comes from the Medieval Latin word vitula,
meaning stringed instrument;this word is also believed to be the source of the Germanic
\"fiddle\".The violin, while it has ancient origins, acquired most of its modern characteristics in
16th-century Italy, with some further modifications occurring in the 18th and 19th centuries.
Violinists and collectors particularly prize the instruments made by the Gasparo da Sal.
the element is phosphorus (P) Solution .pdf
Phosphorus is the element being discussed. The document states that the element is phosphorus and repeats that the element is phosphorus.
Step1 False ; An elemental sample consists of one.pdf
An elemental sample consists of only one kind of atom. The statement "An elemental sample consists of one kind of atoms" is false because an elemental sample contains only one type of atom, not multiple kinds of atoms. The solution confirms that the first step is false.
solution or homogeneous mixture Solution .pdf
A solution is a homogeneous mixture that is made up of two or more substances. The substances that make up a solution are evenly distributed throughout and cannot be easily separated. A solution forms a single phase rather than being made up of separate components.
PV ^ n = constant using pv = N R T we get P ^ .pdf
PV ^ n = constant using pv = N* R * T we get P ^ 1-n * T ^ n = contant T1= 150
T1= 293 P2= 600 T2 = 393 k P1/P2 = (T2/T1 ) ^ n/1-n 1/4 = ( 1.34)^ n/1-n taking log on both
side log( 1/4) = n/1-n * ( log (1.34) n/1-n =-4.736 n= 1.267
Solution
PV ^ n = constant using pv = N* R * T we get P ^ 1-n * T ^ n = contant T1= 150
T1= 293 P2= 600 T2 = 393 k P1/P2 = (T2/T1 ) ^ n/1-n 1/4 = ( 1.34)^ n/1-n taking log on both
side log( 1/4) = n/1-n * ( log (1.34) n/1-n =-4.736 n= 1.267.
methyl-m-nitrobenzoate is more polar than methyl .pdf
methyl-m-nitrobenzoate is more polar than methyl benzoate. There is a large
separation of charge in the nitro group (between the nitrogen and oxygens). This leads to a
higher level of polarity, and the compound containing the nitro is more polar. Hope this helps!
Solution
methyl-m-nitrobenzoate is more polar than methyl benzoate. There is a large
separation of charge in the nitro group (between the nitrogen and oxygens). This leads to a
higher level of polarity, and the compound containing the nitro is more polar. Hope this helps!.
L(f(t)) = 6s^4 - 2s^3 .pdf
This document provides the Laplace transform of a function f(t) which is equal to 6/s^4 - 2/s^3. The Laplace transform of the function f(t) is given as L(f(t)) = 6/s^4 - 2/s^3.
It was a 19th century geologist who first recogni.pdf
It was a 19th century geologist who first recognized periodicity in the physical
properties of the elements. Alexandre Beguyer de Chancourtois (1820-1886), professor of
geology at the School of Mines in Paris, published in 1862 a list of all the known elements. The
list was constructed as a helical graph wrapped around a cylinder--elements with similar
properties occupied positions on the same vertical line of cylinder (the list also included some
ions and compounds). Using geological terms and published without the diagram, de
Chancourtois ideas were completely ignored until the work of Mendeleev. it is used to
distinguish elements by their properties
Solution
It was a 19th century geologist who first recognized periodicity in the physical
properties of the elements. Alexandre Beguyer de Chancourtois (1820-1886), professor of
geology at the School of Mines in Paris, published in 1862 a list of all the known elements. The
list was constructed as a helical graph wrapped around a cylinder--elements with similar
properties occupied positions on the same vertical line of cylinder (the list also included some
ions and compounds). Using geological terms and published without the diagram, de
Chancourtois ideas were completely ignored until the work of Mendeleev. it is used to
distinguish elements by their properties.
here the solution is 0.015 M HCl so we have HCl -.pdf
here the solution is 0.015 M HCl so we have HCl --> H+ + Cl- now, [H+] = 0.0015
M so we have pH = -log[H+] = - log [ 0.0015] = 2.82 and [H3O +]
= [H2O][H+] = 0.0015 * 1.4 x 10-14 = 2.1 x 10-17 M
Solution
here the solution is 0.015 M HCl so we have HCl --> H+ + Cl- now, [H+] = 0.0015
M so we have pH = -log[H+] = - log [ 0.0015] = 2.82 and [H3O +]
= [H2O][H+] = 0.0015 * 1.4 x 10-14 = 2.1 x 10-17 M.
H2SO4 + 2 NaOH = Na2SO4 + 2 H2O moles of NaOH = .pdf
H2SO4 + 2 NaOH = Na2SO4 + 2 H2O moles of NaOH = 2.6 * .45 = 1.17 So
moles of Na2So4 formed = 1.17/2 weight = 1.17/2 * 142 = 83.07 g
Solution
H2SO4 + 2 NaOH = Na2SO4 + 2 H2O moles of NaOH = 2.6 * .45 = 1.17 So
moles of Na2So4 formed = 1.17/2 weight = 1.17/2 * 142 = 83.07 g.
D) H2SO4, heat followed by H2, Pt .pdf
This procedure involves heating sulfuric acid followed by adding hydrogen gas in the presence of a platinum catalyst. The overall reaction is producing something, but the summary does not provide enough context to identify the reactants or products.
CCl4 4. molecular CaBr2 3. ionic RbF 3. ionic .pdf
CCl4: 4. molecular CaBr2: 3. ionic RbF: 3. ionic Kr:1. nonbonding atomic Ni: 2.
metallic atomic
Solution
CCl4: 4. molecular CaBr2: 3. ionic RbF: 3. ionic Kr:1. nonbonding atomic Ni: 2.
metallic atomic.
NO3-(aq) + 2H+(aq) + e- --- NO2(g) + H2O(l) Ered = +0.80 VCu(s) -.pdf
NO3-(aq) + 2H+(aq) + e- ---> NO2(g) + H2O(l) Ered = +0.80 V
Cu(s) ---> Cu2+(aq) + 2e- Eox = -0.34 v
Ecell = Eox + Ered = 0.80 - 0.34 = + 0.46 V
Solution
NO3-(aq) + 2H+(aq) + e- ---> NO2(g) + H2O(l) Ered = +0.80 V
Cu(s) ---> Cu2+(aq) + 2e- Eox = -0.34 v
Ecell = Eox + Ered = 0.80 - 0.34 = + 0.46 V.
internet protocol stack contains 5 layersi)application layerii)t.pdf
internet protocol stack contains 5 layers
i)application layer
ii)transport layer
iii)network layer
iv)link layer
v)physical layer
a) Route selection is done in network layer
b)Provide user services such as e-mail and file transfer is done in application layer
c) Mechanical and electrical interface is done in physical layer
Solution
internet protocol stack contains 5 layers
i)application layer
ii)transport layer
iii)network layer
iv)link layer
v)physical layer
a) Route selection is done in network layer
b)Provide user services such as e-mail and file transfer is done in application layer
c) Mechanical and electrical interface is done in physical layer.
in water, HC2H3O2 does not ionize since it is a weak acid CsOH, ho.pdf
in water, HC2H3O2 does not ionize since it is a weak acid
CsOH, however, dissociates.
Molecular: HC2H3O2(aq) + CsOH(aq) --> CsC2H3O2(aq) + H2O(l)
Ionic: HC2H3O2(aq) + Cs+(aq) + OH-(aq) --> Cs+(aq) + C2H3O2-(aq) + H2O(l)
Net Ionic: HC2H3O2(aq) + OH-(aq) --> C2H3O2-(aq) + H2O(l)
Molecular fomula is where all compound are written as you would see them in reagent bottles.
Ionic equation is where all compounds that dissociate in water are written as ions if in aqueous
environment
Net ionic equation is same as ionic equation, but species identical on both sides of equation
(spectator ions) are not written.
Hope it helps :D
Solution
in water, HC2H3O2 does not ionize since it is a weak acid
CsOH, however, dissociates.
Molecular: HC2H3O2(aq) + CsOH(aq) --> CsC2H3O2(aq) + H2O(l)
Ionic: HC2H3O2(aq) + Cs+(aq) + OH-(aq) --> Cs+(aq) + C2H3O2-(aq) + H2O(l)
Net Ionic: HC2H3O2(aq) + OH-(aq) --> C2H3O2-(aq) + H2O(l)
Molecular fomula is where all compound are written as you would see them in reagent bottles.
Ionic equation is where all compounds that dissociate in water are written as ions if in aqueous
environment
Net ionic equation is same as ionic equation, but species identical on both sides of equation
(spectator ions) are not written.
Hope it helps :D.
In general, ligands are viewed as donating electrons andelectrostati.pdf
In general, ligands are viewed as donating electrons andelectrostatic molecules to the central
atom. Bonding is oftendescribed using the formalisms of molecular orbital theory. Ingeneral,
electron pairs occupy the HOMO (Highest Occupied MolecularOrbital) of the ligands.
Ligands and metal ions can be ordered in many ways; one rankingsystem focuses on ligand
\'hardness\' (see also hard/soft acid/basetheory). Metal ions preferentially bind certain ligands.
Ingeneral, \'hard\' metal ions prefer weak field ligands, whereas\'soft\' metal ions prefer strong
field ligands. According to themolecular orbital theory, the HOMO of the ligand should have
anenergy that overlaps with the LUMO (Lowest Unoccupied MolecularOrbital) of the metal
preferential. Metal ions bound tostrong-field ligands follow the Aufbau principle, whereas
complexesbound to weak-field ligands follow Hund\'s rule.
Binding of the metal with the ligands results in a set ofmolecular orbitals, where the metal can be
identified with a newHOMO and LUMO (the orbitals defining the properties and reactivityof the
resulting complex) and a certain ordering of the 5d-orbitals (which may be filled, or partially
filled withelectrons). In an octahedral environment, the 5 otherwisedegenerate d-orbitals split in
sets of 2 and 3 orbitals (for a morein depth explanation, see crystal field theory).
The energy difference between these 2 sets of d-orbitals iscalled the splitting parameter, o. The
magnitudeof o is determined by the field-strength of theligand: strong field ligands, by
definition, increaseo more than weak field ligands. Ligands can now besorted according to the
magnitude of o (see thetable below). This ordering of ligands is almost invariable for allmetal
ions and is called spectrochemical series.
For complexes with a tetrahedral surrounding, the d-orbitalsagain split into two sets, but this time
in reverse order:
The energy difference between these 2 sets of d-orbitals is nowcalled t. The magnitude of t
issmaller than for o, because in a tetrahedralcomplex only 4 ligands influence the d-orbitals,
whereas in anoctahedral complex the d-orbitals are influenced by 6 ligands. Whenthe
coordination number is neither octahedral nor tetrahedral, thesplitting becomes correspondingly
more complex. For the purposes ofranking ligands, however, the properties of the
octahedralcomplexes and the resulting o has been of primaryinterest.
Solution
In general, ligands are viewed as donating electrons andelectrostatic molecules to the central
atom. Bonding is oftendescribed using the formalisms of molecular orbital theory. Ingeneral,
electron pairs occupy the HOMO (Highest Occupied MolecularOrbital) of the ligands.
Ligands and metal ions can be ordered in many ways; one rankingsystem focuses on ligand
\'hardness\' (see also hard/soft acid/basetheory). Metal ions preferentially bind certain ligands.
Ingeneral, \'hard\' metal ions prefer weak field ligands, whereas\'soft\' metal ions prefer strong
field ligands. .
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