Which of the folllowing statements about the solubility of 1- propan.pdf
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Which of the folllowing statements about the solubility of 1- propanol, ch3ch2ch2oh, and ethanethiol, ch3ch2sh, are true? A) Ethanethiol is more soluble in water because sulfer has agreater electronegativity value than oxygen. B) Ethanethiol is more soluble in water because it has a lower molecular mass than 1-propanol. C) 1-propanol is more soluble in water because it has a greater molecular mass than ethanethiol. D) 1-Propanol is more soluble in water because oxygen has a greater electronegativity value than sulfer. Solution D) 1-Propanol is more soluble in water because oxygen has a greater electronegativity value than sulfer. \\\ eason propanol has a hydrogen bonding by itself while ethanethiol does not have any hydrogen bonding by itself that is why propanol is more soluble in water. because oxygen has a greater electronegativity value than sulfer. the more hydrogen bonding by itself the more soluble in water!.
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Please help me with these General Biology 1 (Bio 111) questions. You.pdf
Please help me with these General Biology 1 (Bio 111) questions. Your response will be highly
appreciated.
1. Describe the following bonds (use examples): ionic bond, polar covalent bond, nonpolar
covalent bond, and hydrogen bond.
2. What are the properties of water? Describe each property being sure to include the following
terms: hydrophobic, hydrophilic, evaporation, cohesion, adhesion, capillarity, solvent, solute,
and solution.
3. What does the pH scale measure? What is the range on the scale? What values indicate acid,
indicate base and indicate neutral?
4. Describe the following bonds (use examples): ionic bond, polar covalent bond, nonpolar
covalent bond, and hydrogen bond.
5. What are the properties of water? Describe each property being sure to include the following
terms: hydrophobic, hydrophilic, evaporation, cohesion, adhesion, capillarity, solvent, solute,
and solution.
Solution
(1)
Ionic bonds result when complete transfer of valence electron(s) between atoms
Example: NaCl, KCl
Covalent bonds result when valence electrons are shared between atoms.
Polar covalent bonding is a type of chemical bond where a pair of electrons is unequally shared
between two atoms: Example: HCl,
Nonpolar covalent bonds are a type of chemical bond where two atoms share a pair of electrons
with each other. Example: CCl3, H2
A hydrogen bond is the attractive force between the hydrogen attached to an electronegative
atom of one molecule and an electronegative atom of a different molecule
Example: Hydrogen bonding occurs between electronegative oxygen of water with hydrogens of
other water molecules..
Benefits Of Alkaline, Ionized Water By Dr. Hidemitsu Hayashi, M.D.
Water: The chemistry of life. Whenever we attempt to determine whether there is life as we know it on Mars or other planets, scientists first seek to establish whether or not water is present. Why? Because life on earth totally depends on water. A High percentage of living things, both plant and animal are found in water. All life on earth is thought to have arisen from water. The bodies of all living organisms are composed largely of water. About 70 to 90 percent of all organic matter is water. The chemical reactions in all plants and animals that support life take place in a water medium. Water not only provides the medium to make these life sustaining reactions possible, but water itself is often an important reactant or product of these reactions. In short, the chemistry of life is water chemistry.
Analysis of phenols and qualitative tests for phenols
Students of gurunanak college of pharmacy,Nagpur
Recommended
Please help me with these General Biology 1 (Bio 111) questions. You.pdf
Please help me with these General Biology 1 (Bio 111) questions. Your response will be highly
appreciated.
1. Describe the following bonds (use examples): ionic bond, polar covalent bond, nonpolar
covalent bond, and hydrogen bond.
2. What are the properties of water? Describe each property being sure to include the following
terms: hydrophobic, hydrophilic, evaporation, cohesion, adhesion, capillarity, solvent, solute,
and solution.
3. What does the pH scale measure? What is the range on the scale? What values indicate acid,
indicate base and indicate neutral?
4. Describe the following bonds (use examples): ionic bond, polar covalent bond, nonpolar
covalent bond, and hydrogen bond.
5. What are the properties of water? Describe each property being sure to include the following
terms: hydrophobic, hydrophilic, evaporation, cohesion, adhesion, capillarity, solvent, solute,
and solution.
Solution
(1)
Ionic bonds result when complete transfer of valence electron(s) between atoms
Example: NaCl, KCl
Covalent bonds result when valence electrons are shared between atoms.
Polar covalent bonding is a type of chemical bond where a pair of electrons is unequally shared
between two atoms: Example: HCl,
Nonpolar covalent bonds are a type of chemical bond where two atoms share a pair of electrons
with each other. Example: CCl3, H2
A hydrogen bond is the attractive force between the hydrogen attached to an electronegative
atom of one molecule and an electronegative atom of a different molecule
Example: Hydrogen bonding occurs between electronegative oxygen of water with hydrogens of
other water molecules..
Benefits Of Alkaline, Ionized Water By Dr. Hidemitsu Hayashi, M.D.
Water: The chemistry of life. Whenever we attempt to determine whether there is life as we know it on Mars or other planets, scientists first seek to establish whether or not water is present. Why? Because life on earth totally depends on water. A High percentage of living things, both plant and animal are found in water. All life on earth is thought to have arisen from water. The bodies of all living organisms are composed largely of water. About 70 to 90 percent of all organic matter is water. The chemical reactions in all plants and animals that support life take place in a water medium. Water not only provides the medium to make these life sustaining reactions possible, but water itself is often an important reactant or product of these reactions. In short, the chemistry of life is water chemistry.
Analysis of phenols and qualitative tests for phenols
Students of gurunanak college of pharmacy,Nagpur
Benefits of alkaline, ionized water by dr. hidemitsu hayashi, m.d.
Benefits Of Alkaline, Ionized Water By Dr. Hidemitsu Hayashi, M.D.
CCCCCCCCHHHHHHHHH.docx
C
C
C
C
C
C
C
C
H
H
H
HH
HH
HH
HH
HH
HH
H
H
H
C
C
C
C
C
C
C
C
H
H
H
HH
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CH
HH
CH
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H
H
H H H HH C H
HH
C C
C
CC
C C
H
H
C
H
H
H
H
H HH
H
H
H
H
H
H
H
octane - unbranched (straight-chain)
4-ethyl-2-methyloctane - branched
ethylcyclohexane - cyclic
Figure 1 - Unbranched, branched, and
cyclic hydrocarbons.
Experiment #3 - Hydrocarbons
Introduction
Organic chemistry is the chemistry of the
compounds of carbon. Currently over twenty
million compounds have been reported in the
chemical literature; about 90% of them are
organic, ie they contain carbon. The remaining
compounds are called inorganic and are formed
from the other elements, of which there are
about 100. That carbon so dominates
compound formation is a result of the fact that it
is almost unique in its ability to form long
chains with other carbon atoms. [Carbon’s
neighbor in the periodic table, silicon, can do
this but rarely does.] These chains with one
carbon joined to a second and the second
joined to a third, etc., can be branched, ie,
chains of carbon atoms can be attached to
carbons in the original chain. It is also possible
for one carbon in a chain to become bonded to
another carbon in that chain, resulting in a
closed ring of atoms. We call these compounds
cyclic.
Since there are so many organic
compounds it is fortunate that we can organize
them into various groups that have some
similarity to each other. For example, one large
group of organic compounds is known as the
hydrocarbons because members of this group
contain only carbon and hydrogen and no other
elements. Figure 1 shows examples of branched, unbranched and cyclic hydrocarbons.
It is possible to subdivide the hydrocarbon group of compounds based on the
bonding between the carbons. If all the carbon-carbon bonds are single, the compound
is an alkane. If at least one of the carbon-carbon bonds in the compound is a double
bond, and the remaining carbon-carbon bonds are single, the compound is an alkene. If
at least one of the carbon-carbon bonds in the compound is a triple bond, and the
remaining carbon-carbon bonds are single, the compound is an alkyne. If the compound
contains a six carbon ring that has alternating double and single bonds around the ring
(three double and three single), we say that ring, and the compound, is aromatic. An
aromatic ring looks like an alkene with three double bonds because that’s the way we
draw it using Lewis structures. However, the actual bonding in such a ring is
considerably different from that in alkenes and, consequently, many of the chemical
properties are different also. Therefore, we place these compounds in a separate family.
By the way, the term aromatic as used here has nothing to do with fragrance.
Experiment #3 Hydrocarbons Page 2
Hydrocarbons may be saturated or unsaturated. A saturated hydrocarbon is one
that is maxed out in terms of the number of hydrogens that can be present given the
nu ...
Carbon acid modified
acid base indicators, carbon acid, pH scale., carbanions, acids and conjugate bases, reactions of carbon acids, phosphonium ions as carbon acids , Carbon acids in synthesis, Super acid,
Non aqueous titration- Pharmaceutical Analysis
Non- aqueous titrations are those in which the titrations of too weakly acidic or basic substances are carried out using non-aqueous solvents so as to get sharp end point.
Such titrations can also be used for the titration of the substances not soluble in water.
The speed, precision and accuracy of the non-aqueous method are close to those of classical acidimetric and alkalimetric titrations.
First reported successful quantitative titration of organic acid and base in non-aqueous solvent: 1910.
To an understanding of non-aqueous acid base titrimetry the theories of acid and base is very important. The theories are:
Acids are hydrogen containing compounds that dissociates to yield hydrogen ions (H+) when dissolved in water.
Bases are compounds that dissociates to yield hydroxide/hydroxyl ions (OH-) when dissolved in water.
Ethanol (CH3-CH2-OH) is miscible in water because it is able to form .pdf
Ethanol (CH3-CH2-OH) is miscible in water because it is able to form hydrogen bonds with
itself and other molecules. However, its structure only allows it to form 1-2 hydrogen bonds.
This is one reason why even low concentrations of ethanol in solution are lethal for cells. Based
on this information, explain why we can use high concentrations ethanol to precipitate DNA out
of solution. Also, describe predict the effects of increasing concentrations of ethanol in (and
around) a cell on macro-molecular interactions (i.e. on weak bonds). Finally, it is possible to
select for yeast that a tolerant to increased concentrations of ethanol. Give an example of a
physiological change in yeast cells that might make them resistant to ethanol.
Solution
1. DNA is a highly polar molecule as a result of which it is highly soluble in water. Due to its
high polarity, the electrostatic forces between charged particles is much less in aqueous solution.
There is formation of a shell by water molecules around charge particles.
Ethanol is less polar and its addition disrupts the shell formed by water molecule around charged
particles. As a result, there is ionic bond formation between phosphate groups of DNA and
positively charged ions present in the solution. This causes precipitation of DNA.
2. Ethanol disrups the hydrogen bonds between water molecules. This alters the hydrophilic
regions of the plasma membrane, thus altering their structure. There is a resultant dehydration of
cells.
3. Upon exposure to ethanol, yeast produces high amounts of monounsaturated fatty acids in
cellular lipids. This an be a method of developing ethanol tolerance in some strains of yeast..
Which statement is true (more than one may be selected)a) Substan.pdf
Which statement is true? (more than one may be selected)
a) Substances that release protons when they dissolve in water are termed bases and result in a
pH higher than 7.
b) Substances that release protons when they dissolve in water are termed acids and result in a
pH higher than 7.
c) Bases (alkali) are any aqueous environments where the proton content decreases and the pH
increases.
d) Acids are any aqueous environments where the proton content increases and the pH decreases.
e) Substances that release protons when they dissolve in water are termed acids and result in a
pH lower than 7.
Solution
Ans. Which statement is true? (more than one may be selected)
a) Substances that release protons when they dissolve in water are termed bases and result in a
pH higher than 7. Incorrect- it is an acid
b) Substances that release protons when they dissolve in water are termed acids and result in a
pH higher than 7. Incorrect- it is “pH lower than 7”
c) Bases (alkali) are any aqueous environments where the proton content decreases and the pH
increases.
d) Acids are any aqueous environments where the proton content increases and the pH decreases.
e) Substances that release protons when they dissolve in water are termed acids and result in a
pH lower than 7. Correct option
Correct option: E
Explanation: Acid is a molecule or ion that donates proton.
An aqueous environment may facilitate acceptance or release of proton to/from a molecule, but
the environment can’t be termed an acid or base because it does not represent a specific molecule
or ion. For example, adding HCl to NaCl solution (say, normal saline) cause donation of proton
from HCl molecule to the solution. Here, NaCl- the aqueous environment, accept a proton- BUT
NaCl solution is NOT a base because it does NOT represent a specific molecule or ion. So, for
being an acid or base- it must be a molecule or ion, NOT an aqueous environment. Though H2O
(water) alone can act as acid or base depending on the acid/base added to it; but the term
‘aqueous environment’ is NOT exclusive for water. So, it can’t be treated as an acid or base.
Therefore, options C, D are incorrect.
pH = - log [H]. Addition of an acid to water (solvent) cause an increase in [H+] ions in water
because acid donates proton to water. Increase in [H+] lowers the pH , i.e. pH is reduced below
7.0; say pH lowered from 7.0 to 4.0. Thus “Substances that release protons when they dissolve in
water are termed acids and result in a pH lower than 7.”
So, options A, B are incorrect..
Write an equation for the height of the point P above the ground as .pdf
Write an equation for the height of the point P above the ground as a function of t, time in
seconds.
h(t) =
Hint: Start by thinking: What is the height of the point P above the center? How is the center
changing?
Solution
height of center at ant time = R + 7(sinA)t = 2 + 7(100/700)t = (2 + t) cm
now height of P wrt center = Rcos(x/2) where x = 7t
height of P = 2+ t + Rcos(7t/2)
= 2 + t + 2cos(7t/2)
7t/2 is value in radian. oon putting t in second
for t= 0,(starting point)
height = 2 + 2cos0 = 4 cm
at the end, at = 100 cm,
h = 2 + 100 + 2cos(350) = 101.433 cm.
Wild-type (WT) mice are irradiated, destroying all of their bone marr.pdf
Wild-type (WT) mice are irradiated, destroying all of their bone marrow cells including CD34+
hematopoietic stem cells (HSC). CD34+ HSC is isolated from (1) an untreated wild-type
syngeneic control mouse, and (2) various immune gene mutant (-/- null) donor mice. The CD34+
HSC from the different mouse genotypes is injected into the wild-type irradiated recipient mice.
Predict the reconstitution of the hematopoietic system in each of the recipient mice (+ or -).
Solution
Answer -
Hematopoietic cells reconstituted in recipient
CD34+HSC
Doner GenotypeErythrocytes macrophages T cells B cells
Wild Type mouse + + + +
CD19(-/-)mouse + + + -
Rag1/2(-/-)mouse + + - -
erythropoietin EPO(-/-) - + + +
AIRE(-/-)mouse + + + +
explaination -Wild type mouse has normal CD34+HSC which is responsible to diffentiate into
all type of immune cells in bone marrow. Whwn this CD34+HSC from wild type mouse injected
into recipient mouse its bone marrow produce all type of immune cells.
CD19 is a B cell lymphocyte antigen It is a protein which is coded by CD19 gene. it is found on
the surface of B cells and helps in its proliferation and differenciation.CD34+HSC from doner
who lack CD19 gene when injected into recipient mouse,It will be unable to produce B cell.
Rag 1/2 also known as recombination activating gene codes a protein which paly an important
role in recombination of the gene of immunoglobuline and T cell receptor marker.two gene
products of recombination activating gene are Rag 1 and Rag 2 protein.These two rag protein are
essential to the generation of mature Bcells and Tcells. .CD34+Hsc from doner who lack the
gene codes gor Rag1/2 protein when injected to the recipient mouse, recipient mousewill be
unable to produce mature B and T cells.
Erythropoietin EPO is a protein harmone secreated by kidney in responce to less
oxygen/Heamoglobine in blood.erythropoietin stimulates bone marrow to produce more Red
blood cells (erythrocytes) which contain Heamoglobine ( an Oxygen carrier) and supply required
amount of oxygen to the body by mean of blood flow through the body. when CD34+HSC from
doner lack the gene codes for Erythropoietin injected into recipient mouse,it bone marrow will
be unable to produce erythrocytes.
AIRE is an auto immune regulator protein encoded by AIRE gene.It controles a mechanism that
prevent T cells (an important cell for immune response) from attacking the self body antigen.T
cells recognizes antigen which are expresed on the surface of antigen presenting cell by T cell
receptors present on T cells and produce Immune responce against that antigen. In autoimmune
diseases T cell Recognize self antigen and produce immune response against self antigen which
leads to sever problem in self body.AIRE protein play an important role to controle these kind of
autoimmune disorders by the removel of T cell through apoptosis while T cells are still in
thymus.
When CD34+HSC from doner WHICh lake the gene for AIRE protein injected into recipient
mouse.Reci.
More Related Content
Similar to Which of the folllowing statements about the solubility of 1- propan.pdf
Benefits of alkaline, ionized water by dr. hidemitsu hayashi, m.d.
Benefits Of Alkaline, Ionized Water By Dr. Hidemitsu Hayashi, M.D.
CCCCCCCCHHHHHHHHH.docx
C
C
C
C
C
C
C
C
H
H
H
HH
HH
HH
HH
HH
HH
H
H
H
C
C
C
C
C
C
C
C
H
H
H
HH
HH
HH
CH
HH
CH
H
H
H
H H H HH C H
HH
C C
C
CC
C C
H
H
C
H
H
H
H
H HH
H
H
H
H
H
H
H
octane - unbranched (straight-chain)
4-ethyl-2-methyloctane - branched
ethylcyclohexane - cyclic
Figure 1 - Unbranched, branched, and
cyclic hydrocarbons.
Experiment #3 - Hydrocarbons
Introduction
Organic chemistry is the chemistry of the
compounds of carbon. Currently over twenty
million compounds have been reported in the
chemical literature; about 90% of them are
organic, ie they contain carbon. The remaining
compounds are called inorganic and are formed
from the other elements, of which there are
about 100. That carbon so dominates
compound formation is a result of the fact that it
is almost unique in its ability to form long
chains with other carbon atoms. [Carbon’s
neighbor in the periodic table, silicon, can do
this but rarely does.] These chains with one
carbon joined to a second and the second
joined to a third, etc., can be branched, ie,
chains of carbon atoms can be attached to
carbons in the original chain. It is also possible
for one carbon in a chain to become bonded to
another carbon in that chain, resulting in a
closed ring of atoms. We call these compounds
cyclic.
Since there are so many organic
compounds it is fortunate that we can organize
them into various groups that have some
similarity to each other. For example, one large
group of organic compounds is known as the
hydrocarbons because members of this group
contain only carbon and hydrogen and no other
elements. Figure 1 shows examples of branched, unbranched and cyclic hydrocarbons.
It is possible to subdivide the hydrocarbon group of compounds based on the
bonding between the carbons. If all the carbon-carbon bonds are single, the compound
is an alkane. If at least one of the carbon-carbon bonds in the compound is a double
bond, and the remaining carbon-carbon bonds are single, the compound is an alkene. If
at least one of the carbon-carbon bonds in the compound is a triple bond, and the
remaining carbon-carbon bonds are single, the compound is an alkyne. If the compound
contains a six carbon ring that has alternating double and single bonds around the ring
(three double and three single), we say that ring, and the compound, is aromatic. An
aromatic ring looks like an alkene with three double bonds because that’s the way we
draw it using Lewis structures. However, the actual bonding in such a ring is
considerably different from that in alkenes and, consequently, many of the chemical
properties are different also. Therefore, we place these compounds in a separate family.
By the way, the term aromatic as used here has nothing to do with fragrance.
Experiment #3 Hydrocarbons Page 2
Hydrocarbons may be saturated or unsaturated. A saturated hydrocarbon is one
that is maxed out in terms of the number of hydrogens that can be present given the
nu ...
Carbon acid modified
acid base indicators, carbon acid, pH scale., carbanions, acids and conjugate bases, reactions of carbon acids, phosphonium ions as carbon acids , Carbon acids in synthesis, Super acid,
Non aqueous titration- Pharmaceutical Analysis
Non- aqueous titrations are those in which the titrations of too weakly acidic or basic substances are carried out using non-aqueous solvents so as to get sharp end point.
Such titrations can also be used for the titration of the substances not soluble in water.
The speed, precision and accuracy of the non-aqueous method are close to those of classical acidimetric and alkalimetric titrations.
First reported successful quantitative titration of organic acid and base in non-aqueous solvent: 1910.
To an understanding of non-aqueous acid base titrimetry the theories of acid and base is very important. The theories are:
Acids are hydrogen containing compounds that dissociates to yield hydrogen ions (H+) when dissolved in water.
Bases are compounds that dissociates to yield hydroxide/hydroxyl ions (OH-) when dissolved in water.
Ethanol (CH3-CH2-OH) is miscible in water because it is able to form .pdf
Ethanol (CH3-CH2-OH) is miscible in water because it is able to form hydrogen bonds with
itself and other molecules. However, its structure only allows it to form 1-2 hydrogen bonds.
This is one reason why even low concentrations of ethanol in solution are lethal for cells. Based
on this information, explain why we can use high concentrations ethanol to precipitate DNA out
of solution. Also, describe predict the effects of increasing concentrations of ethanol in (and
around) a cell on macro-molecular interactions (i.e. on weak bonds). Finally, it is possible to
select for yeast that a tolerant to increased concentrations of ethanol. Give an example of a
physiological change in yeast cells that might make them resistant to ethanol.
Solution
1. DNA is a highly polar molecule as a result of which it is highly soluble in water. Due to its
high polarity, the electrostatic forces between charged particles is much less in aqueous solution.
There is formation of a shell by water molecules around charge particles.
Ethanol is less polar and its addition disrupts the shell formed by water molecule around charged
particles. As a result, there is ionic bond formation between phosphate groups of DNA and
positively charged ions present in the solution. This causes precipitation of DNA.
2. Ethanol disrups the hydrogen bonds between water molecules. This alters the hydrophilic
regions of the plasma membrane, thus altering their structure. There is a resultant dehydration of
cells.
3. Upon exposure to ethanol, yeast produces high amounts of monounsaturated fatty acids in
cellular lipids. This an be a method of developing ethanol tolerance in some strains of yeast..
Which statement is true (more than one may be selected)a) Substan.pdf
Which statement is true? (more than one may be selected)
a) Substances that release protons when they dissolve in water are termed bases and result in a
pH higher than 7.
b) Substances that release protons when they dissolve in water are termed acids and result in a
pH higher than 7.
c) Bases (alkali) are any aqueous environments where the proton content decreases and the pH
increases.
d) Acids are any aqueous environments where the proton content increases and the pH decreases.
e) Substances that release protons when they dissolve in water are termed acids and result in a
pH lower than 7.
Solution
Ans. Which statement is true? (more than one may be selected)
a) Substances that release protons when they dissolve in water are termed bases and result in a
pH higher than 7. Incorrect- it is an acid
b) Substances that release protons when they dissolve in water are termed acids and result in a
pH higher than 7. Incorrect- it is “pH lower than 7”
c) Bases (alkali) are any aqueous environments where the proton content decreases and the pH
increases.
d) Acids are any aqueous environments where the proton content increases and the pH decreases.
e) Substances that release protons when they dissolve in water are termed acids and result in a
pH lower than 7. Correct option
Correct option: E
Explanation: Acid is a molecule or ion that donates proton.
An aqueous environment may facilitate acceptance or release of proton to/from a molecule, but
the environment can’t be termed an acid or base because it does not represent a specific molecule
or ion. For example, adding HCl to NaCl solution (say, normal saline) cause donation of proton
from HCl molecule to the solution. Here, NaCl- the aqueous environment, accept a proton- BUT
NaCl solution is NOT a base because it does NOT represent a specific molecule or ion. So, for
being an acid or base- it must be a molecule or ion, NOT an aqueous environment. Though H2O
(water) alone can act as acid or base depending on the acid/base added to it; but the term
‘aqueous environment’ is NOT exclusive for water. So, it can’t be treated as an acid or base.
Therefore, options C, D are incorrect.
pH = - log [H]. Addition of an acid to water (solvent) cause an increase in [H+] ions in water
because acid donates proton to water. Increase in [H+] lowers the pH , i.e. pH is reduced below
7.0; say pH lowered from 7.0 to 4.0. Thus “Substances that release protons when they dissolve in
water are termed acids and result in a pH lower than 7.”
So, options A, B are incorrect..
Similar to Which of the folllowing statements about the solubility of 1- propan.pdf (20)
Benefits of alkaline, ionized water by dr. hidemitsu hayashi, m.d.
Benefits of alkaline, ionized water by dr. hidemitsu hayashi, m.d.
Chapter-Acids-Bases-and-Salts-Important-questions-Free-pdf-notes.pdf
Chapter-Acids-Bases-and-Salts-Important-questions-Free-pdf-notes.pdf
Ethanol (CH3-CH2-OH) is miscible in water because it is able to form .pdf
Ethanol (CH3-CH2-OH) is miscible in water because it is able to form .pdf
Which statement is true (more than one may be selected)a) Substan.pdf
Which statement is true (more than one may be selected)a) Substan.pdf
More from herminaherman
Write an equation for the height of the point P above the ground as .pdf
Write an equation for the height of the point P above the ground as a function of t, time in
seconds.
h(t) =
Hint: Start by thinking: What is the height of the point P above the center? How is the center
changing?
Solution
height of center at ant time = R + 7(sinA)t = 2 + 7(100/700)t = (2 + t) cm
now height of P wrt center = Rcos(x/2) where x = 7t
height of P = 2+ t + Rcos(7t/2)
= 2 + t + 2cos(7t/2)
7t/2 is value in radian. oon putting t in second
for t= 0,(starting point)
height = 2 + 2cos0 = 4 cm
at the end, at = 100 cm,
h = 2 + 100 + 2cos(350) = 101.433 cm.
Wild-type (WT) mice are irradiated, destroying all of their bone marr.pdf
Wild-type (WT) mice are irradiated, destroying all of their bone marrow cells including CD34+
hematopoietic stem cells (HSC). CD34+ HSC is isolated from (1) an untreated wild-type
syngeneic control mouse, and (2) various immune gene mutant (-/- null) donor mice. The CD34+
HSC from the different mouse genotypes is injected into the wild-type irradiated recipient mice.
Predict the reconstitution of the hematopoietic system in each of the recipient mice (+ or -).
Solution
Answer -
Hematopoietic cells reconstituted in recipient
CD34+HSC
Doner GenotypeErythrocytes macrophages T cells B cells
Wild Type mouse + + + +
CD19(-/-)mouse + + + -
Rag1/2(-/-)mouse + + - -
erythropoietin EPO(-/-) - + + +
AIRE(-/-)mouse + + + +
explaination -Wild type mouse has normal CD34+HSC which is responsible to diffentiate into
all type of immune cells in bone marrow. Whwn this CD34+HSC from wild type mouse injected
into recipient mouse its bone marrow produce all type of immune cells.
CD19 is a B cell lymphocyte antigen It is a protein which is coded by CD19 gene. it is found on
the surface of B cells and helps in its proliferation and differenciation.CD34+HSC from doner
who lack CD19 gene when injected into recipient mouse,It will be unable to produce B cell.
Rag 1/2 also known as recombination activating gene codes a protein which paly an important
role in recombination of the gene of immunoglobuline and T cell receptor marker.two gene
products of recombination activating gene are Rag 1 and Rag 2 protein.These two rag protein are
essential to the generation of mature Bcells and Tcells. .CD34+Hsc from doner who lack the
gene codes gor Rag1/2 protein when injected to the recipient mouse, recipient mousewill be
unable to produce mature B and T cells.
Erythropoietin EPO is a protein harmone secreated by kidney in responce to less
oxygen/Heamoglobine in blood.erythropoietin stimulates bone marrow to produce more Red
blood cells (erythrocytes) which contain Heamoglobine ( an Oxygen carrier) and supply required
amount of oxygen to the body by mean of blood flow through the body. when CD34+HSC from
doner lack the gene codes for Erythropoietin injected into recipient mouse,it bone marrow will
be unable to produce erythrocytes.
AIRE is an auto immune regulator protein encoded by AIRE gene.It controles a mechanism that
prevent T cells (an important cell for immune response) from attacking the self body antigen.T
cells recognizes antigen which are expresed on the surface of antigen presenting cell by T cell
receptors present on T cells and produce Immune responce against that antigen. In autoimmune
diseases T cell Recognize self antigen and produce immune response against self antigen which
leads to sever problem in self body.AIRE protein play an important role to controle these kind of
autoimmune disorders by the removel of T cell through apoptosis while T cells are still in
thymus.
When CD34+HSC from doner WHICh lake the gene for AIRE protein injected into recipient
mouse.Reci.
Which of the following is considered to be a distributed denial of s.pdf
Which of the following is considered to be a distributed denial of service attack because it comes
from numerous sources?a.Smurf
Solution
The Distributed denial of service (DDOS) is an attack where the attacker repeatedly pings the
server from different computers/servers to restrict the server from handling the requests of the
legitimate users. This way all the requests of the authorized users will get be declined.
The SYN flood attack and the death ping send multiple requests to the server from a single
source and the sequence prediction attack is used to analyze the traffic to identify the data
packets.
The smurf attack uses a broadcast network to send multiple requests to the server from different
sources. Hence, the correct options is a. Smurf..
What type of mutation that impacts splicing is a significant contrib.pdf
What type of mutation that impacts splicing is a significant contributor to human disease?
Solution
Disruptions in splicing are major contibutor of human diseases.Most genes in higher eukaryotes
are composed of introns (non coding segments) and exons (coding segments). The majority of
human intron removals ae catalyzed by a large and dynamic ribonucleoprotein (RNP) complex
called the spliceosome.This process involves two sequential transesterification reactions which
ligate the exons and release the intron as a lariat.
Alternate splicing which allows for the productioin of many proteins from a single genomic
locous.This greatly augments the repertorie of the protein.This greatly augments the repertoire of
proteins that can be produced by a given gene and plays an important role in
evolution,development and disease..
What data indicate that all three germ layers are specified in the b.pdf
What data indicate that all three germ layers are specified in the blastula? What are the
differences between dorsal and ventral mesodermal derivatives and what cellular interactions are
required for their specification?
Solution
Three germ layers of amphibians are specified in the blastula is determined by isolating these
tissues in vitro, and they are able to form into specific germ layers. The animal pole cap cells
make ectoderm, marginal region cells make mesoderm, and vegetal cells make endoderm.
Theyare specified but not determined. It is also interesting to note that if animal cap cells are
place co-cultured with vegetal cells, the animal cap cells with become mesoderm. This indicates
that vegetal cells induce other cells to form mesoderm. Experiment paired animal cap cells in 4
different sections of vegetal blastomeres to see if they induce different dorsal-ventral
mesodermal fates. Result showed that different sections of vegetal blastomere have specific
inductive capacities, which is crucial for dorsal-ventral mesoderm determination. Difference
between dorsal and ventral mesoderm derivatives: dorsal mesoderm is the notochord and somite.
In all bilaterian animals, the mesoderm is one of the three primary germ layers in the very early
embryo. The other two layers are the ectoderm (outside layer) and endoderm (inside layer), with
the mesoderm as the middle layer between them.
The mesoderm forms mesenchyme, mesothelium, non-epithelial blood cells and coelomocytes.
Some of the mesoderm derivatives include the muscle (smooth, cardiac and skeletal), the
muscles of the tongue (occipital somites), the pharyngeal arches muscle (muscles of mastication,
muscles of facial expressions), connective tissue, dermis and subcutaneous layer of the skin
At mid-blastula two signaling centers are present on the dorsal side: The prospective
neuroectoderm expresses bone morphogenetic protein (BMP) antagonists, and the future dorsal
endoderm secretes Nodal-related mesoderm-inducing factors. When dorsal mesoderm is formed
at gastrula, a cocktail of growth factor antagonists is secreted by the Spemann organizer and
further patterns the embryo. A ventral gastrula signaling center opposes the actions of the dorsal
organizer, and another set of secreted antagonists is produced ventrally under the control of
BMP4. The early dorsal -Catenin signal inhibits BMP expression at the transcriptional level and
promotes expression of secreted BMP antagonists in the prospective central nervous system
(CNS). In the absence of mesoderm, expression of Chordin and Noggin in ectoderm is required
for anterior CNS formation. FGF (fibroblast growth factor) and IGF (insulin-like growth factor)
signals are also potent neural inducers. Neural induction by anti-BMPs such as Chordin requires
mitogen-activated protein kinase (MAPK) activation mediated by FGF and IGF. These multiple
signals can be integrated at the level of Smad1. Phosphorylation by BMP receptor stimulates
Smad1 transcrip.
Unlike RNA, DNA containsA. adenine.B. uracil.C. phosphate grou.pdf
Unlike RNA, DNA contains
A. adenine.
B. uracil.
C. phosphate groups.
D. thymine.
Solution
Answer is: Thymine
Explanation:
Base Pairing in DNA: A-T (Adenine-Thymine), G-C (Guanine-Cytosine)
Base Pairing in RNA: A-U (Adenine-Uracil), G-C (Guanine-Cytosine)
The phosphate group is present in both RNA and DNA..
Use provided HTML file.Create JavaScript file to validate user i.pdf
Use provided HTML file.
Create *JavaScript file* to validate user input and process user input.
Show the conformation window for Valid Submission.
Title is required field.
Description is required field.
City is required field.
Rate this photo => positive numeric value 1 – 10
User must provide date and time.
User must accept the software license.
User must select “Creative Commons Type
Use JavaScript to display User information.
//HTML
Chapter 6Photo Details
Title
Description
Continent
Choose continentAfricaAsiaEuropeNorth AmericaSouth America
Country
Choose countryCanadaMexicoUnited States
City
CalgaryMontrealTorontoVancouver
Copyright?
All rights reserved
Creative Commons
Creative Commons Types
Attribution
Noncommercial
No Derivative Works
Share Alike
I accept the software license
Rate this photo:
Color Collection:
Date Taken:
Time Taken:
//CSS
/* general text formatting */
h1, h2, h3, nav, footer {
font-family: Georgia, Cambria, \"Times New Roman\", serif;
}
body {
font-family: \"Lucida Sans\", Verdana, Arial, sans-serif;
font-size: 85%;
}
table {
border: collapse;
border-spacing: 0;
width; 90%;
margin: 0 auto;
}
table tbody td{
/* border: 1pt solid #95BEF0; */
line-height: 1.5em;
vertical-align: top;
padding: 0.5em 0.75em;
}
legend {
background-color: #EBF4FB ;
margin: 0 auto;
width: 90%;
padding: 0.25em;
text-align: center;
font-weight: bold;
font-size: 100%;
}
fieldset {
margin: 1em auto;
background-color: #FAFCFF;
width: 60%;
}
form p {
margin-top: 0.5em;
}
.box {
border: 1pt solid #95BEF0;
padding: 0.5em;
margin-bottom: 0.4em;
}
.rectangle {
background-color: #EBF4FB;
padding: 0.5em;
}
.centered {
text-align: center;
}
.rounded, .rounded:hover {
border: 1px solid #172d6e;
border-bottom: 1px solid #0e1d45;
border-radius: 5px;
text-align: center;
color: white;
background-color: #8c9cbf;
padding: 0.5em 0 0.5em 0;
margin: 0.3em;
width: 7em;
-webkit-box-shadow: inset 0 1px 0 0 #72b9eb, 0 1px 4px 0 #b3b3b3;
box-shadow: inset 0 1px 0 0 #72b9eb, 0 1px 4px 0 #b3b3b3;
}
.rounded:hover {
background-color: #7f8dad;
}
Solution
script.js
/**************format.css**************************************/.
Two four-sided dice are rolled. One die has the letters A through D..pdf
Two four-sided dice are rolled. One die has the letters A through D. The other has the numbers 1
through 4.
Express the event “a roll of the dice will contain the letter A or B” as a set (show each of the
possible combinations).
What is the probability that a roll will contain an A or a B?
Solution
The set
S = { A1,A2,A3,A4,B1,B2,B3,B4,C1,C2,C3,C4,D1D,D3D4}
Let event M =a roll of the dice will contain the letter A or B
M = {A1,A2,A3,A4,B1,B2,B3,B4}
Probability = 8/16 = 1/2.
The job of a CODEC (CoderDecoder) is to Convert an analog voice si.pdf
The job of a CODEC (Coder/Decoder) is to: Convert an analog voice signal into a digital signal
Convert a digital signal into an analog voice signal Convert a computer signal into a voice signal
Both a and b
Solution
Option c is true, Since the job of a codec is to encode a stream of data a signal for storage and
transmission, possibly in encrypted form, and the process is revered in decoder function for
playback or editing. Codec are used in videoconferencing, streaming media, and video editing
applications..
Question 4 Constructive retirement is when A. P buys Ss common stoc.pdf
Question 4 Constructive retirement is when A. P buys S\'s common stock O B. P buys S\'s bonds
payable O C. P retires bonds payable. O D. P retires shares of its own stock.
Solution
If a company buys back its own share, which is treasury stock,and if the management of the
company decides to retire them permanently, its called constructive retirement.
So when P retires shares of its own stock, it results in constructive retirement .
Option D is correct..
Private and Public Health are working together through many program .pdf
Private and Public Health are working together through many program initiatives to promote
cost-effective, quality, and compassionate patient care and population health. Under the
Affordable Care Act (2010), and the patient-centered medical home model (PCMH) and
programs such as the All-Inclusive Care for the Elderly (PACE), how are health professionals in
both private and public health working together to meet the critical needs of the elderly
populations?
Solution
Let us start with PACE. It is a program helping people meet their healthcare needs without going
to a hospital or nursing home. A team of healthcare professionals work together to get this done.
PACE organizations work towards providing healthcare and services through their specialists so
that the needs of the elderly are met. the team of specialists come from different locations and
services. Some of them come from Public health organizations and some from private. However,
they all work together towards a common cause of serving the elderly. Thus PACE is an ideal
platform where both private and public health professionals work together to serve the elderly.
PCMH today is transforming the way primary care is managed and delivered. This model meets
a majority of each individual\'s health care needs in a comprehensive manner. Such a
comprehensive coverage would mean that a large team of specialists is involved including
doctors, nurses, pharmacists, assistants and so on. PCMH model is more relationship oriented. It
involves creating a relationship with the patient, understanding them and their families,
respecting their individual needs. This requires a lot of coordination across the healthcare
system, from hospitals to nursing homes to home health care. PCMH model delivers its services
around the clock. this requires the availability of services and specialist at every time. This is
possible through their pool of both private and public health professionals. Services are provided
in the shortest time by the available specialists, both private and public. Thus the PCMH model
also shows a way through which public and private health work together to serve the elderly.
Under the Affordable Care Act and programs like PACE and PCMH, a large pool of specialists
is created who jointly serve the elderly. Such a pool would not be possible if it is made by only
private or public healthcare professionals. Thus, the creation of these pools necessitates the
involvement of both private and public healthcare professionals. The creation of large pools acts
as a strong catalyst in bringing all types of health professionals towards the common goal of
serving the elderly and also the nation at large..
Name the two standards that are supported by major DBMSs for distrib.pdf
Name the two standards that are supported by major DBMSs for distributing data? Describe the
standard set of protocols developed by ISO and ANSI for remote access to database.
Solution
The two standards that are supported by major DBMS\'s Are ODMG 3.0 and JDO .
ODMG 3.0 stands for Object Data Management Group (ODMG) and JDO stands for java data
objects .
Remote database access (RDA) is the standard protocol for database access .
this standard has been developed by ISO and ANSI.
RDA provides protocol for eastablishing remote connection between database client and server.
The client is acting on behalf of an application program while the server is interfacing to a
process that controls data transfers to and from a database.
The RDA Communication Element converts RDA service requests into underlying ACSE and
TP service requests as part of an open systems interconnection..
Material Manufacturing Name 3 ways that the properties of thermosett.pdf
Material Manufacturing Name 3 ways that the properties of thermosetting polymers differ from
those of thermoplastics.
Solution
Thermoplastics can be melted and reshaped over and over again, while thermosetting plastics can
only be heated and shaped once
Thermoset are More resistant to high temperatures than thermoplastics
Thermoset have highly flexible design.
Mean can be computed for variables with ordinal-level measurements..pdf
Mean can be computed for variables with ordinal-level measurements.
True or False?
Solution
Sol)
False
Mean can not be computed for variables with ordinal-level measurements. we can find median
for ordinal measurement.
Ex: Gender, Intelligence and taste. These are qualitative variables so for these type of variables
we can not find mean. we use median.
Let u_1, u_2 be finite dimensional subspaces of a vector space V. Sho.pdf
Let u_1, u_2 be finite dimensional subspaces of a vector space V. Show that the sum u_1 + u_2
is direct if dim(u_1 +u_2) = dim(u_1) + dim (u_2)
Solution
we know that
dim(U1+U2)= dim(U1)+dim(U2)-dim(U1 intersection U2)
given that dim(U1+U2)= dim(U1)+dim(U2)
i.e dim(U1 intersection U2) =0
U1 intersection U2 = {0}
hence sum U1+U2 is direct.
IV. Which compound would have the higher water solubility, tetrahydro.pdf
IV. Which compound would have the higher water solubility, tetrahydropyran (THP) or ethyl
propyl ether? Explain using chemical drawings and no more than 2 sentences. (10 points) THP
ethyl propyl ether
Solution
Cyclic ether is miscible than acyclic ether because the oxygen atom of cyclic ether are more
exposed towards hydrogen bonding. Hence THP (cyclic) is more soluble than ethyl propyl ether
(acyclic)..
Interseting discussion topic,want to chime inDescribe a symbol .pdf
Interseting discussion topic,
want to chime in?
Describe a symbol in your life. It can be a religious one or not. But I want to know something in
your life that is communal (that is to say, it takes place with other people) or personal (you go
through it alone) that points \"beyond itself\". Especially good topics are symbols that you share
with friends, family, teams, scout troops, or whatever - just a meaningful group (important
places, annual parties, hand signs, code words, handshakes, etc).
I want you to deeply diagnose the thing - even if you never have before: What are the motions
involved? Do they have a significance? What is the connection between sign and symbolized?
What feeling or event is it meant to evoke? Just describe each and every element that is
consistently involved and explain its significance... Is there a \"mythology\" behind it? (that is to
say, a key founding story that\'s interpretive, rather than exact, history)
Solution
hand signs:
it involves acceptance, sometime ignorance, sometime victory, some time hapiness. Yes a lot. It
is beyond the language one speaks. It has grat power to communicate. I don\'t know of
mythology but yes its there..
IN JAVA Write a program to create a binary data file named RandomInt.pdf
IN JAVA Write a program to create a binary data file named RandomInt.dat that consists of 10
random integer numbers using DataOutputStream. Use DataInputStream to read the integers
from RandomInt.dat just created and print them on the console in the same program.
Solution
DataInputStreamTest.java
import java.io.DataInputStream;
import java.io.DataOutputStream;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.FileOutputStream;
import java.io.IOException;
import java.util.Random;
public class DataInputStreamTest {
public static void main(String[] args) {
String strFilePath = \"D:\\\\RandomInt.dat\";
Random r = new Random();
try
{
FileOutputStream outputStr = new FileOutputStream(strFilePath);
DataOutputStream outputStrData = new DataOutputStream(outputStr);
for(int i=0; i<10; i++){
outputStrData.writeInt(r.nextInt(100)+1);
outputStrData.writeChar(\'\\t\');
}
outputStrData.flush();
outputStrData.close();
outputStr.close();
FileInputStream inputStr = new FileInputStream(strFilePath);
DataInputStream inputStrData = new DataInputStream(inputStr);
int count = 0;
while(count<10){
int i = inputStrData.readInt();
inputStrData.readChar();
count++;
System.out.println( i);
}
outputStrData.flush();
inputStrData.close();
outputStrData.close();
}
catch(FileNotFoundException fe)
{
System.out.println(\"FileNotFoundException : \" + fe);
}
catch(IOException ioe)
{
System.out.println(\"IOException : \" + ioe);
}
}
}
Output:
65
8
95
69
1
77
16
97
85
1.
I wanna add the shape creator like rectangular , square , circle etc.pdf
I wanna add the shape creator like rectangular , square , circle etc shape creator in this program. I
tried many ways to do it but couldnt figure out the right way. Please help me do this. this is a
painter program and I want to add shape creator for an extra point.. Thanks..
//Painter.java
import java.awt.*;
import java.awt.event.*;
import java.util.ArrayList;
import javax.swing.*;
public class Painter extends JFrame
{
private JButton blueButton, redButton, greenButton,eraserButton,clearButton;
private Container myCP;
public Paint panel;
private JPanel mainContainer;
public Painter()
{
mainContainer = new JPanel(new BorderLayout());
mainContainer.setBackground(Color.WHITE);
JPanel panel2 = new JPanel();
panel2.setLayout(new FlowLayout());
blueButton = new JButton(\"Blue\");
blueButton.setPreferredSize(new Dimension(75,50));
panel2.add(blueButton);
blueButton.addActionListener(new
ButtonListener());
redButton = new JButton(\"Red\");
redButton.setPreferredSize(new Dimension(75,50));
panel2.add(redButton);
redButton.addActionListener(new ButtonListener());
greenButton = new JButton(\"Green\");
greenButton.setPreferredSize(new Dimension(75,50));
panel2.add(greenButton);
greenButton.addActionListener(new ButtonListener());
eraserButton = new JButton(\"Eraser\");
eraserButton.setPreferredSize(new Dimension(75,50));
panel2.add(eraserButton);
eraserButton.addActionListener(new ButtonListener());
clearButton = new JButton(\"Clear\");
clearButton.setPreferredSize(new Dimension(75,50));
panel2.add(clearButton);
clearButton.addActionListener(new ButtonListener());
panel = new Paint();
mainContainer.add(panel2, BorderLayout.NORTH);
mainContainer.add(panel, BorderLayout.CENTER);
add(mainContainer);
setVisible(true);
}
private class ButtonListener implements ActionListener
{
public void actionPerformed(ActionEvent e)
{
if (e.getSource() == blueButton)
{
panel.addCoordinateList(new ColorPoints(Color.BLUE,
new Point()));
}
if (e.getSource() == redButton)
{
panel.addCoordinateList(new ColorPoints(Color.RED,
new Point()));
}
if (e.getSource() == greenButton)
{
panel.addCoordinateList(new
ColorPoints(Color.GREEN,
new Point()));
}
if (e.getSource() == eraserButton)
{
panel.addCoordinateList(new ColorPoints(panel.getBackground(),
new Point()));
}
if (e.getSource() == clearButton)
{
panel.setCoordinateList(new ArrayList());
repaint();
}
}
}
public static void main(String args[])
{
Painter myAppF = new Painter();
myAppF.setTitle(\"Painter\");
myAppF.setVisible(true);
myAppF.setSize(500, 500);
myAppF.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
System.out.println(\"Programmed By Aashutosh Bajgain and Sujan Gautam!\");
}
}
//Paint.java
import java.awt.*;
import java.awt.event.*;
import java.util.ArrayList;
import javax.swing.*;
public class Paint extends JPanel
{
private boolean lineBeingDrawn;
private Point startingPoint;
private ArrayList coordinateList;
private Color selectedColor;
private ColorPoints pwc;
public Paint()
{
lineBeingDrawn = false;
startingPoint .
I keep on get a redefinition error and an undefined error.Customer.pdf
I keep on get a redefinition error and an undefined error.
Customer List
(1) CustomerList() and ~CustomerList() - default constructor and destructor.
(2) bool addStore(Store*s) - Add an instance of store to the linked list. Return true if successful.
(3) Store *removeStore(int ID) - Locate a Store in the list if it exists, remove and return it.
(4) Store *getStore(int ID) - Locate a Store in the list and return a pointer to it.
(5) bool updateStore(int ID, char *name, char *addr, char *city, char *st, char *zip) - Locate a
store in the list then update all its information. Return true if successful.
(6) void printStoreInfo() - Iterate over the list of store and have each print itself.
Employee Record
(1) Add a pointer to a class of type CustomerList called m_pCustomerList.
(2) Add a function CustomerList *getCustomerList to return the pointer to the
EmployeeRecord\'s CustomerList object.
-------------------------------------------------------------
CustomerList.h
#pragma once;
#include
#include \"Store.h\"
class CustomerList
{
public:
Store *m_pHead;
CustomerList();
~CustomerList();
bool addStore(Store *s);
Store *removeStore(int ID);
Store *getStore(int ID);
bool updateStore(int ID, char *name, char *addr, char *city, char *st, char *zip);
void printStoresInfo();
};
----------------------------------------------------------------------------
CustomerList.cpp
#include
#include \"CustomerList.h\"
#include \"Store.h\"
using namespace std;
CustomerList::CustomerList()
{
// Default constructor
}
CustomerList::~CustomerList()
{
// Destructor
}
bool CustomerList:: addStore(Store *s)
{
//creating a new instance
s = new Store();
if(s==NULL)
return true;
}
Store *CustomerList::removeStore(int ID)
{
Store *temp, *back;
temp = m_pHead;
back = NULL;
while((temp != NULL)&&(ID != temp ->getStoreID()))
{
back=temp;
temp=temp->m_pNext;
}
if(temp==NULL)
return NULL;
else if(back==NULL)
{
m_pHead = m_pHead ->m_pNext;
return temp;
}
else
{
back -> m_pNext = temp-> m_pNext;
return temp;
}
return NULL;
}
Store *CustomerList::getStore(int ID)
{
Store *temp;
temp = m_pHead;
while((temp != NULL) && (ID != temp ->getStoreID()))
{
temp = temp->m_pNext;
}
if(temp == NULL)
return NULL;
else
{
Store *retStore = new Store;
*retStore = *temp;
retStore->m_pNext = NULL;
return retStore;
}
}
bool CustomerList::updateStore(int ID, char *name, char *addr, char *city, char *st, char *zip)
{
return false;
}
void CustomerList::printStoresInfo()
{
Store *temp;
cout << \"
=====================================================================
================ \" << endl;
if(m_pHead== NULL)
cout << \" The List is currently empty.\ \" ;
else
{
temp = m_pHead;
while(temp != NULL)
{
cout << temp->m_pNext << endl;
}
}
}
--------------------------------------------------------------------------------
Store.h
#pragma once;
#include
#include
using namespace std;
class Store
{
private:
int m_iStoreID;
char m_sStoreName[64];
char m_sAddress[64];
char m_sCity[32];
char m_sState[32];
char m_sZip[.
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