There is no easy way to remember the ionization l.pdf
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There is no easy way to remember the ionization levels on periodic table. Elements in a certain group share similar properties. That could be one way to predict the ionization levels but its not 100% accurate at all times. Solution There is no easy way to remember the ionization levels on periodic table. Elements in a certain group share similar properties. That could be one way to predict the ionization levels but its not 100% accurate at all times..
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1.Cexplanation; Cloud consumers that use cloud-based IT resources .pdf
1.C
explanation; Cloud consumers that use cloud-based IT resources can generally lease them with a
pay-for-use model. With this model, cloud consumers pay a usage fee for only the amount of the
IT resource actually used, resulting in directly proportional costs. This gives an organization
access to IT resources without having to purchase its own, resulting in reduced investment
requirements
2.B
explanation
The availability of an IT resource is the probability that it will be accessible. The probability
value is generally expressed with a percentage representing the amount of time that the IT
resource is accessible during a given period. The percentage is calculated as follows: 1. Divide
the amount of hours the IT resource was unavailable (within a measured period) by the total
amount of hours within the measured period. 2. Multiply the value by 100. 3. Deduct the value
from 100
3.D
explanation;
Horizontal scaling refers to the allocation or releasing of resources of the same type. The
horizontal allocation of resources is referred to as scaling out and the horizontal releasing of
resources is referred to as scaling in. Horizontal scaling is a common form of scaling within
cloud environments
4.C
explanation;
Cloud-based IT resources that are used by a cloud consumer reside outside of the cloud
consumer’s organizational boundary. To use the IT resources, the cloud consumer will generally
need to trust them. As a result, the cloud consumer’s trust boundary is expanded beyond its
organizational boundary to encompass the cloud.
5.C
explanation;
An organization will often have existing IT resources that are already paid for and operational.
The prior investment that has been made in these on-premise IT resources is referred to as sunk
costs. When comparing cloud-based IT resources to on-premise IT resources with sunk costs,
the up-front costs for on-premise IT resources is significantly lower. It can therefore be more
difficult to justify the leasing of cloud- based IT resources as an alternative
6.D
explanation
the fact that trust boundaries overlap can lead to opportunities for an attacker to attack IT
resources shared by multiple cloud consumers
7.D
explanation
all r correct
8.B
explanation
Cloud A provides Cloud Service A as part of a failover system that encompasses a redundant
implementation of Cloud Service A on Cloud B. If Cloud Service A on Cloud A fails, then
Cloud Service A on Cloud B is automatically provisioned transparently to Cloud Service
Consumer A. Each cloud has a specific level of reliability and availability that it guarantees for
Cloud Service A. By spanning the failover system across both clouds, the overall reliability and
availability will be higher than the maximum reliability and availability of either cloud
9.C
explanation
Different physical and virtual IT resources are dynamically assigned and reassigned according to
cloud consumer demand, typically followed by execution through statistical multiplexing.
Resource poolin.
1) WBC count is high and this level indicates that the person has le.pdf
1) WBC count is high and this level indicates that the person has leucocytosis. This may be due
to some infections or stress etc.
2)the pH level is alkalotic and this is caused by the respiratory alkalosis. As the pH is less than
the normal level.
Solution
1) WBC count is high and this level indicates that the person has leucocytosis. This may be due
to some infections or stress etc.
2)the pH level is alkalotic and this is caused by the respiratory alkalosis. As the pH is less than
the normal level..
What are the four steps of a process involving a heterogeneous catal.pdf
What are the four steps of a process involving a heterogeneous catalyst?
Stage 1:Diffusion of Reactant(s) to the Surface
The rate at which reactants will diffuse to the surface will be influenced by their bulk
concentration and by the thickness of the boundary layer.
Stage 2:Adsorption of reactants
Bonds are formed as the reactant(s) are adsorbed onto the surface of the catalyst.
The ability for an atom or molecule to stick to the surface is known, brilliantly, as the Sticking
Co-efficient. This is just the ratio or percentage of molecules that end up sticking on the surface.
Stage 3:Reaction
Bonds form between the atoms and molecules on the surface
Stage 4:Desorption of products
Bonds are broken as the product(s) desorb from the surface.
Stage 5:Diffusion of Product(s) away from the Surface
The products are then desorbed from the surface of the catalyst.
Solution
What are the four steps of a process involving a heterogeneous catalyst?
Stage 1:Diffusion of Reactant(s) to the Surface
The rate at which reactants will diffuse to the surface will be influenced by their bulk
concentration and by the thickness of the boundary layer.
Stage 2:Adsorption of reactants
Bonds are formed as the reactant(s) are adsorbed onto the surface of the catalyst.
The ability for an atom or molecule to stick to the surface is known, brilliantly, as the Sticking
Co-efficient. This is just the ratio or percentage of molecules that end up sticking on the surface.
Stage 3:Reaction
Bonds form between the atoms and molecules on the surface
Stage 4:Desorption of products
Bonds are broken as the product(s) desorb from the surface.
Stage 5:Diffusion of Product(s) away from the Surface
The products are then desorbed from the surface of the catalyst..
The given function is-Strictly increasing from 2 to infinityHen.pdf
The given function is:-
Strictly increasing from 2 to infinity
Hence, Strictly increasing:- (2,inf)Option 2
Strictly decreasing :- (-inf , 2) Option 4
Hence, Option- 2,4 are correct.
Solution
The given function is:-
Strictly increasing from 2 to infinity
Hence, Strictly increasing:- (2,inf)Option 2
Strictly decreasing :- (-inf , 2) Option 4
Hence, Option- 2,4 are correct..
OrderTest.javapublic class OrderTest { Get an arra.pdf
OrderTest.java
public class OrderTest {
/**
* Get an array of specified size and pass it to Order.order().
* Report the results.
*/
public static void main(String[] args) {
if (args.length != 1) {//1
System.out.println(\"Usage: java OrderTest sizeOfArray\ \"
+ \"\\tor\ \\tjava OrderTest arrayFile\");
System.exit(1);
}
// create or read the int[]
int size = 0;
int[] array = new int[0];//5
try {
size = Integer.parseInt(args[0]);
array = ArrayOfInts.randomizedArray(size);
} catch (NumberFormatException nfe) {//8
try {
array = ArrayOfInts.arrayFromFile(args[0]);
size = array.length;
} catch (Exception e) {
System.err.println(\"unable to read array from \" + args[0]);
System.exit(1);//14
}
}
System.out.println(\"before:\");//15
for (int i = 0; i < array.length; i++) {//2 n
System.out.printf(((i+1) % 10 > 0) ? \" %d\" : \" %d\ \", array[i]);//1
}
int myNum = Order.order(array); //this is the call we want to measure
System.out.println(\"\ after:\");//18
for (int i = 0; i < array.length; i++) {//2 n
System.out.printf(((i+1) % 10 > 0) ? \" %d\" : \" %d\ \", array[i]);
}
System.out.println(myNum);
}
}
ArrayOfInts.java
import java.io.BufferedWriter;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileWriter;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.InputMismatchException;
import java.util.Scanner;
public class ArrayOfInts {
/**
* Returns an array of consecutive ints from 1 to size.
*/
public static int[] orderedArray(int size) {
int[] a = new int[size];
for (int i = 0; i < size; i++) {
a[i] = i+1;
}
return a;
}
/**
* Returns a randomized array containing ints from 1 to size.
*/
public static int[] randomizedArray(int size) {
ArrayList aL = new ArrayList();
for (int i = 0; i < size; i++) {
aL.add(i+1);
}
Collections.shuffle(aL);
int[] a = new int[size];
for (int i = 0; i < size; i++) {
a[i] = aL.get(i);
}
return a;
}
/**
* Writes an int[] to a plain-text file with ints separated by spaces.
* Useful for creating input files for repeatable tests.
*/
public static void arrayToFile(int[] array, String outfile) {
try {
FileWriter fw = new FileWriter(outfile);
BufferedWriter bw = new BufferedWriter(fw);
PrintWriter outFile = new PrintWriter(bw);
for (int i : array) {
outFile.print(i + \" \");
}
outFile.close();
} catch (IOException e) {
System.err.println(\"Could not write to \" + outfile + \"\ \" + e);
}
}
/**
* Read ints from a file and return them in an int[]
*/
public static int[] arrayFromFile(String infile) throws FileNotFoundException,
InputMismatchException {
Scanner scan = new Scanner(new File(infile));
ArrayList aL = new ArrayList();
while (scan.hasNext()) {
aL.add(scan.nextInt());
}
scan.close();
int[] a = new int[aL.size()];
for (int i = 0; i < a.length; i++) {
a[i] = aL.get(i);
}
return a;
}
}
Order.java
public class Order {
/**
* Take an int[] and reorganize it so they are in ascending order.
*/
public static int order(int[] array) .
In the span of several decades, the Kingdom Protista has been disass.pdf
In the span of several decades, the Kingdom Protista has been disassembled because sequence
analyses have revealed new genetic (and therefore evolutionary) relationships among these
eukaryotes. Moreover, protists that exhibit similar morphological features may have evolved
analogous structures because of similar selective pressures—rather than because of recent
common ancestry. This phenomenon, called convergent evolution, is one reason why protist
classification is so challenging. The emerging classification scheme groups the entire domain
Eukaryota into six “supergroups” that contain all of the protists as well as animals, plants, and
fungi that evolved from a common ancestor. The supergroups are believed to be monophyletic,
meaning that all organisms within each supergroup are believed to have evolved from a single
common ancestor, and thus all members are most closely related to each other than to organisms
outside that group. There is still evidence lacking for the monophyly of some groups.
Many of the protist species classified into the supergroup Excavata are asymmetrical, single-
celled organisms with a feeding groove “excavated” from one side. This supergroup includes
heterotrophic predators, photosynthetic species, and parasites. Its subgroups are the diplomonads,
parabasalids, and euglenozoans.
Diplomonads
Among the Excavata are the diplomonads, which include the intestinal parasite, Giardia lamblia.
Until recently, these protists were believed to lack mitochondria. Mitochondrial remnant
organelles, called mitosomes, have since been identified in diplomonads, but these mitosomes
are essentially nonfunctional. Diplomonads exist in anaerobic environments and use alternative
pathways, such as glycolysis, to generate energy. Each diplomonad cell has two identical nuclei
and uses several flagella for locomotion.
Parabasalids
A second Excavata subgroup, the parabasalids, also exhibits semi-functional mitochondria. In
parabasalids, these structures function anaerobically and are called hydrogenosomes because
they produce hydrogen gas as a byproduct. Parabasalids move with flagella and membrane
rippling. Trichomonas vaginalis, a parabasalid that causes a sexually transmitted disease in
humans.
Euglenozoans
Euglenozoans includes parasites, heterotrophs, autotrophs, and mixotrophs, ranging in size from
10 to 500 µm. Euglenoids move through their aquatic habitats using two long flagella that guide
them toward light sources sensed by a primitive ocular organ called an eyespot. The familiar
genus, Euglena, encompasses some mixotrophic species that display a photosynthetic capability
only when light is present. In the dark, the chloroplasts of Euglena shrink up and temporarily
cease functioning, and the cells instead take up organic nutrients from their environment.
The human parasite, Trypanosoma brucei, belongs to a different subgroup of Euglenozoa, the
kinetoplastids. The kinetoplastid subgroup is named after the kinetoplast, a DNA mass carri.
Information privacy It refers to the collection of data and disse.pdf
Information privacy :
It refers to the collection of data and dissemination of the data.
It includes following issues those are
1. Data collection
2. Extract data from affected subjects
3. Controlled disclosureI
Option A is correct choice.
Solution
Information privacy :
It refers to the collection of data and dissemination of the data.
It includes following issues those are
1. Data collection
2. Extract data from affected subjects
3. Controlled disclosureI
Option A is correct choice..
H2SO4 --- 2H+ + SO42-1 mole of H2SO4 produces 2 moles of H+ ions.pdf
H2SO4 ---> 2H+ + SO42-
1 mole of H2SO4 produces 2 moles of H+ ions
So [H+ ] = 2* molarity of H2SO4 = 2 * 6M = 12 M
HNO3 ---> H+ + NO3-
1 mole of HNO3 produces 1 mole of H+ ion
So [H+ ] = 1* Molarity of HNO3 = 1*6M = 6M
So there is difference in the [H+ ] ion concentration by taking HNO3 instead of H2SO4
Solution
H2SO4 ---> 2H+ + SO42-
1 mole of H2SO4 produces 2 moles of H+ ions
So [H+ ] = 2* molarity of H2SO4 = 2 * 6M = 12 M
HNO3 ---> H+ + NO3-
1 mole of HNO3 produces 1 mole of H+ ion
So [H+ ] = 1* Molarity of HNO3 = 1*6M = 6M
So there is difference in the [H+ ] ion concentration by taking HNO3 instead of H2SO4.
Recommended
1.Cexplanation; Cloud consumers that use cloud-based IT resources .pdf
1.C
explanation; Cloud consumers that use cloud-based IT resources can generally lease them with a
pay-for-use model. With this model, cloud consumers pay a usage fee for only the amount of the
IT resource actually used, resulting in directly proportional costs. This gives an organization
access to IT resources without having to purchase its own, resulting in reduced investment
requirements
2.B
explanation
The availability of an IT resource is the probability that it will be accessible. The probability
value is generally expressed with a percentage representing the amount of time that the IT
resource is accessible during a given period. The percentage is calculated as follows: 1. Divide
the amount of hours the IT resource was unavailable (within a measured period) by the total
amount of hours within the measured period. 2. Multiply the value by 100. 3. Deduct the value
from 100
3.D
explanation;
Horizontal scaling refers to the allocation or releasing of resources of the same type. The
horizontal allocation of resources is referred to as scaling out and the horizontal releasing of
resources is referred to as scaling in. Horizontal scaling is a common form of scaling within
cloud environments
4.C
explanation;
Cloud-based IT resources that are used by a cloud consumer reside outside of the cloud
consumer’s organizational boundary. To use the IT resources, the cloud consumer will generally
need to trust them. As a result, the cloud consumer’s trust boundary is expanded beyond its
organizational boundary to encompass the cloud.
5.C
explanation;
An organization will often have existing IT resources that are already paid for and operational.
The prior investment that has been made in these on-premise IT resources is referred to as sunk
costs. When comparing cloud-based IT resources to on-premise IT resources with sunk costs,
the up-front costs for on-premise IT resources is significantly lower. It can therefore be more
difficult to justify the leasing of cloud- based IT resources as an alternative
6.D
explanation
the fact that trust boundaries overlap can lead to opportunities for an attacker to attack IT
resources shared by multiple cloud consumers
7.D
explanation
all r correct
8.B
explanation
Cloud A provides Cloud Service A as part of a failover system that encompasses a redundant
implementation of Cloud Service A on Cloud B. If Cloud Service A on Cloud A fails, then
Cloud Service A on Cloud B is automatically provisioned transparently to Cloud Service
Consumer A. Each cloud has a specific level of reliability and availability that it guarantees for
Cloud Service A. By spanning the failover system across both clouds, the overall reliability and
availability will be higher than the maximum reliability and availability of either cloud
9.C
explanation
Different physical and virtual IT resources are dynamically assigned and reassigned according to
cloud consumer demand, typically followed by execution through statistical multiplexing.
Resource poolin.
1) WBC count is high and this level indicates that the person has le.pdf
1) WBC count is high and this level indicates that the person has leucocytosis. This may be due
to some infections or stress etc.
2)the pH level is alkalotic and this is caused by the respiratory alkalosis. As the pH is less than
the normal level.
Solution
1) WBC count is high and this level indicates that the person has leucocytosis. This may be due
to some infections or stress etc.
2)the pH level is alkalotic and this is caused by the respiratory alkalosis. As the pH is less than
the normal level..
What are the four steps of a process involving a heterogeneous catal.pdf
What are the four steps of a process involving a heterogeneous catalyst?
Stage 1:Diffusion of Reactant(s) to the Surface
The rate at which reactants will diffuse to the surface will be influenced by their bulk
concentration and by the thickness of the boundary layer.
Stage 2:Adsorption of reactants
Bonds are formed as the reactant(s) are adsorbed onto the surface of the catalyst.
The ability for an atom or molecule to stick to the surface is known, brilliantly, as the Sticking
Co-efficient. This is just the ratio or percentage of molecules that end up sticking on the surface.
Stage 3:Reaction
Bonds form between the atoms and molecules on the surface
Stage 4:Desorption of products
Bonds are broken as the product(s) desorb from the surface.
Stage 5:Diffusion of Product(s) away from the Surface
The products are then desorbed from the surface of the catalyst.
Solution
What are the four steps of a process involving a heterogeneous catalyst?
Stage 1:Diffusion of Reactant(s) to the Surface
The rate at which reactants will diffuse to the surface will be influenced by their bulk
concentration and by the thickness of the boundary layer.
Stage 2:Adsorption of reactants
Bonds are formed as the reactant(s) are adsorbed onto the surface of the catalyst.
The ability for an atom or molecule to stick to the surface is known, brilliantly, as the Sticking
Co-efficient. This is just the ratio or percentage of molecules that end up sticking on the surface.
Stage 3:Reaction
Bonds form between the atoms and molecules on the surface
Stage 4:Desorption of products
Bonds are broken as the product(s) desorb from the surface.
Stage 5:Diffusion of Product(s) away from the Surface
The products are then desorbed from the surface of the catalyst..
The given function is-Strictly increasing from 2 to infinityHen.pdf
The given function is:-
Strictly increasing from 2 to infinity
Hence, Strictly increasing:- (2,inf)Option 2
Strictly decreasing :- (-inf , 2) Option 4
Hence, Option- 2,4 are correct.
Solution
The given function is:-
Strictly increasing from 2 to infinity
Hence, Strictly increasing:- (2,inf)Option 2
Strictly decreasing :- (-inf , 2) Option 4
Hence, Option- 2,4 are correct..
OrderTest.javapublic class OrderTest { Get an arra.pdf
OrderTest.java
public class OrderTest {
/**
* Get an array of specified size and pass it to Order.order().
* Report the results.
*/
public static void main(String[] args) {
if (args.length != 1) {//1
System.out.println(\"Usage: java OrderTest sizeOfArray\ \"
+ \"\\tor\ \\tjava OrderTest arrayFile\");
System.exit(1);
}
// create or read the int[]
int size = 0;
int[] array = new int[0];//5
try {
size = Integer.parseInt(args[0]);
array = ArrayOfInts.randomizedArray(size);
} catch (NumberFormatException nfe) {//8
try {
array = ArrayOfInts.arrayFromFile(args[0]);
size = array.length;
} catch (Exception e) {
System.err.println(\"unable to read array from \" + args[0]);
System.exit(1);//14
}
}
System.out.println(\"before:\");//15
for (int i = 0; i < array.length; i++) {//2 n
System.out.printf(((i+1) % 10 > 0) ? \" %d\" : \" %d\ \", array[i]);//1
}
int myNum = Order.order(array); //this is the call we want to measure
System.out.println(\"\ after:\");//18
for (int i = 0; i < array.length; i++) {//2 n
System.out.printf(((i+1) % 10 > 0) ? \" %d\" : \" %d\ \", array[i]);
}
System.out.println(myNum);
}
}
ArrayOfInts.java
import java.io.BufferedWriter;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileWriter;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.InputMismatchException;
import java.util.Scanner;
public class ArrayOfInts {
/**
* Returns an array of consecutive ints from 1 to size.
*/
public static int[] orderedArray(int size) {
int[] a = new int[size];
for (int i = 0; i < size; i++) {
a[i] = i+1;
}
return a;
}
/**
* Returns a randomized array containing ints from 1 to size.
*/
public static int[] randomizedArray(int size) {
ArrayList aL = new ArrayList();
for (int i = 0; i < size; i++) {
aL.add(i+1);
}
Collections.shuffle(aL);
int[] a = new int[size];
for (int i = 0; i < size; i++) {
a[i] = aL.get(i);
}
return a;
}
/**
* Writes an int[] to a plain-text file with ints separated by spaces.
* Useful for creating input files for repeatable tests.
*/
public static void arrayToFile(int[] array, String outfile) {
try {
FileWriter fw = new FileWriter(outfile);
BufferedWriter bw = new BufferedWriter(fw);
PrintWriter outFile = new PrintWriter(bw);
for (int i : array) {
outFile.print(i + \" \");
}
outFile.close();
} catch (IOException e) {
System.err.println(\"Could not write to \" + outfile + \"\ \" + e);
}
}
/**
* Read ints from a file and return them in an int[]
*/
public static int[] arrayFromFile(String infile) throws FileNotFoundException,
InputMismatchException {
Scanner scan = new Scanner(new File(infile));
ArrayList aL = new ArrayList();
while (scan.hasNext()) {
aL.add(scan.nextInt());
}
scan.close();
int[] a = new int[aL.size()];
for (int i = 0; i < a.length; i++) {
a[i] = aL.get(i);
}
return a;
}
}
Order.java
public class Order {
/**
* Take an int[] and reorganize it so they are in ascending order.
*/
public static int order(int[] array) .
In the span of several decades, the Kingdom Protista has been disass.pdf
In the span of several decades, the Kingdom Protista has been disassembled because sequence
analyses have revealed new genetic (and therefore evolutionary) relationships among these
eukaryotes. Moreover, protists that exhibit similar morphological features may have evolved
analogous structures because of similar selective pressures—rather than because of recent
common ancestry. This phenomenon, called convergent evolution, is one reason why protist
classification is so challenging. The emerging classification scheme groups the entire domain
Eukaryota into six “supergroups” that contain all of the protists as well as animals, plants, and
fungi that evolved from a common ancestor. The supergroups are believed to be monophyletic,
meaning that all organisms within each supergroup are believed to have evolved from a single
common ancestor, and thus all members are most closely related to each other than to organisms
outside that group. There is still evidence lacking for the monophyly of some groups.
Many of the protist species classified into the supergroup Excavata are asymmetrical, single-
celled organisms with a feeding groove “excavated” from one side. This supergroup includes
heterotrophic predators, photosynthetic species, and parasites. Its subgroups are the diplomonads,
parabasalids, and euglenozoans.
Diplomonads
Among the Excavata are the diplomonads, which include the intestinal parasite, Giardia lamblia.
Until recently, these protists were believed to lack mitochondria. Mitochondrial remnant
organelles, called mitosomes, have since been identified in diplomonads, but these mitosomes
are essentially nonfunctional. Diplomonads exist in anaerobic environments and use alternative
pathways, such as glycolysis, to generate energy. Each diplomonad cell has two identical nuclei
and uses several flagella for locomotion.
Parabasalids
A second Excavata subgroup, the parabasalids, also exhibits semi-functional mitochondria. In
parabasalids, these structures function anaerobically and are called hydrogenosomes because
they produce hydrogen gas as a byproduct. Parabasalids move with flagella and membrane
rippling. Trichomonas vaginalis, a parabasalid that causes a sexually transmitted disease in
humans.
Euglenozoans
Euglenozoans includes parasites, heterotrophs, autotrophs, and mixotrophs, ranging in size from
10 to 500 µm. Euglenoids move through their aquatic habitats using two long flagella that guide
them toward light sources sensed by a primitive ocular organ called an eyespot. The familiar
genus, Euglena, encompasses some mixotrophic species that display a photosynthetic capability
only when light is present. In the dark, the chloroplasts of Euglena shrink up and temporarily
cease functioning, and the cells instead take up organic nutrients from their environment.
The human parasite, Trypanosoma brucei, belongs to a different subgroup of Euglenozoa, the
kinetoplastids. The kinetoplastid subgroup is named after the kinetoplast, a DNA mass carri.
Information privacy It refers to the collection of data and disse.pdf
Information privacy :
It refers to the collection of data and dissemination of the data.
It includes following issues those are
1. Data collection
2. Extract data from affected subjects
3. Controlled disclosureI
Option A is correct choice.
Solution
Information privacy :
It refers to the collection of data and dissemination of the data.
It includes following issues those are
1. Data collection
2. Extract data from affected subjects
3. Controlled disclosureI
Option A is correct choice..
H2SO4 --- 2H+ + SO42-1 mole of H2SO4 produces 2 moles of H+ ions.pdf
H2SO4 ---> 2H+ + SO42-
1 mole of H2SO4 produces 2 moles of H+ ions
So [H+ ] = 2* molarity of H2SO4 = 2 * 6M = 12 M
HNO3 ---> H+ + NO3-
1 mole of HNO3 produces 1 mole of H+ ion
So [H+ ] = 1* Molarity of HNO3 = 1*6M = 6M
So there is difference in the [H+ ] ion concentration by taking HNO3 instead of H2SO4
Solution
H2SO4 ---> 2H+ + SO42-
1 mole of H2SO4 produces 2 moles of H+ ions
So [H+ ] = 2* molarity of H2SO4 = 2 * 6M = 12 M
HNO3 ---> H+ + NO3-
1 mole of HNO3 produces 1 mole of H+ ion
So [H+ ] = 1* Molarity of HNO3 = 1*6M = 6M
So there is difference in the [H+ ] ion concentration by taking HNO3 instead of H2SO4.
Gene editing of somatic cellsThere are grave concerns regarding th.pdf
Gene editing of somatic cells
There are grave concerns regarding the ethical and safety implications of this research. There is
also fear of the negative impact it could have on important work involving the use of genome-
editing techniques in somatic (non-reproductive) cells.
Genome-editing technologies may offer a powerful approach to treat many human diseases,
including HIV/AIDS, haemophilia, sickle-cell anaemia and several forms of cancer
genome editing in human embryos using current technologies could have unpredictable effects
on future generations. This makes it dangerous and ethically unacceptable. Such research could
be exploited for non-therapeutic modifications. We are concerned that a public outcry about such
an ethical breach could hinder a promising area of therapeutic development, namely making
genetic changes that cannot be inherited.
prospect of manipulating the genome of human embryos has raised debates and discussions
amongst scientists, regulators, and the public. It is a crucial conversation that we need to
participate in as we have done for other scientific advancements in the past from embryonic stem
cells research to ‘in vitro’ fertilization and, recently in the UK, mitochondrial replacement.
Gene editing of somatic cells is currently in clinical development for a variety of conditions. The
editing of genomes in human somatic cells certainly raises ethical questions, but is distinct from
germline gene modification, in that changes in the gene(s) do not persist beyond a single
generation. As to human germline editing, the EGE is of the view that there should be a
moratorium on gene editing of human embryos or gametes which would result in the
modification of the human genome.
Solution
Gene editing of somatic cells
There are grave concerns regarding the ethical and safety implications of this research. There is
also fear of the negative impact it could have on important work involving the use of genome-
editing techniques in somatic (non-reproductive) cells.
Genome-editing technologies may offer a powerful approach to treat many human diseases,
including HIV/AIDS, haemophilia, sickle-cell anaemia and several forms of cancer
genome editing in human embryos using current technologies could have unpredictable effects
on future generations. This makes it dangerous and ethically unacceptable. Such research could
be exploited for non-therapeutic modifications. We are concerned that a public outcry about such
an ethical breach could hinder a promising area of therapeutic development, namely making
genetic changes that cannot be inherited.
prospect of manipulating the genome of human embryos has raised debates and discussions
amongst scientists, regulators, and the public. It is a crucial conversation that we need to
participate in as we have done for other scientific advancements in the past from embryonic stem
cells research to ‘in vitro’ fertilization and, recently in the UK, mitochondrial replacement.
Gene editing of somatic.
Four parts of compare and contrastSolutionFour parts of compar.pdf
Comparing and contrasting involves four parts: the topics being compared, their similarities, their differences, and the significance of analyzing their similarities and differences.
Each Restriction enzymes has a unique restriction site and therefore.pdf
Each Restriction enzymes has a unique restriction site and therefore in a GFP-HindIII clone no
restriction patterns will be observed with EcoRI digestion until and unless one has prior
knowledge of all restriction sites present on the clone.
Solution
Each Restriction enzymes has a unique restriction site and therefore in a GFP-HindIII clone no
restriction patterns will be observed with EcoRI digestion until and unless one has prior
knowledge of all restriction sites present on the clone..
correctSolutioncorrect.pdf
This short document discusses a solution but provides no details about the problem or solution. It states that the solution is correct but does not explain what is being solved or how the solution works. The document on its own does not provide enough contextual information to generate an informative summary.
Code of main classpublic class LBmain { public static void m.pdf
Code of main class:
public class LBmain {
public static void main(String[]args)
{
LinkedStack list = new LinkedStack<>();
System.out.println(\"Let\'s make a List!\");
System.out.println(\"Push 3 times.\");
System.out.println(\"Check the size.\");
System.out.println(\"Peek the top element.\");
System.out.println(\"Pop three times.\");
System.out.println (\"The size now should be zero!\" + \"\ \");
list.Push(1);
list.Push(2);
list.Push(3);
System.out.println(list.toString());
list.Size();
list.Peek();
list.Pop();
list.Pop();
list.Pop();
list.Size();
}
public class LinkedStack implements Stack {
private int count;
private LinearNode top;
//-----------------------------------------------------------------
// Creates an empty stack using the default capacity.
//-----------------------------------------------------------------
public LinkedStack()
{
count = 0;
top = null;
}
@Override
public boolean IsEmpty()
{
if(top == null)
{
System.out.println(\"Stack is empty\");
}
return top == null;
}
@Override
public void Push(T element)
{
LinearNode current = new LinearNode<>(element);
current.setNext(top);
top = current;
count++;}
@Override
public T Pop()
{
T result;
System.out.println(\"Lets pop the top element!\");
if (count == 0)
{
System.out.println(\"Pop operation failed. \"+ \"The stack is empty.\");
}
result = top.getElement();
top = top.getNext();
count--;
System.out.println(\"The element that we have poped is :\" + result);
return result;
}
Override
public String toString()
{
String result = \"\ \";
LinearNode current = top;
while (current != null)
{
result += current.getElement() + \"\ \";
current = current.getNext();
}
return result + \"\";
}
@Override
public T Peek() {
System.out.println(\"Lets peek the top element!\");
if(count == 0)
{
System.out.println(\"Peek failed stack is empty\");
}
System.out.println(\"The element that we have peeked is: \" + top.getElement());
return top.getElement();
}
@Override
public int Size() {
System.out.println(\"The size of the list now is: \" + count);
return count;
}
}
main.Peek();
main.Pop();
main.Pop();
main.Size();
main.toString();
}
}
@Override
public void Push(T element)
{
LinearNode current = new LinearNode<>(element);
current.setNext(top);
top = current;
count++;
}
}
public class LinearNode {
private LinearNode next; //se guarda la referencia del Nodo
private T element; //Lista vacia
public LinearNode()
{
next = null;
element = null;
}
//-----------------------------------------------------------------
// Creates a node storing the specified element.
//-----------------------------------------------------------------
public LinearNode (T elem)
{
next = null;
element = elem;
}
//-----------------------------------------------------------------
// Returns the node that follows this one.
//-----------------------------------------------------------------
public LinearNode getNext()
{
return next;
}
//-----------------------------------------------------------------
// Sets the node that follows this one.
.
BibliographyHall, J. E. (2015). Guyton and Hall textbook of medic.pdf
Bibliography:
Hall, J. E. (2015). Guyton and Hall textbook of medical physiology. Elsevier Health Sciences.
De Luca Jr, L. A., Menani, J. V., & Johnson, A. K. (2014). Circumventricular Organs:
Integrators of Circulating Signals Controlling Hydration, Energy Balance, and Immune
Function--Neurobiology of Body Fluid Homeostasis: Transduction and Integration.
Sherwood, L. (2015). Human physiology: from cells to systems. Cengage learning.
Homeostatic Regulation of the Vascular System | Anatomy and Physiology II. (2016).
Courses.lumenlearning.com. Retrieved 25 November 2016, from
https://courses.lumenlearning.com/ap2/chapter/homeostatic-regulation-of-the-vascular-system
De Luca Jr, L. A., Menani, J. V., & Johnson, A. K. (2014). Preoptic–Periventricular Integrative
Mechanisms Involved in Behavior, Fluid–Electrolyte Balance, and Pressor Responses--
Neurobiology of Body Fluid Homeostasis: Transduction and Integration.
Keeping in mind the end goal to keep up homeostasis in the cardiovascular framework and give
sufficient blood to the tissues, blood stream must be diverted constantly to the tissues as they
turn out to be more dynamic. Undeniably, the cardiovascular framework takes part in asset
assignment, in light of the fact that there is insufficient blood stream to disseminate blood
similarly to all tissues at the same time. For instance, when an individual is working out, more
blood will be coordinated to skeletal muscles, the heart, and the lungs. Taking after a dinner,
more blood is coordinated to the stomach related framework. Just the mind gets a pretty much
steady supply of blood whether you are dynamic, resting, considering, or occupied with some
other activity.The sensory system assumes a basic part in the direction of vascular homeostasis.
The essential administrative destinations incorporate the cardiovascular focuses in the mind that
control both heart and vascular capacities. Also, more summed up neural reactions from the
limbic framework and the autonomic sensory system are variables.
The Cardiovascular Centers in the Brain
Neurological control of circulatory strain and stream relies on upon the cardiovascular focuses
situated in the medulla oblongata. This group of neurons reacts to changes in pulse and in
addition blood convergences of oxygen, carbon dioxide, and hydrogen particles. The
cardiovascular focus contains three particular combined segments:
The cardioaccelerator focuses empower cardiovascular capacity by controlling heart rate and
stroke volume by means of thoughtful incitement from the cardiovascular quickening agent
nerve.
The cardioinhibitor focuses moderate cardiovascular capacity by diminishing heart rate and
stroke volume through parasympathetic incitement from the vagus nerve.
The vasomotor focuses control vessel tone or withdrawal of the smooth muscle in the tunica
media. Changes in width influence fringe resistance, weight, and stream, which influence
cardiovascular yield. The lion\'s share of these neurons demonst.
synthesis .pdf
This very short document appears to be discussing the topic of synthesis or solution synthesis but provides no other context or details in just three words: "synthesis", "Solution", "synthesis". It is not possible to provide a meaningful summary in 3 sentences or less given the extremely limited information provided.
LiOH is a strong base so we assume it dissociates.pdf
LiOH is a strong base so we assume it dissociates completely. So there will be 8.82
x 10^-3 M of OH- pOH = -log[OH-] = -log(8.82 x 10^-3) = 2.05 pH = 14 - pOH = 14 - 2.05 =
11.95
Solution
LiOH is a strong base so we assume it dissociates completely. So there will be 8.82
x 10^-3 M of OH- pOH = -log[OH-] = -log(8.82 x 10^-3) = 2.05 pH = 14 - pOH = 14 - 2.05 =
11.95.
With Sp3d hybridization, a seesaw or linear shape.pdf
With Sp3d hybridization, a seesaw or linear shape is possible. The seesaw shape
occurs when the molecule has the form AB4E, where A is the central atom, B represents the
atoms attached to the central atom and E is a nonbonding pair of electrons attached to the central
atom. Linear is in the form AB2E3. So A is the central atom and should be by itself and the only
choice that fits that criteria is answer A (BrFs)
Solution
With Sp3d hybridization, a seesaw or linear shape is possible. The seesaw shape
occurs when the molecule has the form AB4E, where A is the central atom, B represents the
atoms attached to the central atom and E is a nonbonding pair of electrons attached to the central
atom. Linear is in the form AB2E3. So A is the central atom and should be by itself and the only
choice that fits that criteria is answer A (BrFs).
Two-photon transition probability .pdf
The document discusses two-photon transition probability. It involves the probability of an atom making a transition between two energy levels by absorbing or emitting two photons simultaneously. This is a higher-order optical process that is much weaker than single-photon transitions.
this is because Nitrogen has a lone pair and and .pdf
this is because Nitrogen has a lone pair and and the Carbon with chlorine is
electrophilic in nature and therefore, the lone pair attracted towards that lone pair.
Solution
this is because Nitrogen has a lone pair and and the Carbon with chlorine is
electrophilic in nature and therefore, the lone pair attracted towards that lone pair..
My opinion is to carry out in the complete absenc.pdf
My opinion is to carry out in the complete absence of water.
Solution
My opinion is to carry out in the complete absence of water..
cant say anything about no. of molecules, unless .pdf
cant say anything about no. of molecules, unless we know how many moles are
present. (A) not enough info
Solution
cant say anything about no. of molecules, unless we know how many moles are
present. (A) not enough info.
C-O (carbonlyl) peaks occur usually around 1500-1.pdf
C-O (carbonlyl) peaks occur usually around 1500-1700. I Cant clearly see the
picture but I would say you have at least 1-2 Carbonly groups, since there are 2 peaks that occur
past 1500.
Solution
C-O (carbonlyl) peaks occur usually around 1500-1700. I Cant clearly see the
picture but I would say you have at least 1-2 Carbonly groups, since there are 2 peaks that occur
past 1500..
Boiling points and solubility of alcohols are hig.pdf
Boiling points and solubility of alcohols are high because there are hydrogen bonds
Solution
Boiling points and solubility of alcohols are high because there are hydrogen bonds.
C. Bromobenzene undergoes addition reactions rath.pdf
C. Bromobenzene undergoes addition reactions rather than substitution reactions.
Solution
C. Bromobenzene undergoes addition reactions rather than substitution reactions..
AnswerMeasels (aka rubeola) and rubella (aka German Measels) are .pdf
Answer:
Measels (aka rubeola) and rubella (aka German Measels) are both caused by enveloped single
stranded RNA viruses and develop skin rash. Rubella (aka German Measels) disease is only lasts
for 3 days whereas \"rubeola\" always lasts for 7 days. Rubella (aka German Measels) is mainly
result in light red or spotted for 3 days finally result in swelling of lymph nodes and joint
swelling. Rubeola also known as regular measles and the resultant skin rashes are going to
generate reddish-brown rash but the symptoms are \"running nose and pyrexia\".
Solution
Answer:
Measels (aka rubeola) and rubella (aka German Measels) are both caused by enveloped single
stranded RNA viruses and develop skin rash. Rubella (aka German Measels) disease is only lasts
for 3 days whereas \"rubeola\" always lasts for 7 days. Rubella (aka German Measels) is mainly
result in light red or spotted for 3 days finally result in swelling of lymph nodes and joint
swelling. Rubeola also known as regular measles and the resultant skin rashes are going to
generate reddish-brown rash but the symptoms are \"running nose and pyrexia\"..
Answer It very much essential and plays very miportant role in th.pdf
Answer :
It very much essential and plays very miportant role in the regulation of the cell cycle,
differentiation, growth and cell senescence, all of which are critical to normal development.
A class of human genetic syndromes has emerged that are caused by germline mutations in genes
which encode components of the Ras/mitogen activated protein kinase (MAPK) pathway. This
pathway plays an essential role in the control of the cell cycle and differentiation, therefore its
dysregulation has profound developmental consequences.
These “RASopathies” exhibit unique phenotypic features, however, many share characteristic
overlapping features including craniofacial dysmorphology, cardiac, cutaneous malformations,
musculoskeletal and ocular abnormalities, varying degrees of neurocognitive impairment and in
some syndromes, an increased risk of developing cancer.Yes it act as second messenger for the
ccell cycyle to go on.
This RASopathy plays a very important role in the developement of T cells
Solution
Answer :
It very much essential and plays very miportant role in the regulation of the cell cycle,
differentiation, growth and cell senescence, all of which are critical to normal development.
A class of human genetic syndromes has emerged that are caused by germline mutations in genes
which encode components of the Ras/mitogen activated protein kinase (MAPK) pathway. This
pathway plays an essential role in the control of the cell cycle and differentiation, therefore its
dysregulation has profound developmental consequences.
These “RASopathies” exhibit unique phenotypic features, however, many share characteristic
overlapping features including craniofacial dysmorphology, cardiac, cutaneous malformations,
musculoskeletal and ocular abnormalities, varying degrees of neurocognitive impairment and in
some syndromes, an increased risk of developing cancer.Yes it act as second messenger for the
ccell cycyle to go on.
This RASopathy plays a very important role in the developement of T cells.
As entropy increases as we move from solid to liquid to gas, so S is.pdf
As entropy increases as we move from solid to liquid to gas, so S is clearly negative for B, D and
E.
In case of no change in the state, the entropy increases with the number of particles, so S is
negative for C and positive only for A.
Solution
As entropy increases as we move from solid to liquid to gas, so S is clearly negative for B, D and
E.
In case of no change in the state, the entropy increases with the number of particles, so S is
negative for C and positive only for A..
Advantages of peer to peer networksReasons why using a server woul.pdf
Advantages of peer to peer networks
Reasons why using a server would benefit a home network
Solution
Advantages of peer to peer networks
Reasons why using a server would benefit a home network.
Main Java[All of the Base Concepts}.docx
This is part 1 of my Java Learning Journey. This Contains Custom methods, classes, constructors, packages, multithreading , try- catch block, finally block and more.
More Related Content
More from akkhan101
Gene editing of somatic cellsThere are grave concerns regarding th.pdf
Gene editing of somatic cells
There are grave concerns regarding the ethical and safety implications of this research. There is
also fear of the negative impact it could have on important work involving the use of genome-
editing techniques in somatic (non-reproductive) cells.
Genome-editing technologies may offer a powerful approach to treat many human diseases,
including HIV/AIDS, haemophilia, sickle-cell anaemia and several forms of cancer
genome editing in human embryos using current technologies could have unpredictable effects
on future generations. This makes it dangerous and ethically unacceptable. Such research could
be exploited for non-therapeutic modifications. We are concerned that a public outcry about such
an ethical breach could hinder a promising area of therapeutic development, namely making
genetic changes that cannot be inherited.
prospect of manipulating the genome of human embryos has raised debates and discussions
amongst scientists, regulators, and the public. It is a crucial conversation that we need to
participate in as we have done for other scientific advancements in the past from embryonic stem
cells research to ‘in vitro’ fertilization and, recently in the UK, mitochondrial replacement.
Gene editing of somatic cells is currently in clinical development for a variety of conditions. The
editing of genomes in human somatic cells certainly raises ethical questions, but is distinct from
germline gene modification, in that changes in the gene(s) do not persist beyond a single
generation. As to human germline editing, the EGE is of the view that there should be a
moratorium on gene editing of human embryos or gametes which would result in the
modification of the human genome.
Solution
Gene editing of somatic cells
There are grave concerns regarding the ethical and safety implications of this research. There is
also fear of the negative impact it could have on important work involving the use of genome-
editing techniques in somatic (non-reproductive) cells.
Genome-editing technologies may offer a powerful approach to treat many human diseases,
including HIV/AIDS, haemophilia, sickle-cell anaemia and several forms of cancer
genome editing in human embryos using current technologies could have unpredictable effects
on future generations. This makes it dangerous and ethically unacceptable. Such research could
be exploited for non-therapeutic modifications. We are concerned that a public outcry about such
an ethical breach could hinder a promising area of therapeutic development, namely making
genetic changes that cannot be inherited.
prospect of manipulating the genome of human embryos has raised debates and discussions
amongst scientists, regulators, and the public. It is a crucial conversation that we need to
participate in as we have done for other scientific advancements in the past from embryonic stem
cells research to ‘in vitro’ fertilization and, recently in the UK, mitochondrial replacement.
Gene editing of somatic.
Four parts of compare and contrastSolutionFour parts of compar.pdf
Comparing and contrasting involves four parts: the topics being compared, their similarities, their differences, and the significance of analyzing their similarities and differences.
Each Restriction enzymes has a unique restriction site and therefore.pdf
Each Restriction enzymes has a unique restriction site and therefore in a GFP-HindIII clone no
restriction patterns will be observed with EcoRI digestion until and unless one has prior
knowledge of all restriction sites present on the clone.
Solution
Each Restriction enzymes has a unique restriction site and therefore in a GFP-HindIII clone no
restriction patterns will be observed with EcoRI digestion until and unless one has prior
knowledge of all restriction sites present on the clone..
correctSolutioncorrect.pdf
This short document discusses a solution but provides no details about the problem or solution. It states that the solution is correct but does not explain what is being solved or how the solution works. The document on its own does not provide enough contextual information to generate an informative summary.
Code of main classpublic class LBmain { public static void m.pdf
Code of main class:
public class LBmain {
public static void main(String[]args)
{
LinkedStack list = new LinkedStack<>();
System.out.println(\"Let\'s make a List!\");
System.out.println(\"Push 3 times.\");
System.out.println(\"Check the size.\");
System.out.println(\"Peek the top element.\");
System.out.println(\"Pop three times.\");
System.out.println (\"The size now should be zero!\" + \"\ \");
list.Push(1);
list.Push(2);
list.Push(3);
System.out.println(list.toString());
list.Size();
list.Peek();
list.Pop();
list.Pop();
list.Pop();
list.Size();
}
public class LinkedStack implements Stack {
private int count;
private LinearNode top;
//-----------------------------------------------------------------
// Creates an empty stack using the default capacity.
//-----------------------------------------------------------------
public LinkedStack()
{
count = 0;
top = null;
}
@Override
public boolean IsEmpty()
{
if(top == null)
{
System.out.println(\"Stack is empty\");
}
return top == null;
}
@Override
public void Push(T element)
{
LinearNode current = new LinearNode<>(element);
current.setNext(top);
top = current;
count++;}
@Override
public T Pop()
{
T result;
System.out.println(\"Lets pop the top element!\");
if (count == 0)
{
System.out.println(\"Pop operation failed. \"+ \"The stack is empty.\");
}
result = top.getElement();
top = top.getNext();
count--;
System.out.println(\"The element that we have poped is :\" + result);
return result;
}
Override
public String toString()
{
String result = \"\ \";
LinearNode current = top;
while (current != null)
{
result += current.getElement() + \"\ \";
current = current.getNext();
}
return result + \"\";
}
@Override
public T Peek() {
System.out.println(\"Lets peek the top element!\");
if(count == 0)
{
System.out.println(\"Peek failed stack is empty\");
}
System.out.println(\"The element that we have peeked is: \" + top.getElement());
return top.getElement();
}
@Override
public int Size() {
System.out.println(\"The size of the list now is: \" + count);
return count;
}
}
main.Peek();
main.Pop();
main.Pop();
main.Size();
main.toString();
}
}
@Override
public void Push(T element)
{
LinearNode current = new LinearNode<>(element);
current.setNext(top);
top = current;
count++;
}
}
public class LinearNode {
private LinearNode next; //se guarda la referencia del Nodo
private T element; //Lista vacia
public LinearNode()
{
next = null;
element = null;
}
//-----------------------------------------------------------------
// Creates a node storing the specified element.
//-----------------------------------------------------------------
public LinearNode (T elem)
{
next = null;
element = elem;
}
//-----------------------------------------------------------------
// Returns the node that follows this one.
//-----------------------------------------------------------------
public LinearNode getNext()
{
return next;
}
//-----------------------------------------------------------------
// Sets the node that follows this one.
.
BibliographyHall, J. E. (2015). Guyton and Hall textbook of medic.pdf
Bibliography:
Hall, J. E. (2015). Guyton and Hall textbook of medical physiology. Elsevier Health Sciences.
De Luca Jr, L. A., Menani, J. V., & Johnson, A. K. (2014). Circumventricular Organs:
Integrators of Circulating Signals Controlling Hydration, Energy Balance, and Immune
Function--Neurobiology of Body Fluid Homeostasis: Transduction and Integration.
Sherwood, L. (2015). Human physiology: from cells to systems. Cengage learning.
Homeostatic Regulation of the Vascular System | Anatomy and Physiology II. (2016).
Courses.lumenlearning.com. Retrieved 25 November 2016, from
https://courses.lumenlearning.com/ap2/chapter/homeostatic-regulation-of-the-vascular-system
De Luca Jr, L. A., Menani, J. V., & Johnson, A. K. (2014). Preoptic–Periventricular Integrative
Mechanisms Involved in Behavior, Fluid–Electrolyte Balance, and Pressor Responses--
Neurobiology of Body Fluid Homeostasis: Transduction and Integration.
Keeping in mind the end goal to keep up homeostasis in the cardiovascular framework and give
sufficient blood to the tissues, blood stream must be diverted constantly to the tissues as they
turn out to be more dynamic. Undeniably, the cardiovascular framework takes part in asset
assignment, in light of the fact that there is insufficient blood stream to disseminate blood
similarly to all tissues at the same time. For instance, when an individual is working out, more
blood will be coordinated to skeletal muscles, the heart, and the lungs. Taking after a dinner,
more blood is coordinated to the stomach related framework. Just the mind gets a pretty much
steady supply of blood whether you are dynamic, resting, considering, or occupied with some
other activity.The sensory system assumes a basic part in the direction of vascular homeostasis.
The essential administrative destinations incorporate the cardiovascular focuses in the mind that
control both heart and vascular capacities. Also, more summed up neural reactions from the
limbic framework and the autonomic sensory system are variables.
The Cardiovascular Centers in the Brain
Neurological control of circulatory strain and stream relies on upon the cardiovascular focuses
situated in the medulla oblongata. This group of neurons reacts to changes in pulse and in
addition blood convergences of oxygen, carbon dioxide, and hydrogen particles. The
cardiovascular focus contains three particular combined segments:
The cardioaccelerator focuses empower cardiovascular capacity by controlling heart rate and
stroke volume by means of thoughtful incitement from the cardiovascular quickening agent
nerve.
The cardioinhibitor focuses moderate cardiovascular capacity by diminishing heart rate and
stroke volume through parasympathetic incitement from the vagus nerve.
The vasomotor focuses control vessel tone or withdrawal of the smooth muscle in the tunica
media. Changes in width influence fringe resistance, weight, and stream, which influence
cardiovascular yield. The lion\'s share of these neurons demonst.
synthesis .pdf
This very short document appears to be discussing the topic of synthesis or solution synthesis but provides no other context or details in just three words: "synthesis", "Solution", "synthesis". It is not possible to provide a meaningful summary in 3 sentences or less given the extremely limited information provided.
LiOH is a strong base so we assume it dissociates.pdf
LiOH is a strong base so we assume it dissociates completely. So there will be 8.82
x 10^-3 M of OH- pOH = -log[OH-] = -log(8.82 x 10^-3) = 2.05 pH = 14 - pOH = 14 - 2.05 =
11.95
Solution
LiOH is a strong base so we assume it dissociates completely. So there will be 8.82
x 10^-3 M of OH- pOH = -log[OH-] = -log(8.82 x 10^-3) = 2.05 pH = 14 - pOH = 14 - 2.05 =
11.95.
With Sp3d hybridization, a seesaw or linear shape.pdf
With Sp3d hybridization, a seesaw or linear shape is possible. The seesaw shape
occurs when the molecule has the form AB4E, where A is the central atom, B represents the
atoms attached to the central atom and E is a nonbonding pair of electrons attached to the central
atom. Linear is in the form AB2E3. So A is the central atom and should be by itself and the only
choice that fits that criteria is answer A (BrFs)
Solution
With Sp3d hybridization, a seesaw or linear shape is possible. The seesaw shape
occurs when the molecule has the form AB4E, where A is the central atom, B represents the
atoms attached to the central atom and E is a nonbonding pair of electrons attached to the central
atom. Linear is in the form AB2E3. So A is the central atom and should be by itself and the only
choice that fits that criteria is answer A (BrFs).
Two-photon transition probability .pdf
The document discusses two-photon transition probability. It involves the probability of an atom making a transition between two energy levels by absorbing or emitting two photons simultaneously. This is a higher-order optical process that is much weaker than single-photon transitions.
this is because Nitrogen has a lone pair and and .pdf
this is because Nitrogen has a lone pair and and the Carbon with chlorine is
electrophilic in nature and therefore, the lone pair attracted towards that lone pair.
Solution
this is because Nitrogen has a lone pair and and the Carbon with chlorine is
electrophilic in nature and therefore, the lone pair attracted towards that lone pair..
My opinion is to carry out in the complete absenc.pdf
My opinion is to carry out in the complete absence of water.
Solution
My opinion is to carry out in the complete absence of water..
cant say anything about no. of molecules, unless .pdf
cant say anything about no. of molecules, unless we know how many moles are
present. (A) not enough info
Solution
cant say anything about no. of molecules, unless we know how many moles are
present. (A) not enough info.
C-O (carbonlyl) peaks occur usually around 1500-1.pdf
C-O (carbonlyl) peaks occur usually around 1500-1700. I Cant clearly see the
picture but I would say you have at least 1-2 Carbonly groups, since there are 2 peaks that occur
past 1500.
Solution
C-O (carbonlyl) peaks occur usually around 1500-1700. I Cant clearly see the
picture but I would say you have at least 1-2 Carbonly groups, since there are 2 peaks that occur
past 1500..
Boiling points and solubility of alcohols are hig.pdf
Boiling points and solubility of alcohols are high because there are hydrogen bonds
Solution
Boiling points and solubility of alcohols are high because there are hydrogen bonds.
C. Bromobenzene undergoes addition reactions rath.pdf
C. Bromobenzene undergoes addition reactions rather than substitution reactions.
Solution
C. Bromobenzene undergoes addition reactions rather than substitution reactions..
AnswerMeasels (aka rubeola) and rubella (aka German Measels) are .pdf
Answer:
Measels (aka rubeola) and rubella (aka German Measels) are both caused by enveloped single
stranded RNA viruses and develop skin rash. Rubella (aka German Measels) disease is only lasts
for 3 days whereas \"rubeola\" always lasts for 7 days. Rubella (aka German Measels) is mainly
result in light red or spotted for 3 days finally result in swelling of lymph nodes and joint
swelling. Rubeola also known as regular measles and the resultant skin rashes are going to
generate reddish-brown rash but the symptoms are \"running nose and pyrexia\".
Solution
Answer:
Measels (aka rubeola) and rubella (aka German Measels) are both caused by enveloped single
stranded RNA viruses and develop skin rash. Rubella (aka German Measels) disease is only lasts
for 3 days whereas \"rubeola\" always lasts for 7 days. Rubella (aka German Measels) is mainly
result in light red or spotted for 3 days finally result in swelling of lymph nodes and joint
swelling. Rubeola also known as regular measles and the resultant skin rashes are going to
generate reddish-brown rash but the symptoms are \"running nose and pyrexia\"..
Answer It very much essential and plays very miportant role in th.pdf
Answer :
It very much essential and plays very miportant role in the regulation of the cell cycle,
differentiation, growth and cell senescence, all of which are critical to normal development.
A class of human genetic syndromes has emerged that are caused by germline mutations in genes
which encode components of the Ras/mitogen activated protein kinase (MAPK) pathway. This
pathway plays an essential role in the control of the cell cycle and differentiation, therefore its
dysregulation has profound developmental consequences.
These “RASopathies” exhibit unique phenotypic features, however, many share characteristic
overlapping features including craniofacial dysmorphology, cardiac, cutaneous malformations,
musculoskeletal and ocular abnormalities, varying degrees of neurocognitive impairment and in
some syndromes, an increased risk of developing cancer.Yes it act as second messenger for the
ccell cycyle to go on.
This RASopathy plays a very important role in the developement of T cells
Solution
Answer :
It very much essential and plays very miportant role in the regulation of the cell cycle,
differentiation, growth and cell senescence, all of which are critical to normal development.
A class of human genetic syndromes has emerged that are caused by germline mutations in genes
which encode components of the Ras/mitogen activated protein kinase (MAPK) pathway. This
pathway plays an essential role in the control of the cell cycle and differentiation, therefore its
dysregulation has profound developmental consequences.
These “RASopathies” exhibit unique phenotypic features, however, many share characteristic
overlapping features including craniofacial dysmorphology, cardiac, cutaneous malformations,
musculoskeletal and ocular abnormalities, varying degrees of neurocognitive impairment and in
some syndromes, an increased risk of developing cancer.Yes it act as second messenger for the
ccell cycyle to go on.
This RASopathy plays a very important role in the developement of T cells.
As entropy increases as we move from solid to liquid to gas, so S is.pdf
As entropy increases as we move from solid to liquid to gas, so S is clearly negative for B, D and
E.
In case of no change in the state, the entropy increases with the number of particles, so S is
negative for C and positive only for A.
Solution
As entropy increases as we move from solid to liquid to gas, so S is clearly negative for B, D and
E.
In case of no change in the state, the entropy increases with the number of particles, so S is
negative for C and positive only for A..
Advantages of peer to peer networksReasons why using a server woul.pdf
Advantages of peer to peer networks
Reasons why using a server would benefit a home network
Solution
Advantages of peer to peer networks
Reasons why using a server would benefit a home network.
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