Their physical properties are so dofferent just because theintermole.pdf
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Their physical properties are so dofferent just because theintermolecular or interatomic forces are so different. Solution Their physical properties are so dofferent just because theintermolecular or interatomic forces are so different..
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y = 22Solutiony = 22.pdf
This short document defines the variable y and assigns it the value of 22. No further calculations or explanations are provided.
what ways are quorum sensing and the two component signaling system .pdf
what ways are quorum sensing and the two component signaling system in bacteria similar?
First we will see What is quorum sensing (QS). QS is a cell to cell communication in bacteria
that involve signaling molecules or autoinducers (AIs).
AIs come in different forms depending on bacterial species.
As one would expect, the molecular mechanism of QS is dependent on species, but there are four
main characteristics that govern all QS systems:
At low bacterial cell density (LBCD), AIs diffuse away and their cellular concentration is
approximately the same as the environment.
At high bacterial cell density (HBCD), there is a cumulative production of AIs, which leads to a
local high (environment) concentration. This is detected by the cells which in turn trigger
different kinds of responses.
AIs are detected by receptors that are either transmembrane molecules or are present in the
cytoplasm.
Detection of AIs leads to further production of AIs (feed forward loop) in addition to activating
other genes.
Qouram sensing and two component signaling system are inter connected and we will check how
it is connected: Cell-density-dependent gene expression appears to be widely spread in bacteria.
This quorum-sensing phenomenon has been well established in Gram-negative bacteria, where
N-acyl homoserine lactones are the diffusible communication molecules that modulate cell-
density-dependent phenotypes. Similarly, a variety of processes are known to be regulated in a
cell-density- or growth-phase-dependent manner in Gram-positive bacteria. Examples of such
quorum-sensing modes in Gram-positive bacteria are the development of genetic competence in
Bacillus subtilis andStreptococcus pneumoniae, the virulence response in Staphylococcus aureus,
and the production of antimicrobial peptides by several species of Gram-positive bacteria
including lactic acid bacteria. Cell-density-dependent regulatory modes in these systems appear
to follow a common theme, in which the signal molecule is a post-translationally processed
peptide that is secreted by a dedicated ATP-binding-cassette exporter. This secreted peptide
pheromone functions as the input signal for a specific sensor component of a two-component
signal-transduction system. Moreover, genetic linkage of the common elements involved results
in autoregulation of peptide-pheromone production.
What advantages would an enzyme coupled receptor have over a G-coupled receptor?
To understand the advantage of Enzyme-linked receptors over G-Protein Linked Receptors, we
have to see how these both types of receptors function in cellular level.
G-Protein Linked Receptors
G-protein-linked receptors bind a ligand and activate a membrane protein called a G-protein. The
activated G-protein then interacts with either an ion channel or an enzyme in the membrane. All
G-protein-linked receptors have seven transmembrane domains, but each receptor has its own
specific extracellular domain and G-protein-binding site.
Cell signaling using G.
Average.javaimport java.util.Scanner; import java.util.Arra.pdf
// Average.java
import java.util.Scanner;
import java.util.ArrayList;
public class Average
{
private double mean;
private int [] data;
public Average()
{
Scanner scan = new Scanner(System.in);
//Create an array with 5 elements.
data = new int[5];
//input the data into the array
for (int i = 0; i < data.length; i++)
{
System.out.print(\"Enter your score for test #\" + (i +1) + \": \");
data[i] = scan.nextInt();
}
}
public void calculateMean()
{
double sum = 0; //Accumulator
//calculate the sum of the test scores in the data array.
for (int i = 0; i < data.length; i++)
sum += data[i];
mean = (Double)sum/data.length;
}
public String toString()
{
String result = \"\ The data stored in array: \";
for (int i = 0; i < data.length; i++)
result = result + data[i] + \" \";
result = result + \"\ \" + \"Average: \" + mean + \"\ \";
return result;
}
}
// AverageDriver.java
public class AverageDriver
{
public static void main(String[] args)
{
Average avg = new Average();
avg.calculateMean();
System.out.println(avg.toString());
}
}
/*
output:
Enter your score for test #1: 87
Enter your score for test #2: 90
Enter your score for test #3: 92
Enter your score for test #4: 77
Enter your score for test #5: 89
The data stored in array: 87 90 92 77 89
Average: 87.0
*/
Solution
// Average.java
import java.util.Scanner;
import java.util.ArrayList;
public class Average
{
private double mean;
private int [] data;
public Average()
{
Scanner scan = new Scanner(System.in);
//Create an array with 5 elements.
data = new int[5];
//input the data into the array
for (int i = 0; i < data.length; i++)
{
System.out.print(\"Enter your score for test #\" + (i +1) + \": \");
data[i] = scan.nextInt();
}
}
public void calculateMean()
{
double sum = 0; //Accumulator
//calculate the sum of the test scores in the data array.
for (int i = 0; i < data.length; i++)
sum += data[i];
mean = (Double)sum/data.length;
}
public String toString()
{
String result = \"\ The data stored in array: \";
for (int i = 0; i < data.length; i++)
result = result + data[i] + \" \";
result = result + \"\ \" + \"Average: \" + mean + \"\ \";
return result;
}
}
// AverageDriver.java
public class AverageDriver
{
public static void main(String[] args)
{
Average avg = new Average();
avg.calculateMean();
System.out.println(avg.toString());
}
}
/*
output:
Enter your score for test #1: 87
Enter your score for test #2: 90
Enter your score for test #3: 92
Enter your score for test #4: 77
Enter your score for test #5: 89
The data stored in array: 87 90 92 77 89
Average: 87.0
*/.
C. 0.20 M NaCl Solution C..pdf
This document describes a 0.20 M NaCl solution. A 0.20 M NaCl solution contains 0.20 moles of sodium chloride dissolved in 1 liter of water. This concentration of sodium chloride solution is commonly used in chemistry experiments and labs.
yep in water which is higly polar will dissolve i.pdf
yep in water which is higly polar will dissolve ionic NaCl but not nonpolar steric
acid yes
Solution
yep in water which is higly polar will dissolve ionic NaCl but not nonpolar steric
acid yes.
When thinking of E or Z, the first thing that com.pdf
When thinking of E or Z, the first thing that comes to mind is the double bond of
the alkene. From there, you look at the atom attach to the double bond of the alkene. There are 4
atoms: 1) Carbon (top left) : bonded to 1 carbon and 2 hydrogen 2) Carbon (bottom left) : bonded
to 2 Oxygen [treat double bond here as two bonds] and 1 Hydrogen 3) Carbon (top right) :
bonded to 2 carbon and 1 hydrogen 4) Hydrogen (bottom right) : bonded to nothing. On each
side of the double bond, rank the priority (with respect to molecular mass) Bottom left >> Top
left Top right >> bottom right since the two groups are trans each other, this is an E isomer
Solution
When thinking of E or Z, the first thing that comes to mind is the double bond of
the alkene. From there, you look at the atom attach to the double bond of the alkene. There are 4
atoms: 1) Carbon (top left) : bonded to 1 carbon and 2 hydrogen 2) Carbon (bottom left) : bonded
to 2 Oxygen [treat double bond here as two bonds] and 1 Hydrogen 3) Carbon (top right) :
bonded to 2 carbon and 1 hydrogen 4) Hydrogen (bottom right) : bonded to nothing. On each
side of the double bond, rank the priority (with respect to molecular mass) Bottom left >> Top
left Top right >> bottom right since the two groups are trans each other, this is an E isomer.
Since it contains oxygen double bonded to a carbo.pdf
Since it contains oxygen double bonded to a carbon atom , it may beCaboxylic acid
or carbonyl compound. So the characteristic IRabsorption band is aroud 1700 cm -1
Solution
Since it contains oxygen double bonded to a carbon atom , it may beCaboxylic acid
or carbonyl compound. So the characteristic IRabsorption band is aroud 1700 cm -1.
Recommended
y = 22Solutiony = 22.pdf
This short document defines the variable y and assigns it the value of 22. No further calculations or explanations are provided.
what ways are quorum sensing and the two component signaling system .pdf
what ways are quorum sensing and the two component signaling system in bacteria similar?
First we will see What is quorum sensing (QS). QS is a cell to cell communication in bacteria
that involve signaling molecules or autoinducers (AIs).
AIs come in different forms depending on bacterial species.
As one would expect, the molecular mechanism of QS is dependent on species, but there are four
main characteristics that govern all QS systems:
At low bacterial cell density (LBCD), AIs diffuse away and their cellular concentration is
approximately the same as the environment.
At high bacterial cell density (HBCD), there is a cumulative production of AIs, which leads to a
local high (environment) concentration. This is detected by the cells which in turn trigger
different kinds of responses.
AIs are detected by receptors that are either transmembrane molecules or are present in the
cytoplasm.
Detection of AIs leads to further production of AIs (feed forward loop) in addition to activating
other genes.
Qouram sensing and two component signaling system are inter connected and we will check how
it is connected: Cell-density-dependent gene expression appears to be widely spread in bacteria.
This quorum-sensing phenomenon has been well established in Gram-negative bacteria, where
N-acyl homoserine lactones are the diffusible communication molecules that modulate cell-
density-dependent phenotypes. Similarly, a variety of processes are known to be regulated in a
cell-density- or growth-phase-dependent manner in Gram-positive bacteria. Examples of such
quorum-sensing modes in Gram-positive bacteria are the development of genetic competence in
Bacillus subtilis andStreptococcus pneumoniae, the virulence response in Staphylococcus aureus,
and the production of antimicrobial peptides by several species of Gram-positive bacteria
including lactic acid bacteria. Cell-density-dependent regulatory modes in these systems appear
to follow a common theme, in which the signal molecule is a post-translationally processed
peptide that is secreted by a dedicated ATP-binding-cassette exporter. This secreted peptide
pheromone functions as the input signal for a specific sensor component of a two-component
signal-transduction system. Moreover, genetic linkage of the common elements involved results
in autoregulation of peptide-pheromone production.
What advantages would an enzyme coupled receptor have over a G-coupled receptor?
To understand the advantage of Enzyme-linked receptors over G-Protein Linked Receptors, we
have to see how these both types of receptors function in cellular level.
G-Protein Linked Receptors
G-protein-linked receptors bind a ligand and activate a membrane protein called a G-protein. The
activated G-protein then interacts with either an ion channel or an enzyme in the membrane. All
G-protein-linked receptors have seven transmembrane domains, but each receptor has its own
specific extracellular domain and G-protein-binding site.
Cell signaling using G.
Average.javaimport java.util.Scanner; import java.util.Arra.pdf
// Average.java
import java.util.Scanner;
import java.util.ArrayList;
public class Average
{
private double mean;
private int [] data;
public Average()
{
Scanner scan = new Scanner(System.in);
//Create an array with 5 elements.
data = new int[5];
//input the data into the array
for (int i = 0; i < data.length; i++)
{
System.out.print(\"Enter your score for test #\" + (i +1) + \": \");
data[i] = scan.nextInt();
}
}
public void calculateMean()
{
double sum = 0; //Accumulator
//calculate the sum of the test scores in the data array.
for (int i = 0; i < data.length; i++)
sum += data[i];
mean = (Double)sum/data.length;
}
public String toString()
{
String result = \"\ The data stored in array: \";
for (int i = 0; i < data.length; i++)
result = result + data[i] + \" \";
result = result + \"\ \" + \"Average: \" + mean + \"\ \";
return result;
}
}
// AverageDriver.java
public class AverageDriver
{
public static void main(String[] args)
{
Average avg = new Average();
avg.calculateMean();
System.out.println(avg.toString());
}
}
/*
output:
Enter your score for test #1: 87
Enter your score for test #2: 90
Enter your score for test #3: 92
Enter your score for test #4: 77
Enter your score for test #5: 89
The data stored in array: 87 90 92 77 89
Average: 87.0
*/
Solution
// Average.java
import java.util.Scanner;
import java.util.ArrayList;
public class Average
{
private double mean;
private int [] data;
public Average()
{
Scanner scan = new Scanner(System.in);
//Create an array with 5 elements.
data = new int[5];
//input the data into the array
for (int i = 0; i < data.length; i++)
{
System.out.print(\"Enter your score for test #\" + (i +1) + \": \");
data[i] = scan.nextInt();
}
}
public void calculateMean()
{
double sum = 0; //Accumulator
//calculate the sum of the test scores in the data array.
for (int i = 0; i < data.length; i++)
sum += data[i];
mean = (Double)sum/data.length;
}
public String toString()
{
String result = \"\ The data stored in array: \";
for (int i = 0; i < data.length; i++)
result = result + data[i] + \" \";
result = result + \"\ \" + \"Average: \" + mean + \"\ \";
return result;
}
}
// AverageDriver.java
public class AverageDriver
{
public static void main(String[] args)
{
Average avg = new Average();
avg.calculateMean();
System.out.println(avg.toString());
}
}
/*
output:
Enter your score for test #1: 87
Enter your score for test #2: 90
Enter your score for test #3: 92
Enter your score for test #4: 77
Enter your score for test #5: 89
The data stored in array: 87 90 92 77 89
Average: 87.0
*/.
C. 0.20 M NaCl Solution C..pdf
This document describes a 0.20 M NaCl solution. A 0.20 M NaCl solution contains 0.20 moles of sodium chloride dissolved in 1 liter of water. This concentration of sodium chloride solution is commonly used in chemistry experiments and labs.
yep in water which is higly polar will dissolve i.pdf
yep in water which is higly polar will dissolve ionic NaCl but not nonpolar steric
acid yes
Solution
yep in water which is higly polar will dissolve ionic NaCl but not nonpolar steric
acid yes.
When thinking of E or Z, the first thing that com.pdf
When thinking of E or Z, the first thing that comes to mind is the double bond of
the alkene. From there, you look at the atom attach to the double bond of the alkene. There are 4
atoms: 1) Carbon (top left) : bonded to 1 carbon and 2 hydrogen 2) Carbon (bottom left) : bonded
to 2 Oxygen [treat double bond here as two bonds] and 1 Hydrogen 3) Carbon (top right) :
bonded to 2 carbon and 1 hydrogen 4) Hydrogen (bottom right) : bonded to nothing. On each
side of the double bond, rank the priority (with respect to molecular mass) Bottom left >> Top
left Top right >> bottom right since the two groups are trans each other, this is an E isomer
Solution
When thinking of E or Z, the first thing that comes to mind is the double bond of
the alkene. From there, you look at the atom attach to the double bond of the alkene. There are 4
atoms: 1) Carbon (top left) : bonded to 1 carbon and 2 hydrogen 2) Carbon (bottom left) : bonded
to 2 Oxygen [treat double bond here as two bonds] and 1 Hydrogen 3) Carbon (top right) :
bonded to 2 carbon and 1 hydrogen 4) Hydrogen (bottom right) : bonded to nothing. On each
side of the double bond, rank the priority (with respect to molecular mass) Bottom left >> Top
left Top right >> bottom right since the two groups are trans each other, this is an E isomer.
Since it contains oxygen double bonded to a carbo.pdf
Since it contains oxygen double bonded to a carbon atom , it may beCaboxylic acid
or carbonyl compound. So the characteristic IRabsorption band is aroud 1700 cm -1
Solution
Since it contains oxygen double bonded to a carbon atom , it may beCaboxylic acid
or carbonyl compound. So the characteristic IRabsorption band is aroud 1700 cm -1.
salicylic acid alcohol (R-OH) and carboxylic aci.pdf
salicylic acid: alcohol (R-OH) and carboxylic acid (R-COOH) acetylsalicylic acid
(aspirin): carboxylic acid and ester (R-COOR)
Solution
salicylic acid: alcohol (R-OH) and carboxylic acid (R-COOH) acetylsalicylic acid
(aspirin): carboxylic acid and ester (R-COOR).
no it doesnot .pdf
This very short document does not seem to provide any clear information that can be summarized in 3 sentences or less. It contains the phrase "no it doesnot" repeated but no other context or details.
It would be Ruthenium (II) Sulfide Ruthenium is .pdf
It would be Ruthenium (II) Sulfide Ruthenium is a transition metal, meaning that it
can have multiple oxidation numbers. Sulfur always takes the charge of -2, so from that we see
that there is a 1 to 1 relation between Ru and S in the empirical formula, so that means that
Ruthenium has the opposite charge, +2.
Solution
It would be Ruthenium (II) Sulfide Ruthenium is a transition metal, meaning that it
can have multiple oxidation numbers. Sulfur always takes the charge of -2, so from that we see
that there is a 1 to 1 relation between Ru and S in the empirical formula, so that means that
Ruthenium has the opposite charge, +2..
ketones do not give tollens test.. hence no react.pdf
ketones do not give tollens test.. hence no reaction
Solution
ketones do not give tollens test.. hence no reaction.
for IONIC the metal name followed by nonmetal io.pdf
for IONIC: the metal name followed by nonmetal ion (-ide) for ex: NaCl is Sodium
Chloride or MgS is Magnesium Sulfide
Solution
for IONIC: the metal name followed by nonmetal ion (-ide) for ex: NaCl is Sodium
Chloride or MgS is Magnesium Sulfide.
pH = 4-log2.4 = 3.62 pOH 14 - pH = 14-3.62 = 10.3.pdf
pH = 4-log2.4 = 3.62 pOH 14 - pH = 14-3.62 = 10.38 concentration of OH- = 10^(-
14) / (2.4*10^(-4)) = 4.167* 10^(-11)
Solution
pH = 4-log2.4 = 3.62 pOH 14 - pH = 14-3.62 = 10.38 concentration of OH- = 10^(-
14) / (2.4*10^(-4)) = 4.167* 10^(-11).
This suggest that dissociation of FeCl3 is 100 .pdf
This suggest that dissociation of FeCl3 is 100 % to form 4 ions (1Fe+3Cl). Other
dissociation also may be possible like FeCl3---->(FeCl2) + Cl- FeCl3---->(FeCl)2+ + 2Cl- so
number of ions is averaged considering dissociation constants which is 3.4
Solution
This suggest that dissociation of FeCl3 is 100 % to form 4 ions (1Fe+3Cl). Other
dissociation also may be possible like FeCl3---->(FeCl2) + Cl- FeCl3---->(FeCl)2+ + 2Cl- so
number of ions is averaged considering dissociation constants which is 3.4.
Total probability of dot received P = (37)(78).pdf
Total probability of dot received P = (3/7)*(7/8)+(4/7)*(1/8)=0.4464 probability
that dot sentP1 = 3/7=0.4285 so P(dot sent | dot received).= P1/P=0.4285/.4464=0.96
Solution
Total probability of dot received P = (3/7)*(7/8)+(4/7)*(1/8)=0.4464 probability
that dot sentP1 = 3/7=0.4285 so P(dot sent | dot received).= P1/P=0.4285/.4464=0.96.
Environmental contaminants of recent concern are .pdf
Environmental contaminants of recent concern are pharmaceuticals, estrogens and
other endocrine disrupting chemicals (EDC) such as degradation products of surfactants, algal
and cyanobacterial toxins, disinfection by-products (DBPs) and metalloids. In addition,
pesticides (especially their transformation products), microorganisms, and humic substances
(HS), in their function as vehicles for contaminants and as precursors for by-products in water
treatment, traditionally play an important role. The present status of the application of LC-MS
techniques for these water constituents are discussed and examples of application are given.
Solid-phase extraction with various non-selective materials in combination with liquid
chromatography (LC) on reversed-phase columns have been the most widely used methods for
sample preconcentration and separation for different compound classes like pesticides,
pharmaceuticals or estrogens. Electrospray ionization (ESI) and atmospheric pressure ionization
(APCI) are the most frequently used ionization techniques for polar and ionic compounds, as
well as for less polar non-ionic ones. The facilities of LC-MS have been successfully
demonstrated for different compound classes. Polar compounds from pharmaceuticals used as
betablockers, iodinated X-ray contrast media, or estrogens have been determined without
derivatization down to ultratrace concentrations. LC-MS can be viewed as a prerequisite for the
determination of algal and cyanobacterial toxins and the homologues and oligomers of
alkylphenol ethoxylates and their metabolites. Tandem mass spectrometric techniques and the
use of diagnostic ions reveal their usefulness for compound-class specific screening and
unknown identification, and are also valid for the analysis of pesticides and especially for their
transformation products.
Solution
Environmental contaminants of recent concern are pharmaceuticals, estrogens and
other endocrine disrupting chemicals (EDC) such as degradation products of surfactants, algal
and cyanobacterial toxins, disinfection by-products (DBPs) and metalloids. In addition,
pesticides (especially their transformation products), microorganisms, and humic substances
(HS), in their function as vehicles for contaminants and as precursors for by-products in water
treatment, traditionally play an important role. The present status of the application of LC-MS
techniques for these water constituents are discussed and examples of application are given.
Solid-phase extraction with various non-selective materials in combination with liquid
chromatography (LC) on reversed-phase columns have been the most widely used methods for
sample preconcentration and separation for different compound classes like pesticides,
pharmaceuticals or estrogens. Electrospray ionization (ESI) and atmospheric pressure ionization
(APCI) are the most frequently used ionization techniques for polar and ionic compounds, as
well as for less polar non-ionic ones. The facilities of .
C-H polar covalent bond Solution .pdf
The document discusses C-H polar covalent bonds. C-H bonds have some ionic character due to the difference in electronegativity between carbon and hydrogen. Hydrogen is more electronegative so it pulls the electrons in the bond slightly toward itself, giving the bond some polar character though it remains primarily covalent.
To insert an element at given position in arrayInsert(stack s1, s.pdf
To insert an element at given position in array:
Insert(stack s1, stack s2, int value, int pos)
//Inserts a value into the array at specified position.
//Assume s1 is a stack which holds the values, and s2 is an empty stack.
while(!s1.empty()) //Till the stack is empty.
s2.push(s1.pop()) //Pop element from s1, and push it in s2.
//By now, all elements are placed in reverse order from s1 to s2.
for i = 1 to pos - 1 //Till the pos-1
s1.push(s2.pop()) //Remove elements from s2, and push them to s1.
s1.push(value) //Now push your element in specified position pos.
while(!s2.empty()) //Push all the remaining elements in s2 to s1.
s1.push(s2.pop())
To delete an element at given position in array:
Delete(stack s1, stack s2, int pos)
//Deletes a value into the array at specified position.
//Assume s1 is a stack which holds the values, and s2 is an empty stack.
while(!s1.empty()) //Till the stack is empty.
s2.push(s1.pop()) //Pop element from s1, and push it in s2.
//By now, all elements are placed in reverse order from s1 to s2.
for i = 1 to pos - 1 //Till the pos-1
s1.push(s2.pop()) //Remove elements from s2, and push them to s1.
s1.pop(value) //Now pop your element in specified position pos.
while(!s2.empty()) //Push all the remaining elements in s2 to s1.
s1.push(s2.pop())
To delete a given element in array:
DeleteVal(stack s1, stack s2, int value)
//Deletes a value from the array.
//Assume s1 is a stack which holds the values, and s2 is an empty stack.
while(!s1.empty()) //Till the stack is empty.
s2.push(s1.pop()) //Pop element from s1, and push it in s2.
//By now, all elements are placed in reverse order from s1 to s2.
ele = s2.peek();
while(ele != value || !s2.empty()) //Till you found the element to be deleted.
s1.push(s2.pop()) //Remove elements from s2, and push them to s1.
ele = s2.peek()
s2.pop(value) //If element is found, delete it.
while(!s2.empty()) //Push all the remaining elements in s2 to s1.
s1.push(s2.pop())
If you have any queries with this, just get back to me.
Solution
To insert an element at given position in array:
Insert(stack s1, stack s2, int value, int pos)
//Inserts a value into the array at specified position.
//Assume s1 is a stack which holds the values, and s2 is an empty stack.
while(!s1.empty()) //Till the stack is empty.
s2.push(s1.pop()) //Pop element from s1, and push it in s2.
//By now, all elements are placed in reverse order from s1 to s2.
for i = 1 to pos - 1 //Till the pos-1
s1.push(s2.pop()) //Remove elements from s2, and push them to s1.
s1.push(value) //Now push your element in specified position pos.
while(!s2.empty()) //Push all the remaining elements in s2 to s1.
s1.push(s2.pop())
To delete an element at given position in array:
Delete(stack s1, stack s2, int pos)
//Deletes a value into the array at specified position.
//Assume s1 is a stack which holds the values, and s2 is an empty stack.
while(!s1.empty()) //Till the stack is empty.
s2.push(s1.pop()) //Pop element from s1, and push it in s2.
//By .
Chemical weathering is the breakdown of rock by c.pdf
Chemical weathering is the breakdown of rock by chemical reaction. In this process
the minerals within the rock are changed into particles that can be easily carried away. Air and
water are both involved in many complex chemical reactions. The minerals in igneous rocks may
be unstable under normal atmospheric conditions, those formed at higher temperatures being
more readily attacked than those formed at lower temperatures. Igneous rocks are commonly
attacked by water, particularly acid or alkaline solutions, and all of the common igneous rock
forming minerals (with the exception of quartz, which is very resistant) are changed in this way
into clay minerals and chemicals in solution. Rock particles in the form of clay, silt, sand, and
gravel are transported by the agents of erosion (usually water, and less frequently, ice and wind)
to new locations and redeposited in layers, generally at a lower elevation. These agents reduce
the size of the particles, sort them by size, and then deposit them in new locations. The sediments
dropped by streams and rivers form alluvial fans, flood plains, deltas, and on the bottom of lakes
and the sea floor. The wind may move large amounts of sand and other smaller particles.
Glaciers transport and deposit great quantities of usually unsorted rock material as till. These
deposited particles eventually become compacted and cemented together, forming clastic
sedimentary rocks. Such rocks contain inert minerals that resist mechanical and chemical
breakdown, such as quartz. Quartz is one of the most mechanically and chemically resistant
minerals. Highly weathered sediments can contain several heavy and stable minerals, best
illustrated by the ZTR index.
Solution
Chemical weathering is the breakdown of rock by chemical reaction. In this process
the minerals within the rock are changed into particles that can be easily carried away. Air and
water are both involved in many complex chemical reactions. The minerals in igneous rocks may
be unstable under normal atmospheric conditions, those formed at higher temperatures being
more readily attacked than those formed at lower temperatures. Igneous rocks are commonly
attacked by water, particularly acid or alkaline solutions, and all of the common igneous rock
forming minerals (with the exception of quartz, which is very resistant) are changed in this way
into clay minerals and chemicals in solution. Rock particles in the form of clay, silt, sand, and
gravel are transported by the agents of erosion (usually water, and less frequently, ice and wind)
to new locations and redeposited in layers, generally at a lower elevation. These agents reduce
the size of the particles, sort them by size, and then deposit them in new locations. The sediments
dropped by streams and rivers form alluvial fans, flood plains, deltas, and on the bottom of lakes
and the sea floor. The wind may move large amounts of sand and other smaller particles.
Glaciers transport and deposit great quantities of.
bu puting your values in formula You need not to.pdf
bu puting your values in formula You need not to be concerned about the exact
mechanism of the reaction. First, try writing out the balanced equation. 4Li + O2 --->2Li2O So
which element was oxidized? Li does not have a charge while LiO is an ionic compound in
which electrons are \"shared\" according to electronegativity. In this case oxygen is much more
electronegative and it essentially obtains two electrons from Li through bonding. Now write out
the half reactions. Oxygen is reduced. O2 + 4e- ---> 2O (with a 2- charge) Li is oxidized. 4Li--
->4e- + 4Li (with a 1+ charge) *So overall O2 + 4Li + 4e- ---> 2O (2-) + 4Li (1+) + 4e- A
third reaction is also important. The products of the latter two steps combine to form 2Li2O. 2O
(2-) + 4Li (1+) ---> 2Li2O Combine this with the above equation and you will return to the
original balanced equation.
Solution
bu puting your values in formula You need not to be concerned about the exact
mechanism of the reaction. First, try writing out the balanced equation. 4Li + O2 --->2Li2O So
which element was oxidized? Li does not have a charge while LiO is an ionic compound in
which electrons are \"shared\" according to electronegativity. In this case oxygen is much more
electronegative and it essentially obtains two electrons from Li through bonding. Now write out
the half reactions. Oxygen is reduced. O2 + 4e- ---> 2O (with a 2- charge) Li is oxidized. 4Li--
->4e- + 4Li (with a 1+ charge) *So overall O2 + 4Li + 4e- ---> 2O (2-) + 4Li (1+) + 4e- A
third reaction is also important. The products of the latter two steps combine to form 2Li2O. 2O
(2-) + 4Li (1+) ---> 2Li2O Combine this with the above equation and you will return to the
original balanced equation..
Please post the figure.SolutionPlease post the figure..pdf
The document requests that a figure be posted but does not provide any context or details about the figure. It repeats the phrase "Please post the figure" twice without any additional information about the figure, its purpose, or why it should be posted. The summary is unable to extract any meaningful high-level information from the given document.
option B,C,D is true. gantt chart is used for project scheduling,tra.pdf
option B,C,D is true. gantt chart is used for project scheduling,tracking. it mainky uses the
technique of crtical path method.
Solution
option B,C,D is true. gantt chart is used for project scheduling,tracking. it mainky uses the
technique of crtical path method..
The mole is a unit of measurement for the amount .pdf
The mole is a unit of measurement for the amount of substance or chemical amount.
It is one of the base units in the International System of Units, and has the unit symbol mol. n
ionic compound is a chemical compound in which ions are held together in a lattice structure by
ionic bonds. Usually, the positively charged portion consists of metal cations and the negatively
charged portion is an anion or polyatomic ion. Ions in ionic compounds are held together by the
electrostatic force between oppositely charged bodies. Ionic compounds have a high melting and
boiling point, and they are hard and very brittle.
Solution
The mole is a unit of measurement for the amount of substance or chemical amount.
It is one of the base units in the International System of Units, and has the unit symbol mol. n
ionic compound is a chemical compound in which ions are held together in a lattice structure by
ionic bonds. Usually, the positively charged portion consists of metal cations and the negatively
charged portion is an anion or polyatomic ion. Ions in ionic compounds are held together by the
electrostatic force between oppositely charged bodies. Ionic compounds have a high melting and
boiling point, and they are hard and very brittle..
Ideally birds evolved from Dinosaurs, but they are closely related t.pdf
Ideally birds evolved from Dinosaurs, but they are closely related to even reptiles. Biologists use
2 types of classification system:
According to the Linnaen system, organisms are grouped on the basis of their characteristics
without considering their ancestry. Based on this a reptile is an animal that is ectothermic and
consists of scales, and thus birds cannot be reptiles.
While in Phylogenetics,organisms are grouped only on the basis of their ancestry, and
characteristics are only used to discover this ancestry. So based on this a reptile is any animal
descending from reptiles, and birds (including mammals) would be considered as reptiles.
Looking back at history, the 1st group of reptiles evolved around 300 million years ago.
Around 40 million years later, a group of reptiles called therapsids branched off, that became
modern mammals.
Another group of reptiles split off in the next 120 million years, and one branch called the
dinosaurs was very successful. Dinosaurs was only distantly related to modern snakes, lizards,
and turtles, groups that had split off at different times.
But 65 million years ago there was a massive extinction event, and all dinosaurs were killed
except for a single group of feathered dinosaurs. These evolved over the next 65 million years
into modern birds. So birds aren\'t just closely related to dinosaurs, they really are dinosaurs!
This is what most people mean when they say that birds are reptiles, although technically
according to the phylogenetic system mammals are also reptiles.
Birds and reptiles are both vertebrates and are alike in ways:
1. They both lay eggs which have a protective shell. This is an adaptation to prevent drying-out,
necessary in evolution before vertebrates could lay their eggs on land.
2. Both birds and reptiles have lungs and thus breathe air.
3. Both birds and reptiles have scales on their skin. In birds, most of the body is now covered
with feathers, which are modified scales. You can still see the scales on birds\' legs.
4. It is thought that birds evolved from reptiles. These probably were able to fly as well.
Birds and reptiles differ in ways such as:
While most reptiles are ectothermic (\"cold blooded\"), birds are endothermic (\"warm blooded).
Thus birds have mechanisms like mammals that enable them to maintain a body temperature that
is independent of their surroundings. Their feathers and their high metabolic rate help keep them
warm in cold environments, for example.
While reptiles have a 3-chambered heart, birds have a 4-chambered heart. This allows them to
be more efficient, as the deoxygenated blood from the tissues is completely separated from the
oxygenated blood from the lungs.While reptiles have teeth, birds do not. They grind their food in
an organ called the gizzard.
Solution
Ideally birds evolved from Dinosaurs, but they are closely related to even reptiles. Biologists use
2 types of classification system:
According to the Linnaen system, organisms are grouped on th.
Introduction Termination of a project is inevitable, but how it .pdf
Introduction
Termination of a project is inevitable, but how it is terminated and when may have a profound
and long lasting impact on the organization and its employees. The success of future projects
may depend on not only the success of past ones, but also on how unsuccessful projects were
treated by the organization and its stakeholders. Firms have the option of initiating a variety of
entrepreneurial projects with varying degrees of risk. If an organization chooses to accept greater
risks, it should avoid penalizing members of projects that turn out to be unsuccessful. If team
members believe they will be penalized for participating in unsuccessful projects, they will be
less willing to terminate failed projects and may become risk adverse.
This research contains information about the project life cycle and the importance of continually
monitoring a project to determine if it is meeting the objectives established at the outset. We
have identified and categorized external and internal factors that influence the success or failure
of projects. The relative importance of each factor varies by organization and project type.
Organizing a project\'s termination process is especially important when it has failed, because of
the lasting impact on future projects as well as the organization\'s image. Including project team
members in the termination process will increase their loyalty and commitment, not only to the
organization but also to the success of future projects. At the end of a project a post-audit report
will be prepared that summarizes the project and provides recommendations for similar projects
in the future. Lastly, as a project is closed down or completed it is important that senior
management recognize the contributions of the project team.
The Project Life Cycle
There are several stages in the life cycle of a project: (1) project selection, (2) planning, (3)
execution, and (4) termination (Ruhl, 1988). The first phase, project selection, will vary among
firms. Each project must be evaluated to determine which is the best use of corporate funds. Each
will have different risks, benefits, and costs, making the selection very difficult. The final
decision should be based on the project\'s financial return and how well it assists the organization
in achieving its long-run strategic objectives.
Once a selection has been made, formal plans must be developed. The importance of thorough
project planning cannot be overemphasized. The objective of this process should be to develop a
master plan that details how each asset of the organization will be used to accomplish the
project\'s goals. Thorough and aggressive planning will also increase the team\'s commitment to
success. The two most important components of the master plan are the project budget and the
master schedule, which are developed from a detailed list of specific project tasks. The master
plan should include measures for evaluating the progress of the project as well as guideli.
In inline heapsort we use only single array throughout the program t.pdf
In inline heapsort we use only single array throughout the program to decrease the running time,
where as, in simple heap sort we use external data structure memory usage of more than O(1) at
a time.
Please consider the below inline heap program which prints out the running time, which is lesser
to the the simple heap sort.
//program
class heapsort
{
private final int array[]; // The array being sorted
heapsort(int array[], int N)
{
this.array = array;
int Last = N-1;
// its children are at elements 2*I+1 and 2*I+2.
// phase 1: form heap
// Construct heap bottom-up, starting with small trees just above leaves
// and coalescing into larger trees near the root.
for( int Top = Last/2; Top >= 0; Top-- )
{
adjust(Top, Last);
}
// phase 2: use heap to sort
// Move top element (largest) out of heap, swapping with last element
// and changing the heap boundary, until only one element remains.
while( Last > 0 )
{
swap(0, Last);
adjust(0, --Last);
}
}
/**
* adjust(Top, Last) adjusts the tree between Top and Last
**/
void adjust(int Top, int Last)
{
int TopVal = array[Top]; // Set aside top of heap
int Parent, Child;
for( Parent = Top; ; Parent = Child ) // Iterate down through tree
{
Child = 2*Parent+1; // Child means left child
if( Child > Last )
break; // Left child non-existent
if( Child+1 <= Last // Right child exists
&& array[Child] < array[Child+1] ) // and right child is larger
Child++; // Child is the larger child
if( TopVal >= array[Child] )
break; // Location for TopVal found
array[Parent] = array[Child]; // Move larger child up in tree
}
array[Parent] = TopVal; // Install TopVal in place
}
//swapping
void swap(int i, int j)
{
int temp = array[i];
array[i] = array[j];
array[j] = temp;
}
public static void main(String[] args)
{
int N = 0; // number of elements in array
int array[] =
{43,65,12,89,11,5,1,88,123,999,233,18,553,332,144,98,452,7663,111,23,654,8735,998};
//initialising array
N=array.length;
System.err.println(\"Sorting started\");
long startTime = System.nanoTime();
new heapsort(array, N); // sorting
long endTime = System.nanoTime();
long totalTime = endTime - startTime;
for( int i = 0; i < N; i++ )
{
System.out.print(array[i] + \" \");
}
System.out.println();
System.out.println(\"Time in nanos\"+totalTime);
}
}
//runn the above program for inline heap sort.
Solution
In inline heapsort we use only single array throughout the program to decrease the running time,
where as, in simple heap sort we use external data structure memory usage of more than O(1) at
a time.
Please consider the below inline heap program which prints out the running time, which is lesser
to the the simple heap sort.
//program
class heapsort
{
private final int array[]; // The array being sorted
heapsort(int array[], int N)
{
this.array = array;
int Last = N-1;
// its children are at elements 2*I+1 and 2*I+2.
// phase 1: form heap
// Construct heap bottom-up, starting with small trees just above leaves
// and coalescing into larger trees near the root.
for( int T.
I think that y=42SolutionI think that y=42.pdf
This very short document states that the author thinks y=42 but provides no context, explanation or reasoning for this statement. It is an unsupported claim that y equals 42 repeated twice with no further information given.
DRUGS AND ITS classification slide share
Any substance (other than food) that is used to prevent, diagnose, treat, or relieve symptoms of a
disease or abnormal condition
ISO/IEC 27001, ISO/IEC 42001, and GDPR: Best Practices for Implementation and...
Denis is a dynamic and results-driven Chief Information Officer (CIO) with a distinguished career spanning information systems analysis and technical project management. With a proven track record of spearheading the design and delivery of cutting-edge Information Management solutions, he has consistently elevated business operations, streamlined reporting functions, and maximized process efficiency.
Certified as an ISO/IEC 27001: Information Security Management Systems (ISMS) Lead Implementer, Data Protection Officer, and Cyber Risks Analyst, Denis brings a heightened focus on data security, privacy, and cyber resilience to every endeavor.
His expertise extends across a diverse spectrum of reporting, database, and web development applications, underpinned by an exceptional grasp of data storage and virtualization technologies. His proficiency in application testing, database administration, and data cleansing ensures seamless execution of complex projects.
What sets Denis apart is his comprehensive understanding of Business and Systems Analysis technologies, honed through involvement in all phases of the Software Development Lifecycle (SDLC). From meticulous requirements gathering to precise analysis, innovative design, rigorous development, thorough testing, and successful implementation, he has consistently delivered exceptional results.
Throughout his career, he has taken on multifaceted roles, from leading technical project management teams to owning solutions that drive operational excellence. His conscientious and proactive approach is unwavering, whether he is working independently or collaboratively within a team. His ability to connect with colleagues on a personal level underscores his commitment to fostering a harmonious and productive workplace environment.
Date: May 29, 2024
Tags: Information Security, ISO/IEC 27001, ISO/IEC 42001, Artificial Intelligence, GDPR
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salicylic acid alcohol (R-OH) and carboxylic aci.pdf
salicylic acid: alcohol (R-OH) and carboxylic acid (R-COOH) acetylsalicylic acid
(aspirin): carboxylic acid and ester (R-COOR)
Solution
salicylic acid: alcohol (R-OH) and carboxylic acid (R-COOH) acetylsalicylic acid
(aspirin): carboxylic acid and ester (R-COOR).
no it doesnot .pdf
This very short document does not seem to provide any clear information that can be summarized in 3 sentences or less. It contains the phrase "no it doesnot" repeated but no other context or details.
It would be Ruthenium (II) Sulfide Ruthenium is .pdf
It would be Ruthenium (II) Sulfide Ruthenium is a transition metal, meaning that it
can have multiple oxidation numbers. Sulfur always takes the charge of -2, so from that we see
that there is a 1 to 1 relation between Ru and S in the empirical formula, so that means that
Ruthenium has the opposite charge, +2.
Solution
It would be Ruthenium (II) Sulfide Ruthenium is a transition metal, meaning that it
can have multiple oxidation numbers. Sulfur always takes the charge of -2, so from that we see
that there is a 1 to 1 relation between Ru and S in the empirical formula, so that means that
Ruthenium has the opposite charge, +2..
ketones do not give tollens test.. hence no react.pdf
ketones do not give tollens test.. hence no reaction
Solution
ketones do not give tollens test.. hence no reaction.
for IONIC the metal name followed by nonmetal io.pdf
for IONIC: the metal name followed by nonmetal ion (-ide) for ex: NaCl is Sodium
Chloride or MgS is Magnesium Sulfide
Solution
for IONIC: the metal name followed by nonmetal ion (-ide) for ex: NaCl is Sodium
Chloride or MgS is Magnesium Sulfide.
pH = 4-log2.4 = 3.62 pOH 14 - pH = 14-3.62 = 10.3.pdf
pH = 4-log2.4 = 3.62 pOH 14 - pH = 14-3.62 = 10.38 concentration of OH- = 10^(-
14) / (2.4*10^(-4)) = 4.167* 10^(-11)
Solution
pH = 4-log2.4 = 3.62 pOH 14 - pH = 14-3.62 = 10.38 concentration of OH- = 10^(-
14) / (2.4*10^(-4)) = 4.167* 10^(-11).
This suggest that dissociation of FeCl3 is 100 .pdf
This suggest that dissociation of FeCl3 is 100 % to form 4 ions (1Fe+3Cl). Other
dissociation also may be possible like FeCl3---->(FeCl2) + Cl- FeCl3---->(FeCl)2+ + 2Cl- so
number of ions is averaged considering dissociation constants which is 3.4
Solution
This suggest that dissociation of FeCl3 is 100 % to form 4 ions (1Fe+3Cl). Other
dissociation also may be possible like FeCl3---->(FeCl2) + Cl- FeCl3---->(FeCl)2+ + 2Cl- so
number of ions is averaged considering dissociation constants which is 3.4.
Total probability of dot received P = (37)(78).pdf
Total probability of dot received P = (3/7)*(7/8)+(4/7)*(1/8)=0.4464 probability
that dot sentP1 = 3/7=0.4285 so P(dot sent | dot received).= P1/P=0.4285/.4464=0.96
Solution
Total probability of dot received P = (3/7)*(7/8)+(4/7)*(1/8)=0.4464 probability
that dot sentP1 = 3/7=0.4285 so P(dot sent | dot received).= P1/P=0.4285/.4464=0.96.
Environmental contaminants of recent concern are .pdf
Environmental contaminants of recent concern are pharmaceuticals, estrogens and
other endocrine disrupting chemicals (EDC) such as degradation products of surfactants, algal
and cyanobacterial toxins, disinfection by-products (DBPs) and metalloids. In addition,
pesticides (especially their transformation products), microorganisms, and humic substances
(HS), in their function as vehicles for contaminants and as precursors for by-products in water
treatment, traditionally play an important role. The present status of the application of LC-MS
techniques for these water constituents are discussed and examples of application are given.
Solid-phase extraction with various non-selective materials in combination with liquid
chromatography (LC) on reversed-phase columns have been the most widely used methods for
sample preconcentration and separation for different compound classes like pesticides,
pharmaceuticals or estrogens. Electrospray ionization (ESI) and atmospheric pressure ionization
(APCI) are the most frequently used ionization techniques for polar and ionic compounds, as
well as for less polar non-ionic ones. The facilities of LC-MS have been successfully
demonstrated for different compound classes. Polar compounds from pharmaceuticals used as
betablockers, iodinated X-ray contrast media, or estrogens have been determined without
derivatization down to ultratrace concentrations. LC-MS can be viewed as a prerequisite for the
determination of algal and cyanobacterial toxins and the homologues and oligomers of
alkylphenol ethoxylates and their metabolites. Tandem mass spectrometric techniques and the
use of diagnostic ions reveal their usefulness for compound-class specific screening and
unknown identification, and are also valid for the analysis of pesticides and especially for their
transformation products.
Solution
Environmental contaminants of recent concern are pharmaceuticals, estrogens and
other endocrine disrupting chemicals (EDC) such as degradation products of surfactants, algal
and cyanobacterial toxins, disinfection by-products (DBPs) and metalloids. In addition,
pesticides (especially their transformation products), microorganisms, and humic substances
(HS), in their function as vehicles for contaminants and as precursors for by-products in water
treatment, traditionally play an important role. The present status of the application of LC-MS
techniques for these water constituents are discussed and examples of application are given.
Solid-phase extraction with various non-selective materials in combination with liquid
chromatography (LC) on reversed-phase columns have been the most widely used methods for
sample preconcentration and separation for different compound classes like pesticides,
pharmaceuticals or estrogens. Electrospray ionization (ESI) and atmospheric pressure ionization
(APCI) are the most frequently used ionization techniques for polar and ionic compounds, as
well as for less polar non-ionic ones. The facilities of .
C-H polar covalent bond Solution .pdf
The document discusses C-H polar covalent bonds. C-H bonds have some ionic character due to the difference in electronegativity between carbon and hydrogen. Hydrogen is more electronegative so it pulls the electrons in the bond slightly toward itself, giving the bond some polar character though it remains primarily covalent.
To insert an element at given position in arrayInsert(stack s1, s.pdf
To insert an element at given position in array:
Insert(stack s1, stack s2, int value, int pos)
//Inserts a value into the array at specified position.
//Assume s1 is a stack which holds the values, and s2 is an empty stack.
while(!s1.empty()) //Till the stack is empty.
s2.push(s1.pop()) //Pop element from s1, and push it in s2.
//By now, all elements are placed in reverse order from s1 to s2.
for i = 1 to pos - 1 //Till the pos-1
s1.push(s2.pop()) //Remove elements from s2, and push them to s1.
s1.push(value) //Now push your element in specified position pos.
while(!s2.empty()) //Push all the remaining elements in s2 to s1.
s1.push(s2.pop())
To delete an element at given position in array:
Delete(stack s1, stack s2, int pos)
//Deletes a value into the array at specified position.
//Assume s1 is a stack which holds the values, and s2 is an empty stack.
while(!s1.empty()) //Till the stack is empty.
s2.push(s1.pop()) //Pop element from s1, and push it in s2.
//By now, all elements are placed in reverse order from s1 to s2.
for i = 1 to pos - 1 //Till the pos-1
s1.push(s2.pop()) //Remove elements from s2, and push them to s1.
s1.pop(value) //Now pop your element in specified position pos.
while(!s2.empty()) //Push all the remaining elements in s2 to s1.
s1.push(s2.pop())
To delete a given element in array:
DeleteVal(stack s1, stack s2, int value)
//Deletes a value from the array.
//Assume s1 is a stack which holds the values, and s2 is an empty stack.
while(!s1.empty()) //Till the stack is empty.
s2.push(s1.pop()) //Pop element from s1, and push it in s2.
//By now, all elements are placed in reverse order from s1 to s2.
ele = s2.peek();
while(ele != value || !s2.empty()) //Till you found the element to be deleted.
s1.push(s2.pop()) //Remove elements from s2, and push them to s1.
ele = s2.peek()
s2.pop(value) //If element is found, delete it.
while(!s2.empty()) //Push all the remaining elements in s2 to s1.
s1.push(s2.pop())
If you have any queries with this, just get back to me.
Solution
To insert an element at given position in array:
Insert(stack s1, stack s2, int value, int pos)
//Inserts a value into the array at specified position.
//Assume s1 is a stack which holds the values, and s2 is an empty stack.
while(!s1.empty()) //Till the stack is empty.
s2.push(s1.pop()) //Pop element from s1, and push it in s2.
//By now, all elements are placed in reverse order from s1 to s2.
for i = 1 to pos - 1 //Till the pos-1
s1.push(s2.pop()) //Remove elements from s2, and push them to s1.
s1.push(value) //Now push your element in specified position pos.
while(!s2.empty()) //Push all the remaining elements in s2 to s1.
s1.push(s2.pop())
To delete an element at given position in array:
Delete(stack s1, stack s2, int pos)
//Deletes a value into the array at specified position.
//Assume s1 is a stack which holds the values, and s2 is an empty stack.
while(!s1.empty()) //Till the stack is empty.
s2.push(s1.pop()) //Pop element from s1, and push it in s2.
//By .
Chemical weathering is the breakdown of rock by c.pdf
Chemical weathering is the breakdown of rock by chemical reaction. In this process
the minerals within the rock are changed into particles that can be easily carried away. Air and
water are both involved in many complex chemical reactions. The minerals in igneous rocks may
be unstable under normal atmospheric conditions, those formed at higher temperatures being
more readily attacked than those formed at lower temperatures. Igneous rocks are commonly
attacked by water, particularly acid or alkaline solutions, and all of the common igneous rock
forming minerals (with the exception of quartz, which is very resistant) are changed in this way
into clay minerals and chemicals in solution. Rock particles in the form of clay, silt, sand, and
gravel are transported by the agents of erosion (usually water, and less frequently, ice and wind)
to new locations and redeposited in layers, generally at a lower elevation. These agents reduce
the size of the particles, sort them by size, and then deposit them in new locations. The sediments
dropped by streams and rivers form alluvial fans, flood plains, deltas, and on the bottom of lakes
and the sea floor. The wind may move large amounts of sand and other smaller particles.
Glaciers transport and deposit great quantities of usually unsorted rock material as till. These
deposited particles eventually become compacted and cemented together, forming clastic
sedimentary rocks. Such rocks contain inert minerals that resist mechanical and chemical
breakdown, such as quartz. Quartz is one of the most mechanically and chemically resistant
minerals. Highly weathered sediments can contain several heavy and stable minerals, best
illustrated by the ZTR index.
Solution
Chemical weathering is the breakdown of rock by chemical reaction. In this process
the minerals within the rock are changed into particles that can be easily carried away. Air and
water are both involved in many complex chemical reactions. The minerals in igneous rocks may
be unstable under normal atmospheric conditions, those formed at higher temperatures being
more readily attacked than those formed at lower temperatures. Igneous rocks are commonly
attacked by water, particularly acid or alkaline solutions, and all of the common igneous rock
forming minerals (with the exception of quartz, which is very resistant) are changed in this way
into clay minerals and chemicals in solution. Rock particles in the form of clay, silt, sand, and
gravel are transported by the agents of erosion (usually water, and less frequently, ice and wind)
to new locations and redeposited in layers, generally at a lower elevation. These agents reduce
the size of the particles, sort them by size, and then deposit them in new locations. The sediments
dropped by streams and rivers form alluvial fans, flood plains, deltas, and on the bottom of lakes
and the sea floor. The wind may move large amounts of sand and other smaller particles.
Glaciers transport and deposit great quantities of.
bu puting your values in formula You need not to.pdf
bu puting your values in formula You need not to be concerned about the exact
mechanism of the reaction. First, try writing out the balanced equation. 4Li + O2 --->2Li2O So
which element was oxidized? Li does not have a charge while LiO is an ionic compound in
which electrons are \"shared\" according to electronegativity. In this case oxygen is much more
electronegative and it essentially obtains two electrons from Li through bonding. Now write out
the half reactions. Oxygen is reduced. O2 + 4e- ---> 2O (with a 2- charge) Li is oxidized. 4Li--
->4e- + 4Li (with a 1+ charge) *So overall O2 + 4Li + 4e- ---> 2O (2-) + 4Li (1+) + 4e- A
third reaction is also important. The products of the latter two steps combine to form 2Li2O. 2O
(2-) + 4Li (1+) ---> 2Li2O Combine this with the above equation and you will return to the
original balanced equation.
Solution
bu puting your values in formula You need not to be concerned about the exact
mechanism of the reaction. First, try writing out the balanced equation. 4Li + O2 --->2Li2O So
which element was oxidized? Li does not have a charge while LiO is an ionic compound in
which electrons are \"shared\" according to electronegativity. In this case oxygen is much more
electronegative and it essentially obtains two electrons from Li through bonding. Now write out
the half reactions. Oxygen is reduced. O2 + 4e- ---> 2O (with a 2- charge) Li is oxidized. 4Li--
->4e- + 4Li (with a 1+ charge) *So overall O2 + 4Li + 4e- ---> 2O (2-) + 4Li (1+) + 4e- A
third reaction is also important. The products of the latter two steps combine to form 2Li2O. 2O
(2-) + 4Li (1+) ---> 2Li2O Combine this with the above equation and you will return to the
original balanced equation..
Please post the figure.SolutionPlease post the figure..pdf
The document requests that a figure be posted but does not provide any context or details about the figure. It repeats the phrase "Please post the figure" twice without any additional information about the figure, its purpose, or why it should be posted. The summary is unable to extract any meaningful high-level information from the given document.
option B,C,D is true. gantt chart is used for project scheduling,tra.pdf
option B,C,D is true. gantt chart is used for project scheduling,tracking. it mainky uses the
technique of crtical path method.
Solution
option B,C,D is true. gantt chart is used for project scheduling,tracking. it mainky uses the
technique of crtical path method..
The mole is a unit of measurement for the amount .pdf
The mole is a unit of measurement for the amount of substance or chemical amount.
It is one of the base units in the International System of Units, and has the unit symbol mol. n
ionic compound is a chemical compound in which ions are held together in a lattice structure by
ionic bonds. Usually, the positively charged portion consists of metal cations and the negatively
charged portion is an anion or polyatomic ion. Ions in ionic compounds are held together by the
electrostatic force between oppositely charged bodies. Ionic compounds have a high melting and
boiling point, and they are hard and very brittle.
Solution
The mole is a unit of measurement for the amount of substance or chemical amount.
It is one of the base units in the International System of Units, and has the unit symbol mol. n
ionic compound is a chemical compound in which ions are held together in a lattice structure by
ionic bonds. Usually, the positively charged portion consists of metal cations and the negatively
charged portion is an anion or polyatomic ion. Ions in ionic compounds are held together by the
electrostatic force between oppositely charged bodies. Ionic compounds have a high melting and
boiling point, and they are hard and very brittle..
Ideally birds evolved from Dinosaurs, but they are closely related t.pdf
Ideally birds evolved from Dinosaurs, but they are closely related to even reptiles. Biologists use
2 types of classification system:
According to the Linnaen system, organisms are grouped on the basis of their characteristics
without considering their ancestry. Based on this a reptile is an animal that is ectothermic and
consists of scales, and thus birds cannot be reptiles.
While in Phylogenetics,organisms are grouped only on the basis of their ancestry, and
characteristics are only used to discover this ancestry. So based on this a reptile is any animal
descending from reptiles, and birds (including mammals) would be considered as reptiles.
Looking back at history, the 1st group of reptiles evolved around 300 million years ago.
Around 40 million years later, a group of reptiles called therapsids branched off, that became
modern mammals.
Another group of reptiles split off in the next 120 million years, and one branch called the
dinosaurs was very successful. Dinosaurs was only distantly related to modern snakes, lizards,
and turtles, groups that had split off at different times.
But 65 million years ago there was a massive extinction event, and all dinosaurs were killed
except for a single group of feathered dinosaurs. These evolved over the next 65 million years
into modern birds. So birds aren\'t just closely related to dinosaurs, they really are dinosaurs!
This is what most people mean when they say that birds are reptiles, although technically
according to the phylogenetic system mammals are also reptiles.
Birds and reptiles are both vertebrates and are alike in ways:
1. They both lay eggs which have a protective shell. This is an adaptation to prevent drying-out,
necessary in evolution before vertebrates could lay their eggs on land.
2. Both birds and reptiles have lungs and thus breathe air.
3. Both birds and reptiles have scales on their skin. In birds, most of the body is now covered
with feathers, which are modified scales. You can still see the scales on birds\' legs.
4. It is thought that birds evolved from reptiles. These probably were able to fly as well.
Birds and reptiles differ in ways such as:
While most reptiles are ectothermic (\"cold blooded\"), birds are endothermic (\"warm blooded).
Thus birds have mechanisms like mammals that enable them to maintain a body temperature that
is independent of their surroundings. Their feathers and their high metabolic rate help keep them
warm in cold environments, for example.
While reptiles have a 3-chambered heart, birds have a 4-chambered heart. This allows them to
be more efficient, as the deoxygenated blood from the tissues is completely separated from the
oxygenated blood from the lungs.While reptiles have teeth, birds do not. They grind their food in
an organ called the gizzard.
Solution
Ideally birds evolved from Dinosaurs, but they are closely related to even reptiles. Biologists use
2 types of classification system:
According to the Linnaen system, organisms are grouped on th.
Introduction Termination of a project is inevitable, but how it .pdf
Introduction
Termination of a project is inevitable, but how it is terminated and when may have a profound
and long lasting impact on the organization and its employees. The success of future projects
may depend on not only the success of past ones, but also on how unsuccessful projects were
treated by the organization and its stakeholders. Firms have the option of initiating a variety of
entrepreneurial projects with varying degrees of risk. If an organization chooses to accept greater
risks, it should avoid penalizing members of projects that turn out to be unsuccessful. If team
members believe they will be penalized for participating in unsuccessful projects, they will be
less willing to terminate failed projects and may become risk adverse.
This research contains information about the project life cycle and the importance of continually
monitoring a project to determine if it is meeting the objectives established at the outset. We
have identified and categorized external and internal factors that influence the success or failure
of projects. The relative importance of each factor varies by organization and project type.
Organizing a project\'s termination process is especially important when it has failed, because of
the lasting impact on future projects as well as the organization\'s image. Including project team
members in the termination process will increase their loyalty and commitment, not only to the
organization but also to the success of future projects. At the end of a project a post-audit report
will be prepared that summarizes the project and provides recommendations for similar projects
in the future. Lastly, as a project is closed down or completed it is important that senior
management recognize the contributions of the project team.
The Project Life Cycle
There are several stages in the life cycle of a project: (1) project selection, (2) planning, (3)
execution, and (4) termination (Ruhl, 1988). The first phase, project selection, will vary among
firms. Each project must be evaluated to determine which is the best use of corporate funds. Each
will have different risks, benefits, and costs, making the selection very difficult. The final
decision should be based on the project\'s financial return and how well it assists the organization
in achieving its long-run strategic objectives.
Once a selection has been made, formal plans must be developed. The importance of thorough
project planning cannot be overemphasized. The objective of this process should be to develop a
master plan that details how each asset of the organization will be used to accomplish the
project\'s goals. Thorough and aggressive planning will also increase the team\'s commitment to
success. The two most important components of the master plan are the project budget and the
master schedule, which are developed from a detailed list of specific project tasks. The master
plan should include measures for evaluating the progress of the project as well as guideli.
In inline heapsort we use only single array throughout the program t.pdf
In inline heapsort we use only single array throughout the program to decrease the running time,
where as, in simple heap sort we use external data structure memory usage of more than O(1) at
a time.
Please consider the below inline heap program which prints out the running time, which is lesser
to the the simple heap sort.
//program
class heapsort
{
private final int array[]; // The array being sorted
heapsort(int array[], int N)
{
this.array = array;
int Last = N-1;
// its children are at elements 2*I+1 and 2*I+2.
// phase 1: form heap
// Construct heap bottom-up, starting with small trees just above leaves
// and coalescing into larger trees near the root.
for( int Top = Last/2; Top >= 0; Top-- )
{
adjust(Top, Last);
}
// phase 2: use heap to sort
// Move top element (largest) out of heap, swapping with last element
// and changing the heap boundary, until only one element remains.
while( Last > 0 )
{
swap(0, Last);
adjust(0, --Last);
}
}
/**
* adjust(Top, Last) adjusts the tree between Top and Last
**/
void adjust(int Top, int Last)
{
int TopVal = array[Top]; // Set aside top of heap
int Parent, Child;
for( Parent = Top; ; Parent = Child ) // Iterate down through tree
{
Child = 2*Parent+1; // Child means left child
if( Child > Last )
break; // Left child non-existent
if( Child+1 <= Last // Right child exists
&& array[Child] < array[Child+1] ) // and right child is larger
Child++; // Child is the larger child
if( TopVal >= array[Child] )
break; // Location for TopVal found
array[Parent] = array[Child]; // Move larger child up in tree
}
array[Parent] = TopVal; // Install TopVal in place
}
//swapping
void swap(int i, int j)
{
int temp = array[i];
array[i] = array[j];
array[j] = temp;
}
public static void main(String[] args)
{
int N = 0; // number of elements in array
int array[] =
{43,65,12,89,11,5,1,88,123,999,233,18,553,332,144,98,452,7663,111,23,654,8735,998};
//initialising array
N=array.length;
System.err.println(\"Sorting started\");
long startTime = System.nanoTime();
new heapsort(array, N); // sorting
long endTime = System.nanoTime();
long totalTime = endTime - startTime;
for( int i = 0; i < N; i++ )
{
System.out.print(array[i] + \" \");
}
System.out.println();
System.out.println(\"Time in nanos\"+totalTime);
}
}
//runn the above program for inline heap sort.
Solution
In inline heapsort we use only single array throughout the program to decrease the running time,
where as, in simple heap sort we use external data structure memory usage of more than O(1) at
a time.
Please consider the below inline heap program which prints out the running time, which is lesser
to the the simple heap sort.
//program
class heapsort
{
private final int array[]; // The array being sorted
heapsort(int array[], int N)
{
this.array = array;
int Last = N-1;
// its children are at elements 2*I+1 and 2*I+2.
// phase 1: form heap
// Construct heap bottom-up, starting with small trees just above leaves
// and coalescing into larger trees near the root.
for( int T.
I think that y=42SolutionI think that y=42.pdf
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