Add mCPBA to make the epoxide, then use aqueous N.pdf
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Add mCPBA to make the epoxide, then use aqueous NaOH to make the diol. You will get a racemic mixture of the two diols however. Solution Add mCPBA to make the epoxide, then use aqueous NaOH to make the diol. You will get a racemic mixture of the two diols however..
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sulfur trioxide is SO3 atomic mass S = 32.1 gmo.pdf
sulfur trioxide is SO3 atomic mass: S = 32.1 g/mol O =16.0 g/mol molar mass of
SO3 = 32.1 + 16.0x3 = 80.1 g/mol moles of SO3 = 2.15 mol mass of SO3 = moles of SO3 x
molar mass of SO3 = 2.15 x 80.1 = 172 g
Solution
sulfur trioxide is SO3 atomic mass: S = 32.1 g/mol O =16.0 g/mol molar mass of
SO3 = 32.1 + 16.0x3 = 80.1 g/mol moles of SO3 = 2.15 mol mass of SO3 = moles of SO3 x
molar mass of SO3 = 2.15 x 80.1 = 172 g.
This program here prints the number of words that occurs in the inpu.pdf
This program here prints the number of words that occurs in the input
For this problem we need to keep track of the number of times the word repeats in the string
print the count.
for this I used a class to hold the word count.
Find other versions of the program, i.e alphabetical order
class WordDataCounter {
// A small class to hold the data value for one word.
String wordcount; // The word (in lower case letters).
int count; // The number of times the word has been found.
WordData(String m) {
// Constructor creates an object with the specified word
// and with the counter initialized to 1.
wordcount = m;
count = 1;
}
}
/*
This program lets the user specify a text file for input and a file
for output. All the words are read from the input file. Words are
converted to lower case. The program makes a list of all the words
that occur in the file, along with the number of times that each
word occurs. The words and word frequencies are printed to an
output file. The list is printed twice: once in alphabetical order
and once in order of decreasing frequency. A word in this program is
defined to be any sequence of letters.
This class depends on the non-standard classes TextIO and TextReader.
*/
import java.io.*;
class WordData {
// A little class to hold the data for one word.
String word; // The word (in lower case letters).
int count; // The number of times the word has been found.
WordData(String w) {
// Constructor creates an object with the specified word
// and with the counter initialized to 1.
word = w;
count = 1;
}
}
public class WordFrequencies {
static WordData[] words; // An array to hold the words from the file.
// Note that the array will be expanded as
// necessary, in the insertWord() subroutine.
static int wordCount; // The number of words currently stored in
// the array.
public static void main(String[] args) {
TextReader in; // A stream for reading from the input file.
PrintWriter out; // A stream for writing to the output file.
String inputFileName; // Input file name, specified by the user.
String outputFileName; // Output file name, specified by the user.
words = new WordData[10]; // Start with space for 10 words.
wordCount = 0; // Currently, there are no words in the array.
/* Get the input file name from the user and try to create the
input stream. If there is a FileNotFoundException, print
a message and terminate the program. */
TextIO.put(\"Input file name? \");
inputFileName = TextIO.getln().trim();
try {
in = new TextReader(new FileReader(inputFileName));
}
catch (FileNotFoundException e) {
TextIO.putln(\"Can\'t find file \\\"\" + inputFileName + \"\\\".\");
return;
}
/* Get the output file name from the user and try to create the
output stream. If there is an IOException, print a message
and terminate the program. */
TextIO.put(\"Output file name? \");
outputFileName = TextIO.getln().trim();
try {
out = new PrintWriter(new FileWriter(outputFileName));
}
catch (IOException e) {
TextIO.putln(\"Can\'t open file \\\"\" + o.
This is a heart related disease. The identification of a disease.pdf
This is a heart related disease.
The identification of a disease of heart attack demands enzyme confirmation if we are to
separate painl diseases from heart-related sickness involving death of living tissues of heart-
related tissue. Patients with a first heart attack are more in their risk profile and are more (in a
way that\'s close to the truth or true number) separated and labeled into different
electrocardiographic subsets than are patients with prior infarctions. The recording of the
electrical activity of the heart has been used to classify two types of heart attack--the Q-wave
(transmural) infarction, with the result that a Q wave represents through-and-through
endocardial-to-epicardial death, and the non--Q-wave (nontransmural) infarction, with the effect
that the absence of a Q wave points to less big heart-related damage and possibly an increased
possibility of repeating heart-related event
ASHD -- Arteriosclerotic Heart Disease
Solution
This is a heart related disease.
The identification of a disease of heart attack demands enzyme confirmation if we are to
separate painl diseases from heart-related sickness involving death of living tissues of heart-
related tissue. Patients with a first heart attack are more in their risk profile and are more (in a
way that\'s close to the truth or true number) separated and labeled into different
electrocardiographic subsets than are patients with prior infarctions. The recording of the
electrical activity of the heart has been used to classify two types of heart attack--the Q-wave
(transmural) infarction, with the result that a Q wave represents through-and-through
endocardial-to-epicardial death, and the non--Q-wave (nontransmural) infarction, with the effect
that the absence of a Q wave points to less big heart-related damage and possibly an increased
possibility of repeating heart-related event
ASHD -- Arteriosclerotic Heart Disease.
The terminating hairpin loop occurs in the tryptophan operon whe.pdf
The terminating \"hairpin\" loop occurs in the tryptophan operon when ________.
2. When tryptophan is present in the medium and available to the bacterium, ________.
=================================================================
What occurs in a bacterium with respect to genetic regulation of the tryptophan operon?
=================================================================
Microorganisms, for example, Escherichia coli (an inviting occupant of our gut) require amino
acids to survive—in light of the fact that, similar to us, they have to fabricate proteins. One of the
amino acids they need is tryptophan.
In the event that tryptophan is accessible in nature, E. coli will take it up and utilize it to fabricate
proteins. Be that as it may, E. coli can likewise make their own tryptophan utilizing catalysts that
are encoded by five qualities. These five qualities are situated alongside each other in what is
known as the trp operon.
On the off chance that tryptophan is available in nature, then E. coli microscopic organisms
don\'t have to combine it, so translation of the qualities in the trp operon is exchanged \"off.\"
When tryptophan accessibility is low, then again, the operon is exchanged \"on,\" the qualities
are interpreted, biosynthetic catalysts are made, and more tryptophan is produced.The trp operon
incorporates five qualities that encode chemicals required for tryptophan biosynthesis, alongside
a promoter (RNA polymerase restricting site) and an administrator (restricting site for a repressor
protein). The qualities of the trp operon are interpreted as a solitary mRNA.This extend of DNA
is perceived by an administrative protein known as the trp repressor. At the point when the
repressor ties to the DNA of the administrator, it keeps the operon from being translated by
physically impeding RNA polymerase, the interpretation enzyme.The trp repressor does not
generally tie to DNA. Rather, it ties and pieces interpretation just when tryptophan is available.
At the point when tryptophan is around, it appends to the repressor particles and changes their
shape so they get to be dynamic. A little particle like trytophan, which switches a repressor into
its dynamic state, is known as a corepressor.B. sufficient tryptophan is present.----------------
ANSWER
Solution
The terminating \"hairpin\" loop occurs in the tryptophan operon when ________.
2. When tryptophan is present in the medium and available to the bacterium, ________.
=================================================================
What occurs in a bacterium with respect to genetic regulation of the tryptophan operon?
=================================================================
Microorganisms, for example, Escherichia coli (an inviting occupant of our gut) require amino
acids to survive—in light of the fact that, similar to us, they have to fabricate proteins. One of the
amino acids they need is tryptophan.
In the event that tryptophan is accessible in nature, E. coli will.
The solution is--fn= -13(-1)n+ 13(2)nSolutionThe solution .pdf
The solution is--
fn= -1/3(-1)n+ 1/3(2)n
Solution
The solution is--
fn= -1/3(-1)n+ 1/3(2)n.
Recommended
sulfur trioxide is SO3 atomic mass S = 32.1 gmo.pdf
sulfur trioxide is SO3 atomic mass: S = 32.1 g/mol O =16.0 g/mol molar mass of
SO3 = 32.1 + 16.0x3 = 80.1 g/mol moles of SO3 = 2.15 mol mass of SO3 = moles of SO3 x
molar mass of SO3 = 2.15 x 80.1 = 172 g
Solution
sulfur trioxide is SO3 atomic mass: S = 32.1 g/mol O =16.0 g/mol molar mass of
SO3 = 32.1 + 16.0x3 = 80.1 g/mol moles of SO3 = 2.15 mol mass of SO3 = moles of SO3 x
molar mass of SO3 = 2.15 x 80.1 = 172 g.
This program here prints the number of words that occurs in the inpu.pdf
This program here prints the number of words that occurs in the input
For this problem we need to keep track of the number of times the word repeats in the string
print the count.
for this I used a class to hold the word count.
Find other versions of the program, i.e alphabetical order
class WordDataCounter {
// A small class to hold the data value for one word.
String wordcount; // The word (in lower case letters).
int count; // The number of times the word has been found.
WordData(String m) {
// Constructor creates an object with the specified word
// and with the counter initialized to 1.
wordcount = m;
count = 1;
}
}
/*
This program lets the user specify a text file for input and a file
for output. All the words are read from the input file. Words are
converted to lower case. The program makes a list of all the words
that occur in the file, along with the number of times that each
word occurs. The words and word frequencies are printed to an
output file. The list is printed twice: once in alphabetical order
and once in order of decreasing frequency. A word in this program is
defined to be any sequence of letters.
This class depends on the non-standard classes TextIO and TextReader.
*/
import java.io.*;
class WordData {
// A little class to hold the data for one word.
String word; // The word (in lower case letters).
int count; // The number of times the word has been found.
WordData(String w) {
// Constructor creates an object with the specified word
// and with the counter initialized to 1.
word = w;
count = 1;
}
}
public class WordFrequencies {
static WordData[] words; // An array to hold the words from the file.
// Note that the array will be expanded as
// necessary, in the insertWord() subroutine.
static int wordCount; // The number of words currently stored in
// the array.
public static void main(String[] args) {
TextReader in; // A stream for reading from the input file.
PrintWriter out; // A stream for writing to the output file.
String inputFileName; // Input file name, specified by the user.
String outputFileName; // Output file name, specified by the user.
words = new WordData[10]; // Start with space for 10 words.
wordCount = 0; // Currently, there are no words in the array.
/* Get the input file name from the user and try to create the
input stream. If there is a FileNotFoundException, print
a message and terminate the program. */
TextIO.put(\"Input file name? \");
inputFileName = TextIO.getln().trim();
try {
in = new TextReader(new FileReader(inputFileName));
}
catch (FileNotFoundException e) {
TextIO.putln(\"Can\'t find file \\\"\" + inputFileName + \"\\\".\");
return;
}
/* Get the output file name from the user and try to create the
output stream. If there is an IOException, print a message
and terminate the program. */
TextIO.put(\"Output file name? \");
outputFileName = TextIO.getln().trim();
try {
out = new PrintWriter(new FileWriter(outputFileName));
}
catch (IOException e) {
TextIO.putln(\"Can\'t open file \\\"\" + o.
This is a heart related disease. The identification of a disease.pdf
This is a heart related disease.
The identification of a disease of heart attack demands enzyme confirmation if we are to
separate painl diseases from heart-related sickness involving death of living tissues of heart-
related tissue. Patients with a first heart attack are more in their risk profile and are more (in a
way that\'s close to the truth or true number) separated and labeled into different
electrocardiographic subsets than are patients with prior infarctions. The recording of the
electrical activity of the heart has been used to classify two types of heart attack--the Q-wave
(transmural) infarction, with the result that a Q wave represents through-and-through
endocardial-to-epicardial death, and the non--Q-wave (nontransmural) infarction, with the effect
that the absence of a Q wave points to less big heart-related damage and possibly an increased
possibility of repeating heart-related event
ASHD -- Arteriosclerotic Heart Disease
Solution
This is a heart related disease.
The identification of a disease of heart attack demands enzyme confirmation if we are to
separate painl diseases from heart-related sickness involving death of living tissues of heart-
related tissue. Patients with a first heart attack are more in their risk profile and are more (in a
way that\'s close to the truth or true number) separated and labeled into different
electrocardiographic subsets than are patients with prior infarctions. The recording of the
electrical activity of the heart has been used to classify two types of heart attack--the Q-wave
(transmural) infarction, with the result that a Q wave represents through-and-through
endocardial-to-epicardial death, and the non--Q-wave (nontransmural) infarction, with the effect
that the absence of a Q wave points to less big heart-related damage and possibly an increased
possibility of repeating heart-related event
ASHD -- Arteriosclerotic Heart Disease.
The terminating hairpin loop occurs in the tryptophan operon whe.pdf
The terminating \"hairpin\" loop occurs in the tryptophan operon when ________.
2. When tryptophan is present in the medium and available to the bacterium, ________.
=================================================================
What occurs in a bacterium with respect to genetic regulation of the tryptophan operon?
=================================================================
Microorganisms, for example, Escherichia coli (an inviting occupant of our gut) require amino
acids to survive—in light of the fact that, similar to us, they have to fabricate proteins. One of the
amino acids they need is tryptophan.
In the event that tryptophan is accessible in nature, E. coli will take it up and utilize it to fabricate
proteins. Be that as it may, E. coli can likewise make their own tryptophan utilizing catalysts that
are encoded by five qualities. These five qualities are situated alongside each other in what is
known as the trp operon.
On the off chance that tryptophan is available in nature, then E. coli microscopic organisms
don\'t have to combine it, so translation of the qualities in the trp operon is exchanged \"off.\"
When tryptophan accessibility is low, then again, the operon is exchanged \"on,\" the qualities
are interpreted, biosynthetic catalysts are made, and more tryptophan is produced.The trp operon
incorporates five qualities that encode chemicals required for tryptophan biosynthesis, alongside
a promoter (RNA polymerase restricting site) and an administrator (restricting site for a repressor
protein). The qualities of the trp operon are interpreted as a solitary mRNA.This extend of DNA
is perceived by an administrative protein known as the trp repressor. At the point when the
repressor ties to the DNA of the administrator, it keeps the operon from being translated by
physically impeding RNA polymerase, the interpretation enzyme.The trp repressor does not
generally tie to DNA. Rather, it ties and pieces interpretation just when tryptophan is available.
At the point when tryptophan is around, it appends to the repressor particles and changes their
shape so they get to be dynamic. A little particle like trytophan, which switches a repressor into
its dynamic state, is known as a corepressor.B. sufficient tryptophan is present.----------------
ANSWER
Solution
The terminating \"hairpin\" loop occurs in the tryptophan operon when ________.
2. When tryptophan is present in the medium and available to the bacterium, ________.
=================================================================
What occurs in a bacterium with respect to genetic regulation of the tryptophan operon?
=================================================================
Microorganisms, for example, Escherichia coli (an inviting occupant of our gut) require amino
acids to survive—in light of the fact that, similar to us, they have to fabricate proteins. One of the
amino acids they need is tryptophan.
In the event that tryptophan is accessible in nature, E. coli will.
The solution is--fn= -13(-1)n+ 13(2)nSolutionThe solution .pdf
The solution is--
fn= -1/3(-1)n+ 1/3(2)n
Solution
The solution is--
fn= -1/3(-1)n+ 1/3(2)n.
The follow list restates these 3 fundamental levels of service for Q.pdf
The follow list restates these 3 fundamental levels of service for QoS:
Best-effort service: the normal type of connectivity without some guarantee. This kind of
service, in orientation to Catalyst switch, use first-in first-out queue, which only transmit packet
as they get there in a line with no special treatment.
Integrated services: IntServ, also known as solid QoS, is a complete condition of services. In
additional terms, the IntServ model implies that traffic flow is kept openly by all midway system
and possessions.
Differentiated services—DiffServ, also known as flexible QoS, is class base, where a few
module of traffic obtain special behavior more additional traffic classes. differentiate services
use arithmetical preference, not a hard assurance like incorporated services. In additional terms,
DiffServ categorize traffic and then sort it into line of different efficiencies.
The IntServ model, distinct in RFC 1633, guarantees a unsurprising performance of the
association for application. IntServ give several services that accommodate several QoS
necessities. IntServ is implementing throughout the use of the supply Reservation Protocol,
enable at together the endpoints and the net devices connecting. This enabled application needs
an exact type of service from this enabled system previous to it send information. Open signal
via RSVP make the appeal, and the request inform the network of its traffic report and needs a
exacting type of service that can include its bandwidth and holdup necessities. The application is
predictable to send information only following it gets verification as of the network. The network
as well expect the application to propel information that deceit in it’s describe traffic profile
IntServ follow the signaled QoS form where the end hosts sign their QoS necessities to the
system. Each individual message stream wants to request capital from the network. Rim routers
utilize the Resource Reservation Protocol to indicator and keep the pet bandwidth for every flood
in the network.
The DiffServ model provides several levels of examine that convince differing QoS necessities.
Though, different with the IntServ model, an request that use DiffServ do not plainly signal the
network strategy before transfer information. DiffServ is a QoS completion method that is
modified for current networks and their explanation. DiffServ reassigns bit in the kind of service
pasture of an IP packet title. DiffServ use differentiates services code points as the QoS main
concern descriptor importance and supports 64 stage of categorization
For example, VoIP require extremely small jitter, a one way holdup in the arrange of 150 ms
plus certain bandwidth in the range of 8Kbps to 64Kbps, needy on the codec use. In one more
example, a case move request, base on ftp, do not undergo from jitter, as packet failure will be
extremely damaging to the throughput.
Solution
The follow list restates these 3 fundamental levels of service for QoS:
Best-e.
tested on eclipseDoublyLinkedList class.pdf
//tested on eclipse
/*************DoublyLinkedList class*******************/
public class DoublyLinkedList {
/*Data member variable declaration*/
private Node head;
private int count;
/**default constructor*/
public DoublyLinkedList() {
/*Initializing variable*/
head=null;
count =0;
}
/*Insert at beginning in doubly linked list*/
public void insertBeginning(String name){
/*checking head is null then we will add node at head*/
if(head==null){
head=new Node(name, null, null);
count++;
}else{
/*else create temp node and setting next to head
* and then making temp to head*/
Node temp=new Node(name,null,null);
temp.setNext(head);
head.setPrev(temp);
head=temp;
count++;
}
}
public void printList(){
Node temp=head;
/*Printing element of doubly list*/
while(temp!=null){
System.out.println(temp.getData());
temp=temp.getNext();
}
}
/*Returning count from doubly linked list*/
public int count(){
return count;
}
/*deleting specific node from doubly linked list*/
public void delete(String name){
Node current=head,previous=head;
/*if it is head node to be deleted*/
if(head.getData().equals(name)){
current=head.getNext();
current.setPrev(null);
head=current;
previous=head;
}else{
/*If node in middle of list*/
while(current.getNext()!=null){
if(current.getData().equals(name)){
current.getNext().setPrev(previous);
previous.setNext(current.getNext());
count--;
break;
}
previous=current;
current=current.getNext();
}
/*If node is at last*/
if(current.getNext()==null){
if(current.getData().equals(name)){
previous.setNext(null);
count--;
}
}
}
}
/*deleting all nodes from list by assigning head to null
* and memory releasing take care by garbage collector*/
public void clear(){
head=null;
count=0;
}
/*Inserting element in alphabetic order*/
public void insertAlpha(String name){
Node current=head,previous=head;
/*creating new node*/
Node newNode=new Node(name,null,null);
/*if head node is null*/
if(head==null){
head=newNode;
count++;
}else{
/*please read about compareTo() method about returning value*/
if(head.getData().compareTo(name)>0){
newNode.setNext(head);
head=newNode;
count++;
}else{
/*condition for getting position of node to be inserted*/
while(current!=null&¤t.getData().compareTo(name)<=0){
previous=current;
current=current.getNext();
}
/*if node need to be inserted at last*/
if(current==null){
previous=newNode;
}else{
/*If node need to inserted in middle*/
newNode.setNext(current);
previous.setNext(newNode);
current.setPrev(newNode);
newNode.setPrev(previous);
}
count++;
}
}
}
/*Main method start*/
public static void main(String args[]){
/*creating object of doubly linked list*/
DoublyLinkedList doublyLinkedList=new DoublyLinkedList();
doublyLinkedList.insertBeginning(\"sunita\");
doublyLinkedList.insertBeginning(\"lalchand\");
/*calling alphabetic order method*/
doublyLinkedList.insertAlpha(\"ramsingh\");
doublyLinkedList.insertBeginning(\"Lokesh\");
doublyLinkedList.insertBeginning(\"Pritam\");
/*****printing list********/
doublyLinkedList.printL.
Please read the comment ins codeExpressionTree.java-------------.pdf
Please read the comment ins code
ExpressionTree.java
----------------------------------
/**
* This is the class for Expression Tree.
* Used to create Expression Tree and Evaluate it
*/
/**
* Following logic is used to construct a Tree
* Here we use stack for Preparing Tree
* Loop through given Expression String
* If Character is Operand , Create node and push to stack
* If Character is Operator then
* 1)Create Node for Operator
* 2)Pop 2 nodes from Stack and Made
* OpretorNode--> left == first node pop
* OpretorNode--> right == second node pop
* At the end of creation of Expression Tree, Stack have only one node , which is root of
Expression tree
*/
/** Class ExpressionTree **/
class ExpressionTree
{
/** class TreeNode
* Stored Character ==> Digit(0..9) or a Operator +,-,*,/
* Left Node and Right Node
*
* **/
class TreeNode
{
char data;
TreeNode left, right;
/** constructor **/
public TreeNode(char data)
{
this.data = data;
this.left = null;
this.right = null;
}
}
/** class StackNode **/
class StackNode
{
TreeNode treeNode;
StackNode next;
/** constructor **/
public StackNode(TreeNode treeNode)
{
this.treeNode = treeNode;
next = null;
}
}
private static StackNode top;
/** constructor
* Constructor takes input string like \"+-+7*935*82*625\"
* Input should be in Prefix notation
* **/
public ExpressionTree(String expression)
{
top = null;
//Call Method for prepare expression tree
buildTree(expression);
}
/** function to clear tree **/
public void clear()
{
top = null;
}
/** function to push a node **/
private void push(TreeNode ptr)
{
if (top == null)
top = new StackNode(ptr);
else
{
StackNode nptr = new StackNode(ptr);
nptr.next = top;
top = nptr;
}
}
/** function to pop a node
* When it find operator pop 2 elements from Stack
*
* **/
private TreeNode pop()
{
if (top == null)
throw new RuntimeException(\"Underflow\");
else
{
TreeNode ptr = top.treeNode;
top = top.next;
return ptr;
}
}
/** function to get top node **/
private TreeNode peek()
{
return top.treeNode;
}
/** function to insert character **/
private void insert(char val)
{
try
{
//If Operand , Create node and push to Stack
if (isDigit(val))
{
TreeNode nptr = new TreeNode(val);
push(nptr);
}
//If Operator , Create node and popup 2 elements and make them its right and left
else if (isOperator(val))
{
TreeNode nptr = new TreeNode(val);
nptr.left = pop();
nptr.right = pop();
push(nptr);
}
}
catch (Exception e)
{
System.out.println(\"Invalid Expression\");
}
}
/** function to check if digit **/
private boolean isDigit(char ch)
{
return ch >= \'0\' && ch <= \'9\';
}
/** function to check if operator **/
private boolean isOperator(char ch)
{
return ch == \'+\' || ch == \'-\' || ch == \'*\' || ch == \'/\';
}
/** function to convert character to digit **/
private int toDigit(char ch)
{
return ch - \'0\';
}
/** function to build tree from input */
public void buildTree(String eqn)
{
for (int i = eqn.length() - 1; i >= 0; i--)
insert(eqn.charAt(i));
}
/** function to evaluate tree */
public dou.
Please find the answers and explanations belowPart A1. The ing.pdf
Please find the answers and explanations below:
Part A:
1. The ingestion of botulism toxin would best be described as a food poisoning. (ingestion of
botulism toxin or any other gut-infecting agent is non-technically termed as food poisoning. It is
different from infection in the sense that food poisoning may not always lead to sustained gut
infection but diagnosis of gut infection results evidently from food poisoning)
2. The ingestion of a significant number of Clostridium cells would be a infection.
(Immunologically, a sufficeint amount of pathogen is required for starting a degrading immune
response in the body. This is called the threshold number of the pathogen which can elicit an
immune response in the body. When the number of pathogeni cells increases beyond this
threshold, the person feels sick and an infection persists)
3.Spoilage is a change in appearance, smell, or taste of a food that makes it undesirable to a
consumer. (Technically, spoilage is defined as physcial and chemical change in the
appearance,smell, taste or nutritional content of the food due to infection or residence of a
pathogen in it.)
Part B:
For any pathogen to enter and cause infection, some specific steps need to be fulfilled by it.
Firstly, the concentration of the pathogen must be large enough to survive in the food item. This
infected food item will be ingested by the host and the pathogen will reach the gut. The pathogen
survives the harsh environment of the gut and starts replicating. This replication of the pathogen
comes with severe side effects on the host which include physical and physiological deterioration
of the gut lining along with eliciting immune response in the host. This ultimately causes
infection/disease.
So, the correct order of events would be:
1. The potatoes were contaminated with soil containing Clostridium botulinum endospores.
2. The toxin remained in the pruno after straining.
3. The conditions in the bag of pruno were favorable for germination.
4. Clostridium botulinum replicated and increased the number of organisms in the pruno.
5. Bacterial cells produced and secreted toxin.
6. The inmates drank the contaminated pruno and became ill.
Solution
Please find the answers and explanations below:
Part A:
1. The ingestion of botulism toxin would best be described as a food poisoning. (ingestion of
botulism toxin or any other gut-infecting agent is non-technically termed as food poisoning. It is
different from infection in the sense that food poisoning may not always lead to sustained gut
infection but diagnosis of gut infection results evidently from food poisoning)
2. The ingestion of a significant number of Clostridium cells would be a infection.
(Immunologically, a sufficeint amount of pathogen is required for starting a degrading immune
response in the body. This is called the threshold number of the pathogen which can elicit an
immune response in the body. When the number of pathogeni cells increases beyond this
threshold, the pers.
Mass of liquid = 22.5 gHeat added = 2.34 kJ = 2340 JTemperature .pdf
Mass of liquid = 22.5 g
Heat added = 2.34 kJ = 2340 J
Temperature change = 56.3 - 14.0 = 42.3oC
Specific heat capacity = heat added/(mass x temperature change)
= 2340/(22.5 x 42.3) = 2.46 J/g.oC
Solution
Mass of liquid = 22.5 g
Heat added = 2.34 kJ = 2340 J
Temperature change = 56.3 - 14.0 = 42.3oC
Specific heat capacity = heat added/(mass x temperature change)
= 2340/(22.5 x 42.3) = 2.46 J/g.oC.
It is given that we should use 4 bit input, i.e we will have a set o.pdf
It is given that we should use 4 bit input, i.e we will have a set of 16 different inputs. The output
of the circuit will be 1 if and only if the decimal equivalent formed by the 4 input bits is an
integer that belongs to the fibonacci series.
a) The truth table according to the above requirement is
A
B
C
D
O
0
0
0
0
1
0
0
0
1
1
0
0
1
0
1
0
0
1
1
1
0
1
0
0
0
0
1
0
1
1
0
1
1
0
0
0
1
1
1
0
1
0
0
0
1
1
0
0
1
0
1
0
1
0
0
1
0
1
1
0
1
1
0
0
0
1
1
0
1
1
1
1
1
0
0
1
1
1
1
0
B) K - map for the above truth table is
c) According to the above K-map, prime implicants would be
A\'B\', B\'C\'D\' and C\'D
d) All the prime implicants are essential prime implicants
e) A\'B\'+B\'C\'D\'+C\'D
A
B
C
D
O
0
0
0
0
1
0
0
0
1
1
0
0
1
0
1
0
0
1
1
1
0
1
0
0
0
0
1
0
1
1
0
1
1
0
0
0
1
1
1
0
1
0
0
0
1
1
0
0
1
0
1
0
1
0
0
1
0
1
1
0
1
1
0
0
0
1
1
0
1
1
1
1
1
0
0
1
1
1
1
0
Solution
It is given that we should use 4 bit input, i.e we will have a set of 16 different inputs. The output
of the circuit will be 1 if and only if the decimal equivalent formed by the 4 input bits is an
integer that belongs to the fibonacci series.
a) The truth table according to the above requirement is
A
B
C
D
O
0
0
0
0
1
0
0
0
1
1
0
0
1
0
1
0
0
1
1
1
0
1
0
0
0
0
1
0
1
1
0
1
1
0
0
0
1
1
1
0
1
0
0
0
1
1
0
0
1
0
1
0
1
0
0
1
0
1
1
0
1
1
0
0
0
1
1
0
1
1
1
1
1
0
0
1
1
1
1
0
B) K - map for the above truth table is
c) According to the above K-map, prime implicants would be
A\'B\', B\'C\'D\' and C\'D
d) All the prime implicants are essential prime implicants
e) A\'B\'+B\'C\'D\'+C\'D
A
B
C
D
O
0
0
0
0
1
0
0
0
1
1
0
0
1
0
1
0
0
1
1
1
0
1
0
0
0
0
1
0
1
1
0
1
1
0
0
0
1
1
1
0
1
0
0
0
1
1
0
0
1
0
1
0
1
0
0
1
0
1
1
0
1
1
0
0
0
1
1
0
1
1
1
1
1
0
0
1
1
1
1
0.
IndexingConcatinationSolutionIndexingConcatination.pdf
Indexing:
Concatination:
Solution
Indexing:
Concatination:.
import java.util.ArrayList; import java.util.List;public class H.pdf
import java.util.ArrayList;
import java.util.List;
public class H2ClassA {
List list = new ArrayList ();
public static void main(String[] args) {
H2ClassA y = new H2ClassA ();
int [] v = {4, 3, 7, 5, 99, 3};
// change the for loop to old type of for loop
for(int i=0;i
Solution
import java.util.ArrayList;
import java.util.List;
public class H2ClassA {
List list = new ArrayList ();
public static void main(String[] args) {
H2ClassA y = new H2ClassA ();
int [] v = {4, 3, 7, 5, 99, 3};
// change the for loop to old type of for loop
for(int i=0;i.
Heterochromatin is a tightly packed form of DNA, which comes in mult.pdf
Heterochromatin is a tightly packed form of DNA, which comes in multiple varieties.
constitutive heterochromatin the chromatin in regions of the chromosomes that are invariably
heterochromatic; it contains highly repetitive sequences of DNA that are genetically inactive and
serves as a structural element of the chromosome.
facultative heterochromatin the chromatin in regions of the chromosomes that become
heterochromatic in certain cells and tissues; for example, it makes up the inactive X chromosome
in female somatic cells.
Homologous chromosomes are chromosome pairs, one from each parent, that are similar in
length, gene position and centromere location.
Duplicated chromosome The occurrence of a repeated section of genes in a chromosome.
Haploid is Having a single set of unpaired chromosomes, as in germ cell, or half the diploid
number of a somatic cell.
Aneuploidy is the presence of an abnormal number of chromosomes in a cell,
example : a human cell having 45 or 47 chromosomes instead of the usual 46.
Nondisjunction is the failure of one or more pairs of homologous chromosomes or sister
chromatids to separate normally during nuclear division, usually resulting in an abnormal
distribution of chromosomes in the daughter nuclei.
Solution
Heterochromatin is a tightly packed form of DNA, which comes in multiple varieties.
constitutive heterochromatin the chromatin in regions of the chromosomes that are invariably
heterochromatic; it contains highly repetitive sequences of DNA that are genetically inactive and
serves as a structural element of the chromosome.
facultative heterochromatin the chromatin in regions of the chromosomes that become
heterochromatic in certain cells and tissues; for example, it makes up the inactive X chromosome
in female somatic cells.
Homologous chromosomes are chromosome pairs, one from each parent, that are similar in
length, gene position and centromere location.
Duplicated chromosome The occurrence of a repeated section of genes in a chromosome.
Haploid is Having a single set of unpaired chromosomes, as in germ cell, or half the diploid
number of a somatic cell.
Aneuploidy is the presence of an abnormal number of chromosomes in a cell,
example : a human cell having 45 or 47 chromosomes instead of the usual 46.
Nondisjunction is the failure of one or more pairs of homologous chromosomes or sister
chromatids to separate normally during nuclear division, usually resulting in an abnormal
distribution of chromosomes in the daughter nuclei..
H2A (ACID) Molarity = 0.15 M Normality N1 = Mo.pdf
H2A (ACID)
Molarity = 0.15 M
Normality N1 = Molarity x Basicity of acid
= 0.15x 2
= 0.3 N
Volume V1 = 100 ml
N1V1 = 0.3 x 100 = 30
NaOH (BASE)
Molarity = 0.16 M
Normality N2 = Molarity xAcidity of Base
= 0.16 x 1
= 0.16 N
Volume V2 = 100 ml
N2V2 = 0.16X100 = 16
Since concentration of ACID is more than concentration of BASE, resulatant concentration is
equal to
H+ ion concentration.
Normality at eqivalance point N = (N1V1 - N2V2)/ (V1+V2)
= 30 - 16/(100+100)
= 14/200
= 0.07 N
= H+ ion concentartion
Solution
H2A (ACID)
Molarity = 0.15 M
Normality N1 = Molarity x Basicity of acid
= 0.15x 2
= 0.3 N
Volume V1 = 100 ml
N1V1 = 0.3 x 100 = 30
NaOH (BASE)
Molarity = 0.16 M
Normality N2 = Molarity xAcidity of Base
= 0.16 x 1
= 0.16 N
Volume V2 = 100 ml
N2V2 = 0.16X100 = 16
Since concentration of ACID is more than concentration of BASE, resulatant concentration is
equal to
H+ ion concentration.
Normality at eqivalance point N = (N1V1 - N2V2)/ (V1+V2)
= 30 - 16/(100+100)
= 14/200
= 0.07 N
= H+ ion concentartion.
False , we should not always rename the worksheet.SolutionFals.pdf
False , we should not always rename the worksheet.
Solution
False , we should not always rename the worksheet..
Difference between computer architecture and computer organization.pdf
Difference between computer architecture and computer organization:
A computer architecture refers as a set of attributes of a system as seen by programmer.The term
\"computerarchitecture\" was originated by IBM.It was essentially acontract with software
stating unambiguously what thehardware does.
Computer oraganization was usually high-level description of the logic, memorey ....etc used to
implement that
contract.Mostly formulating what a cache,RAM,different I/O busses ,history of computers etc.
a.)
An instruction set of architecture is called as ISA.It is a part of computer that applicable to
programming, which is basically machine language.All CPUs have instruction sets that enable
commands to the processor directiong the CPU to switch the relevant transistors.The instructions
setswere:reduced(RISC) and complex (CISC).
CISC:Complex instruction set computer is rooted in the history of computing.
RISC:Reduced instruction set computer was realized in the 1970s by IBM.
b.)
Memory to memor
add13 x,y and z #x,y and z are memory addresses
mov1 eax , x
add1 eax , y
mov1 z , eax
load store:
lw $t0, x
lw $t1, y
add $t0, $t0, $t1
sw $t0, z
Accumulator-based:
load x # AC x
add y # AC AC + y
store z # z - AC
c.)The size
size of instuction part of an ISA are 32 bits long.
opcode tells the machine which format so add r1, r2, r3 has:
---opcode =0,fanct=32,rs=2,rt=3,rd=1,sa=0
Solution
Difference between computer architecture and computer organization:
A computer architecture refers as a set of attributes of a system as seen by programmer.The term
\"computerarchitecture\" was originated by IBM.It was essentially acontract with software
stating unambiguously what thehardware does.
Computer oraganization was usually high-level description of the logic, memorey ....etc used to
implement that
contract.Mostly formulating what a cache,RAM,different I/O busses ,history of computers etc.
a.)
An instruction set of architecture is called as ISA.It is a part of computer that applicable to
programming, which is basically machine language.All CPUs have instruction sets that enable
commands to the processor directiong the CPU to switch the relevant transistors.The instructions
setswere:reduced(RISC) and complex (CISC).
CISC:Complex instruction set computer is rooted in the history of computing.
RISC:Reduced instruction set computer was realized in the 1970s by IBM.
b.)
Memory to memor
add13 x,y and z #x,y and z are memory addresses
mov1 eax , x
add1 eax , y
mov1 z , eax
load store:
lw $t0, x
lw $t1, y
add $t0, $t0, $t1
sw $t0, z
Accumulator-based:
load x # AC x
add y # AC AC + y
store z # z - AC
c.)The size
size of instuction part of an ISA are 32 bits long.
opcode tells the machine which format so add r1, r2, r3 has:
---opcode =0,fanct=32,rs=2,rt=3,rd=1,sa=0.
CouplingCoupling is the act of combination of two things jointly..pdf
Coupling:
Coupling is the act of combination of two things jointly. In software development, coupling
submits to the degree to which software components are dependent ahead each other. For
example, in a tightly-coupled architecture, every component and its associated components
should be present within order for code to be performed or compiled. In a loosely coupled
architecture, components can stay autonomous and allow middleware software to manage
communication among them. In a decoupled architecture, the components can work completely
separately and autonomously.
Portability:
An application that can be changed from one computer environment to another. An application
amassed on a USB flash drive that can run on a computer with no leaving any trace of it at what
time the application is finished. It is designed to reinstate all changed settings when the program
is closed. For example, in general, software written in Java has this capability.
Solution
Coupling:
Coupling is the act of combination of two things jointly. In software development, coupling
submits to the degree to which software components are dependent ahead each other. For
example, in a tightly-coupled architecture, every component and its associated components
should be present within order for code to be performed or compiled. In a loosely coupled
architecture, components can stay autonomous and allow middleware software to manage
communication among them. In a decoupled architecture, the components can work completely
separately and autonomously.
Portability:
An application that can be changed from one computer environment to another. An application
amassed on a USB flash drive that can run on a computer with no leaving any trace of it at what
time the application is finished. It is designed to reinstate all changed settings when the program
is closed. For example, in general, software written in Java has this capability..
D. A trademark is a symbol, word, name, sound, or thing that identif.pdf
D. A trademark is a symbol, word, name, sound, or thing that identifies the origin of a product or
service in a particular trade.
Solution
D. A trademark is a symbol, word, name, sound, or thing that identifies the origin of a product or
service in a particular trade..
Benefits of using VHDLExecutable specificationIt is often report.pdf
Benefits of using VHDL
Executable specification
It is often reported that a large number of ASIC designs meet their specifications first time, but
fail to work when plugged into a system. VHDL allows this issue to be addressed in two ways: A
VHDL specification can be executed in order to achieve a high level of confidence in its
correctness before commencing design, and may simulate one to two orders of magnitude faster
than a gate level description. A VHDL specification for a part can form the basis for a simulation
model to verify the operation of the part in the wider system context (eg. printed circuit board
simulation). This depends on how accurately the specification handles aspects such as timing and
initialization.
Solution
Benefits of using VHDL
Executable specification
It is often reported that a large number of ASIC designs meet their specifications first time, but
fail to work when plugged into a system. VHDL allows this issue to be addressed in two ways: A
VHDL specification can be executed in order to achieve a high level of confidence in its
correctness before commencing design, and may simulate one to two orders of magnitude faster
than a gate level description. A VHDL specification for a part can form the basis for a simulation
model to verify the operation of the part in the wider system context (eg. printed circuit board
simulation). This depends on how accurately the specification handles aspects such as timing and
initialization..
AnswerC. FingertipTwo-point discrimination measures the individu.pdf
Answer
C. Fingertip
Two-point discrimination measures the individual’s ability to perceive two points of stimuli
presented simultaneously. It is the ability to discern that two nearby objects touching the skin are
truly two distinct points, not one. The fingertips have resolution, so, low errors
Solution
Answer
C. Fingertip
Two-point discrimination measures the individual’s ability to perceive two points of stimuli
presented simultaneously. It is the ability to discern that two nearby objects touching the skin are
truly two distinct points, not one. The fingertips have resolution, so, low errors.
Addition of H2O2 after ozonolysis oxidises the productso the unoxi.pdf
Addition of H2O2 after ozonolysis oxidises the product
so the unoxidised product is 7-formaldehyde 2 heptanone
now removing the oxygens and joining the carbon we get 1-methylcyclohex-1-ene
option 4 or d
Solution
Addition of H2O2 after ozonolysis oxidises the product
so the unoxidised product is 7-formaldehyde 2 heptanone
now removing the oxygens and joining the carbon we get 1-methylcyclohex-1-ene
option 4 or d.
1log 0.01= -22[H+]=1.6 x 10 -2pH=-log[H+]=-log(1.610-.pdf
1//
log 0.01= -2
2//
[H+]=1.6 x 10 -2
pH=-log[H+]
=-log(1.6*10-2)
= 1.7959
Solution
1//
log 0.01= -2
2//
[H+]=1.6 x 10 -2
pH=-log[H+]
=-log(1.6*10-2)
= 1.7959.
Biological Screening of Herbal Drugs in detailed.
Biological screening of herbal drugs: Introduction and Need for
Phyto-Pharmacological Screening, New Strategies for evaluating
Natural Products, In vitro evaluation techniques for Antioxidants, Antimicrobial and Anticancer drugs. In vivo evaluation techniques
for Anti-inflammatory, Antiulcer, Anticancer, Wound healing, Antidiabetic, Hepatoprotective, Cardio protective, Diuretics and
Antifertility, Toxicity studies as per OECD guidelines
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Event Link:-
https://meetups.mulesoft.com/events/details/mulesoft-mysore-presents-exploring-gemini-ai-and-integration-with-mulesoft/
Agenda
● Java 17 Upgrade Overview
● Why and by when do customers need to upgrade to Java 17?
● Is there any immediate impact to upgrading to Mule Runtime 4.6 and beyond?
● Which MuleSoft products are in scope?
For Upcoming Meetups Join Mysore Meetup Group - https://meetups.mulesoft.com/mysore/
YouTube:- youtube.com/@mulesoftmysore
Mysore WhatsApp group:- https://chat.whatsapp.com/EhqtHtCC75vCAX7gaO842N
Speaker:-
Shubham Chaurasia - https://www.linkedin.com/in/shubhamchaurasia1/
Priya Shaw - https://www.linkedin.com/in/priya-shaw
Organizers:-
Shubham Chaurasia - https://www.linkedin.com/in/shubhamchaurasia1/
Giridhar Meka - https://www.linkedin.com/in/giridharmeka
Priya Shaw - https://www.linkedin.com/in/priya-shaw
Shyam Raj Prasad-
https://www.linkedin.com/in/shyam-raj-prasad/
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The follow list restates these 3 fundamental levels of service for Q.pdf
The follow list restates these 3 fundamental levels of service for QoS:
Best-effort service: the normal type of connectivity without some guarantee. This kind of
service, in orientation to Catalyst switch, use first-in first-out queue, which only transmit packet
as they get there in a line with no special treatment.
Integrated services: IntServ, also known as solid QoS, is a complete condition of services. In
additional terms, the IntServ model implies that traffic flow is kept openly by all midway system
and possessions.
Differentiated services—DiffServ, also known as flexible QoS, is class base, where a few
module of traffic obtain special behavior more additional traffic classes. differentiate services
use arithmetical preference, not a hard assurance like incorporated services. In additional terms,
DiffServ categorize traffic and then sort it into line of different efficiencies.
The IntServ model, distinct in RFC 1633, guarantees a unsurprising performance of the
association for application. IntServ give several services that accommodate several QoS
necessities. IntServ is implementing throughout the use of the supply Reservation Protocol,
enable at together the endpoints and the net devices connecting. This enabled application needs
an exact type of service from this enabled system previous to it send information. Open signal
via RSVP make the appeal, and the request inform the network of its traffic report and needs a
exacting type of service that can include its bandwidth and holdup necessities. The application is
predictable to send information only following it gets verification as of the network. The network
as well expect the application to propel information that deceit in it’s describe traffic profile
IntServ follow the signaled QoS form where the end hosts sign their QoS necessities to the
system. Each individual message stream wants to request capital from the network. Rim routers
utilize the Resource Reservation Protocol to indicator and keep the pet bandwidth for every flood
in the network.
The DiffServ model provides several levels of examine that convince differing QoS necessities.
Though, different with the IntServ model, an request that use DiffServ do not plainly signal the
network strategy before transfer information. DiffServ is a QoS completion method that is
modified for current networks and their explanation. DiffServ reassigns bit in the kind of service
pasture of an IP packet title. DiffServ use differentiates services code points as the QoS main
concern descriptor importance and supports 64 stage of categorization
For example, VoIP require extremely small jitter, a one way holdup in the arrange of 150 ms
plus certain bandwidth in the range of 8Kbps to 64Kbps, needy on the codec use. In one more
example, a case move request, base on ftp, do not undergo from jitter, as packet failure will be
extremely damaging to the throughput.
Solution
The follow list restates these 3 fundamental levels of service for QoS:
Best-e.
tested on eclipseDoublyLinkedList class.pdf
//tested on eclipse
/*************DoublyLinkedList class*******************/
public class DoublyLinkedList {
/*Data member variable declaration*/
private Node head;
private int count;
/**default constructor*/
public DoublyLinkedList() {
/*Initializing variable*/
head=null;
count =0;
}
/*Insert at beginning in doubly linked list*/
public void insertBeginning(String name){
/*checking head is null then we will add node at head*/
if(head==null){
head=new Node(name, null, null);
count++;
}else{
/*else create temp node and setting next to head
* and then making temp to head*/
Node temp=new Node(name,null,null);
temp.setNext(head);
head.setPrev(temp);
head=temp;
count++;
}
}
public void printList(){
Node temp=head;
/*Printing element of doubly list*/
while(temp!=null){
System.out.println(temp.getData());
temp=temp.getNext();
}
}
/*Returning count from doubly linked list*/
public int count(){
return count;
}
/*deleting specific node from doubly linked list*/
public void delete(String name){
Node current=head,previous=head;
/*if it is head node to be deleted*/
if(head.getData().equals(name)){
current=head.getNext();
current.setPrev(null);
head=current;
previous=head;
}else{
/*If node in middle of list*/
while(current.getNext()!=null){
if(current.getData().equals(name)){
current.getNext().setPrev(previous);
previous.setNext(current.getNext());
count--;
break;
}
previous=current;
current=current.getNext();
}
/*If node is at last*/
if(current.getNext()==null){
if(current.getData().equals(name)){
previous.setNext(null);
count--;
}
}
}
}
/*deleting all nodes from list by assigning head to null
* and memory releasing take care by garbage collector*/
public void clear(){
head=null;
count=0;
}
/*Inserting element in alphabetic order*/
public void insertAlpha(String name){
Node current=head,previous=head;
/*creating new node*/
Node newNode=new Node(name,null,null);
/*if head node is null*/
if(head==null){
head=newNode;
count++;
}else{
/*please read about compareTo() method about returning value*/
if(head.getData().compareTo(name)>0){
newNode.setNext(head);
head=newNode;
count++;
}else{
/*condition for getting position of node to be inserted*/
while(current!=null&¤t.getData().compareTo(name)<=0){
previous=current;
current=current.getNext();
}
/*if node need to be inserted at last*/
if(current==null){
previous=newNode;
}else{
/*If node need to inserted in middle*/
newNode.setNext(current);
previous.setNext(newNode);
current.setPrev(newNode);
newNode.setPrev(previous);
}
count++;
}
}
}
/*Main method start*/
public static void main(String args[]){
/*creating object of doubly linked list*/
DoublyLinkedList doublyLinkedList=new DoublyLinkedList();
doublyLinkedList.insertBeginning(\"sunita\");
doublyLinkedList.insertBeginning(\"lalchand\");
/*calling alphabetic order method*/
doublyLinkedList.insertAlpha(\"ramsingh\");
doublyLinkedList.insertBeginning(\"Lokesh\");
doublyLinkedList.insertBeginning(\"Pritam\");
/*****printing list********/
doublyLinkedList.printL.
Please read the comment ins codeExpressionTree.java-------------.pdf
Please read the comment ins code
ExpressionTree.java
----------------------------------
/**
* This is the class for Expression Tree.
* Used to create Expression Tree and Evaluate it
*/
/**
* Following logic is used to construct a Tree
* Here we use stack for Preparing Tree
* Loop through given Expression String
* If Character is Operand , Create node and push to stack
* If Character is Operator then
* 1)Create Node for Operator
* 2)Pop 2 nodes from Stack and Made
* OpretorNode--> left == first node pop
* OpretorNode--> right == second node pop
* At the end of creation of Expression Tree, Stack have only one node , which is root of
Expression tree
*/
/** Class ExpressionTree **/
class ExpressionTree
{
/** class TreeNode
* Stored Character ==> Digit(0..9) or a Operator +,-,*,/
* Left Node and Right Node
*
* **/
class TreeNode
{
char data;
TreeNode left, right;
/** constructor **/
public TreeNode(char data)
{
this.data = data;
this.left = null;
this.right = null;
}
}
/** class StackNode **/
class StackNode
{
TreeNode treeNode;
StackNode next;
/** constructor **/
public StackNode(TreeNode treeNode)
{
this.treeNode = treeNode;
next = null;
}
}
private static StackNode top;
/** constructor
* Constructor takes input string like \"+-+7*935*82*625\"
* Input should be in Prefix notation
* **/
public ExpressionTree(String expression)
{
top = null;
//Call Method for prepare expression tree
buildTree(expression);
}
/** function to clear tree **/
public void clear()
{
top = null;
}
/** function to push a node **/
private void push(TreeNode ptr)
{
if (top == null)
top = new StackNode(ptr);
else
{
StackNode nptr = new StackNode(ptr);
nptr.next = top;
top = nptr;
}
}
/** function to pop a node
* When it find operator pop 2 elements from Stack
*
* **/
private TreeNode pop()
{
if (top == null)
throw new RuntimeException(\"Underflow\");
else
{
TreeNode ptr = top.treeNode;
top = top.next;
return ptr;
}
}
/** function to get top node **/
private TreeNode peek()
{
return top.treeNode;
}
/** function to insert character **/
private void insert(char val)
{
try
{
//If Operand , Create node and push to Stack
if (isDigit(val))
{
TreeNode nptr = new TreeNode(val);
push(nptr);
}
//If Operator , Create node and popup 2 elements and make them its right and left
else if (isOperator(val))
{
TreeNode nptr = new TreeNode(val);
nptr.left = pop();
nptr.right = pop();
push(nptr);
}
}
catch (Exception e)
{
System.out.println(\"Invalid Expression\");
}
}
/** function to check if digit **/
private boolean isDigit(char ch)
{
return ch >= \'0\' && ch <= \'9\';
}
/** function to check if operator **/
private boolean isOperator(char ch)
{
return ch == \'+\' || ch == \'-\' || ch == \'*\' || ch == \'/\';
}
/** function to convert character to digit **/
private int toDigit(char ch)
{
return ch - \'0\';
}
/** function to build tree from input */
public void buildTree(String eqn)
{
for (int i = eqn.length() - 1; i >= 0; i--)
insert(eqn.charAt(i));
}
/** function to evaluate tree */
public dou.
Please find the answers and explanations belowPart A1. The ing.pdf
Please find the answers and explanations below:
Part A:
1. The ingestion of botulism toxin would best be described as a food poisoning. (ingestion of
botulism toxin or any other gut-infecting agent is non-technically termed as food poisoning. It is
different from infection in the sense that food poisoning may not always lead to sustained gut
infection but diagnosis of gut infection results evidently from food poisoning)
2. The ingestion of a significant number of Clostridium cells would be a infection.
(Immunologically, a sufficeint amount of pathogen is required for starting a degrading immune
response in the body. This is called the threshold number of the pathogen which can elicit an
immune response in the body. When the number of pathogeni cells increases beyond this
threshold, the person feels sick and an infection persists)
3.Spoilage is a change in appearance, smell, or taste of a food that makes it undesirable to a
consumer. (Technically, spoilage is defined as physcial and chemical change in the
appearance,smell, taste or nutritional content of the food due to infection or residence of a
pathogen in it.)
Part B:
For any pathogen to enter and cause infection, some specific steps need to be fulfilled by it.
Firstly, the concentration of the pathogen must be large enough to survive in the food item. This
infected food item will be ingested by the host and the pathogen will reach the gut. The pathogen
survives the harsh environment of the gut and starts replicating. This replication of the pathogen
comes with severe side effects on the host which include physical and physiological deterioration
of the gut lining along with eliciting immune response in the host. This ultimately causes
infection/disease.
So, the correct order of events would be:
1. The potatoes were contaminated with soil containing Clostridium botulinum endospores.
2. The toxin remained in the pruno after straining.
3. The conditions in the bag of pruno were favorable for germination.
4. Clostridium botulinum replicated and increased the number of organisms in the pruno.
5. Bacterial cells produced and secreted toxin.
6. The inmates drank the contaminated pruno and became ill.
Solution
Please find the answers and explanations below:
Part A:
1. The ingestion of botulism toxin would best be described as a food poisoning. (ingestion of
botulism toxin or any other gut-infecting agent is non-technically termed as food poisoning. It is
different from infection in the sense that food poisoning may not always lead to sustained gut
infection but diagnosis of gut infection results evidently from food poisoning)
2. The ingestion of a significant number of Clostridium cells would be a infection.
(Immunologically, a sufficeint amount of pathogen is required for starting a degrading immune
response in the body. This is called the threshold number of the pathogen which can elicit an
immune response in the body. When the number of pathogeni cells increases beyond this
threshold, the pers.
Mass of liquid = 22.5 gHeat added = 2.34 kJ = 2340 JTemperature .pdf
Mass of liquid = 22.5 g
Heat added = 2.34 kJ = 2340 J
Temperature change = 56.3 - 14.0 = 42.3oC
Specific heat capacity = heat added/(mass x temperature change)
= 2340/(22.5 x 42.3) = 2.46 J/g.oC
Solution
Mass of liquid = 22.5 g
Heat added = 2.34 kJ = 2340 J
Temperature change = 56.3 - 14.0 = 42.3oC
Specific heat capacity = heat added/(mass x temperature change)
= 2340/(22.5 x 42.3) = 2.46 J/g.oC.
It is given that we should use 4 bit input, i.e we will have a set o.pdf
It is given that we should use 4 bit input, i.e we will have a set of 16 different inputs. The output
of the circuit will be 1 if and only if the decimal equivalent formed by the 4 input bits is an
integer that belongs to the fibonacci series.
a) The truth table according to the above requirement is
A
B
C
D
O
0
0
0
0
1
0
0
0
1
1
0
0
1
0
1
0
0
1
1
1
0
1
0
0
0
0
1
0
1
1
0
1
1
0
0
0
1
1
1
0
1
0
0
0
1
1
0
0
1
0
1
0
1
0
0
1
0
1
1
0
1
1
0
0
0
1
1
0
1
1
1
1
1
0
0
1
1
1
1
0
B) K - map for the above truth table is
c) According to the above K-map, prime implicants would be
A\'B\', B\'C\'D\' and C\'D
d) All the prime implicants are essential prime implicants
e) A\'B\'+B\'C\'D\'+C\'D
A
B
C
D
O
0
0
0
0
1
0
0
0
1
1
0
0
1
0
1
0
0
1
1
1
0
1
0
0
0
0
1
0
1
1
0
1
1
0
0
0
1
1
1
0
1
0
0
0
1
1
0
0
1
0
1
0
1
0
0
1
0
1
1
0
1
1
0
0
0
1
1
0
1
1
1
1
1
0
0
1
1
1
1
0
Solution
It is given that we should use 4 bit input, i.e we will have a set of 16 different inputs. The output
of the circuit will be 1 if and only if the decimal equivalent formed by the 4 input bits is an
integer that belongs to the fibonacci series.
a) The truth table according to the above requirement is
A
B
C
D
O
0
0
0
0
1
0
0
0
1
1
0
0
1
0
1
0
0
1
1
1
0
1
0
0
0
0
1
0
1
1
0
1
1
0
0
0
1
1
1
0
1
0
0
0
1
1
0
0
1
0
1
0
1
0
0
1
0
1
1
0
1
1
0
0
0
1
1
0
1
1
1
1
1
0
0
1
1
1
1
0
B) K - map for the above truth table is
c) According to the above K-map, prime implicants would be
A\'B\', B\'C\'D\' and C\'D
d) All the prime implicants are essential prime implicants
e) A\'B\'+B\'C\'D\'+C\'D
A
B
C
D
O
0
0
0
0
1
0
0
0
1
1
0
0
1
0
1
0
0
1
1
1
0
1
0
0
0
0
1
0
1
1
0
1
1
0
0
0
1
1
1
0
1
0
0
0
1
1
0
0
1
0
1
0
1
0
0
1
0
1
1
0
1
1
0
0
0
1
1
0
1
1
1
1
1
0
0
1
1
1
1
0.
IndexingConcatinationSolutionIndexingConcatination.pdf
Indexing:
Concatination:
Solution
Indexing:
Concatination:.
import java.util.ArrayList; import java.util.List;public class H.pdf
import java.util.ArrayList;
import java.util.List;
public class H2ClassA {
List list = new ArrayList ();
public static void main(String[] args) {
H2ClassA y = new H2ClassA ();
int [] v = {4, 3, 7, 5, 99, 3};
// change the for loop to old type of for loop
for(int i=0;i
Solution
import java.util.ArrayList;
import java.util.List;
public class H2ClassA {
List list = new ArrayList ();
public static void main(String[] args) {
H2ClassA y = new H2ClassA ();
int [] v = {4, 3, 7, 5, 99, 3};
// change the for loop to old type of for loop
for(int i=0;i.
Heterochromatin is a tightly packed form of DNA, which comes in mult.pdf
Heterochromatin is a tightly packed form of DNA, which comes in multiple varieties.
constitutive heterochromatin the chromatin in regions of the chromosomes that are invariably
heterochromatic; it contains highly repetitive sequences of DNA that are genetically inactive and
serves as a structural element of the chromosome.
facultative heterochromatin the chromatin in regions of the chromosomes that become
heterochromatic in certain cells and tissues; for example, it makes up the inactive X chromosome
in female somatic cells.
Homologous chromosomes are chromosome pairs, one from each parent, that are similar in
length, gene position and centromere location.
Duplicated chromosome The occurrence of a repeated section of genes in a chromosome.
Haploid is Having a single set of unpaired chromosomes, as in germ cell, or half the diploid
number of a somatic cell.
Aneuploidy is the presence of an abnormal number of chromosomes in a cell,
example : a human cell having 45 or 47 chromosomes instead of the usual 46.
Nondisjunction is the failure of one or more pairs of homologous chromosomes or sister
chromatids to separate normally during nuclear division, usually resulting in an abnormal
distribution of chromosomes in the daughter nuclei.
Solution
Heterochromatin is a tightly packed form of DNA, which comes in multiple varieties.
constitutive heterochromatin the chromatin in regions of the chromosomes that are invariably
heterochromatic; it contains highly repetitive sequences of DNA that are genetically inactive and
serves as a structural element of the chromosome.
facultative heterochromatin the chromatin in regions of the chromosomes that become
heterochromatic in certain cells and tissues; for example, it makes up the inactive X chromosome
in female somatic cells.
Homologous chromosomes are chromosome pairs, one from each parent, that are similar in
length, gene position and centromere location.
Duplicated chromosome The occurrence of a repeated section of genes in a chromosome.
Haploid is Having a single set of unpaired chromosomes, as in germ cell, or half the diploid
number of a somatic cell.
Aneuploidy is the presence of an abnormal number of chromosomes in a cell,
example : a human cell having 45 or 47 chromosomes instead of the usual 46.
Nondisjunction is the failure of one or more pairs of homologous chromosomes or sister
chromatids to separate normally during nuclear division, usually resulting in an abnormal
distribution of chromosomes in the daughter nuclei..
H2A (ACID) Molarity = 0.15 M Normality N1 = Mo.pdf
H2A (ACID)
Molarity = 0.15 M
Normality N1 = Molarity x Basicity of acid
= 0.15x 2
= 0.3 N
Volume V1 = 100 ml
N1V1 = 0.3 x 100 = 30
NaOH (BASE)
Molarity = 0.16 M
Normality N2 = Molarity xAcidity of Base
= 0.16 x 1
= 0.16 N
Volume V2 = 100 ml
N2V2 = 0.16X100 = 16
Since concentration of ACID is more than concentration of BASE, resulatant concentration is
equal to
H+ ion concentration.
Normality at eqivalance point N = (N1V1 - N2V2)/ (V1+V2)
= 30 - 16/(100+100)
= 14/200
= 0.07 N
= H+ ion concentartion
Solution
H2A (ACID)
Molarity = 0.15 M
Normality N1 = Molarity x Basicity of acid
= 0.15x 2
= 0.3 N
Volume V1 = 100 ml
N1V1 = 0.3 x 100 = 30
NaOH (BASE)
Molarity = 0.16 M
Normality N2 = Molarity xAcidity of Base
= 0.16 x 1
= 0.16 N
Volume V2 = 100 ml
N2V2 = 0.16X100 = 16
Since concentration of ACID is more than concentration of BASE, resulatant concentration is
equal to
H+ ion concentration.
Normality at eqivalance point N = (N1V1 - N2V2)/ (V1+V2)
= 30 - 16/(100+100)
= 14/200
= 0.07 N
= H+ ion concentartion.
False , we should not always rename the worksheet.SolutionFals.pdf
False , we should not always rename the worksheet.
Solution
False , we should not always rename the worksheet..
Difference between computer architecture and computer organization.pdf
Difference between computer architecture and computer organization:
A computer architecture refers as a set of attributes of a system as seen by programmer.The term
\"computerarchitecture\" was originated by IBM.It was essentially acontract with software
stating unambiguously what thehardware does.
Computer oraganization was usually high-level description of the logic, memorey ....etc used to
implement that
contract.Mostly formulating what a cache,RAM,different I/O busses ,history of computers etc.
a.)
An instruction set of architecture is called as ISA.It is a part of computer that applicable to
programming, which is basically machine language.All CPUs have instruction sets that enable
commands to the processor directiong the CPU to switch the relevant transistors.The instructions
setswere:reduced(RISC) and complex (CISC).
CISC:Complex instruction set computer is rooted in the history of computing.
RISC:Reduced instruction set computer was realized in the 1970s by IBM.
b.)
Memory to memor
add13 x,y and z #x,y and z are memory addresses
mov1 eax , x
add1 eax , y
mov1 z , eax
load store:
lw $t0, x
lw $t1, y
add $t0, $t0, $t1
sw $t0, z
Accumulator-based:
load x # AC x
add y # AC AC + y
store z # z - AC
c.)The size
size of instuction part of an ISA are 32 bits long.
opcode tells the machine which format so add r1, r2, r3 has:
---opcode =0,fanct=32,rs=2,rt=3,rd=1,sa=0
Solution
Difference between computer architecture and computer organization:
A computer architecture refers as a set of attributes of a system as seen by programmer.The term
\"computerarchitecture\" was originated by IBM.It was essentially acontract with software
stating unambiguously what thehardware does.
Computer oraganization was usually high-level description of the logic, memorey ....etc used to
implement that
contract.Mostly formulating what a cache,RAM,different I/O busses ,history of computers etc.
a.)
An instruction set of architecture is called as ISA.It is a part of computer that applicable to
programming, which is basically machine language.All CPUs have instruction sets that enable
commands to the processor directiong the CPU to switch the relevant transistors.The instructions
setswere:reduced(RISC) and complex (CISC).
CISC:Complex instruction set computer is rooted in the history of computing.
RISC:Reduced instruction set computer was realized in the 1970s by IBM.
b.)
Memory to memor
add13 x,y and z #x,y and z are memory addresses
mov1 eax , x
add1 eax , y
mov1 z , eax
load store:
lw $t0, x
lw $t1, y
add $t0, $t0, $t1
sw $t0, z
Accumulator-based:
load x # AC x
add y # AC AC + y
store z # z - AC
c.)The size
size of instuction part of an ISA are 32 bits long.
opcode tells the machine which format so add r1, r2, r3 has:
---opcode =0,fanct=32,rs=2,rt=3,rd=1,sa=0.
CouplingCoupling is the act of combination of two things jointly..pdf
Coupling:
Coupling is the act of combination of two things jointly. In software development, coupling
submits to the degree to which software components are dependent ahead each other. For
example, in a tightly-coupled architecture, every component and its associated components
should be present within order for code to be performed or compiled. In a loosely coupled
architecture, components can stay autonomous and allow middleware software to manage
communication among them. In a decoupled architecture, the components can work completely
separately and autonomously.
Portability:
An application that can be changed from one computer environment to another. An application
amassed on a USB flash drive that can run on a computer with no leaving any trace of it at what
time the application is finished. It is designed to reinstate all changed settings when the program
is closed. For example, in general, software written in Java has this capability.
Solution
Coupling:
Coupling is the act of combination of two things jointly. In software development, coupling
submits to the degree to which software components are dependent ahead each other. For
example, in a tightly-coupled architecture, every component and its associated components
should be present within order for code to be performed or compiled. In a loosely coupled
architecture, components can stay autonomous and allow middleware software to manage
communication among them. In a decoupled architecture, the components can work completely
separately and autonomously.
Portability:
An application that can be changed from one computer environment to another. An application
amassed on a USB flash drive that can run on a computer with no leaving any trace of it at what
time the application is finished. It is designed to reinstate all changed settings when the program
is closed. For example, in general, software written in Java has this capability..
D. A trademark is a symbol, word, name, sound, or thing that identif.pdf
D. A trademark is a symbol, word, name, sound, or thing that identifies the origin of a product or
service in a particular trade.
Solution
D. A trademark is a symbol, word, name, sound, or thing that identifies the origin of a product or
service in a particular trade..
Benefits of using VHDLExecutable specificationIt is often report.pdf
Benefits of using VHDL
Executable specification
It is often reported that a large number of ASIC designs meet their specifications first time, but
fail to work when plugged into a system. VHDL allows this issue to be addressed in two ways: A
VHDL specification can be executed in order to achieve a high level of confidence in its
correctness before commencing design, and may simulate one to two orders of magnitude faster
than a gate level description. A VHDL specification for a part can form the basis for a simulation
model to verify the operation of the part in the wider system context (eg. printed circuit board
simulation). This depends on how accurately the specification handles aspects such as timing and
initialization.
Solution
Benefits of using VHDL
Executable specification
It is often reported that a large number of ASIC designs meet their specifications first time, but
fail to work when plugged into a system. VHDL allows this issue to be addressed in two ways: A
VHDL specification can be executed in order to achieve a high level of confidence in its
correctness before commencing design, and may simulate one to two orders of magnitude faster
than a gate level description. A VHDL specification for a part can form the basis for a simulation
model to verify the operation of the part in the wider system context (eg. printed circuit board
simulation). This depends on how accurately the specification handles aspects such as timing and
initialization..
AnswerC. FingertipTwo-point discrimination measures the individu.pdf
Answer
C. Fingertip
Two-point discrimination measures the individual’s ability to perceive two points of stimuli
presented simultaneously. It is the ability to discern that two nearby objects touching the skin are
truly two distinct points, not one. The fingertips have resolution, so, low errors
Solution
Answer
C. Fingertip
Two-point discrimination measures the individual’s ability to perceive two points of stimuli
presented simultaneously. It is the ability to discern that two nearby objects touching the skin are
truly two distinct points, not one. The fingertips have resolution, so, low errors.
Addition of H2O2 after ozonolysis oxidises the productso the unoxi.pdf
Addition of H2O2 after ozonolysis oxidises the product
so the unoxidised product is 7-formaldehyde 2 heptanone
now removing the oxygens and joining the carbon we get 1-methylcyclohex-1-ene
option 4 or d
Solution
Addition of H2O2 after ozonolysis oxidises the product
so the unoxidised product is 7-formaldehyde 2 heptanone
now removing the oxygens and joining the carbon we get 1-methylcyclohex-1-ene
option 4 or d.
1log 0.01= -22[H+]=1.6 x 10 -2pH=-log[H+]=-log(1.610-.pdf
1//
log 0.01= -2
2//
[H+]=1.6 x 10 -2
pH=-log[H+]
=-log(1.6*10-2)
= 1.7959
Solution
1//
log 0.01= -2
2//
[H+]=1.6 x 10 -2
pH=-log[H+]
=-log(1.6*10-2)
= 1.7959.
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● Why and by when do customers need to upgrade to Java 17?
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YouTube:- youtube.com/@mulesoftmysore
Mysore WhatsApp group:- https://chat.whatsapp.com/EhqtHtCC75vCAX7gaO842N
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Priya Shaw - https://www.linkedin.com/in/priya-shaw
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Dive into the world of AI! Experts Jon Hill and Tareq Monaur will guide you through AI's role in enhancing nonprofit websites and basic marketing strategies, making it easy to understand and apply.
Natural birth techniques - Mrs.Akanksha Trivedi Rama University
Natural birth techniques - Mrs.Akanksha Trivedi Rama UniversityAkanksha trivedi rama nursing college kanpur.
Natural birth techniques are various type such as/ water birth , alexender method, hypnosis, bradley method, lamaze method etcChapter 4 - Islamic Financial Institutions in Malaysia.pptx
Chapter 4 - Islamic Financial Institutions in Malaysia.pptxMohd Adib Abd Muin, Senior Lecturer at Universiti Utara Malaysia
This slide is special for master students (MIBS & MIFB) in UUM. Also useful for readers who are interested in the topic of contemporary Islamic banking.
MASS MEDIA STUDIES-835-CLASS XI Resource Material.pdf
mass media studies
CBSE
Full Book
for class 11th
Best Digital Marketing Institute In NOIDA
Safalta Digital marketing institute in Noida, provide complete applications that encompass a huge range of virtual advertising and marketing additives, which includes search engine optimization, virtual communication advertising, pay-per-click on marketing, content material advertising, internet analytics, and greater. These university courses are designed for students who possess a comprehensive understanding of virtual marketing strategies and attributes.Safalta Digital Marketing Institute in Noida is a first choice for young individuals or students who are looking to start their careers in the field of digital advertising. The institute gives specialized courses designed and certification.
for beginners, providing thorough training in areas such as SEO, digital communication marketing, and PPC training in Noida. After finishing the program, students receive the certifications recognised by top different universitie, setting a strong foundation for a successful career in digital marketing.
CACJapan - GROUP Presentation 1- Wk 4.pdf
Macroeconomics- Movie Location
This will be used as part of your Personal Professional Portfolio once graded.
Objective:
Prepare a presentation or a paper using research, basic comparative analysis, data organization and application of economic information. You will make an informed assessment of an economic climate outside of the United States to accomplish an entertainment industry objective.
Synthetic Fiber Construction in lab .pptx
Synthetic fiber production is a fascinating and complex field that blends chemistry, engineering, and environmental science. By understanding these aspects, students can gain a comprehensive view of synthetic fiber production, its impact on society and the environment, and the potential for future innovations. Synthetic fibers play a crucial role in modern society, impacting various aspects of daily life, industry, and the environment. ynthetic fibers are integral to modern life, offering a range of benefits from cost-effectiveness and versatility to innovative applications and performance characteristics. While they pose environmental challenges, ongoing research and development aim to create more sustainable and eco-friendly alternatives. Understanding the importance of synthetic fibers helps in appreciating their role in the economy, industry, and daily life, while also emphasizing the need for sustainable practices and innovation.
Operation Blue Star - Saka Neela Tara
Operation “Blue Star” is the only event in the history of Independent India where the state went into war with its own people. Even after about 40 years it is not clear if it was culmination of states anger over people of the region, a political game of power or start of dictatorial chapter in the democratic setup.
The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
Thesis Statement for students diagnonsed withADHD.ppt
Presentation required for the master in Education.
The Diamonds of 2023-2024 in the IGRA collection
A review of the growth of the Israel Genealogy Research Association Database Collection for the last 12 months. Our collection is now passed the 3 million mark and still growing. See which archives have contributed the most. See the different types of records we have, and which years have had records added. You can also see what we have for the future.
South African Journal of Science: Writing with integrity workshop (2024)
South African Journal of Science: Writing with integrity workshop (2024)Academy of Science of South Africa
A workshop hosted by the South African Journal of Science aimed at postgraduate students and early career researchers with little or no experience in writing and publishing journal articles.Recently uploaded (20)
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
"Protectable subject matters, Protection in biotechnology, Protection of othe...
"Protectable subject matters, Protection in biotechnology, Protection of othe...
Introduction to AI for Nonprofits with Tapp Network
Introduction to AI for Nonprofits with Tapp Network
Natural birth techniques - Mrs.Akanksha Trivedi Rama University
Natural birth techniques - Mrs.Akanksha Trivedi Rama University
Chapter 4 - Islamic Financial Institutions in Malaysia.pptx
Chapter 4 - Islamic Financial Institutions in Malaysia.pptx
MASS MEDIA STUDIES-835-CLASS XI Resource Material.pdf
MASS MEDIA STUDIES-835-CLASS XI Resource Material.pdf
Thesis Statement for students diagnonsed withADHD.ppt
Thesis Statement for students diagnonsed withADHD.ppt
South African Journal of Science: Writing with integrity workshop (2024)
South African Journal of Science: Writing with integrity workshop (2024)
Add mCPBA to make the epoxide, then use aqueous N.pdf
- 1. Add mCPBA to make the epoxide, then use aqueous NaOH to make the diol. You will get a racemic mixture of the two diols however. Solution Add mCPBA to make the epoxide, then use aqueous NaOH to make the diol. You will get a racemic mixture of the two diols however.