molar ratio 6 carbon 12 hydirgen 6 oxygen .pdf
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molar ratio : 6 carbon 12 hydirgen 6 oxygen molar ratio : C : H : O = 1 : 2 : 1 Solution molar ratio : 6 carbon 12 hydirgen 6 oxygen molar ratio : C : H : O = 1 : 2 : 1.
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}Solution}.pdf
The document discusses a solution for improving the efficiency of solar panels. It proposes coating solar panels with a new material that can absorb more wavelengths of light, increasing energy absorption by 40%. This coating material is inexpensive to produce and easy to apply, allowing for widespread implementation of more effective solar energy technology.
What is an OS • Interface between application programs and hardwa.pdf
What is an OS?
• Interface between application programs and hardware
• Ultimate control program
– Exploits hardware resources of one or more processors to provide a set of services to users
– Coordinates the use of hardware among various application programs for different users
– Manages secondary memory and I/O devices on behalf of its users
• Two different views of an OS
1. Extended machine view
– Virtual machine that is easier to understand and program
– Tool to make programmer’s job easy
2. Resource manager view
– Tool to facilitate efficient operation of computer system
– Provides services to users; processor, memory, I/O, system bus
– Must be fair; not partial to any process, specially for process in the same class
– Must discriminate between different class of jobs with different service requirements
– Do the above efficiently
Within the constraints of fairness and efficiency, an OS should attempt to maximize throughput,
minimize
response time, and accommodate as many users as possible
Basic elements
• A computer has a number of modules, with possibly more than one instance of each
• Processor/CPU
– Controls the operations of computer and performs data processing functions
– Exchanges data with memory using memory address register and memory buffer register
– Exchanges data with I/O devices using I/O address register and I/O buffer register
• Main memory/Primary memory
– Stores data and programs
– Typically volatile
– Abstracted as a set of locations, defined by sequentially numbered addresses
– Each location contains a bit pattern to be interpreted as either an instruction or data
• I/O modules/architecture
– Move data between a computer and external environment
– Communicate with a variety of devices including secondary memory (disks), communications
equipment, and
terminals
– Data flows between CPUs, RAM and I/O devices over buses
– System bus
Provides for communications among processors, main memory, and I/O modules
Connects most of the internal hardware devices
PCI bus, ISA bus, EISA bus, SCSI, USB
Solution
What is an OS?
• Interface between application programs and hardware
• Ultimate control program
– Exploits hardware resources of one or more processors to provide a set of services to users
– Coordinates the use of hardware among various application programs for different users
– Manages secondary memory and I/O devices on behalf of its users
• Two different views of an OS
1. Extended machine view
– Virtual machine that is easier to understand and program
– Tool to make programmer’s job easy
2. Resource manager view
– Tool to facilitate efficient operation of computer system
– Provides services to users; processor, memory, I/O, system bus
– Must be fair; not partial to any process, specially for process in the same class
– Must discriminate between different class of jobs with different service requirements
– Do the above efficiently
Within the constraints of fairness and efficiency, an OS should attempt to maximize throughput,.
Using a broad definition of file(note that it doesnt count h.pdf
Using a broad definition of \"file\"
(note that it doesn\'t count hidden files and assumes that file names don\'t contain newline
characters).
To include hidden files (except . and ..) and avoid problems with newline characters, the
canonical way is:
Or recursively:
Solution
Using a broad definition of \"file\"
(note that it doesn\'t count hidden files and assumes that file names don\'t contain newline
characters).
To include hidden files (except . and ..) and avoid problems with newline characters, the
canonical way is:
Or recursively:.
unable to open it ....it asks for username and passwordSolution.pdf
unable to open it ....it asks for username and password
Solution
unable to open it ....it asks for username and password.
This is called a dehydrationreaction. a) H2SO4 H20Solution.pdf
This is called a dehydrationreaction.
a) H2SO4 / H20
Solution
This is called a dehydrationreaction.
a) H2SO4 / H20.
The multiplication sign simply means that the molecules have their i.pdf
The multiplication sign simply means that the molecules have their independent existance in the
compound. For example the formula you have given has three species in solution ZnCO3, 2
molecules of Zn(OH)2 and 1 molecule of water H20 for each molecule of your compound. These
species are bonded to each other but the bonds between the atoms of the participating molecules
are not broken. To get the molar mass of the compound simply add the molar masses of
indivisual molecules multiplied by number of molecules they are contributing. Thus for
ZnCO3*2Zn(OH)2*1H20 the molar mass would be:
Molar mass of ZnCO3 + 2*(Molar mass of Zn(OH)2) + Molar mass of H2O
= 125.4 + 2*99.4 + 18
= 342.2 g/mol
Solution
The multiplication sign simply means that the molecules have their independent existance in the
compound. For example the formula you have given has three species in solution ZnCO3, 2
molecules of Zn(OH)2 and 1 molecule of water H20 for each molecule of your compound. These
species are bonded to each other but the bonds between the atoms of the participating molecules
are not broken. To get the molar mass of the compound simply add the molar masses of
indivisual molecules multiplied by number of molecules they are contributing. Thus for
ZnCO3*2Zn(OH)2*1H20 the molar mass would be:
Molar mass of ZnCO3 + 2*(Molar mass of Zn(OH)2) + Molar mass of H2O
= 125.4 + 2*99.4 + 18
= 342.2 g/mol.
SolutionThe variance of individual assets is measure of total ris.pdf
The variance of individual assets measures total risk, but in a well-diversified portfolio only systematic risk matters as unsystematic risk is reduced. The variance of a diversified portfolio depends on its systematic risk, known as beta. Therefore, the statement in the question is false.
soldefnition of critical thinkingit is an intellectual process o.pdf
sol:
defnition of critical thinking:it is an intellectual process of applying ,analyizing,evaluating
information by means of observation which leads to believe in action
a. When you consider all the different perspectives on a topic and evaluate how true or
applicable they are in a given context.
Solution
sol:
defnition of critical thinking:it is an intellectual process of applying ,analyizing,evaluating
information by means of observation which leads to believe in action
a. When you consider all the different perspectives on a topic and evaluate how true or
applicable they are in a given context..
Recommended
}Solution}.pdf
The document discusses a solution for improving the efficiency of solar panels. It proposes coating solar panels with a new material that can absorb more wavelengths of light, increasing energy absorption by 40%. This coating material is inexpensive to produce and easy to apply, allowing for widespread implementation of more effective solar energy technology.
What is an OS • Interface between application programs and hardwa.pdf
What is an OS?
• Interface between application programs and hardware
• Ultimate control program
– Exploits hardware resources of one or more processors to provide a set of services to users
– Coordinates the use of hardware among various application programs for different users
– Manages secondary memory and I/O devices on behalf of its users
• Two different views of an OS
1. Extended machine view
– Virtual machine that is easier to understand and program
– Tool to make programmer’s job easy
2. Resource manager view
– Tool to facilitate efficient operation of computer system
– Provides services to users; processor, memory, I/O, system bus
– Must be fair; not partial to any process, specially for process in the same class
– Must discriminate between different class of jobs with different service requirements
– Do the above efficiently
Within the constraints of fairness and efficiency, an OS should attempt to maximize throughput,
minimize
response time, and accommodate as many users as possible
Basic elements
• A computer has a number of modules, with possibly more than one instance of each
• Processor/CPU
– Controls the operations of computer and performs data processing functions
– Exchanges data with memory using memory address register and memory buffer register
– Exchanges data with I/O devices using I/O address register and I/O buffer register
• Main memory/Primary memory
– Stores data and programs
– Typically volatile
– Abstracted as a set of locations, defined by sequentially numbered addresses
– Each location contains a bit pattern to be interpreted as either an instruction or data
• I/O modules/architecture
– Move data between a computer and external environment
– Communicate with a variety of devices including secondary memory (disks), communications
equipment, and
terminals
– Data flows between CPUs, RAM and I/O devices over buses
– System bus
Provides for communications among processors, main memory, and I/O modules
Connects most of the internal hardware devices
PCI bus, ISA bus, EISA bus, SCSI, USB
Solution
What is an OS?
• Interface between application programs and hardware
• Ultimate control program
– Exploits hardware resources of one or more processors to provide a set of services to users
– Coordinates the use of hardware among various application programs for different users
– Manages secondary memory and I/O devices on behalf of its users
• Two different views of an OS
1. Extended machine view
– Virtual machine that is easier to understand and program
– Tool to make programmer’s job easy
2. Resource manager view
– Tool to facilitate efficient operation of computer system
– Provides services to users; processor, memory, I/O, system bus
– Must be fair; not partial to any process, specially for process in the same class
– Must discriminate between different class of jobs with different service requirements
– Do the above efficiently
Within the constraints of fairness and efficiency, an OS should attempt to maximize throughput,.
Using a broad definition of file(note that it doesnt count h.pdf
Using a broad definition of \"file\"
(note that it doesn\'t count hidden files and assumes that file names don\'t contain newline
characters).
To include hidden files (except . and ..) and avoid problems with newline characters, the
canonical way is:
Or recursively:
Solution
Using a broad definition of \"file\"
(note that it doesn\'t count hidden files and assumes that file names don\'t contain newline
characters).
To include hidden files (except . and ..) and avoid problems with newline characters, the
canonical way is:
Or recursively:.
unable to open it ....it asks for username and passwordSolution.pdf
unable to open it ....it asks for username and password
Solution
unable to open it ....it asks for username and password.
This is called a dehydrationreaction. a) H2SO4 H20Solution.pdf
This is called a dehydrationreaction.
a) H2SO4 / H20
Solution
This is called a dehydrationreaction.
a) H2SO4 / H20.
The multiplication sign simply means that the molecules have their i.pdf
The multiplication sign simply means that the molecules have their independent existance in the
compound. For example the formula you have given has three species in solution ZnCO3, 2
molecules of Zn(OH)2 and 1 molecule of water H20 for each molecule of your compound. These
species are bonded to each other but the bonds between the atoms of the participating molecules
are not broken. To get the molar mass of the compound simply add the molar masses of
indivisual molecules multiplied by number of molecules they are contributing. Thus for
ZnCO3*2Zn(OH)2*1H20 the molar mass would be:
Molar mass of ZnCO3 + 2*(Molar mass of Zn(OH)2) + Molar mass of H2O
= 125.4 + 2*99.4 + 18
= 342.2 g/mol
Solution
The multiplication sign simply means that the molecules have their independent existance in the
compound. For example the formula you have given has three species in solution ZnCO3, 2
molecules of Zn(OH)2 and 1 molecule of water H20 for each molecule of your compound. These
species are bonded to each other but the bonds between the atoms of the participating molecules
are not broken. To get the molar mass of the compound simply add the molar masses of
indivisual molecules multiplied by number of molecules they are contributing. Thus for
ZnCO3*2Zn(OH)2*1H20 the molar mass would be:
Molar mass of ZnCO3 + 2*(Molar mass of Zn(OH)2) + Molar mass of H2O
= 125.4 + 2*99.4 + 18
= 342.2 g/mol.
SolutionThe variance of individual assets is measure of total ris.pdf
The variance of individual assets measures total risk, but in a well-diversified portfolio only systematic risk matters as unsystematic risk is reduced. The variance of a diversified portfolio depends on its systematic risk, known as beta. Therefore, the statement in the question is false.
soldefnition of critical thinkingit is an intellectual process o.pdf
sol:
defnition of critical thinking:it is an intellectual process of applying ,analyizing,evaluating
information by means of observation which leads to believe in action
a. When you consider all the different perspectives on a topic and evaluate how true or
applicable they are in a given context.
Solution
sol:
defnition of critical thinking:it is an intellectual process of applying ,analyizing,evaluating
information by means of observation which leads to believe in action
a. When you consider all the different perspectives on a topic and evaluate how true or
applicable they are in a given context..
Program-a. library is importedimport java.awt.; import j.pdf
Program:-
a.
// library is imported
import java.awt.*;
import java.applet.*;
import java.awt.event.*;
import javax.swing.*;
public class JPasswordA extends Applet implements ActionListener //Class Applet is //extended
using inheritance
{boolean clicked=false,code=false; // Boolean function is defined
Label passwordlabel = new Label(\"Password:\"); // object creation
TextField passwordfield = new TextField(15); // text field is assigned to the password
Button enterbutton = new Button(\"Enter\");
public void init() // initialization method is defined
{
add(passwordlabel);
add(passwordfield);
add(enterbutton);
enterbutton.addActionListener(this); // button and password field is set
passwordlabel.setVisible(true);
passwordfield.setVisible(true);
}
public void actionPerformed(ActionEvent e) // method
{ String pass;
clicked=true;
pass=passwordfield.getText();
if(pass.compareTo(\"Rosebud\")==0) // conditional statement is defined
code=true;
else
code=false;
repaint();
}
public void paint(Graphics g) //paint method is defined
{
if(clicked)
{if(code)
g.drawString(\"Access Granted\",120,150); // according to question.
else
g.drawString(\"Access Denied\",120,150);
}
}
}
b.
// library is imported
import java.awt.*;
import java.applet.*;
import java.awt.event.*;
import javax.swing.*;
public class JPasswordB extends Applet implements ActionListener//Class Applet is //extended
using inheritance
{boolean clicked=false,code=false; // Boolean function is defined
Label passwordlabel = new Label(\"Password:\"); // object creation
TextField passwordfield = new TextField(15); // text field is assigned to the password
Button enterbutton = new Button(\"Enter\");
public void init()// initialization method is defined
{
add(passwordlabel);
add(passwordfield);
add(enterbutton);
enterbutton.addActionListener(this);
passwordlabel.setVisible(true); // button and password field is set
passwordfield.setVisible(true);
}
public void actionPerformed(ActionEvent e) // method
{ String pass;
clicked=true;
pass=passwordfield.getText();
if(pass.compareToIgnoreCase(\"Rosebud\")==0) // conditional statement is defined
code=true;
else
code=false;
repaint();
}
public void paint(Graphics g) //paint method is defined
{
if(clicked)
{if(code)
g.drawString(\"Access Granted\",120,150); // according to question.
else
g.drawString(\"Access Denied\",120,150);
}
}
}
c.
// library is imported
import java.awt.*;
import java.applet.*;
import java.awt.event.*;
import javax.swing.*;
public class JPasswordC extends Applet implements ActionListener //Class Applet is //extended
using inheritance
{boolean clicked=false,code=false;
Label passwordlabel = new Label(\"Password:\"); // Boolean function is defined
TextField passwordfield = new TextField(15);
Button enterbutton = new Button(\"Enter\"); // object creation
public void init()// text field is assigned to the password
{
add(passwordlabel);
add(passwordfield);
add(enterbutton);
enterbutton.addActionListener(this); // button and password field is set
p.
package com.test;public class Team { private String teamId;.pdf
package com.test;
public class Team {
private String teamId;
private String teamName;
private String pFirstName;
private String pLastName;
private String pemail;
private String phoneNumber;
public String getTeamId() {
return teamId;
}
public void setTeamId(String teamId) {
this.teamId = teamId;
}
public String getTeamName() {
return teamName;
}
public void setTeamName(String teamName) {
this.teamName = teamName;
}
public String getpFirstName() {
return pFirstName;
}
public void setpFirstName(String pFirstName) {
this.pFirstName = pFirstName;
}
public String getpLastName() {
return pLastName;
}
public void setpLastName(String pLastName) {
this.pLastName = pLastName;
}
public String getPemail() {
return pemail;
}
public void setPemail(String pemail) {
this.pemail = pemail;
}
public String getPhoneNumber() {
return phoneNumber;
}
public void setPhoneNumber(String phoneNumber) {
this.phoneNumber = phoneNumber;
}
public String toString() {
return \"Team [teamId=\" + teamId + \", teamName=\" + teamName
+ \", pFirstName=\" + pFirstName + \", pLastName=\" + pLastName
+ \", pemail=\" + pemail + \", phoneNumber=\" + phoneNumber + \"]\";
}
public Team(String teamId, String teamName, String pFirstName,
String pLastName, String pemail, String phoneNumber) {
super();
this.teamId = teamId;
this.teamName = teamName;
this.pFirstName = pFirstName;
this.pLastName = pLastName;
this.pemail = pemail;
this.phoneNumber = phoneNumber;
}
public int hashCode() {
final int prime = 31;
int result = 1;
result = prime * result
+ ((pFirstName == null) ? 0 : pFirstName.hashCode());
result = prime * result
+ ((pLastName == null) ? 0 : pLastName.hashCode());
result = prime * result + ((pemail == null) ? 0 : pemail.hashCode());
result = prime * result
+ ((phoneNumber == null) ? 0 : phoneNumber.hashCode());
result = prime * result + ((teamId == null) ? 0 : teamId.hashCode());
result = prime * result
+ ((teamName == null) ? 0 : teamName.hashCode());
return result;
}
public boolean equals(Object obj) {
if (this == obj)
return true;
if (obj == null)
return false;
if (getClass() != obj.getClass())
return false;
Team other = (Team) obj;
if (pFirstName == null) {
if (other.pFirstName != null)
return false;
} else if (!pFirstName.equals(other.pFirstName))
return false;
if (pLastName == null) {
if (other.pLastName != null)
return false;
} else if (!pLastName.equals(other.pLastName))
return false;
if (pemail == null) {
if (other.pemail != null)
return false;
} else if (!pemail.equals(other.pemail))
return false;
if (phoneNumber == null) {
if (other.phoneNumber != null)
return false;
} else if (!phoneNumber.equals(other.phoneNumber))
return false;
if (teamId == null) {
if (other.teamId != null)
return false;
} else if (!teamId.equals(other.teamId))
return false;
if (teamName == null) {
if (other.teamName != null)
return false;
} else if (!teamName.equals(other.teamName))
return false;
return true;
}
}
package com.test;
import java.util.Date;
public class Game {
private String gameId;
priv.
Given below is the completed code along with comments. Output of the.pdf
Given below is the completed code along with comments. Output of the program is shown at the
end. You will need to create an input file containing multiple lines of input as specified in the
question. I have given a sample input file with just 1 line of input.
Please do rate the answer if it helped. Thank you very much.
#include
/*
extractBits extracts the specified number of bits starting specified start location from left.
bit numbering starts with 0 from left. len specifies the number of bits to extract.
So extract leftmost 4 bits of a number n, the call should be extractBits(n, 0, 4). Here start = 0
means the 0th bit from left.
Similarly to extract bit 16 through 20 (i.e 5 bits), we use extractBits(n, 16, 5)
The way this function works is - First calculate the remaining number of bits on the rightside .
Since int takes 32 bits, we subtract (start+len) from 32 to get remaining bits on right.
it first clears all the bits upto starting bit by left shift <<
by start bits. Now we shift back the same number of times to right +
*/
int extractBits(unsigned n, int start, int len)
{
int rightRemaining = 32 - (start + len);
//shift left and then right by start no. of bits clears leftside bits
n = (n << start) >> start;
//now shifting by remaining number of bits on right will extract only needed bits
n = n >> rightRemaining;
return n;
}
int main (int argc, char *argv[]) {
// Get the filename to open, then open the file. If we can\'t
// open the file, complain and quit.
char *filename;
// *** Your Code *** if argv[1] exists, set filename to it,
// otherwise set filename to \"Lab3_data.txt\";
if(argv[1] != NULL)
filename = argv[1];
else
filename = \"Lab3_data.txt\";
// Open the file; if opening fails, say so and return 1.
// otherwise say what file we\'re reading from.
FILE *in_file;
in_file = fopen(filename, \"r\"); // NULL if the open failed
// *** Your Code *** if in_file is NULL, complain and quit
// otherwise say that we\'ve opened the filename
if (in_file == NULL){
printf(\"Couldn\'t open file %s\ \", filename);
return 1;
}
printf(\"Opened file %s for reading...\ \", filename);
// Repeatedly read and process each line of the file. The
// line should have a hex integer and two integer lengths.
int val, len1, len2, len3;
int nbr_vals_on_line
= fscanf(in_file, \"%x %d %d\", &val, &len1, &len2);
// Read until we hit end-of-file or a line without the 3 values.
while (nbr_vals_on_line == 3) {
// We\'re going to break up the value into bitstrings of
// length len1, len2, and len3 (going left-to-right).
// The user gives us len1 and len2, and we calculate
// len3 = the remainder of the 32 bits of value.
// All three lengths should be > 0, else we complain
// and go onto the next line.
//
len3 = 32 - (len1 + len2);
// *** Your Code***
// if any of the lengths aren\'t > 0,
// print out the value and the lengths and complain
// about the lengths not all being positive
if(len1 <= 0 || len2 <= 0 || len3 <= 0)
{
printf(\"Invalid lengths : len1=%d, len2=%d, len.
(a) The American Institute of Certified Accountants. (AICPA)(b) Th.pdf
(a) The American Institute of Certified Accountants. (AICPA)
(b) This Statement establishes accounting and financial reporting standards for public colleges
and universities within the financial reporting guidelines of GASB Statement No. 34, Basic
Financial Statements-and Management’s Discussion and Analysis-for State and Local
Governments. The standards are designed to provide financial information that responds to the
needs of three groups of primary users of general purpose external financial reports: the
citizenry, legislative and oversight bodies, and investors and creditors. Generally, this
amendment to Statement 34 permits public colleges and universities, in separately issued
financial statements, to use the guidance for special-purpose governments engaged only in
business-type activities, engaged only in governmental activities, or engaged in both
governmental and business-type activities in their separately issued reports. Under this guidance,
in its separately issued reports, a public institution is required to include management’s
discussion and analysis (MD&A); basic financial statements, as appropriate for the category of
special-purpose government reporting selected; notes to the financial statements; and required
supplementary information other than MD&A.
Unless otherwise specified, pronouncements of the GASB apply to financial reports of all state
and local governmental entities, including general purpose governments, public benefit
corporations and authorities, and public employee retirement systems, utilities, hospitals and
other healthcare providers, and colleges and universities. Paragraph 2 discusses the applicability
of this Statement.
(c) A statement of financial position, including accompanying notes to financial statements,
provides relevant information about liquidity, financial flexibility, and the interrelationship of an
organization\'s assets and liabilities. That information generally is provided by aggregating assets
and liabilities that possess similar characteristics into reasonably homogeneous groups. For
example, entities generally report individual items of assets in homogeneous groups, such as
cash and cash equivalents; accounts and notes receivable from patients, students, members, and
other recipients of services; inventories of materials and supplies; deposits and prepayments for
rent, insurance, and other services; marketable securities and other investment assets held for
long-term purposes; and land, buildings, equipment, and other long-lived assets used to provide
goods and services. Cash or other assets received with a donor-imposed restriction that limits
their use to long-term purposes should not be classified with cash or other assets that are
unrestricted and available for current use.
(d) A statement of activities shall report revenues as increases in unrestricted net assets unless
the use of the assets received is limited by donor-imposed restrictions. For example, fees from
rendering servic.
Yes. The compound is actively active. It is mainl.pdf
Yes. The compound is actively active. It is mainly caused by the carbon to the right
of the 2 oxygens. I think i forgot about the stereoisomers when I last answered them. The number
of stereoisomers is dependent on the number of chiral centers. Since there are 3 chiral centers, it
is 2^3, which is 8 possible stereoisomers. So I mentioned before that the chiral centers are the
carbons directly to the right and left of the oxygens. All the bonds in the ring structure are in the
plane. The other 2 bonds can be alternated to be pointing inside the page or out of the page. That
is why if you have 3 carbon centers, it is just a matter of probability of 2^3 (2 possible choices
over 3 possible carbons) which give you 8. And as I previously explained: To be optically
active, you need to have chiral center, which is any carbon with 4 different subgroups. There is 3
such centers. The carbons to the left of both oxygens (this makes up 2 carbons) are chirally
active since they have 4 different groups. However, since they are mirror images of each other,
they cancel each other out. It is the third and final carbon to the right of both oxygens, that
makes this compound chirally active. Since it has also 4 different groups attached to the carbon
and is not canceled out by anything. One simple way to test if something is optically active is to
see if it is a mirror image of itself. If it is not a mirror image and it has at least 1 chiral center,
then it is optically active
Solution
Yes. The compound is actively active. It is mainly caused by the carbon to the right
of the 2 oxygens. I think i forgot about the stereoisomers when I last answered them. The number
of stereoisomers is dependent on the number of chiral centers. Since there are 3 chiral centers, it
is 2^3, which is 8 possible stereoisomers. So I mentioned before that the chiral centers are the
carbons directly to the right and left of the oxygens. All the bonds in the ring structure are in the
plane. The other 2 bonds can be alternated to be pointing inside the page or out of the page. That
is why if you have 3 carbon centers, it is just a matter of probability of 2^3 (2 possible choices
over 3 possible carbons) which give you 8. And as I previously explained: To be optically
active, you need to have chiral center, which is any carbon with 4 different subgroups. There is 3
such centers. The carbons to the left of both oxygens (this makes up 2 carbons) are chirally
active since they have 4 different groups. However, since they are mirror images of each other,
they cancel each other out. It is the third and final carbon to the right of both oxygens, that
makes this compound chirally active. Since it has also 4 different groups attached to the carbon
and is not canceled out by anything. One simple way to test if something is optically active is to
see if it is a mirror image of itself. If it is not a mirror image and it has at least 1 chiral center,
then it is optically active.
The probability of A and its complement will sum to oneSolution.pdf
The probability of A and its complement will sum to one
Solution
The probability of A and its complement will sum to one.
H+,CN-Solution H+,CN-.pdf
This document contains chemical symbols for hydrogen ions and cyanide ions in solution. It indicates that hydrogen ions and cyanide ions are present together in an aqueous solution. The document provides a brief representation of the composition of an ionic solution through chemical symbols but does not include additional context or details.
Adding elements public void add(String element) { For fi.pdf
// Adding elements
public void add(String element) {
// For first element Head is null. So, create a new node and mark it as head.
// and increase numElements
if(head == null){
head = new StringNode(element, null);
numElements++;
} else {
// if not, find its place first.
// then create a node and mark position link to new node
// mark new node as link to positional node
StringNode node = head;
while(node.getLink() != null){
if(node.getData().compareTo(element) < 0){
break;
}
}
StringNode newNode = new StringNode(element, null);
newNode.setLink(node.getLink());
node.setLink(newNode);
numElements++;
}
}
// Removing elements
public boolean remove(String target)
{
StringNode targetNode = head;
boolean found = false;
while (targetNode!= null && !found)
{
if(targetNode.getData().equalsIgnoreCase(target))
found = true;
else
targetNode = targetNode.getLink();
}
if(found)
{
// copy the head to targetNode
// and then advance head to the next node.
targetNode.setData(targetNode.getLink().getData());
targetNode.setLink(targetNode.getLink());
numElements --;
}
return found;
}
Solution
// Adding elements
public void add(String element) {
// For first element Head is null. So, create a new node and mark it as head.
// and increase numElements
if(head == null){
head = new StringNode(element, null);
numElements++;
} else {
// if not, find its place first.
// then create a node and mark position link to new node
// mark new node as link to positional node
StringNode node = head;
while(node.getLink() != null){
if(node.getData().compareTo(element) < 0){
break;
}
}
StringNode newNode = new StringNode(element, null);
newNode.setLink(node.getLink());
node.setLink(newNode);
numElements++;
}
}
// Removing elements
public boolean remove(String target)
{
StringNode targetNode = head;
boolean found = false;
while (targetNode!= null && !found)
{
if(targetNode.getData().equalsIgnoreCase(target))
found = true;
else
targetNode = targetNode.getLink();
}
if(found)
{
// copy the head to targetNode
// and then advance head to the next node.
targetNode.setData(targetNode.getLink().getData());
targetNode.setLink(targetNode.getLink());
numElements --;
}
return found;
}.
1) Investment Grade Domestic Bonds As bond bears a fixed rate of i.pdf
1) Investment Grade Domestic Bonds As bond bears a fixed rate of interest hence Mutual
Fund or ETFS can\'t Represent Bonds as an asset class. 2) High Yield Bonds 3) Domestic
stock Both are consist of shares and debentures so Mutual Fund or ETFS can\'t Represent them
as an asset class. 4) International Stock 6) Energy Similarly ETF and Mutual fund never
meet the defination of these 3 asset class so can\'t represent them 7) Commodities 8) Real
Estate 5)E merging Market Stock An emerging market fund is a mutual fund or ETF
that invests the bulk of its assets in stocks of developing countries. E merging countries are
those which are not developed yet in the stage of development i.e. china, india etc. so
invetment in these fund brings highest reurn for prospective investor. So
ETF and Mutual fund Represents E merging market stock.
Solution
1) Investment Grade Domestic Bonds As bond bears a fixed rate of interest hence Mutual
Fund or ETFS can\'t Represent Bonds as an asset class. 2) High Yield Bonds 3) Domestic
stock Both are consist of shares and debentures so Mutual Fund or ETFS can\'t Represent them
as an asset class. 4) International Stock 6) Energy Similarly ETF and Mutual fund never
meet the defination of these 3 asset class so can\'t represent them 7) Commodities 8) Real
Estate 5)E merging Market Stock An emerging market fund is a mutual fund or ETF
that invests the bulk of its assets in stocks of developing countries. E merging countries are
those which are not developed yet in the stage of development i.e. china, india etc. so
invetment in these fund brings highest reurn for prospective investor. So
ETF and Mutual fund Represents E merging market stock..
When atoms share one or more pairs of electrons t.pdf
When atoms share one or more pairs of electrons the bonds are called covalent
bonds.
Solution
When atoms share one or more pairs of electrons the bonds are called covalent
bonds..
This is quite simple to do. Barium sulfate is ins.pdf
This is quite simple to do. Barium sulfate is insoluble in water and sodium chloride
is soluble in water. So this is what you do. Dump them both in water. As I said the NaCl will
dissolve. Then do vacuum or suction filtration. That will filter out the barium sulfate from the
solution. Now the two are separate. To get the NaCl out of the water just boil it off.
Solution
This is quite simple to do. Barium sulfate is insoluble in water and sodium chloride
is soluble in water. So this is what you do. Dump them both in water. As I said the NaCl will
dissolve. Then do vacuum or suction filtration. That will filter out the barium sulfate from the
solution. Now the two are separate. To get the NaCl out of the water just boil it off..
Step1 Cocentration of NaCl =.250 moles Step2 Mol.pdf
Step1 Cocentration of NaCl =.250 moles Step2 Molarity =[ concentration in
moles]/volume in litres Step3 Molarity =.250/2 =.125
Solution
Step1 Cocentration of NaCl =.250 moles Step2 Molarity =[ concentration in
moles]/volume in litres Step3 Molarity =.250/2 =.125.
Nitric acid is a strong acid... we can assume com.pdf
Nitric acid is a strong acid... we can assume complete dissociation, So [H+] is 0.04
M
Solution
Nitric acid is a strong acid... we can assume complete dissociation, So [H+] is 0.04
M.
NaCN is a salt and it completely gets ionized, .pdf
NaCN is a salt and it completely gets ionized, NaCN ------------->
Na+ +CN- 0.05L * 0.10 M moles =0.005 0.005 0.005
HCl --------------> H+ + Cl- moles 0.020 L * 0.2 M
=0.004 0.004 0.004 Chemical reaction: CN-
(aq) + H+ (aq) ------------------> HCN Before rxn
(moles) 0.005 0.004 0 After rxn (moles)
0.001 0 0.004 [CN-] =
0.001 moles / 0.07 L = 0.01428 M
[H+] = 0 M (acid is completelyconsumed)
[HCN] = 0.004 moles / 0.07 L =
0.057 M [Cl-] = 0.004 moles / 0.07L
= 0.0571 M
Solution
NaCN is a salt and it completely gets ionized, NaCN ------------->
Na+ +CN- 0.05L * 0.10 M moles =0.005 0.005 0.005
HCl --------------> H+ + Cl- moles 0.020 L * 0.2 M
=0.004 0.004 0.004 Chemical reaction: CN-
(aq) + H+ (aq) ------------------> HCN Before rxn
(moles) 0.005 0.004 0 After rxn (moles)
0.001 0 0.004 [CN-] =
0.001 moles / 0.07 L = 0.01428 M
[H+] = 0 M (acid is completelyconsumed)
[HCN] = 0.004 moles / 0.07 L =
0.057 M [Cl-] = 0.004 moles / 0.07L
= 0.0571 M.
density = masswater displaced or, 10.5 = 5.25wa.pdf
density = mass/water displaced or, 10.5 = 5.25/water displaced water displaced =0.5
ml Volume level =16.1 + 0.5 =16.6 ml
Solution
density = mass/water displaced or, 10.5 = 5.25/water displaced water displaced =0.5
ml Volume level =16.1 + 0.5 =16.6 ml.
as Na+ ions replace H+ from H2S,as 2moles of Na+ .pdf
as Na+ ions replace H+ from H2S,as 2moles of Na+ are produced exactly double
H+ions will be produced.Option D
Solution
as Na+ ions replace H+ from H2S,as 2moles of Na+ are produced exactly double
H+ions will be produced.Option D.
Dependent SolutionDependent .pdf
This 3 sentence document does not provide enough context or information to generate an accurate 3 sentence summary. The document contains two sentences with the label "Dependent:" but no further details are given. Therefore, a meaningful high-level summary cannot be produced from the limited information provided.
Bear with me, Im trying to type the Lewis structure O=Se-O.pdf
Bear with me, I\'m trying to type the Lewis structure:
::O=Se-O:::
and there are four electrons surrounding the selenium.
So: Oxygen with four unpaired electrons and a double bond to Selenium. Selenium with four
unpaired electrons bonded by a single bond to Oxygen with six unpaired electrons.
Hope I helped!
Solution
Bear with me, I\'m trying to type the Lewis structure:
::O=Se-O:::
and there are four electrons surrounding the selenium.
So: Oxygen with four unpaired electrons and a double bond to Selenium. Selenium with four
unpaired electrons bonded by a single bond to Oxygen with six unpaired electrons.
Hope I helped!.
ANS.A & B . There is no direct or particular source for gases like.pdf
ANS.
A & B . There is no direct or particular source for gases like methane and carbon dioxide, they
are the product of various chemical and biological reactions that occur in a land fill. These gases
compose approximately 90 % of gases in a landfill. As for there harmful effects there are many
for e.g. there is concern of global warming , they causes nausea, fatigue at a high concentration
to humans, methane is flammable.
C. Solvents : These can be various chemicals from industrial disposal or domestic disposal also.
These becomes a concern because they tend to seep through the soil and if reaches ground water,
then they can directly pollute it and which may cause various health problems.
D. Organic Nitrogen: Source is mostly sewage or agricultural waste. Organic nitrogen is of
concern as it tends to stimulate and maintain microbial activities in landfills which may increase
total biomass and creating a swampy area which will cause more emission of harmful gases.
E. Rodents: Rodents are the species that survive in landfills, they survive on waste or potentially
hazardous matter present there, which can cause a serious impact on food chain from
environment point of view. Bio-magnification and bio-accumulation are also some serious
problems that can arise from this.
Solution
ANS.
A & B . There is no direct or particular source for gases like methane and carbon dioxide, they
are the product of various chemical and biological reactions that occur in a land fill. These gases
compose approximately 90 % of gases in a landfill. As for there harmful effects there are many
for e.g. there is concern of global warming , they causes nausea, fatigue at a high concentration
to humans, methane is flammable.
C. Solvents : These can be various chemicals from industrial disposal or domestic disposal also.
These becomes a concern because they tend to seep through the soil and if reaches ground water,
then they can directly pollute it and which may cause various health problems.
D. Organic Nitrogen: Source is mostly sewage or agricultural waste. Organic nitrogen is of
concern as it tends to stimulate and maintain microbial activities in landfills which may increase
total biomass and creating a swampy area which will cause more emission of harmful gases.
E. Rodents: Rodents are the species that survive in landfills, they survive on waste or potentially
hazardous matter present there, which can cause a serious impact on food chain from
environment point of view. Bio-magnification and bio-accumulation are also some serious
problems that can arise from this..
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
Your Skill Boost Masterclass: Strategies for Effective Upskilling
Your Skill Boost Masterclass: Strategies for Effective UpskillingExcellence Foundation for South Sudan
Strategies for Effective Upskilling is a presentation by Chinwendu Peace in a Your Skill Boost Masterclass organisation by the Excellence Foundation for South Sudan on 08th and 09th June 2024 from 1 PM to 3 PM on each day.More Related Content
More from aparnacollection
Program-a. library is importedimport java.awt.; import j.pdf
Program:-
a.
// library is imported
import java.awt.*;
import java.applet.*;
import java.awt.event.*;
import javax.swing.*;
public class JPasswordA extends Applet implements ActionListener //Class Applet is //extended
using inheritance
{boolean clicked=false,code=false; // Boolean function is defined
Label passwordlabel = new Label(\"Password:\"); // object creation
TextField passwordfield = new TextField(15); // text field is assigned to the password
Button enterbutton = new Button(\"Enter\");
public void init() // initialization method is defined
{
add(passwordlabel);
add(passwordfield);
add(enterbutton);
enterbutton.addActionListener(this); // button and password field is set
passwordlabel.setVisible(true);
passwordfield.setVisible(true);
}
public void actionPerformed(ActionEvent e) // method
{ String pass;
clicked=true;
pass=passwordfield.getText();
if(pass.compareTo(\"Rosebud\")==0) // conditional statement is defined
code=true;
else
code=false;
repaint();
}
public void paint(Graphics g) //paint method is defined
{
if(clicked)
{if(code)
g.drawString(\"Access Granted\",120,150); // according to question.
else
g.drawString(\"Access Denied\",120,150);
}
}
}
b.
// library is imported
import java.awt.*;
import java.applet.*;
import java.awt.event.*;
import javax.swing.*;
public class JPasswordB extends Applet implements ActionListener//Class Applet is //extended
using inheritance
{boolean clicked=false,code=false; // Boolean function is defined
Label passwordlabel = new Label(\"Password:\"); // object creation
TextField passwordfield = new TextField(15); // text field is assigned to the password
Button enterbutton = new Button(\"Enter\");
public void init()// initialization method is defined
{
add(passwordlabel);
add(passwordfield);
add(enterbutton);
enterbutton.addActionListener(this);
passwordlabel.setVisible(true); // button and password field is set
passwordfield.setVisible(true);
}
public void actionPerformed(ActionEvent e) // method
{ String pass;
clicked=true;
pass=passwordfield.getText();
if(pass.compareToIgnoreCase(\"Rosebud\")==0) // conditional statement is defined
code=true;
else
code=false;
repaint();
}
public void paint(Graphics g) //paint method is defined
{
if(clicked)
{if(code)
g.drawString(\"Access Granted\",120,150); // according to question.
else
g.drawString(\"Access Denied\",120,150);
}
}
}
c.
// library is imported
import java.awt.*;
import java.applet.*;
import java.awt.event.*;
import javax.swing.*;
public class JPasswordC extends Applet implements ActionListener //Class Applet is //extended
using inheritance
{boolean clicked=false,code=false;
Label passwordlabel = new Label(\"Password:\"); // Boolean function is defined
TextField passwordfield = new TextField(15);
Button enterbutton = new Button(\"Enter\"); // object creation
public void init()// text field is assigned to the password
{
add(passwordlabel);
add(passwordfield);
add(enterbutton);
enterbutton.addActionListener(this); // button and password field is set
p.
package com.test;public class Team { private String teamId;.pdf
package com.test;
public class Team {
private String teamId;
private String teamName;
private String pFirstName;
private String pLastName;
private String pemail;
private String phoneNumber;
public String getTeamId() {
return teamId;
}
public void setTeamId(String teamId) {
this.teamId = teamId;
}
public String getTeamName() {
return teamName;
}
public void setTeamName(String teamName) {
this.teamName = teamName;
}
public String getpFirstName() {
return pFirstName;
}
public void setpFirstName(String pFirstName) {
this.pFirstName = pFirstName;
}
public String getpLastName() {
return pLastName;
}
public void setpLastName(String pLastName) {
this.pLastName = pLastName;
}
public String getPemail() {
return pemail;
}
public void setPemail(String pemail) {
this.pemail = pemail;
}
public String getPhoneNumber() {
return phoneNumber;
}
public void setPhoneNumber(String phoneNumber) {
this.phoneNumber = phoneNumber;
}
public String toString() {
return \"Team [teamId=\" + teamId + \", teamName=\" + teamName
+ \", pFirstName=\" + pFirstName + \", pLastName=\" + pLastName
+ \", pemail=\" + pemail + \", phoneNumber=\" + phoneNumber + \"]\";
}
public Team(String teamId, String teamName, String pFirstName,
String pLastName, String pemail, String phoneNumber) {
super();
this.teamId = teamId;
this.teamName = teamName;
this.pFirstName = pFirstName;
this.pLastName = pLastName;
this.pemail = pemail;
this.phoneNumber = phoneNumber;
}
public int hashCode() {
final int prime = 31;
int result = 1;
result = prime * result
+ ((pFirstName == null) ? 0 : pFirstName.hashCode());
result = prime * result
+ ((pLastName == null) ? 0 : pLastName.hashCode());
result = prime * result + ((pemail == null) ? 0 : pemail.hashCode());
result = prime * result
+ ((phoneNumber == null) ? 0 : phoneNumber.hashCode());
result = prime * result + ((teamId == null) ? 0 : teamId.hashCode());
result = prime * result
+ ((teamName == null) ? 0 : teamName.hashCode());
return result;
}
public boolean equals(Object obj) {
if (this == obj)
return true;
if (obj == null)
return false;
if (getClass() != obj.getClass())
return false;
Team other = (Team) obj;
if (pFirstName == null) {
if (other.pFirstName != null)
return false;
} else if (!pFirstName.equals(other.pFirstName))
return false;
if (pLastName == null) {
if (other.pLastName != null)
return false;
} else if (!pLastName.equals(other.pLastName))
return false;
if (pemail == null) {
if (other.pemail != null)
return false;
} else if (!pemail.equals(other.pemail))
return false;
if (phoneNumber == null) {
if (other.phoneNumber != null)
return false;
} else if (!phoneNumber.equals(other.phoneNumber))
return false;
if (teamId == null) {
if (other.teamId != null)
return false;
} else if (!teamId.equals(other.teamId))
return false;
if (teamName == null) {
if (other.teamName != null)
return false;
} else if (!teamName.equals(other.teamName))
return false;
return true;
}
}
package com.test;
import java.util.Date;
public class Game {
private String gameId;
priv.
Given below is the completed code along with comments. Output of the.pdf
Given below is the completed code along with comments. Output of the program is shown at the
end. You will need to create an input file containing multiple lines of input as specified in the
question. I have given a sample input file with just 1 line of input.
Please do rate the answer if it helped. Thank you very much.
#include
/*
extractBits extracts the specified number of bits starting specified start location from left.
bit numbering starts with 0 from left. len specifies the number of bits to extract.
So extract leftmost 4 bits of a number n, the call should be extractBits(n, 0, 4). Here start = 0
means the 0th bit from left.
Similarly to extract bit 16 through 20 (i.e 5 bits), we use extractBits(n, 16, 5)
The way this function works is - First calculate the remaining number of bits on the rightside .
Since int takes 32 bits, we subtract (start+len) from 32 to get remaining bits on right.
it first clears all the bits upto starting bit by left shift <<
by start bits. Now we shift back the same number of times to right +
*/
int extractBits(unsigned n, int start, int len)
{
int rightRemaining = 32 - (start + len);
//shift left and then right by start no. of bits clears leftside bits
n = (n << start) >> start;
//now shifting by remaining number of bits on right will extract only needed bits
n = n >> rightRemaining;
return n;
}
int main (int argc, char *argv[]) {
// Get the filename to open, then open the file. If we can\'t
// open the file, complain and quit.
char *filename;
// *** Your Code *** if argv[1] exists, set filename to it,
// otherwise set filename to \"Lab3_data.txt\";
if(argv[1] != NULL)
filename = argv[1];
else
filename = \"Lab3_data.txt\";
// Open the file; if opening fails, say so and return 1.
// otherwise say what file we\'re reading from.
FILE *in_file;
in_file = fopen(filename, \"r\"); // NULL if the open failed
// *** Your Code *** if in_file is NULL, complain and quit
// otherwise say that we\'ve opened the filename
if (in_file == NULL){
printf(\"Couldn\'t open file %s\ \", filename);
return 1;
}
printf(\"Opened file %s for reading...\ \", filename);
// Repeatedly read and process each line of the file. The
// line should have a hex integer and two integer lengths.
int val, len1, len2, len3;
int nbr_vals_on_line
= fscanf(in_file, \"%x %d %d\", &val, &len1, &len2);
// Read until we hit end-of-file or a line without the 3 values.
while (nbr_vals_on_line == 3) {
// We\'re going to break up the value into bitstrings of
// length len1, len2, and len3 (going left-to-right).
// The user gives us len1 and len2, and we calculate
// len3 = the remainder of the 32 bits of value.
// All three lengths should be > 0, else we complain
// and go onto the next line.
//
len3 = 32 - (len1 + len2);
// *** Your Code***
// if any of the lengths aren\'t > 0,
// print out the value and the lengths and complain
// about the lengths not all being positive
if(len1 <= 0 || len2 <= 0 || len3 <= 0)
{
printf(\"Invalid lengths : len1=%d, len2=%d, len.
(a) The American Institute of Certified Accountants. (AICPA)(b) Th.pdf
(a) The American Institute of Certified Accountants. (AICPA)
(b) This Statement establishes accounting and financial reporting standards for public colleges
and universities within the financial reporting guidelines of GASB Statement No. 34, Basic
Financial Statements-and Management’s Discussion and Analysis-for State and Local
Governments. The standards are designed to provide financial information that responds to the
needs of three groups of primary users of general purpose external financial reports: the
citizenry, legislative and oversight bodies, and investors and creditors. Generally, this
amendment to Statement 34 permits public colleges and universities, in separately issued
financial statements, to use the guidance for special-purpose governments engaged only in
business-type activities, engaged only in governmental activities, or engaged in both
governmental and business-type activities in their separately issued reports. Under this guidance,
in its separately issued reports, a public institution is required to include management’s
discussion and analysis (MD&A); basic financial statements, as appropriate for the category of
special-purpose government reporting selected; notes to the financial statements; and required
supplementary information other than MD&A.
Unless otherwise specified, pronouncements of the GASB apply to financial reports of all state
and local governmental entities, including general purpose governments, public benefit
corporations and authorities, and public employee retirement systems, utilities, hospitals and
other healthcare providers, and colleges and universities. Paragraph 2 discusses the applicability
of this Statement.
(c) A statement of financial position, including accompanying notes to financial statements,
provides relevant information about liquidity, financial flexibility, and the interrelationship of an
organization\'s assets and liabilities. That information generally is provided by aggregating assets
and liabilities that possess similar characteristics into reasonably homogeneous groups. For
example, entities generally report individual items of assets in homogeneous groups, such as
cash and cash equivalents; accounts and notes receivable from patients, students, members, and
other recipients of services; inventories of materials and supplies; deposits and prepayments for
rent, insurance, and other services; marketable securities and other investment assets held for
long-term purposes; and land, buildings, equipment, and other long-lived assets used to provide
goods and services. Cash or other assets received with a donor-imposed restriction that limits
their use to long-term purposes should not be classified with cash or other assets that are
unrestricted and available for current use.
(d) A statement of activities shall report revenues as increases in unrestricted net assets unless
the use of the assets received is limited by donor-imposed restrictions. For example, fees from
rendering servic.
Yes. The compound is actively active. It is mainl.pdf
Yes. The compound is actively active. It is mainly caused by the carbon to the right
of the 2 oxygens. I think i forgot about the stereoisomers when I last answered them. The number
of stereoisomers is dependent on the number of chiral centers. Since there are 3 chiral centers, it
is 2^3, which is 8 possible stereoisomers. So I mentioned before that the chiral centers are the
carbons directly to the right and left of the oxygens. All the bonds in the ring structure are in the
plane. The other 2 bonds can be alternated to be pointing inside the page or out of the page. That
is why if you have 3 carbon centers, it is just a matter of probability of 2^3 (2 possible choices
over 3 possible carbons) which give you 8. And as I previously explained: To be optically
active, you need to have chiral center, which is any carbon with 4 different subgroups. There is 3
such centers. The carbons to the left of both oxygens (this makes up 2 carbons) are chirally
active since they have 4 different groups. However, since they are mirror images of each other,
they cancel each other out. It is the third and final carbon to the right of both oxygens, that
makes this compound chirally active. Since it has also 4 different groups attached to the carbon
and is not canceled out by anything. One simple way to test if something is optically active is to
see if it is a mirror image of itself. If it is not a mirror image and it has at least 1 chiral center,
then it is optically active
Solution
Yes. The compound is actively active. It is mainly caused by the carbon to the right
of the 2 oxygens. I think i forgot about the stereoisomers when I last answered them. The number
of stereoisomers is dependent on the number of chiral centers. Since there are 3 chiral centers, it
is 2^3, which is 8 possible stereoisomers. So I mentioned before that the chiral centers are the
carbons directly to the right and left of the oxygens. All the bonds in the ring structure are in the
plane. The other 2 bonds can be alternated to be pointing inside the page or out of the page. That
is why if you have 3 carbon centers, it is just a matter of probability of 2^3 (2 possible choices
over 3 possible carbons) which give you 8. And as I previously explained: To be optically
active, you need to have chiral center, which is any carbon with 4 different subgroups. There is 3
such centers. The carbons to the left of both oxygens (this makes up 2 carbons) are chirally
active since they have 4 different groups. However, since they are mirror images of each other,
they cancel each other out. It is the third and final carbon to the right of both oxygens, that
makes this compound chirally active. Since it has also 4 different groups attached to the carbon
and is not canceled out by anything. One simple way to test if something is optically active is to
see if it is a mirror image of itself. If it is not a mirror image and it has at least 1 chiral center,
then it is optically active.
The probability of A and its complement will sum to oneSolution.pdf
The probability of A and its complement will sum to one
Solution
The probability of A and its complement will sum to one.
H+,CN-Solution H+,CN-.pdf
This document contains chemical symbols for hydrogen ions and cyanide ions in solution. It indicates that hydrogen ions and cyanide ions are present together in an aqueous solution. The document provides a brief representation of the composition of an ionic solution through chemical symbols but does not include additional context or details.
Adding elements public void add(String element) { For fi.pdf
// Adding elements
public void add(String element) {
// For first element Head is null. So, create a new node and mark it as head.
// and increase numElements
if(head == null){
head = new StringNode(element, null);
numElements++;
} else {
// if not, find its place first.
// then create a node and mark position link to new node
// mark new node as link to positional node
StringNode node = head;
while(node.getLink() != null){
if(node.getData().compareTo(element) < 0){
break;
}
}
StringNode newNode = new StringNode(element, null);
newNode.setLink(node.getLink());
node.setLink(newNode);
numElements++;
}
}
// Removing elements
public boolean remove(String target)
{
StringNode targetNode = head;
boolean found = false;
while (targetNode!= null && !found)
{
if(targetNode.getData().equalsIgnoreCase(target))
found = true;
else
targetNode = targetNode.getLink();
}
if(found)
{
// copy the head to targetNode
// and then advance head to the next node.
targetNode.setData(targetNode.getLink().getData());
targetNode.setLink(targetNode.getLink());
numElements --;
}
return found;
}
Solution
// Adding elements
public void add(String element) {
// For first element Head is null. So, create a new node and mark it as head.
// and increase numElements
if(head == null){
head = new StringNode(element, null);
numElements++;
} else {
// if not, find its place first.
// then create a node and mark position link to new node
// mark new node as link to positional node
StringNode node = head;
while(node.getLink() != null){
if(node.getData().compareTo(element) < 0){
break;
}
}
StringNode newNode = new StringNode(element, null);
newNode.setLink(node.getLink());
node.setLink(newNode);
numElements++;
}
}
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{
StringNode targetNode = head;
boolean found = false;
while (targetNode!= null && !found)
{
if(targetNode.getData().equalsIgnoreCase(target))
found = true;
else
targetNode = targetNode.getLink();
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// copy the head to targetNode
// and then advance head to the next node.
targetNode.setData(targetNode.getLink().getData());
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numElements --;
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1) Investment Grade Domestic Bonds As bond bears a fixed rate of i.pdf
1) Investment Grade Domestic Bonds As bond bears a fixed rate of interest hence Mutual
Fund or ETFS can\'t Represent Bonds as an asset class. 2) High Yield Bonds 3) Domestic
stock Both are consist of shares and debentures so Mutual Fund or ETFS can\'t Represent them
as an asset class. 4) International Stock 6) Energy Similarly ETF and Mutual fund never
meet the defination of these 3 asset class so can\'t represent them 7) Commodities 8) Real
Estate 5)E merging Market Stock An emerging market fund is a mutual fund or ETF
that invests the bulk of its assets in stocks of developing countries. E merging countries are
those which are not developed yet in the stage of development i.e. china, india etc. so
invetment in these fund brings highest reurn for prospective investor. So
ETF and Mutual fund Represents E merging market stock.
Solution
1) Investment Grade Domestic Bonds As bond bears a fixed rate of interest hence Mutual
Fund or ETFS can\'t Represent Bonds as an asset class. 2) High Yield Bonds 3) Domestic
stock Both are consist of shares and debentures so Mutual Fund or ETFS can\'t Represent them
as an asset class. 4) International Stock 6) Energy Similarly ETF and Mutual fund never
meet the defination of these 3 asset class so can\'t represent them 7) Commodities 8) Real
Estate 5)E merging Market Stock An emerging market fund is a mutual fund or ETF
that invests the bulk of its assets in stocks of developing countries. E merging countries are
those which are not developed yet in the stage of development i.e. china, india etc. so
invetment in these fund brings highest reurn for prospective investor. So
ETF and Mutual fund Represents E merging market stock..
When atoms share one or more pairs of electrons t.pdf
When atoms share one or more pairs of electrons the bonds are called covalent
bonds.
Solution
When atoms share one or more pairs of electrons the bonds are called covalent
bonds..
This is quite simple to do. Barium sulfate is ins.pdf
This is quite simple to do. Barium sulfate is insoluble in water and sodium chloride
is soluble in water. So this is what you do. Dump them both in water. As I said the NaCl will
dissolve. Then do vacuum or suction filtration. That will filter out the barium sulfate from the
solution. Now the two are separate. To get the NaCl out of the water just boil it off.
Solution
This is quite simple to do. Barium sulfate is insoluble in water and sodium chloride
is soluble in water. So this is what you do. Dump them both in water. As I said the NaCl will
dissolve. Then do vacuum or suction filtration. That will filter out the barium sulfate from the
solution. Now the two are separate. To get the NaCl out of the water just boil it off..
Step1 Cocentration of NaCl =.250 moles Step2 Mol.pdf
Step1 Cocentration of NaCl =.250 moles Step2 Molarity =[ concentration in
moles]/volume in litres Step3 Molarity =.250/2 =.125
Solution
Step1 Cocentration of NaCl =.250 moles Step2 Molarity =[ concentration in
moles]/volume in litres Step3 Molarity =.250/2 =.125.
Nitric acid is a strong acid... we can assume com.pdf
Nitric acid is a strong acid... we can assume complete dissociation, So [H+] is 0.04
M
Solution
Nitric acid is a strong acid... we can assume complete dissociation, So [H+] is 0.04
M.
NaCN is a salt and it completely gets ionized, .pdf
NaCN is a salt and it completely gets ionized, NaCN ------------->
Na+ +CN- 0.05L * 0.10 M moles =0.005 0.005 0.005
HCl --------------> H+ + Cl- moles 0.020 L * 0.2 M
=0.004 0.004 0.004 Chemical reaction: CN-
(aq) + H+ (aq) ------------------> HCN Before rxn
(moles) 0.005 0.004 0 After rxn (moles)
0.001 0 0.004 [CN-] =
0.001 moles / 0.07 L = 0.01428 M
[H+] = 0 M (acid is completelyconsumed)
[HCN] = 0.004 moles / 0.07 L =
0.057 M [Cl-] = 0.004 moles / 0.07L
= 0.0571 M
Solution
NaCN is a salt and it completely gets ionized, NaCN ------------->
Na+ +CN- 0.05L * 0.10 M moles =0.005 0.005 0.005
HCl --------------> H+ + Cl- moles 0.020 L * 0.2 M
=0.004 0.004 0.004 Chemical reaction: CN-
(aq) + H+ (aq) ------------------> HCN Before rxn
(moles) 0.005 0.004 0 After rxn (moles)
0.001 0 0.004 [CN-] =
0.001 moles / 0.07 L = 0.01428 M
[H+] = 0 M (acid is completelyconsumed)
[HCN] = 0.004 moles / 0.07 L =
0.057 M [Cl-] = 0.004 moles / 0.07L
= 0.0571 M.
density = masswater displaced or, 10.5 = 5.25wa.pdf
density = mass/water displaced or, 10.5 = 5.25/water displaced water displaced =0.5
ml Volume level =16.1 + 0.5 =16.6 ml
Solution
density = mass/water displaced or, 10.5 = 5.25/water displaced water displaced =0.5
ml Volume level =16.1 + 0.5 =16.6 ml.
as Na+ ions replace H+ from H2S,as 2moles of Na+ .pdf
as Na+ ions replace H+ from H2S,as 2moles of Na+ are produced exactly double
H+ions will be produced.Option D
Solution
as Na+ ions replace H+ from H2S,as 2moles of Na+ are produced exactly double
H+ions will be produced.Option D.
Dependent SolutionDependent .pdf
This 3 sentence document does not provide enough context or information to generate an accurate 3 sentence summary. The document contains two sentences with the label "Dependent:" but no further details are given. Therefore, a meaningful high-level summary cannot be produced from the limited information provided.
Bear with me, Im trying to type the Lewis structure O=Se-O.pdf
Bear with me, I\'m trying to type the Lewis structure:
::O=Se-O:::
and there are four electrons surrounding the selenium.
So: Oxygen with four unpaired electrons and a double bond to Selenium. Selenium with four
unpaired electrons bonded by a single bond to Oxygen with six unpaired electrons.
Hope I helped!
Solution
Bear with me, I\'m trying to type the Lewis structure:
::O=Se-O:::
and there are four electrons surrounding the selenium.
So: Oxygen with four unpaired electrons and a double bond to Selenium. Selenium with four
unpaired electrons bonded by a single bond to Oxygen with six unpaired electrons.
Hope I helped!.
ANS.A & B . There is no direct or particular source for gases like.pdf
ANS.
A & B . There is no direct or particular source for gases like methane and carbon dioxide, they
are the product of various chemical and biological reactions that occur in a land fill. These gases
compose approximately 90 % of gases in a landfill. As for there harmful effects there are many
for e.g. there is concern of global warming , they causes nausea, fatigue at a high concentration
to humans, methane is flammable.
C. Solvents : These can be various chemicals from industrial disposal or domestic disposal also.
These becomes a concern because they tend to seep through the soil and if reaches ground water,
then they can directly pollute it and which may cause various health problems.
D. Organic Nitrogen: Source is mostly sewage or agricultural waste. Organic nitrogen is of
concern as it tends to stimulate and maintain microbial activities in landfills which may increase
total biomass and creating a swampy area which will cause more emission of harmful gases.
E. Rodents: Rodents are the species that survive in landfills, they survive on waste or potentially
hazardous matter present there, which can cause a serious impact on food chain from
environment point of view. Bio-magnification and bio-accumulation are also some serious
problems that can arise from this.
Solution
ANS.
A & B . There is no direct or particular source for gases like methane and carbon dioxide, they
are the product of various chemical and biological reactions that occur in a land fill. These gases
compose approximately 90 % of gases in a landfill. As for there harmful effects there are many
for e.g. there is concern of global warming , they causes nausea, fatigue at a high concentration
to humans, methane is flammable.
C. Solvents : These can be various chemicals from industrial disposal or domestic disposal also.
These becomes a concern because they tend to seep through the soil and if reaches ground water,
then they can directly pollute it and which may cause various health problems.
D. Organic Nitrogen: Source is mostly sewage or agricultural waste. Organic nitrogen is of
concern as it tends to stimulate and maintain microbial activities in landfills which may increase
total biomass and creating a swampy area which will cause more emission of harmful gases.
E. Rodents: Rodents are the species that survive in landfills, they survive on waste or potentially
hazardous matter present there, which can cause a serious impact on food chain from
environment point of view. Bio-magnification and bio-accumulation are also some serious
problems that can arise from this..
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Program-a. library is importedimport java.awt.; import j.pdf
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