FeCO3 is an ionic compound, composed of the Fe(II)2+ ion and the CO3.pdf
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FeCO3 is an ionic compound, composed of the Fe(II)2+ ion and the CO32- carbonate ion. Ions tend to dissociate in water, attracted by the partial charges of the molecule. FeCO3 + H2O <---> Fe2+ + CO32- + H2O Water also dissociates into H+ and OH- at a certain equilibrium rate: H2O <---> H+ + OH- Which when combined with an ionic compound is obviously very attractive to the ions in solution... H+ + OH- + Fe2+ + CO32- <----> Fe(OH)2 + H2CO3 Solution FeCO3 is an ionic compound, composed of the Fe(II)2+ ion and the CO32- carbonate ion. Ions tend to dissociate in water, attracted by the partial charges of the molecule. FeCO3 + H2O <---> Fe2+ + CO32- + H2O Water also dissociates into H+ and OH- at a certain equilibrium rate: H2O <---> H+ + OH- Which when combined with an ionic compound is obviously very attractive to the ions in solution... H+ + OH- + Fe2+ + CO32- <----> Fe(OH)2 + H2CO3.
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The water must be boiled to remove traces of dissolved carbon dioxid.pdf
The water must be boiled to remove traces of dissolved carbon dioxide (CO2).
CO2 dissolves in water to form carbonic acid (H2CO3):
CO2 + H2O => H2CO3
H2CO3 is a weak acid and partially ionizes to give H+ ions:
H2CO3 => H+ + HCO3-
HCO3- => H+ + CO32-
The H+ ions can decompose thiosulfate ions and reduce the concentration of sodium thiosulfate
according to the reaction:
2 S2O32- + H+ => HSO3- + S
Solution
The water must be boiled to remove traces of dissolved carbon dioxide (CO2).
CO2 dissolves in water to form carbonic acid (H2CO3):
CO2 + H2O => H2CO3
H2CO3 is a weak acid and partially ionizes to give H+ ions:
H2CO3 => H+ + HCO3-
HCO3- => H+ + CO32-
The H+ ions can decompose thiosulfate ions and reduce the concentration of sodium thiosulfate
according to the reaction:
2 S2O32- + H+ => HSO3- + S.
The effect of acid deposition is basically the ef.pdf
The effect of acid deposition is basically the effect any acid has on metal. It tends to
corrode it. Corrosion is the electrochemical degradation of a material. That means that as the
metal breaks down, it loses electrons to become a different form. Solid iron (Fe) loses electrons
to become a soluble form, Fe(s) --> Fe2+(in solution) + 2e- These electrons are taken up by
oxygen (O2), which reacts with hydrogen ions (H+) to form water. (1/2) O2 + 2H+ + 2e- -->
H2O The overall reaction is , Fe(s) + (1/2) O2 + 2H+ --> Fe2+ + H2O Normally, the H+ can
come from the acid that forms naturally as CO2 dissolves in water. However, acid rain provides
a lot more H+ & makes it much easier for the reaction to run because there is plenty of
everything needed. Once the iron is in solution it can react further to form iron oxide, which is
also known as rusting.
Solution
The effect of acid deposition is basically the effect any acid has on metal. It tends to
corrode it. Corrosion is the electrochemical degradation of a material. That means that as the
metal breaks down, it loses electrons to become a different form. Solid iron (Fe) loses electrons
to become a soluble form, Fe(s) --> Fe2+(in solution) + 2e- These electrons are taken up by
oxygen (O2), which reacts with hydrogen ions (H+) to form water. (1/2) O2 + 2H+ + 2e- -->
H2O The overall reaction is , Fe(s) + (1/2) O2 + 2H+ --> Fe2+ + H2O Normally, the H+ can
come from the acid that forms naturally as CO2 dissolves in water. However, acid rain provides
a lot more H+ & makes it much easier for the reaction to run because there is plenty of
everything needed. Once the iron is in solution it can react further to form iron oxide, which is
also known as rusting..
Chemical reactions of acids and bases
THIS PRESENTATION IS FOR THE STUDENTS STUDYING IN SENIOR CLASSES .IT WILL HELP THE CHILD TO RECALL THE CONTENT IN SHORT TIME IT WILL HELP TO BUILD THE STRONG AND CLEAR CONCEPT KNOWLEDGE.
Recommended
The water must be boiled to remove traces of dissolved carbon dioxid.pdf
The water must be boiled to remove traces of dissolved carbon dioxide (CO2).
CO2 dissolves in water to form carbonic acid (H2CO3):
CO2 + H2O => H2CO3
H2CO3 is a weak acid and partially ionizes to give H+ ions:
H2CO3 => H+ + HCO3-
HCO3- => H+ + CO32-
The H+ ions can decompose thiosulfate ions and reduce the concentration of sodium thiosulfate
according to the reaction:
2 S2O32- + H+ => HSO3- + S
Solution
The water must be boiled to remove traces of dissolved carbon dioxide (CO2).
CO2 dissolves in water to form carbonic acid (H2CO3):
CO2 + H2O => H2CO3
H2CO3 is a weak acid and partially ionizes to give H+ ions:
H2CO3 => H+ + HCO3-
HCO3- => H+ + CO32-
The H+ ions can decompose thiosulfate ions and reduce the concentration of sodium thiosulfate
according to the reaction:
2 S2O32- + H+ => HSO3- + S.
The effect of acid deposition is basically the ef.pdf
The effect of acid deposition is basically the effect any acid has on metal. It tends to
corrode it. Corrosion is the electrochemical degradation of a material. That means that as the
metal breaks down, it loses electrons to become a different form. Solid iron (Fe) loses electrons
to become a soluble form, Fe(s) --> Fe2+(in solution) + 2e- These electrons are taken up by
oxygen (O2), which reacts with hydrogen ions (H+) to form water. (1/2) O2 + 2H+ + 2e- -->
H2O The overall reaction is , Fe(s) + (1/2) O2 + 2H+ --> Fe2+ + H2O Normally, the H+ can
come from the acid that forms naturally as CO2 dissolves in water. However, acid rain provides
a lot more H+ & makes it much easier for the reaction to run because there is plenty of
everything needed. Once the iron is in solution it can react further to form iron oxide, which is
also known as rusting.
Solution
The effect of acid deposition is basically the effect any acid has on metal. It tends to
corrode it. Corrosion is the electrochemical degradation of a material. That means that as the
metal breaks down, it loses electrons to become a different form. Solid iron (Fe) loses electrons
to become a soluble form, Fe(s) --> Fe2+(in solution) + 2e- These electrons are taken up by
oxygen (O2), which reacts with hydrogen ions (H+) to form water. (1/2) O2 + 2H+ + 2e- -->
H2O The overall reaction is , Fe(s) + (1/2) O2 + 2H+ --> Fe2+ + H2O Normally, the H+ can
come from the acid that forms naturally as CO2 dissolves in water. However, acid rain provides
a lot more H+ & makes it much easier for the reaction to run because there is plenty of
everything needed. Once the iron is in solution it can react further to form iron oxide, which is
also known as rusting..
Chemical reactions of acids and bases
THIS PRESENTATION IS FOR THE STUDENTS STUDYING IN SENIOR CLASSES .IT WILL HELP THE CHILD TO RECALL THE CONTENT IN SHORT TIME IT WILL HELP TO BUILD THE STRONG AND CLEAR CONCEPT KNOWLEDGE.
12th Chemistry P-block elements Notes for JEE Main 2015
12th Chemistry P-block elements Notes for JEE Main 2015 by ednexa.com
Ferritin oxidizes Fe2+ to Fe3+ before mineralizing the iron in the fer.docx
Ferritin oxidizes Fe2+ to Fe3+ before mineralizing the iron in the ferritin core.?What is the oxidant for this process, and what is the byproduct produced when the oxidant is reduced?
Solution
Ferritin are a family of natural, highly conserved supramolecular nanostructures designed for sequencing the thousands of iron atoms in a mineralised form.
The main important property of ferritin is the protection of DNA from oxidative damage. In presence of atmospheric oxygen at neutral pH, ferrous ion is readily oxidised to ferric form.
In addition to this hydrogen peroxide (H 2 O 2 ) which is a byproduct of oxygen metabaolism is used as an oxidant and catalyse the generation of highly toxic hydroxyl radicals which damages the proteins, membranes and DNA. So inorder to maintain adequate supply of iron, a molecular mechanism was developed which was utilised by ferritin.\\
Toxification reaction : H 2 O 2 + O 2 - -----> OH - + OH * + O 2
(Fe)
Ferritin consists of 2 subunits H(heavy) and L(light). The H subunit have a dinuclear iron center which consists of A and B binding sites where the fast conversion of Fe +2 to Fe +3 by dioxygen or hydrogen peroxide occurs. The L subunit is used for nucleation of iron core and stores it.
Ferrioxidation reaction : 2Fe +2 + O 2 + 4H 2 O -------> 2FeOOH Core + H 2 O 2 + 4H +
1,2- peroxodi-Fe III intermediate
Minerlization reaction : 4Fe +2 + O 2 + 6H 2 O -------> 4FeOOH Core + 8H +
Detoxification reaction : 4Fe +2 + H 2 O 2 + 2H 2 O -----> 2FeOOH Core + 4H +
Net reaction : Â Â 2Fe +2 + O 2 + 4H 2 O + P --------> (P-(Fe 2 O 2 ) FS +2 ----> (P-(Fe 2 O(OH) 2 ) FS +2 ---->
P + 2FeOOH Core + H 2 O 2 + 4H +
.
The answer is b) water decomposes firstThe standard oxidation pot.pdf
The answer is: b) water decomposes first
The standard oxidation potentials are:
2 F- => F2 + 2 e-, E(oxidation) = -2.87 V
2 H2O => O2 + 4 H+ + 4 e-, E(oxidation) = -1.23 V
Since H2O has a less negative oxidation potential, it is oxidized more easily than F-.
Thus when a solution of F- ions in H2O is electrolyzed, the H2O will decompose first into O2
and H+, while the F- ions remain unreacted.
Solution
The answer is: b) water decomposes first
The standard oxidation potentials are:
2 F- => F2 + 2 e-, E(oxidation) = -2.87 V
2 H2O => O2 + 4 H+ + 4 e-, E(oxidation) = -1.23 V
Since H2O has a less negative oxidation potential, it is oxidized more easily than F-.
Thus when a solution of F- ions in H2O is electrolyzed, the H2O will decompose first into O2
and H+, while the F- ions remain unreacted..
IB Chemistry on Chemical Properties, Oxides and Chlorides of period 3
IB Chemistry on Chemical Properties, Oxides and Chlorides of period 3
Hydrogen
Hydrogen, the most abundant element in the universe and the third most abundant on the surface of the globe.
All you have to know about this inflammable gas.
S = {0,1,2,3,4,5} P(S) be the set of all nonempty subsets of.pdf
S = {0,1,2,3,4,5}
P(S)* be the set of all nonempty subsets of S
a function m: P(S)* -> S by m(H) = the largest element in H
(a) Is m one-to-one? Why or why not?
P(s) having n=2^6-1 element
and S having m= 6 element
function m is not one to one since n>m. (n be less then or equal to m )
M is not one to one since
{1,3} will map to 3
{2,3} will also map tp 3 // by defination of m
two element having same mapping.(IT shd be unique for each element).
Does m map P(S)* onto S? Why or why not?
yes m is onto funtion
max element of P(s) will be in s since P(s) in subset of S only.
therefore for all the element present in s will be mapping of atleat one element present in P(s).
Solution
S = {0,1,2,3,4,5}
P(S)* be the set of all nonempty subsets of S
a function m: P(S)* -> S by m(H) = the largest element in H
(a) Is m one-to-one? Why or why not?
P(s) having n=2^6-1 element
and S having m= 6 element
function m is not one to one since n>m. (n be less then or equal to m )
M is not one to one since
{1,3} will map to 3
{2,3} will also map tp 3 // by defination of m
two element having same mapping.(IT shd be unique for each element).
Does m map P(S)* onto S? Why or why not?
yes m is onto funtion
max element of P(s) will be in s since P(s) in subset of S only.
therefore for all the element present in s will be mapping of atleat one element present in P(s)..
The java program that prompts user to enter a string and .pdf
/*
* The java program that prompts user to
* enter a string and shift key and then
* converts the string to the encoded string
* and prints the encoded string value to console.
* */
//CeaserCipher.java
import java.util.Scanner;
public class CeaserCipher
{
public static void main(String[] args)
{
Scanner scanner=new Scanner(System.in);
String message;
int shift;
System.out.println(\"Enter a message in small letters,a-z\");
message=scanner.nextLine();
System.out.println(\"Enter shift value\");
shift=scanner.nextInt();
//calling encode method with message, shift value
String encodedMsg=encode(message,shift);
//calling WriteMessage
WriteMessage(encodedMsg);
}
/*Method that prints the encoded text to console*/
private static void WriteMessage(String encodedMsg)
{
System.out.println(\"Encoded text :\"+encodedMsg);
}
//Method that takes string and integer value and returns
//the encoded string to calling method
private static String encode(String message, int shift)
{
StringBuffer sb=new StringBuffer(message);
for (int i = 0; i < message.length(); i++)
{
//shift to back to the alphabets
char encodedkey = (char) (\'a\' + (message.charAt(i) -\'a\' + shift) % (\'z\' - \'a\' + 1));
//set character at index,i
sb.setCharAt(i, encodedkey);
}
//return encoded string to calling method
return new String(sb);
}
}
Sample Output:
Enter a message in small letters,a-z
sunday is holiday
Enter shift value
5
Encoded text :xzsifdYnxYmtqnifd
Solution
/*
* The java program that prompts user to
* enter a string and shift key and then
* converts the string to the encoded string
* and prints the encoded string value to console.
* */
//CeaserCipher.java
import java.util.Scanner;
public class CeaserCipher
{
public static void main(String[] args)
{
Scanner scanner=new Scanner(System.in);
String message;
int shift;
System.out.println(\"Enter a message in small letters,a-z\");
message=scanner.nextLine();
System.out.println(\"Enter shift value\");
shift=scanner.nextInt();
//calling encode method with message, shift value
String encodedMsg=encode(message,shift);
//calling WriteMessage
WriteMessage(encodedMsg);
}
/*Method that prints the encoded text to console*/
private static void WriteMessage(String encodedMsg)
{
System.out.println(\"Encoded text :\"+encodedMsg);
}
//Method that takes string and integer value and returns
//the encoded string to calling method
private static String encode(String message, int shift)
{
StringBuffer sb=new StringBuffer(message);
for (int i = 0; i < message.length(); i++)
{
//shift to back to the alphabets
char encodedkey = (char) (\'a\' + (message.charAt(i) -\'a\' + shift) % (\'z\' - \'a\' + 1));
//set character at index,i
sb.setCharAt(i, encodedkey);
}
//return encoded string to calling method
return new String(sb);
}
}
Sample Output:
Enter a message in small letters,a-z
sunday is holiday
Enter shift value
5
Encoded text :xzsifdYnxYmtqnifd.
More Related Content
Similar to FeCO3 is an ionic compound, composed of the Fe(II)2+ ion and the CO3.pdf
12th Chemistry P-block elements Notes for JEE Main 2015
12th Chemistry P-block elements Notes for JEE Main 2015 by ednexa.com
Ferritin oxidizes Fe2+ to Fe3+ before mineralizing the iron in the fer.docx
Ferritin oxidizes Fe2+ to Fe3+ before mineralizing the iron in the ferritin core.?What is the oxidant for this process, and what is the byproduct produced when the oxidant is reduced?
Solution
Ferritin are a family of natural, highly conserved supramolecular nanostructures designed for sequencing the thousands of iron atoms in a mineralised form.
The main important property of ferritin is the protection of DNA from oxidative damage. In presence of atmospheric oxygen at neutral pH, ferrous ion is readily oxidised to ferric form.
In addition to this hydrogen peroxide (H 2 O 2 ) which is a byproduct of oxygen metabaolism is used as an oxidant and catalyse the generation of highly toxic hydroxyl radicals which damages the proteins, membranes and DNA. So inorder to maintain adequate supply of iron, a molecular mechanism was developed which was utilised by ferritin.\\
Toxification reaction : H 2 O 2 + O 2 - -----> OH - + OH * + O 2
(Fe)
Ferritin consists of 2 subunits H(heavy) and L(light). The H subunit have a dinuclear iron center which consists of A and B binding sites where the fast conversion of Fe +2 to Fe +3 by dioxygen or hydrogen peroxide occurs. The L subunit is used for nucleation of iron core and stores it.
Ferrioxidation reaction : 2Fe +2 + O 2 + 4H 2 O -------> 2FeOOH Core + H 2 O 2 + 4H +
1,2- peroxodi-Fe III intermediate
Minerlization reaction : 4Fe +2 + O 2 + 6H 2 O -------> 4FeOOH Core + 8H +
Detoxification reaction : 4Fe +2 + H 2 O 2 + 2H 2 O -----> 2FeOOH Core + 4H +
Net reaction : Â Â 2Fe +2 + O 2 + 4H 2 O + P --------> (P-(Fe 2 O 2 ) FS +2 ----> (P-(Fe 2 O(OH) 2 ) FS +2 ---->
P + 2FeOOH Core + H 2 O 2 + 4H +
.
The answer is b) water decomposes firstThe standard oxidation pot.pdf
The answer is: b) water decomposes first
The standard oxidation potentials are:
2 F- => F2 + 2 e-, E(oxidation) = -2.87 V
2 H2O => O2 + 4 H+ + 4 e-, E(oxidation) = -1.23 V
Since H2O has a less negative oxidation potential, it is oxidized more easily than F-.
Thus when a solution of F- ions in H2O is electrolyzed, the H2O will decompose first into O2
and H+, while the F- ions remain unreacted.
Solution
The answer is: b) water decomposes first
The standard oxidation potentials are:
2 F- => F2 + 2 e-, E(oxidation) = -2.87 V
2 H2O => O2 + 4 H+ + 4 e-, E(oxidation) = -1.23 V
Since H2O has a less negative oxidation potential, it is oxidized more easily than F-.
Thus when a solution of F- ions in H2O is electrolyzed, the H2O will decompose first into O2
and H+, while the F- ions remain unreacted..
IB Chemistry on Chemical Properties, Oxides and Chlorides of period 3
IB Chemistry on Chemical Properties, Oxides and Chlorides of period 3
Hydrogen
Hydrogen, the most abundant element in the universe and the third most abundant on the surface of the globe.
All you have to know about this inflammable gas.
Similar to FeCO3 is an ionic compound, composed of the Fe(II)2+ ion and the CO3.pdf (20)
12th Chemistry P-block elements Notes for JEE Main 2015
12th Chemistry P-block elements Notes for JEE Main 2015
Ferritin oxidizes Fe2+ to Fe3+ before mineralizing the iron in the fer.docx
Ferritin oxidizes Fe2+ to Fe3+ before mineralizing the iron in the fer.docx
New chm-152-unit-3-power-points-sp13-140227172047-phpapp01
New chm-152-unit-3-power-points-sp13-140227172047-phpapp01
The answer is b) water decomposes firstThe standard oxidation pot.pdf
The answer is b) water decomposes firstThe standard oxidation pot.pdf
IB Chemistry on Chemical Properties, Oxides and Chlorides of period 3
IB Chemistry on Chemical Properties, Oxides and Chlorides of period 3
More from DEEPAKSONI562
S = {0,1,2,3,4,5} P(S) be the set of all nonempty subsets of.pdf
S = {0,1,2,3,4,5}
P(S)* be the set of all nonempty subsets of S
a function m: P(S)* -> S by m(H) = the largest element in H
(a) Is m one-to-one? Why or why not?
P(s) having n=2^6-1 element
and S having m= 6 element
function m is not one to one since n>m. (n be less then or equal to m )
M is not one to one since
{1,3} will map to 3
{2,3} will also map tp 3 // by defination of m
two element having same mapping.(IT shd be unique for each element).
Does m map P(S)* onto S? Why or why not?
yes m is onto funtion
max element of P(s) will be in s since P(s) in subset of S only.
therefore for all the element present in s will be mapping of atleat one element present in P(s).
Solution
S = {0,1,2,3,4,5}
P(S)* be the set of all nonempty subsets of S
a function m: P(S)* -> S by m(H) = the largest element in H
(a) Is m one-to-one? Why or why not?
P(s) having n=2^6-1 element
and S having m= 6 element
function m is not one to one since n>m. (n be less then or equal to m )
M is not one to one since
{1,3} will map to 3
{2,3} will also map tp 3 // by defination of m
two element having same mapping.(IT shd be unique for each element).
Does m map P(S)* onto S? Why or why not?
yes m is onto funtion
max element of P(s) will be in s since P(s) in subset of S only.
therefore for all the element present in s will be mapping of atleat one element present in P(s)..
The java program that prompts user to enter a string and .pdf
/*
* The java program that prompts user to
* enter a string and shift key and then
* converts the string to the encoded string
* and prints the encoded string value to console.
* */
//CeaserCipher.java
import java.util.Scanner;
public class CeaserCipher
{
public static void main(String[] args)
{
Scanner scanner=new Scanner(System.in);
String message;
int shift;
System.out.println(\"Enter a message in small letters,a-z\");
message=scanner.nextLine();
System.out.println(\"Enter shift value\");
shift=scanner.nextInt();
//calling encode method with message, shift value
String encodedMsg=encode(message,shift);
//calling WriteMessage
WriteMessage(encodedMsg);
}
/*Method that prints the encoded text to console*/
private static void WriteMessage(String encodedMsg)
{
System.out.println(\"Encoded text :\"+encodedMsg);
}
//Method that takes string and integer value and returns
//the encoded string to calling method
private static String encode(String message, int shift)
{
StringBuffer sb=new StringBuffer(message);
for (int i = 0; i < message.length(); i++)
{
//shift to back to the alphabets
char encodedkey = (char) (\'a\' + (message.charAt(i) -\'a\' + shift) % (\'z\' - \'a\' + 1));
//set character at index,i
sb.setCharAt(i, encodedkey);
}
//return encoded string to calling method
return new String(sb);
}
}
Sample Output:
Enter a message in small letters,a-z
sunday is holiday
Enter shift value
5
Encoded text :xzsifdYnxYmtqnifd
Solution
/*
* The java program that prompts user to
* enter a string and shift key and then
* converts the string to the encoded string
* and prints the encoded string value to console.
* */
//CeaserCipher.java
import java.util.Scanner;
public class CeaserCipher
{
public static void main(String[] args)
{
Scanner scanner=new Scanner(System.in);
String message;
int shift;
System.out.println(\"Enter a message in small letters,a-z\");
message=scanner.nextLine();
System.out.println(\"Enter shift value\");
shift=scanner.nextInt();
//calling encode method with message, shift value
String encodedMsg=encode(message,shift);
//calling WriteMessage
WriteMessage(encodedMsg);
}
/*Method that prints the encoded text to console*/
private static void WriteMessage(String encodedMsg)
{
System.out.println(\"Encoded text :\"+encodedMsg);
}
//Method that takes string and integer value and returns
//the encoded string to calling method
private static String encode(String message, int shift)
{
StringBuffer sb=new StringBuffer(message);
for (int i = 0; i < message.length(); i++)
{
//shift to back to the alphabets
char encodedkey = (char) (\'a\' + (message.charAt(i) -\'a\' + shift) % (\'z\' - \'a\' + 1));
//set character at index,i
sb.setCharAt(i, encodedkey);
}
//return encoded string to calling method
return new String(sb);
}
}
Sample Output:
Enter a message in small letters,a-z
sunday is holiday
Enter shift value
5
Encoded text :xzsifdYnxYmtqnifd.
The answer is b, theres no increase in H. .pdf
The answer is b, theres no increase in H.
Solution
The answer is b, theres no increase in H..
Step1 white ppt with NaCl(aq) indicates Ag+ Step.pdf
Step1 white ppt with NaCl(aq) indicates Ag+ Step2 Passing H2S in acidic salt
solution ; no ppt being formes indicates absence of group II ; Cu+2 and Cd+2 are absent. Step3
Passing H2S in basic condition ; formation of ppt marks presence of group IV Step4 Possible
ions are Ag+; Ni+2 ;Mn+2 ; Na+
Solution
Step1 white ppt with NaCl(aq) indicates Ag+ Step2 Passing H2S in acidic salt
solution ; no ppt being formes indicates absence of group II ; Cu+2 and Cd+2 are absent. Step3
Passing H2S in basic condition ; formation of ppt marks presence of group IV Step4 Possible
ions are Ag+; Ni+2 ;Mn+2 ; Na+.
Natural rubber is a polymer of isoprene units ; e.pdf
Natural rubber is a polymer of isoprene units ; every monomer unit contains a
double bond which undergoes ozonolysis reaction in presence of ozone. Thus polymer chains
are broken.
Solution
Natural rubber is a polymer of isoprene units ; every monomer unit contains a
double bond which undergoes ozonolysis reaction in presence of ozone. Thus polymer chains
are broken..
no double displacement reactions can be mixing tw.pdf
no double displacement reactions can be mixing two soluble salts whose product
can be an insoluble salt
Solution
no double displacement reactions can be mixing two soluble salts whose product
can be an insoluble salt.
using System; using System.Collections.Generic; using System.Lin.pdf
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace ConsoleApplication3
{
class BMI
{
double weight;
double height;
public BMI() //default constructor
{
weight = 0;
height = 0;
}
public BMI(double wt,double ht) //parameterized constructor
{
weight = wt;
height = ht;
}
public void input() //function to input height and weight
{
Console.WriteLine(\"Enter weight in kilograms\");
weight = Convert.ToDouble(Console.ReadLine());
Console.WriteLine(\"Enter height in metres\");
height = Convert.ToDouble(Console.ReadLine());
}
public void BMIstatus() //function to calculate BMI and display status of person
{
double bmi;
bmi = weight / (height * height);
if (bmi<18.5)
Console.WriteLine(\"Person is Underweight\");
else if(bmi>18.5 && bmi<24.9)
Console.WriteLine(\"Person is normal\");
else if(bmi>25 && bmi<29.9)
Console.WriteLine(\"Person is Overweight\");
else if(bmi>=30)
Console.WriteLine(\"Person is Obese\");
}
}
class Program
{
static void Main(string[] args)
{
BMI person1=new BMI();
person1.input(); //call functions
person1.BMIstatus();
Console.ReadLine();
}
}
}
output:
Enter weight in kilograms 70
Enter height in metres 1.57
Person is Overweight
Solution
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace ConsoleApplication3
{
class BMI
{
double weight;
double height;
public BMI() //default constructor
{
weight = 0;
height = 0;
}
public BMI(double wt,double ht) //parameterized constructor
{
weight = wt;
height = ht;
}
public void input() //function to input height and weight
{
Console.WriteLine(\"Enter weight in kilograms\");
weight = Convert.ToDouble(Console.ReadLine());
Console.WriteLine(\"Enter height in metres\");
height = Convert.ToDouble(Console.ReadLine());
}
public void BMIstatus() //function to calculate BMI and display status of person
{
double bmi;
bmi = weight / (height * height);
if (bmi<18.5)
Console.WriteLine(\"Person is Underweight\");
else if(bmi>18.5 && bmi<24.9)
Console.WriteLine(\"Person is normal\");
else if(bmi>25 && bmi<29.9)
Console.WriteLine(\"Person is Overweight\");
else if(bmi>=30)
Console.WriteLine(\"Person is Obese\");
}
}
class Program
{
static void Main(string[] args)
{
BMI person1=new BMI();
person1.input(); //call functions
person1.BMIstatus();
Console.ReadLine();
}
}
}
output:
Enter weight in kilograms 70
Enter height in metres 1.57
Person is Overweight.
Toxicants may move across biological membranes by all the following .pdf
Toxicants may move across biological membranes by all the following mechanisms except
aqueous diffusion. This is because for the diffusion of the toxicant molecules, the aqueous pores
are too small.
Hence, the option a is the correct answer.
Solution
Toxicants may move across biological membranes by all the following mechanisms except
aqueous diffusion. This is because for the diffusion of the toxicant molecules, the aqueous pores
are too small.
Hence, the option a is the correct answer..
time is 18.2yearsdetailed answer i will submit nowSolutionti.pdf
time is 18.2years
detailed answer i will submit now
Solution
time is 18.2years
detailed answer i will submit now.
This is the code for the above 5 public class Input extends JFram.pdf
This is the code for the above 5 :
public class Input extends JFrame{
JRadioButton employee ;
JRadioButton manager;
JPanel panel;
Button button;
public Input(){
button = new Button(\"Enter\");
employee = new JRadioButton(\"Employee\");
manager = new JRadioButton(\"Manager\");
ButtonGroup group = new ButtonGroup();
group.add(employee);
group.add(manager);
employee.setBounds(50,50,80,50);
manager.setBounds(50,100,80,50);
button.setBounds(130, 200,80,25);
panel = new JPanel();
panel.setLayout(null);
add(panel);
panel.add(employee);
panel.add(manager);
panel.add(button);
employee.addActionListener(new ActionListener() {
@Override
public void actionPerformed(ActionEvent e) {
Frame frame = JOptionPane.getFrameForComponent(getParent());
frame.setVisible(true);
frame.setSize(300,300);
}
});
manager.addActionListener(new ActionListener() {
@Override
public void actionPerformed(ActionEvent e) {
}
});
setSize(400,400);
setVisible(true);
setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
}
/*public void actionPerformed(ActionEvent event){
//JOptionPane.showMessageDialog(parent,\"Popup\");
//System.out.print(\"action \");
}*/
public static void main(String[] args) {
new Input();
}
}
Solution
This is the code for the above 5 :
public class Input extends JFrame{
JRadioButton employee ;
JRadioButton manager;
JPanel panel;
Button button;
public Input(){
button = new Button(\"Enter\");
employee = new JRadioButton(\"Employee\");
manager = new JRadioButton(\"Manager\");
ButtonGroup group = new ButtonGroup();
group.add(employee);
group.add(manager);
employee.setBounds(50,50,80,50);
manager.setBounds(50,100,80,50);
button.setBounds(130, 200,80,25);
panel = new JPanel();
panel.setLayout(null);
add(panel);
panel.add(employee);
panel.add(manager);
panel.add(button);
employee.addActionListener(new ActionListener() {
@Override
public void actionPerformed(ActionEvent e) {
Frame frame = JOptionPane.getFrameForComponent(getParent());
frame.setVisible(true);
frame.setSize(300,300);
}
});
manager.addActionListener(new ActionListener() {
@Override
public void actionPerformed(ActionEvent e) {
}
});
setSize(400,400);
setVisible(true);
setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
}
/*public void actionPerformed(ActionEvent event){
//JOptionPane.showMessageDialog(parent,\"Popup\");
//System.out.print(\"action \");
}*/
public static void main(String[] args) {
new Input();
}
}.
Team public class Team { private String teamId; priva.pdf
Team:
public class Team {
private String teamId;
private String name;
private String firstName;
private String lastName;
private String phoneNumber;
private String emailAddress;
Team(String teamId, String name, String firstName, String lastName, String phoneNumber,
String emailAddress) {
this.teamId = teamId;
this.name = name;
this.firstName = firstName;
this.lastName = lastName;
this.phoneNumber = phoneNumber;
this.emailAddress = emailAddress;
}
public String getTeamId() {
return teamId;
}
public void setTeamId(String teamId) {
this.teamId = teamId;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public String getFirstName() {
return firstName;
}
public void setFirstName(String firstName) {
this.firstName = firstName;
}
public String getLastName() {
return lastName;
}
public void setLastName(String lastName) {
this.lastName = lastName;
}
public String getPhoneNumber() {
return phoneNumber;
}
public void setPhoneNumber(String phoneNumber) {
this.phoneNumber = phoneNumber;
}
public String getEmailAddress() {
return emailAddress;
}
public void setEmailAddress(String emailAddress) {
this.emailAddress = emailAddress;
}
public static boolean checkEmailAddress(String emailAddress) {
return emailAddress.contains(\"@\");
}
@Override
public String toString() {
return \"Team [teamId=\" + teamId + \", name=\" + name + \", firstName=\" + firstName +
\", lastName=\" + lastName
+ \", phoneNumber=\" + phoneNumber + \", emailAddress=\" + emailAddress + \"]\";
}
@Override
public int hashCode() {
final int prime = 31;
int result = 1;
result = prime * result + ((emailAddress == null) ? 0 : emailAddress.hashCode());
result = prime * result + ((firstName == null) ? 0 : firstName.hashCode());
result = prime * result + ((lastName == null) ? 0 : lastName.hashCode());
result = prime * result + ((name == null) ? 0 : name.hashCode());
result = prime * result + ((phoneNumber == null) ? 0 : phoneNumber.hashCode());
result = prime * result + ((teamId == null) ? 0 : teamId.hashCode());
return result;
}
@Override
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 (emailAddress == null) {
if (other.emailAddress != null)
return false;
} else if (!emailAddress.equals(other.emailAddress))
return false;
if (firstName == null) {
if (other.firstName != null)
return false;
} else if (!firstName.equals(other.firstName))
return false;
if (lastName == null) {
if (other.lastName != null)
return false;
} else if (!lastName.equals(other.lastName))
return false;
if (name == null) {
if (other.name != null)
return false;
} else if (!name.equals(other.name))
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;
r.
PollinationFertilization1. Pollination is the transfer of po.pdf
Pollination
Fertilization
1. Pollination is the transfer of pollen from the male organ (anther)of a flower to the female
organ (stigma)of a flower.
It is the fusion of male gametes and egg cell in embryo sac situated inside the ovule.
2. It takes place before fertilization
It takes place after pollination
3. Need a vector, insects, water, air like agencies are essential
There is no external medium for this process
4. No pollen tube formation
Pollen tube is formed which transfer male gametes up to egg cell
5. It takes place on outer parts of flower (external mechanism)
It occurs inside the flower (internal mechanism)
Pollination
Fertilization
1. Pollination is the transfer of pollen from the male organ (anther)of a flower to the female
organ (stigma)of a flower.
It is the fusion of male gametes and egg cell in embryo sac situated inside the ovule.
2. It takes place before fertilization
It takes place after pollination
3. Need a vector, insects, water, air like agencies are essential
There is no external medium for this process
4. No pollen tube formation
Pollen tube is formed which transfer male gametes up to egg cell
5. It takes place on outer parts of flower (external mechanism)
It occurs inside the flower (internal mechanism)
Solution
Pollination
Fertilization
1. Pollination is the transfer of pollen from the male organ (anther)of a flower to the female
organ (stigma)of a flower.
It is the fusion of male gametes and egg cell in embryo sac situated inside the ovule.
2. It takes place before fertilization
It takes place after pollination
3. Need a vector, insects, water, air like agencies are essential
There is no external medium for this process
4. No pollen tube formation
Pollen tube is formed which transfer male gametes up to egg cell
5. It takes place on outer parts of flower (external mechanism)
It occurs inside the flower (internal mechanism)
Pollination
Fertilization
1. Pollination is the transfer of pollen from the male organ (anther)of a flower to the female
organ (stigma)of a flower.
It is the fusion of male gametes and egg cell in embryo sac situated inside the ovule.
2. It takes place before fertilization
It takes place after pollination
3. Need a vector, insects, water, air like agencies are essential
There is no external medium for this process
4. No pollen tube formation
Pollen tube is formed which transfer male gametes up to egg cell
5. It takes place on outer parts of flower (external mechanism)
It occurs inside the flower (internal mechanism).
Patient.java package A9.toStudents; public class Patient imple.pdf
//Patient.java
package A9.toStudents;
public class Patient implements Comparable
{
//attributes
private int order;
private String name;
private boolean emergencyCase;
//constructor
public Patient(int order, String name, boolean emergencyCase) {
this.order = order;
this.name = name;
this.emergencyCase = emergencyCase;
}
//compareTo
public int compareTo(Patient other) {
if(this.isEmergencyCase() && !other.isEmergencyCase())
return -1; //place this first
else if(!this.isEmergencyCase() && other.isEmergencyCase())
return 1; //place other first
else //if both are emergency or both are not emergency
return this.getOrder()-other.getOrder(); //place smaller order first
}
//getters and setters
public int getOrder() {
return order;
}
public void setOrder(int order) {
this.order = order;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public boolean isEmergencyCase() {
return emergencyCase;
}
public void setEmergencyCase(boolean emergencyCase) {
this.emergencyCase = emergencyCase;
}
public String toString() {
return (isEmergencyCase()?\"*\":\"\") + name;
}
}
----------------------------------------------------------------------------
//PatientComparator.java
package A9.toStudents;
import java.util.Comparator;
public class PatientComparator implements Comparator
{
public int compare(Patient p1, Patient p2)
{
if(p1.isEmergencyCase() && !p2.isEmergencyCase())
return -1; //place p1 first
else if(!p1.isEmergencyCase() && p2.isEmergencyCase())
return 1; //place p2 first
else //if both are emergency or both are not emergency
return p1.getOrder()-p2.getOrder(); //place smaller order first
}
} //end of the PatientComparator
----------------------------------------------------------------------------
/*Sorter class contains static methods that take ArrayList
and sorts the list using bubble sort,insertion sort and
selection sort */
//Sorter.java
package A9.toStudents;
import java.util.ArrayList;
public class Sorter
{
//Bubble sort
public static void bubbleSort(ArrayList list)
{
for (int c = 0; c < ( list.size() - 1 ); c++)
{
for (int d = 0; d < list.size() - c - 1; d++)
{
if (list.get(d).compareTo(list.get(d+1))>0)
/* For ascending order*/
{
Patient temp= list.get(d);
list.set(d, list.get(d+1));
list.set(d+1, temp);
}
}
}
}
//Selection sort
public static void selectionSort(ArrayList list)
{
for (int i = 0; i < list.size() - 1; i++)
{
int index = i;
for (int j = i + 1; j < list.size(); j++)
if (list.get(j).compareTo(list.get(index))<0)
index = j;
Patient tempPatient = list.get(index);
list.set(index, list.get(i));
list.set(i,tempPatient);
}
}
//Insertion sort
public static void insertionSort(ArrayList list)
{
for (int i = 1; i < list.size(); i++)
{
for(int j = i ; j > 0 ; j--)
{
if(list.get(j).compareTo(list.get(j-1))<0)
{
Patient temp = list.get(j);
list.set(j,list.get(j-1));
list.set(j-1,temp);
}
}
}
}
public static void bubbleSort(ArrayList list,
PatientComparator patientComparator) {
for (int c = 0; c < ( list.s.
Non financial performance indicators of United Utilities Group PLC. .pdf
Non financial performance indicators of United Utilities Group PLC. Are :
1. Customer satisfaction: Improvements in customer satisfaction are measured through Ofwat\'s
SIM mechanism.
Strong performance on Ofwat\'s and environment agency KPI Assessment.
Reinvestment of £280 million has been done to outperform customer servicing.
Below inflation growth in household bills for the period of 10 years from 2010-20.
2. Effective delivery of Capital investment programme:
Capex delivery has further improved. Time cost quality ind3x up to 98%.
Initiated 40£million of transitional investment to aid a smoother and most effective start to
AMP6.
3. Strong Financials.
Operating profits up by £37 million to 641£ million.
RCV Gearing went down by 2%, at 58%.
Final dividend of 24.03 pence per share was in line with policy.
4. Growth in Business Retail:
Growth in developing a range of value added services.
Largest new entrant and Second largest water retailer in Scotland.
Solution
Non financial performance indicators of United Utilities Group PLC. Are :
1. Customer satisfaction: Improvements in customer satisfaction are measured through Ofwat\'s
SIM mechanism.
Strong performance on Ofwat\'s and environment agency KPI Assessment.
Reinvestment of £280 million has been done to outperform customer servicing.
Below inflation growth in household bills for the period of 10 years from 2010-20.
2. Effective delivery of Capital investment programme:
Capex delivery has further improved. Time cost quality ind3x up to 98%.
Initiated 40£million of transitional investment to aid a smoother and most effective start to
AMP6.
3. Strong Financials.
Operating profits up by £37 million to 641£ million.
RCV Gearing went down by 2%, at 58%.
Final dividend of 24.03 pence per share was in line with policy.
4. Growth in Business Retail:
Growth in developing a range of value added services.
Largest new entrant and Second largest water retailer in Scotland..
C) III is correct note the configuration keep un.pdf
C) III is correct note: the configuration keep unchanged in first step the inversion of
configuration in 2nd step (SN2)
Solution
C) III is correct note: the configuration keep unchanged in first step the inversion of
configuration in 2nd step (SN2).
LHS =2y+1RHS=-(-2y-1) =2y+1LHS =RHSy=all the real nu.pdf
LHS =2y+1
RHS=-(-2y-1)
=2y+1
LHS =RHS
y=all the real numbers are the solution
Solution
LHS =2y+1
RHS=-(-2y-1)
=2y+1
LHS =RHS
y=all the real numbers are the solution.
import com.rogue.roguer.command.CommandHandler; import com.rogue.r.pdf
import com.rogue.roguer.command.CommandHandler;
import com.rogue.roguer.configuration.Configuration;
import com.rogue.roguer.lstnr.AFKListener;
import com.rogue.roguer.player.PlayerHandler;
import com.rogue.roguer.runnable.AFKRunnable;
import com.rogue.roguer.runnable.UpdateRunnable;
import java.io.File;
import java.util.logging.Level;
import org.bukkit.Bukkit;
import org.bukkit.plugin.java.JavaPlugin;
public class roguer extends JavaPlugin {
private Configuration cnfgs;
private AFKListener lstnr;
private PlayerHandler phndle;
private CommandHandler chndle;
private int dbgg = 0;
@Override
public void onLoad() {
if (!this.getDataFolder().exists()) {
this.getDataFolder().mkdirs();
}
if (!(new File(getDataFolder(), \"cnfg.yml\").exists())) {
this.saveResource(\"cnfg.yml\", true);
}
this.getLogger().log(Level.INFO, \"Loading cnfgs\");
cnfgs = new Configuration();
cnfgs.loadDefaults();
this.getLogger().log(Level.INFO, \"Running update check\");
this.getLogger().log(Level.INFO, \"This can be disabled in your cnfg.yml\");
if (cnfgs.getBoolean(\"update-check\")) {
Bukkit.getScheduler().runTaskLater(this, new UpdateRunnable(), 1);
}
}
@Override
public void onEnable() {
dbgg = cnfgs.getInt(\"general.dbgg-level\");
if (dbgg > 3) {
dbgg = 3;
}
if (dbgg < 0) {
dbgg = 0;
}
if (dbgg >= 1) {
this.getLogger().log(Level.INFO, \"Debug level set to {0}!\", dbgg);
}
this.getLogger().log(Level.INFO, \"Enabling Listener\");
lstnr = new AFKListener(this);
Bukkit.getPluginManager().registerEvents(lstnr, this);
this.getLogger().log(Level.INFO, \"Enabling Player Handler\");
phndle = new PlayerHandler(this, cnfgs.getInt(\"afk.check-interval\"),
cnfgs.getInt(\"afk.timeout\"));
this.getLogger().log(Level.INFO, \"Enabling Command Handler\");
chndle = new CommandHandler();
long interval = this.getConfig().getInt(\"afk.check-interval\") * 20; // multiplied by 20, due to a
server tick being 1/20th of a second
Bukkit.getServer().getScheduler().runTaskTimer(this, new AFKRunnable(this), interval,
interval);
this.getLogger().log(Level.INFO, \"{0} is enabled!\", this.getName());
}
@Override
public void onDisable() {
this.getLogger().log(Level.INFO, \"{0} is disabled!\", this.getName());
}
public int getDebug() {
return dbgg;
}
public static roguer getPlugin() {
return (roguer) Bukkit.getServer().getPluginManager().getPlugin(\"roguer\");
}
public AFKListener getListener() {
return lstnr;
}
public PlayerHandler getPlayerHandler() {
return phndle;
}
public CommandHandler getCommandHandler() {
return chndle;
}
public Configuration getConfiguration() {
return cnfgs;
}
}
Solution
import com.rogue.roguer.command.CommandHandler;
import com.rogue.roguer.configuration.Configuration;
import com.rogue.roguer.lstnr.AFKListener;
import com.rogue.roguer.player.PlayerHandler;
import com.rogue.roguer.runnable.AFKRunnable;
import com.rogue.roguer.runnable.UpdateRunnable;
import java.io.File;
import java.util.logging.Level;
import org.bukkit.Bukkit;
import org.bukkit.plugin.java.JavaPlugin;
public c.
D the polar covalent bond within methanol cannot form hydrogen bond .pdf
D the polar covalent bond within methanol cannot form hydrogen bond and hense cannot interact
with water molecule hence methyl group hydrophobic in nature.
Solution
D the polar covalent bond within methanol cannot form hydrogen bond and hense cannot interact
with water molecule hence methyl group hydrophobic in nature..
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Shyam Raj Prasad-
https://www.linkedin.com/in/shyam-raj-prasad/
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Let us know about Maruti Suzuki, a successful Joint venture in India.
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Bills have a main role in point of sale procedure. It will help to track sales, handling payments and giving receipts to customers. Bill splitting also has an important role in POS. For example, If some friends come together for dinner and if they want to divide the bill then it is possible by POS bill splitting. This slide will show how to split bills in odoo 17 POS.
GIÁO ÁN DẠY THÊM (KẾ HOẠCH BÀI BUỔI 2) - TIẾNG ANH 8 GLOBAL SUCCESS (2 CỘT) N...
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Unit 8 - Information and Communication Technology (Paper I).pdf
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Students, digital devices and success - Andreas Schleicher - 27 May 2024..pptx
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FeCO3 is an ionic compound, composed of the Fe(II)2+ ion and the CO3.pdf
- 1. FeCO3 is an ionic compound, composed of the Fe(II)2+ ion and the CO32- carbonate ion. Ions tend to dissociate in water, attracted by the partial charges of the molecule. FeCO3 + H2O <---> Fe2+ + CO32- + H2O Water also dissociates into H+ and OH- at a certain equilibrium rate: H2O <---> H+ + OH- Which when combined with an ionic compound is obviously very attractive to the ions in solution... H+ + OH- + Fe2+ + CO32- <----> Fe(OH)2 + H2CO3 Solution FeCO3 is an ionic compound, composed of the Fe(II)2+ ion and the CO32- carbonate ion. Ions tend to dissociate in water, attracted by the partial charges of the molecule. FeCO3 + H2O <---> Fe2+ + CO32- + H2O Water also dissociates into H+ and OH- at a certain equilibrium rate: H2O <---> H+ + OH- Which when combined with an ionic compound is obviously very attractive to the ions in solution... H+ + OH- + Fe2+ + CO32- <----> Fe(OH)2 + H2CO3