Because Fes oxidation state is increasing from +2 to +3, it is an .pdf
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Because Fe\'s oxidation state is increasing from +2 to +3, it is an oxidation reaction. electrons transferred = 1 MnO4-, reduction. O has -2 charge and the molecule itself has -1 charge, so (-2 * 4oxygen molecules) -1 (charge on molecule) = -7charge is balanced by Mn7+ Mn is going through reduction because its change 7+ is reduced to 2+. electrons transferred = 5 -increase in oxidation state is oxidation, decrease is reduction -Oxidation occurs at the anode and reduction at cathode. Reduction half-reaction:__MnO4-_-> Mn2+ +5e- Electrode this reaction occurs at:___cathode_____ Balance elements that are not H or O: Balance O with H2O: MnO4- + 5e- -> Mn2+ + 4H2O Balance H with H+ion: MnO4- + 8H+ + 5e- -> Mn2+ +4H2O Balance Charge: since 5 electrons transferred in this reaction whereas in oxidation reaction only 1 electron transferred, no need to balance this reaction. ( we are going to balance the oxidation reaction.) Oxidation half-reaction:___Fe2+ ->Fe3+ + e- __ Electrode this reaction occurs at: ___anode_ Balance eements that are not H or O: Balance O:no O Balance H: no H Balance Charge: [Fe2+ ->Fe3+ + e- ] x 5 because 5 electrons transferred in reduction reaction, we need to match this up by multiplying 5 to the reaction to balance # of electrons. Multiply one reaction by an integer to get electrons equal: ____5________ Solution Because Fe\'s oxidation state is increasing from +2 to +3, it is an oxidation reaction. electrons transferred = 1 MnO4-, reduction. O has -2 charge and the molecule itself has -1 charge, so (-2 * 4oxygen molecules) -1 (charge on molecule) = -7charge is balanced by Mn7+ Mn is going through reduction because its change 7+ is reduced to 2+. electrons transferred = 5 -increase in oxidation state is oxidation, decrease is reduction -Oxidation occurs at the anode and reduction at cathode. Reduction half-reaction:__MnO4-_-> Mn2+ +5e- Electrode this reaction occurs at:___cathode_____ Balance elements that are not H or O: Balance O with H2O: MnO4- + 5e- -> Mn2+ + 4H2O Balance H with H+ion: MnO4- + 8H+ + 5e- -> Mn2+ +4H2O Balance Charge: since 5 electrons transferred in this reaction whereas in oxidation reaction only 1 electron transferred, no need to balance this reaction. ( we are going to balance the oxidation reaction.) Oxidation half-reaction:___Fe2+ ->Fe3+ + e- __ Electrode this reaction occurs at: ___anode_ Balance eements that are not H or O: Balance O:no O Balance H: no H Balance Charge: [Fe2+ ->Fe3+ + e- ] x 5 because 5 electrons transferred in reduction reaction, we need to match this up by multiplying 5 to the reaction to balance # of electrons. Multiply one reaction by an integer to get electrons equal: ____5________.
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![Because Fe's oxidation state is increasing from +2 to +3, it is an oxidation reaction.
electrons transferred = 1
MnO4-, reduction.
O has -2 charge and the molecule itself has -1 charge, so (-2 * 4oxygen molecules) -1 (charge on
molecule) =
-7charge is balanced by Mn7+
Mn is going through reduction because its change 7+ is reduced to 2+.
electrons transferred = 5
-increase in oxidation state is oxidation, decrease is reduction
-Oxidation occurs at the anode and reduction at cathode.
Reduction half-reaction:__MnO4-_-> Mn2+ +5e- Electrode this reaction occurs
at:___cathode_____
Balance elements that are not H or O:
Balance O with H2O: MnO4- + 5e- -> Mn2+ + 4H2O
Balance H with H+ion: MnO4- + 8H+ + 5e- -> Mn2+ +4H2O
Balance Charge: since 5 electrons transferred in this reaction whereas in oxidation reaction only
1 electron transferred, no need to balance this reaction. ( we are going to balance the oxidation
reaction.)
Oxidation half-reaction:___Fe2+ ->Fe3+ + e- __ Electrode this reaction occurs at: ___anode_
Balance eements that are not H or O:
Balance O:no O
Balance H: no H
Balance Charge: [Fe2+ ->Fe3+ + e- ] x 5
because 5 electrons transferred in reduction reaction, we need to match this up by multiplying 5
to the reaction to balance # of electrons.
Multiply one reaction by an integer to get electrons equal: ____5________
Solution
Because Fe's oxidation state is increasing from +2 to +3, it is an oxidation reaction.
electrons transferred = 1
MnO4-, reduction.
O has -2 charge and the molecule itself has -1 charge, so (-2 * 4oxygen molecules) -1 (charge on
molecule) =](https://image.slidesharecdn.com/becausefesoxidationstateisincreasingfrom2to3itisan-230414005620-f4aa56d0/85/Because-Fes-oxidation-state-is-increasing-from-2-to-3-it-is-an-pdf-1-320.jpg)
![-7charge is balanced by Mn7+
Mn is going through reduction because its change 7+ is reduced to 2+.
electrons transferred = 5
-increase in oxidation state is oxidation, decrease is reduction
-Oxidation occurs at the anode and reduction at cathode.
Reduction half-reaction:__MnO4-_-> Mn2+ +5e- Electrode this reaction occurs
at:___cathode_____
Balance elements that are not H or O:
Balance O with H2O: MnO4- + 5e- -> Mn2+ + 4H2O
Balance H with H+ion: MnO4- + 8H+ + 5e- -> Mn2+ +4H2O
Balance Charge: since 5 electrons transferred in this reaction whereas in oxidation reaction only
1 electron transferred, no need to balance this reaction. ( we are going to balance the oxidation
reaction.)
Oxidation half-reaction:___Fe2+ ->Fe3+ + e- __ Electrode this reaction occurs at: ___anode_
Balance eements that are not H or O:
Balance O:no O
Balance H: no H
Balance Charge: [Fe2+ ->Fe3+ + e- ] x 5
because 5 electrons transferred in reduction reaction, we need to match this up by multiplying 5
to the reaction to balance # of electrons.
Multiply one reaction by an integer to get electrons equal: ____5________](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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Hello!Lets look at the balanced reaction equation starting with th.pdf
Hello!
Lets look at the balanced reaction equation starting with the first item in the equation, Fe2+
If we follow Fe2+ to the product side of the reaction, we see that it becomes Fe3+. In other
words, it goes UP +1 in oxidation state (it is OXIDIZED) and therefore it must have LOST an
electron (since electrons have a negative charge). Since it donated it\'s electron to another atom
in the reaction, we call iron the REDUCING AGENT.
Next, lets look at MnO4- (specifically, lets look at the Mn) which becomes Mn2+ on the product
side of the equation. When bonded with oxygen, Mn has a +7 charge to offset the oxygen\'s
negative charge (Note: each oxygen has a -2 charge times 4 oxygen molecules making a -8
charge. But the overall atom has a -1 charge. We subtract this from oxygen\'s -8 to make -7 total.
So Mn must have a +7 charge to offset this.) Since Mn goes from +7 to +4 it goes DOWN -3 in
oxidation state (it is REDUCED) and therefore it must have GAINED electrons (since electrons
have a negative charge). Since it takes electrons from another atom in the reaction, we call Mn
the OXIDIZING AGENT.
I hope this helps! Please let me know if additional clarification is needed : )
Solution
Hello!
Lets look at the balanced reaction equation starting with the first item in the equation, Fe2+
If we follow Fe2+ to the product side of the reaction, we see that it becomes Fe3+. In other
words, it goes UP +1 in oxidation state (it is OXIDIZED) and therefore it must have LOST an
electron (since electrons have a negative charge). Since it donated it\'s electron to another atom
in the reaction, we call iron the REDUCING AGENT.
Next, lets look at MnO4- (specifically, lets look at the Mn) which becomes Mn2+ on the product
side of the equation. When bonded with oxygen, Mn has a +7 charge to offset the oxygen\'s
negative charge (Note: each oxygen has a -2 charge times 4 oxygen molecules making a -8
charge. But the overall atom has a -1 charge. We subtract this from oxygen\'s -8 to make -7 total.
So Mn must have a +7 charge to offset this.) Since Mn goes from +7 to +4 it goes DOWN -3 in
oxidation state (it is REDUCED) and therefore it must have GAINED electrons (since electrons
have a negative charge). Since it takes electrons from another atom in the reaction, we call Mn
the OXIDIZING AGENT.
I hope this helps! Please let me know if additional clarification is needed : ).
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Hello!Lets look at the balanced reaction equation starting with th.pdf
Hello!
Lets look at the balanced reaction equation starting with the first item in the equation, Fe2+
If we follow Fe2+ to the product side of the reaction, we see that it becomes Fe3+. In other
words, it goes UP +1 in oxidation state (it is OXIDIZED) and therefore it must have LOST an
electron (since electrons have a negative charge). Since it donated it\'s electron to another atom
in the reaction, we call iron the REDUCING AGENT.
Next, lets look at MnO4- (specifically, lets look at the Mn) which becomes Mn2+ on the product
side of the equation. When bonded with oxygen, Mn has a +7 charge to offset the oxygen\'s
negative charge (Note: each oxygen has a -2 charge times 4 oxygen molecules making a -8
charge. But the overall atom has a -1 charge. We subtract this from oxygen\'s -8 to make -7 total.
So Mn must have a +7 charge to offset this.) Since Mn goes from +7 to +4 it goes DOWN -3 in
oxidation state (it is REDUCED) and therefore it must have GAINED electrons (since electrons
have a negative charge). Since it takes electrons from another atom in the reaction, we call Mn
the OXIDIZING AGENT.
I hope this helps! Please let me know if additional clarification is needed : )
Solution
Hello!
Lets look at the balanced reaction equation starting with the first item in the equation, Fe2+
If we follow Fe2+ to the product side of the reaction, we see that it becomes Fe3+. In other
words, it goes UP +1 in oxidation state (it is OXIDIZED) and therefore it must have LOST an
electron (since electrons have a negative charge). Since it donated it\'s electron to another atom
in the reaction, we call iron the REDUCING AGENT.
Next, lets look at MnO4- (specifically, lets look at the Mn) which becomes Mn2+ on the product
side of the equation. When bonded with oxygen, Mn has a +7 charge to offset the oxygen\'s
negative charge (Note: each oxygen has a -2 charge times 4 oxygen molecules making a -8
charge. But the overall atom has a -1 charge. We subtract this from oxygen\'s -8 to make -7 total.
So Mn must have a +7 charge to offset this.) Since Mn goes from +7 to +4 it goes DOWN -3 in
oxidation state (it is REDUCED) and therefore it must have GAINED electrons (since electrons
have a negative charge). Since it takes electrons from another atom in the reaction, we call Mn
the OXIDIZING AGENT.
I hope this helps! Please let me know if additional clarification is needed : ).
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Okay, here are the steps to balance this reaction:
Step 1) Identify oxidizing and reducing agents:
MnO4- is reduced, so it is the oxidizing agent.
MnO4- + 5e- → Mn2+
SO2 is oxidized, so it is the reducing agent.
SO2 → SO42- + 4e-
Step 2) Balance other elements: No need here.
Step 3) Balance O by adding H2O:
MnO4- + 5e- → Mn2+ + 4H2O
SO2 → SO42- + 4e-
Step 4) Balance H by adding H+:
M
HeadPhone.java All the outputs were same because the variabl.pdf
// HeadPhone.java
/* All the outputs were same because the variables declared and defined in HeadPhone class
were static,
they will be updated only once during the program execution.
removing the static keyword from the variables which will be updated during the program
execution will
give the desired results
*/
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private static final int LOW = 1;
private static final int MEDIUM = 2;
private static final int HIGH = 3;
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private boolean pluggedIn = false;
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private String manufacturer = none;
private String headPhoneColor = none;
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private String switchPos= none;
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volume = MEDIUM;
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manufacturer = \"Sony\";
headPhoneColor = \"Gold\";
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pluggedIn = pluggedIn2;
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headPhoneColor = headPhoneColor2;
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volPos = \"MEDIUM\";
}
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volPos = \"LOW\";
}
else {
System.out.println(\"Your entry is invalid, please try again.\");
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return volume;
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public boolean getPluggedIn() {
if(pluggedIn == false) {
onOrOff = \"Off\";
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else {
onOrOff = \"On\";
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return pluggedIn;
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System.out.println(\"Please the the volume (Options are by integer (1 = Low, 2 = Medium, 3 =
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hpObj2.setV.
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// Program to print all combination of size r in an array of size n
#include
void combinationUtil(int arr[], int data[], int start, int end,
int index, int r);
// The main function that prints all combinations of size r
// in arr[] of size n. This function mainly uses combinationUtil()
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index ---> Current index in data[]
r ---> Size of a combination to be printed */
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for (int j=0; j= r-index\" makes sure that including one element
// at index will make a combination with remaining elements
// at remaining positions
for (int i=start; i<=end && end-i+1 >= r-index; i++)
{
data[index] = arr[i];
combinationUtil(arr, data, i+1, end, index+1, r);
}
}
// Driver program to test above functions
int main()
{
int arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9};
int r = 4;
int n = sizeof(arr)/sizeof(arr[0]);
printCombination(arr, n, r);
}
Solution
// Program to print all combination of size r in an array of size n
#include
void combinationUtil(int arr[], int data[], int start, int end,
int index, int r);
// The main function that prints all combinations of size r
// in arr[] of size n. This function mainly uses combinationUtil()
void printCombination(int arr[], int n, int r)
{
// A temporary array to store all combination one by one
int data[r];
// Print all combination using temprary array \'data[]\'
combinationUtil(arr, data, 0, n-1, 0, r);
}
/* arr[] ---> Input Array
data[] ---> Temporary array to store current combination
start & end ---> Staring and Ending indexes in arr[]
index ---> Current index in data[]
r ---> Size of a combination to be printed */
void combinationUtil(int arr[], int data[], int start, int end,
int index, int r)
{
// Current combination is ready to be printed, print it
if (index == r)
{
if (data[0]+data[1]+data[2]+data[3]==24)
{
for (int j=0; j= r-index\" makes sure that including one element
// at index will make a combination with remaining elements
// at remaining positions
for (int i=start; i<=end && end-i+1 >= r-index; i++)
{
data[index] = arr[i];
combinationUtil(arr, data, i+1, end, index+1, r);
}
}
// Driver program to test above functions
int main()
{
int arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9};
int r = 4;
int n = sizeof(arr)/sizeof(arr[0]);
printCombination(arr, n, r);
}.
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Chapter 20 Lecture- Electrochemistry
Let's balance this step-by-step using the half-reaction method:
Oxidation: Ag → Ag+
Reduction: 1/2O2 + H+ → H2O
Balanced half-reactions:
Ag → Ag+
1/2O2 + H+ → H2O
Overall balanced reaction:
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The correct coefficients are:
1, 1/2, 1, 1, 1
The answer is 1.
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**More good stuff available at:
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Okay, here are the steps to balance this reaction:
Step 1) Identify oxidizing and reducing agents:
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SO2 is oxidized, so it is the reducing agent.
SO2 → SO42- + 4e-
Step 2) Balance other elements: No need here.
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MnO4- + 5e- → Mn2+ + 4H2O
SO2 → SO42- + 4e-
Step 4) Balance H by adding H+:
M
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HeadPhone.java All the outputs were same because the variabl.pdf
// HeadPhone.java
/* All the outputs were same because the variables declared and defined in HeadPhone class
were static,
they will be updated only once during the program execution.
removing the static keyword from the variables which will be updated during the program
execution will
give the desired results
*/
public class HeadPhone {
private static final int LOW = 1;
private static final int MEDIUM = 2;
private static final int HIGH = 3;
private int volume = 0;
private boolean pluggedIn = false;
private String none = null;
private String manufacturer = none;
private String headPhoneColor = none;
private String volPos = none;
private String switchPos= none;
private String onOrOff= none;
// Default Headphone Constructor
public HeadPhone() {
volume = MEDIUM;
pluggedIn = false;
manufacturer = \"Sony\";
headPhoneColor = \"Gold\";
}
// Headphone Constructor
public HeadPhone (int volume2, boolean pluggedIn2, String manufacturer2, String
headPhoneColor2) {
volume = volume2;
pluggedIn = pluggedIn2;
manufacturer = manufacturer2;
headPhoneColor = headPhoneColor2;
}
// Setter Methods
// setVolume
public void setVolume(int volume2) {
volume = volume2;
}
// setPluggedIn
public void setPluggedIn(boolean pluggedIn2) {
pluggedIn = pluggedIn2;
}
// setManufacturer
public void setManufacturer(String manufacturer2) {
manufacturer = manufacturer2;
}
// setColor
public void setColor(String headPhoneColor2) {
headPhoneColor = headPhoneColor2;
}
// Getter Methods
// getVolume
public int getVolume() {
if(volume == 3) {
volPos = \"HIGH\";
}
else if (volume == 2) {
volPos = \"MEDIUM\";
}
else if(volume == 1) {
volPos = \"LOW\";
}
else {
System.out.println(\"Your entry is invalid, please try again.\");
}
return volume;
}
//getPluggedIn
public boolean getPluggedIn() {
if(pluggedIn == false) {
onOrOff = \"Off\";
}
else {
onOrOff = \"On\";
}
return pluggedIn;
}
//getManufacturer
public String getManufacturer() {
return manufacturer;
}
//getColor
public String getColor() {
return headPhoneColor;
}
// changeVolume method
public void changeVolume(int volume2) {
setVolume(volume2);
}
// toString method
public String toString() {
String str = \"toString() results: (volume=\" + volume + \", pluggedIn=\" + pluggedIn + \",
Manufacturer=\" + manufacturer + \", HeadPhone Color=\" + headPhoneColor +\")\";
return str;
}
}
// TestHeadPhone.java
import java.util.Scanner;
public class TestHeadPhone {
public static void main(String[] args) {
int row = 4;
HeadPhone [] hfArray = new HeadPhone[row];
HeadPhone hpObj1 = new HeadPhone();
hfArray[0] = hpObj1;
Scanner input = new Scanner(System.in);
for(int i=1; i<4; i++) {
HeadPhone hpObj2 = new HeadPhone();
System.out.println(\"Please enter the data for new headphones:\");
System.out.println(\"Would you like to turn the headphones on? (Type \'true\' for on, Type
\'false\' for off: \");
hpObj2.setPluggedIn(input.nextBoolean());
System.out.println(\"Please the the volume (Options are by integer (1 = Low, 2 = Medium, 3 =
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hpObj2.setV.
Program to print all combination of size r in an array of size n.pdf
// Program to print all combination of size r in an array of size n
#include
void combinationUtil(int arr[], int data[], int start, int end,
int index, int r);
// The main function that prints all combinations of size r
// in arr[] of size n. This function mainly uses combinationUtil()
void printCombination(int arr[], int n, int r)
{
// A temporary array to store all combination one by one
int data[r];
// Print all combination using temprary array \'data[]\'
combinationUtil(arr, data, 0, n-1, 0, r);
}
/* arr[] ---> Input Array
data[] ---> Temporary array to store current combination
start & end ---> Staring and Ending indexes in arr[]
index ---> Current index in data[]
r ---> Size of a combination to be printed */
void combinationUtil(int arr[], int data[], int start, int end,
int index, int r)
{
// Current combination is ready to be printed, print it
if (index == r)
{
if (data[0]+data[1]+data[2]+data[3]==24)
{
for (int j=0; j= r-index\" makes sure that including one element
// at index will make a combination with remaining elements
// at remaining positions
for (int i=start; i<=end && end-i+1 >= r-index; i++)
{
data[index] = arr[i];
combinationUtil(arr, data, i+1, end, index+1, r);
}
}
// Driver program to test above functions
int main()
{
int arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9};
int r = 4;
int n = sizeof(arr)/sizeof(arr[0]);
printCombination(arr, n, r);
}
Solution
// Program to print all combination of size r in an array of size n
#include
void combinationUtil(int arr[], int data[], int start, int end,
int index, int r);
// The main function that prints all combinations of size r
// in arr[] of size n. This function mainly uses combinationUtil()
void printCombination(int arr[], int n, int r)
{
// A temporary array to store all combination one by one
int data[r];
// Print all combination using temprary array \'data[]\'
combinationUtil(arr, data, 0, n-1, 0, r);
}
/* arr[] ---> Input Array
data[] ---> Temporary array to store current combination
start & end ---> Staring and Ending indexes in arr[]
index ---> Current index in data[]
r ---> Size of a combination to be printed */
void combinationUtil(int arr[], int data[], int start, int end,
int index, int r)
{
// Current combination is ready to be printed, print it
if (index == r)
{
if (data[0]+data[1]+data[2]+data[3]==24)
{
for (int j=0; j= r-index\" makes sure that including one element
// at index will make a combination with remaining elements
// at remaining positions
for (int i=start; i<=end && end-i+1 >= r-index; i++)
{
data[index] = arr[i];
combinationUtil(arr, data, i+1, end, index+1, r);
}
}
// Driver program to test above functions
int main()
{
int arr[] = {1, 2, 3, 4, 5, 6, 7, 8, 9};
int r = 4;
int n = sizeof(arr)/sizeof(arr[0]);
printCombination(arr, n, r);
}.
Methanol is soluble in water. methanol is a one carbon organic alco.pdf
Methanol is soluble in water. methanol is a one carbon organic alcohol. The -OH group forms
favorable hydrogen bonds with water, and the hydrophobic portion is small enough that it can
readily dissolve.
Solution
Methanol is soluble in water. methanol is a one carbon organic alcohol. The -OH group forms
favorable hydrogen bonds with water, and the hydrophobic portion is small enough that it can
readily dissolve..
Solubility in water happens when something is pol.pdf
Solubility in water happens when something is polar or charged. Charged is much
easier to understand--it\'s when there\'s an actual positive or negative charge on any given
molecule. Polarity is just like charge, but instead of the electrons being specifically on a given
atom and not on another (like in Na+ and Cl-), they\'re shared in a way such that they spend
more time on one atom vs another (like CH3OH--the electrons spend more time on the O than
they do on either the C or the Hs). How do you know which atoms like to hold onto electrons
more than other atoms? Electronegativity differences. Certain atoms like F, N, and O are really
electronegative, which means they like to hold onto electrons more than other atoms, like, say, C
and H. So, if F were bound to C, the electrons would be on the F more often than the C. On the
other hand, if you have to atoms, like C and H, which have very similar electronegativity values,
they will share the electrons more readily, and the electrons don\'t stay on one or the other atoms
more often. Such molecules are called non-polar, and are, in fact, not soluble in water. Instead,
they\'re fat-soluble, because fat is just a whole bunch of C-H bonds next to each other. So,
looking at the structures of those given vitamins, vitamin A consists mostly of a bunch of C-H
bonds; it\'s likely fat soluble. Vitamin B6 has a bunch of C-O-H bonds and an N, too. It\'s much
more polar, and likely water soluble. Vitamin C has tons of oxygens and is definitely water
soluble. Vitamin E is mostly C-H bonds and not too many O-H or O-C bonds, so it\'s likely fat
soluble. Hope that helps! :)
Solution
Solubility in water happens when something is polar or charged. Charged is much
easier to understand--it\'s when there\'s an actual positive or negative charge on any given
molecule. Polarity is just like charge, but instead of the electrons being specifically on a given
atom and not on another (like in Na+ and Cl-), they\'re shared in a way such that they spend
more time on one atom vs another (like CH3OH--the electrons spend more time on the O than
they do on either the C or the Hs). How do you know which atoms like to hold onto electrons
more than other atoms? Electronegativity differences. Certain atoms like F, N, and O are really
electronegative, which means they like to hold onto electrons more than other atoms, like, say, C
and H. So, if F were bound to C, the electrons would be on the F more often than the C. On the
other hand, if you have to atoms, like C and H, which have very similar electronegativity values,
they will share the electrons more readily, and the electrons don\'t stay on one or the other atoms
more often. Such molecules are called non-polar, and are, in fact, not soluble in water. Instead,
they\'re fat-soluble, because fat is just a whole bunch of C-H bonds next to each other. So,
looking at the structures of those given vitamins, vitamin A consists mostly of a bunch of C-H
bonds; it\'s likely fat soluble. Vi.
moles = 0.144 .pdf
The document provides the value of moles as 0.144. No other information or context is given about what this value represents.
It is used to remove water from the organic layer.pdf
It is used to remove water from the organic layer. During the experiment some
water gets into the organic layer because of the partial miscibility of the organic products in
water. Saturated sodium chloride will pull the water from the organic layer following simple
solute rules. Sat. NaCl wants to become more dilute and salts have a higher affinity for water
than for organic solvents. Therefore removing the bulk of the water in the organic layer.
Solution
It is used to remove water from the organic layer. During the experiment some
water gets into the organic layer because of the partial miscibility of the organic products in
water. Saturated sodium chloride will pull the water from the organic layer following simple
solute rules. Sat. NaCl wants to become more dilute and salts have a higher affinity for water
than for organic solvents. Therefore removing the bulk of the water in the organic layer..
nitrogen is more electronegative than phosphorus .pdf
nitrogen is more electronegative than phosphorus , so nitrogen tends to attracts
more of the electron cloud towards itself. hence nitrogen phosphorous bond in not fully covalent.
hence there is no equal sharing of electrons
Solution
nitrogen is more electronegative than phosphorus , so nitrogen tends to attracts
more of the electron cloud towards itself. hence nitrogen phosphorous bond in not fully covalent.
hence there is no equal sharing of electrons.
ionic equation is 2 Cu+2(aq) + 4Cl-(aq) + 4 K+(aq.pdf
ionic equation is 2 Cu+2(aq) + 4Cl-(aq) + 4 K+(aq)+ 4I-(aq) --> 2 CuI(s) + I2(aq) +
4 K+(aq)+ 4Cl-(aq)
Solution
ionic equation is 2 Cu+2(aq) + 4Cl-(aq) + 4 K+(aq)+ 4I-(aq) --> 2 CuI(s) + I2(aq) +
4 K+(aq)+ 4Cl-(aq).
hydration of alkene. .pdf
The document discusses hydration of alkene. Alkene undergoes addition reaction with water to form an alcohol. Water adds across the carbon-carbon double bond in alkene resulting in a saturated alcohol product.
H2O + NO + NO2 -- 2HNO2 Three molecules collide.pdf
H2O + NO + NO2 --> 2HNO2 Three molecules collide at an instant is rare, thus it
is not an elementary reaction that can occur at a significant rate.
Solution
H2O + NO + NO2 --> 2HNO2 Three molecules collide at an instant is rare, thus it
is not an elementary reaction that can occur at a significant rate..
he coordination compounds , metal complexes or si.pdf
he coordination compounds , metal complexes or simply complexes are compounds
which contain an ion or a central atom, usually a metal, surrounded by a collection of ions or
molecules. The chance discovery dell\'esamminocobalto (III) chloride, CoCl 3 6 NH 3 , due to
Tassaert in 1798 marks the beginning of the chemistry of the complexes. Many of the jobs
immediately after they were made ??with ammonia and complex that resulted were called
metallammine. Subsequently it was found that other amines and other anions such as CN - , NO
2 - , NCS - , Cl - can form complexes with metals. Many compounds were prepared with these
anions and, at first, were called by the name of the chemical that prepared them first.
Subsequently, because of the fact that many of these complexes was colored, was used a scheme
of nomenclature based on the color. Our current knowledge about the nature of the metal
complexes are due to Werner , in 1893 at the age of 26 years suggested that even today is called
coordination theory of Werner. His most important are three postulates: Most of the elements
have two types of meaning: a) primary valence and b) secondary valence. In modern terminology
a) corresponds to the oxidation state and b) the coordination number Each element tends to
satisfy the valence both primary and secondary The secondary valence is directed towards fixed
positions of the space Consider the complex cloroamminocobalto (III). According to the theory
of Werner the first member of the series has the formula Co [(NH 3 ) 6 ] Cl 3 and is represented
as shown in the figure: (i upload for u on this address: Structures
http://i45.tinypic.com/2l2654.gif http://i50.tinypic.com/jaykgy.jpg) The primary valence or
oxidation state of the cobalt (III) is 3. The three ions Cl - primary saturate the valence of cobalt
chloride ions and then the three primary saturate the valence of cobalt negative ions that
neutralize the positive charge of the metal ion affect the primary value.
Solution
he coordination compounds , metal complexes or simply complexes are compounds
which contain an ion or a central atom, usually a metal, surrounded by a collection of ions or
molecules. The chance discovery dell\'esamminocobalto (III) chloride, CoCl 3 6 NH 3 , due to
Tassaert in 1798 marks the beginning of the chemistry of the complexes. Many of the jobs
immediately after they were made ??with ammonia and complex that resulted were called
metallammine. Subsequently it was found that other amines and other anions such as CN - , NO
2 - , NCS - , Cl - can form complexes with metals. Many compounds were prepared with these
anions and, at first, were called by the name of the chemical that prepared them first.
Subsequently, because of the fact that many of these complexes was colored, was used a scheme
of nomenclature based on the color. Our current knowledge about the nature of the metal
complexes are due to Werner , in 1893 at the age of 26 years suggested that even today is called
coordination .
PdCl2 + 2DMSO PdCl22DMSO It forms a complexe..pdf
PdCl2 + 2DMSO ? PdCl2*2DMSO It forms a complexe.
Solution
PdCl2 + 2DMSO ? PdCl2*2DMSO It forms a complexe..
melting point of the mixture decreases when small.pdf
melting point of the mixture decreases when small amount of naphthalene is added
Solution
melting point of the mixture decreases when small amount of naphthalene is added.
Polymers are made by joining small molecules into large ones.But m.pdf
Polymers are made by joining small molecules into large ones.But most of these monomer
molecules are perfectly stable as they are, so they have devised two general methods to make
them react with each other, building up the backbone chain as the reaction proceeds.
This method (also known as step-growth) requires that the monomers possess two or more kinds
of functional groups that are able to react with each other in such a way that parts of these groups
combine to form a small molecule (often H2O) which is eliminated from the two pieces.
2 Additional Polymerisation:
Polymer play a major role in various industry sector and which are used in plastic bags ,water
pipes outer layer of
electric wires clothing and,woven fabric and also other materials.
PLASTIC BAGS:
WATER PIPES
OUTER LAYER OF ELECTRIC WIRES:
Outer layer of insulator is also called as electric insulator.
It is the most commonly used thermoplastic insulator for cables.
It is cheap, durable and widely available.
However, the chlorine in PVC (a halogen) causes the production of thick, toxic, black smoke
when burnt and can be a health hazard in areas where low smoke and toxicity are required
Normal operating temperatures are typically between 75C and 105C (depending on PVC type).
CLOTHING:
Solution
Polymers are made by joining small molecules into large ones.But most of these monomer
molecules are perfectly stable as they are, so they have devised two general methods to make
them react with each other, building up the backbone chain as the reaction proceeds.
This method (also known as step-growth) requires that the monomers possess two or more kinds
of functional groups that are able to react with each other in such a way that parts of these groups
combine to form a small molecule (often H2O) which is eliminated from the two pieces.
2 Additional Polymerisation:
Polymer play a major role in various industry sector and which are used in plastic bags ,water
pipes outer layer of
electric wires clothing and,woven fabric and also other materials.
PLASTIC BAGS:
WATER PIPES
OUTER LAYER OF ELECTRIC WIRES:
Outer layer of insulator is also called as electric insulator.
It is the most commonly used thermoplastic insulator for cables.
It is cheap, durable and widely available.
However, the chlorine in PVC (a halogen) causes the production of thick, toxic, black smoke
when burnt and can be a health hazard in areas where low smoke and toxicity are required
Normal operating temperatures are typically between 75C and 105C (depending on PVC type).
CLOTHING:.
What is Operational ExcellenceAchieving Operational Excellence requ.pdf
What is Operational Excellence?Achieving Operational Excellence requires the successful
implementation of a integrated Business Execution System that effectively and seamlessly
integrates the following four building blocks: 1)Strategy Deployment2)Performance
Management3)Process Excellence and4)High Performance Work Teams.Strategy or Policy
Deployment is the process that aligns and links business strategy and execution. Strategy
Deployment Performance Management High Performance Work Teams Process Excellence
Operational Excellence Business Execution System Strategy Deployment Process
Tools: 1)The Hoshin X-Matrix is a tool that visualizes an organization’sstrategic
objectives,strategic initiatives,key performance indicators,key projects &action items, andhuman
resources requirements in one simple matrix. 2)The Hoshin X-Matrixenables an organization to
easily review the alignment of its strategic objectives, strategicinitiatives, key performance
indicators, key action items and human resources.
Supply chain management:If you manage a small firm that makes a few products or sells a few
services, chances are you will have a small number of suppliers. Supply chain: 1)A firm\'s
supply chain is a network of organization and business processes for procuring raw materials,
transforming these materials into intermediate and finished products, and distributing the
finished products to customers.2)Supply chain software is classified as either software to help
businesses plan their supply chains (supply chain planning) or software to help them execute the
supply chain steps. 3)it manage the flow of products through distribution centers and warehouses
to ensure that products are delivered to the right locations in the most efficient manner.
Customer Relationship Management Systems 1) You were probably heard phrases such as the
customer is always right or the customer comes first. Today these words ring more true than
ever. 2)Because competitive advantage based on an innovative new product or service is often
very short lived, companies are realizing that their only enduring competitive strength may be
their relationships with their customers. 3) Some say that the basis of competition has switched
from who sells the most products and services to whom own the customer, and that customer
relationships represent a firm\'s most valuable asset. 4) Customer relationship management
systems typically provide software and online tools for sales, customer service, and marketing.
5)Partner relationship management (PRM) uses many of the same data, tools, and systems as
customer relationship management to enhance collaboration between a company and its selling
partners.
Enterprise Application: 1)Enterprise application require not only deep seated technological
changes but also fundamental changes in the way the business operates. 2)Companies must make
sweeping changes to their business processes to work with the software. 3)Companies adopting
enterprise applications can .
the name arsenic acid, like phosphoric acid (H3PO4), does notconvey .pdf
the name arsenic acid, like phosphoric acid (H3PO4), does notconvey the structure, but does fit
into a general patternwhereby the ending ‘ic’ denotes a higher or the
highest possible oxidation state (compare nitric acid, sulfuricacid).
HClO: Hypochlorous Acid
Solution
the name arsenic acid, like phosphoric acid (H3PO4), does notconvey the structure, but does fit
into a general patternwhereby the ending ‘ic’ denotes a higher or the
highest possible oxidation state (compare nitric acid, sulfuricacid).
HClO: Hypochlorous Acid.
Sucrose is a molecular compound and does not ionize in solution.KC.pdf
Sucrose is a molecular compound and does not ionize in solution.
KCl is a soluble ionic compound and fully ionizes in solution:
KCl(aq) => K+(aq) + Cl-(aq)
Sucrose and KCl do not interact in solution.
Mass balance equation: [KCl] = [K+]
Charge balance equation: [K+] = [Cl-]
Solution
Sucrose is a molecular compound and does not ionize in solution.
KCl is a soluble ionic compound and fully ionizes in solution:
KCl(aq) => K+(aq) + Cl-(aq)
Sucrose and KCl do not interact in solution.
Mass balance equation: [KCl] = [K+]
Charge balance equation: [K+] = [Cl-].
C. Increasing the temperature of the system .pdf
Increasing the temperature of the system is proposed as a solution. No other details are provided about the system, the current temperature, the proposed increased temperature, or the expected effects of a higher temperature. The summary is limited due to the brevity of the original document.
because Electron Affinity is the amount of energy.pdf
because Electron Affinity is the amount of energy needed to add an electron to an
atom or molecule.. Thus Decreasing its Stabillity and Whenever We have Decrease in Stability
We Have A Negative Energy Released thus energy required and hence negative electron
affinity..The Electron affinity of an atom or molecule is defined as the amount of energy released
when an electron is added to a neutral atom or molecule to form a negative ion. X + e- ? X-
(note: the correct equation in words is X + electron -> X + energy) This property is measured for
atoms and molecules in the gaseous state only, since in the solid or liquid states their energy
levels would be changed by contact with other atoms or molecules. A list of the electron
affinities was used by Robert S. Mulliken to develop an electronegativity scale for atoms, equal
to the average of the electron affinity and ionization potential.Other theoretical concepts that use
electron affinity include electronic chemical potential and chemical hardness. Another example,
a molecule or atom that has a more positive value of electron affinity than another is often called
an electron acceptor and the less positive an electron donor. Together they may undergo charge-
transfer reactions. In solids, the electron affinity is the energy difference between the vacuum
energy and the conduction band minimum. To use electron affinities properly, it is essential to
keep track of sign. For any reaction that releases energy, the change in energy, ?E, has a negative
value and the reaction is called an exothermic process. Electron capture for almost all non-noble
gas atoms involves the release of energy and thus are exothermic. The positive values that are
listed in tables of Eea are amounts or magnitudes. It is the word, released within the definition
energy released that supplies the negative sign. Confusion arises in mistaking Eea for a change in
energy, ?E, in which case the positive values listed in tables would be for an endo- not exo-
thermic process. The relation between the two is, Eea = - ?E(attach). However, if the value
assigned to Eea is negative, the negative sign implies a reversal of direction, and energy is
required to attach an electron. In this case, the electron capture is an endothermic process and the
relationship, Eea = - ?E(attach) is still valid. Negative values typically arise for the capture of a
second electron, but also for the nitrogen atom. The usual expression for calculating Eea when
an electron is attached is Eea = (Einitial - Efinal)attach = - ?E(attach) This expression does
follow the convention ?X = X(final) - X(initial) since - ?E = - (E(final) - E(initial)) = E(initial) -
E(final). Equivalently, electron affinity can also be defined as the amount of energy required to
detach an electron from a singly charged negative ion,i.e. the energy change for the process X- ?
X + e- If the same table is employed for the forward and reverse reactions, without switching
signs, care must be take.
Ques-1Clostredium botulinum, bacteria that cause food poisoning &.pdf
Ques-1:
Clostredium botulinum, bacteria that cause food poisoning & it is the etiological agent in this
outbreak of food poisoning
Ques-2:
This food intoxication because \"botulinum\" is the toxin generated by these bacteria by
putrefying food material and these bacteria is anaerobic. These toxins are made of exotoxins,
endotoxins (tetanus toxin, botulism toxin & cholera toxin) as these are composed of proteins,
lipoppolysaccharides produced through cell wall to act on the host cell immune system. These
are virulent and cause damage to host cells through inflammatory mediators released via
phagocytosis by immune system. I would declare the situation as a sortumme an outbreak, this is
because acute episodes of \"gastrointestinal illness is mainly due to food borne illness caused by
\"clostridium botulinum\". It has observed clear symptoms of disease characterized by nausea,
headache, blurred vision, bronchoconstriction due to stiffen throat & neck
Fever, headache, nausea, and severe respiratory symptoms followed by blurred vision; this is
because botulinum toxin is a neurotoxin that potentially inhibits release of acetylcholine from
nerve endings and leading to drooping eyelid and slurred speech.
Ques-3:
Potato salad was transported in a loose in a pickup truck & it is not covered or packed before
preparing meals therefore anaerobic fermentation of this carbohydrate is the source of
contamination of this bacteria & cause food borne illness
Solution
Ques-1:
Clostredium botulinum, bacteria that cause food poisoning & it is the etiological agent in this
outbreak of food poisoning
Ques-2:
This food intoxication because \"botulinum\" is the toxin generated by these bacteria by
putrefying food material and these bacteria is anaerobic. These toxins are made of exotoxins,
endotoxins (tetanus toxin, botulism toxin & cholera toxin) as these are composed of proteins,
lipoppolysaccharides produced through cell wall to act on the host cell immune system. These
are virulent and cause damage to host cells through inflammatory mediators released via
phagocytosis by immune system. I would declare the situation as a sortumme an outbreak, this is
because acute episodes of \"gastrointestinal illness is mainly due to food borne illness caused by
\"clostridium botulinum\". It has observed clear symptoms of disease characterized by nausea,
headache, blurred vision, bronchoconstriction due to stiffen throat & neck
Fever, headache, nausea, and severe respiratory symptoms followed by blurred vision; this is
because botulinum toxin is a neurotoxin that potentially inhibits release of acetylcholine from
nerve endings and leading to drooping eyelid and slurred speech.
Ques-3:
Potato salad was transported in a loose in a pickup truck & it is not covered or packed before
preparing meals therefore anaerobic fermentation of this carbohydrate is the source of
contamination of this bacteria & cause food borne illness.
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Because Fes oxidation state is increasing from +2 to +3, it is an .pdf
- 1. Because Fe's oxidation state is increasing from +2 to +3, it is an oxidation reaction. electrons transferred = 1 MnO4-, reduction. O has -2 charge and the molecule itself has -1 charge, so (-2 * 4oxygen molecules) -1 (charge on molecule) = -7charge is balanced by Mn7+ Mn is going through reduction because its change 7+ is reduced to 2+. electrons transferred = 5 -increase in oxidation state is oxidation, decrease is reduction -Oxidation occurs at the anode and reduction at cathode. Reduction half-reaction:__MnO4-_-> Mn2+ +5e- Electrode this reaction occurs at:___cathode_____ Balance elements that are not H or O: Balance O with H2O: MnO4- + 5e- -> Mn2+ + 4H2O Balance H with H+ion: MnO4- + 8H+ + 5e- -> Mn2+ +4H2O Balance Charge: since 5 electrons transferred in this reaction whereas in oxidation reaction only 1 electron transferred, no need to balance this reaction. ( we are going to balance the oxidation reaction.) Oxidation half-reaction:___Fe2+ ->Fe3+ + e- __ Electrode this reaction occurs at: ___anode_ Balance eements that are not H or O: Balance O:no O Balance H: no H Balance Charge: [Fe2+ ->Fe3+ + e- ] x 5 because 5 electrons transferred in reduction reaction, we need to match this up by multiplying 5 to the reaction to balance # of electrons. Multiply one reaction by an integer to get electrons equal: ____5________ Solution Because Fe's oxidation state is increasing from +2 to +3, it is an oxidation reaction. electrons transferred = 1 MnO4-, reduction. O has -2 charge and the molecule itself has -1 charge, so (-2 * 4oxygen molecules) -1 (charge on molecule) =
- 2. -7charge is balanced by Mn7+ Mn is going through reduction because its change 7+ is reduced to 2+. electrons transferred = 5 -increase in oxidation state is oxidation, decrease is reduction -Oxidation occurs at the anode and reduction at cathode. Reduction half-reaction:__MnO4-_-> Mn2+ +5e- Electrode this reaction occurs at:___cathode_____ Balance elements that are not H or O: Balance O with H2O: MnO4- + 5e- -> Mn2+ + 4H2O Balance H with H+ion: MnO4- + 8H+ + 5e- -> Mn2+ +4H2O Balance Charge: since 5 electrons transferred in this reaction whereas in oxidation reaction only 1 electron transferred, no need to balance this reaction. ( we are going to balance the oxidation reaction.) Oxidation half-reaction:___Fe2+ ->Fe3+ + e- __ Electrode this reaction occurs at: ___anode_ Balance eements that are not H or O: Balance O:no O Balance H: no H Balance Charge: [Fe2+ ->Fe3+ + e- ] x 5 because 5 electrons transferred in reduction reaction, we need to match this up by multiplying 5 to the reaction to balance # of electrons. Multiply one reaction by an integer to get electrons equal: ____5________