a. formic acid (it has more polar components than cyclohexane, inclu.pdf
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a. formic acid (it has more polar components than cyclohexane, including an OH group, which can form hydrogen bonds with water; this is true of short chain carboxylic acids). Cyclohexane is completely nonpolar. b. Hydrogen iodide because it is an ionic compound, which will dissociate into H+ and I- in water. Furthermore, with the exception of silver/mercury/lead, all iodides are soluble in water. Iodine, or I2, is a solid that offers no means by which to achieve solubility. Its electronegativity difference is 0 (since iodine contains two identical atoms) whereas HI\'s electronegativity difference is much higher, which means it is a very polar molecule that can interact with polar water molecules. c. K2SO4 : all compounds that contain a group I metal (such as K, Na, Li) will be soluble in water because they are ionic and can therefore interact with polar water molecules. SiO2 (aka silica and quartz) is a nonpolar solid that will not interact with polar water molecules. Solution a. formic acid (it has more polar components than cyclohexane, including an OH group, which can form hydrogen bonds with water; this is true of short chain carboxylic acids). Cyclohexane is completely nonpolar. b. Hydrogen iodide because it is an ionic compound, which will dissociate into H+ and I- in water. Furthermore, with the exception of silver/mercury/lead, all iodides are soluble in water. Iodine, or I2, is a solid that offers no means by which to achieve solubility. Its electronegativity difference is 0 (since iodine contains two identical atoms) whereas HI\'s electronegativity difference is much higher, which means it is a very polar molecule that can interact with polar water molecules. c. K2SO4 : all compounds that contain a group I metal (such as K, Na, Li) will be soluble in water because they are ionic and can therefore interact with polar water molecules. SiO2 (aka silica and quartz) is a nonpolar solid that will not interact with polar water molecules..
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Explain how water interacts with the following types of compounds to.pdf
Explain how water interacts with the following types of compounds to influence their water
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1.Compounds that can hydrogen bond
2. Compounds that are polar but cannot hydrogen bond
3. Compounds that are non-polar
4. Compounds that dissociate into cations and anions.
Thank you so much.
Solution
Compounds that can hydrogen bond
Water tends to disassociate into H+ and OH- ions. In this disassociation, the oxygen retains the
electrons and only one of the hydrogens, becoming a negatively charged ion known as
hydroxide. Pure water has the same number (or concentration) of H+ as OH- ions. Acidic
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Explain how water interacts with the following types of compounds to.pdf
Explain how water interacts with the following types of compounds to influence their water
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1.Compounds that can hydrogen bond
2. Compounds that are polar but cannot hydrogen bond
3. Compounds that are non-polar
4. Compounds that dissociate into cations and anions.
Thank you so much.
Solution
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1. Copper is above silver in the activity series. Thus Cu metal will.pdf
1. Copper is above silver in the activity series. Thus Cu metal will displace Ag+ ions from
solution, forming Cu2+ ions and metallic Ag.
Observation: The reddish brown copper metal dissolves and silver metal precipitates out. The
colorless Ag+ solution turns blue due to Cu2+ ions formed.
2. Molecular equation:
Cu(s) + 2 AgNO3(aq) => 2 Ag(s) + Cu(NO3)2(aq)
3. Complete ionic equation (obtained by expanding soluble ions in molecular equation):
Cu(s) + 2 Ag+(aq) + 2 NO3-(aq) => 2 Ag(s) + Cu2+(aq) + 2 NO3-(aq)
4. Net ionic equation (obtained by cancelling common ions in complete ionic equation):
Cu(s) + 2 Ag+(aq) => 2 Ag(s) + Cu2+(aq)
Solution
1. Copper is above silver in the activity series. Thus Cu metal will displace Ag+ ions from
solution, forming Cu2+ ions and metallic Ag.
Observation: The reddish brown copper metal dissolves and silver metal precipitates out. The
colorless Ag+ solution turns blue due to Cu2+ ions formed.
2. Molecular equation:
Cu(s) + 2 AgNO3(aq) => 2 Ag(s) + Cu(NO3)2(aq)
3. Complete ionic equation (obtained by expanding soluble ions in molecular equation):
Cu(s) + 2 Ag+(aq) + 2 NO3-(aq) => 2 Ag(s) + Cu2+(aq) + 2 NO3-(aq)
4. Net ionic equation (obtained by cancelling common ions in complete ionic equation):
Cu(s) + 2 Ag+(aq) => 2 Ag(s) + Cu2+(aq).
using recursive method .pdf
/******************* using recursive method **********************/
/**
* 1.2.1 Java program to calculate the Fibonacci Number using recursion
*/
import java.util.Scanner;
public class Fibo {
public static void main(String args[]){
Scanner scan= new Scanner(System.in); //Scanner object to read from the user
int n;
System.out.println(\"Enter the value of n\");
n=scan.nextInt();
System.out.println(\"The \"+n+\". Fibonacci number is \"+calFibonacci(n+1));// Calling the
recursive method
scan.close();
}
//method to calculate Fibonacci number recursively
public static long calFibonacci(int index){
if (index == 1)
return 0;
if (index == 2)
return 1;
return calFibonacci(index - 1) + calFibonacci(index - 2);
}
}
/** Outputs
* Enter the value of n 25
The 25. Fibonacci number is 75025
*/
/******************************using iterative
method************************************/
/**
* 1.2.2 Java program to calculate the fibonacci number using iterative method
*/
import java.util.Scanner;
public class Fibo2 {
public static void main(String args[]){
Scanner scan= new Scanner(System.in); //Scanner object to read from the user
int n;
System.out.println(\"Enter the value of n\");
n=scan.nextInt();
System.out.println(\"The \"+n+\". Fibonacci number is \"+calFibonacci(n+1)); // Calling the
iterative method to calculate fibonacci number
scan.close();
}
public static long calFibonacci(int index){
int first=0;
int second=1;
for(int i=2;i<=index;i++){
int temp=first;
first=first+second;
second=temp;
}
return first;
}
}
/** Output
*
* Enter the value of n 25
The 25. Fibonacci number is 75025
*
*/
/**************************************************/
The efficient method is the iterative way of calculating the number
Explaination
In recursion method we calculate the the numbers from current index till 0. So for each index we
do calculate the fibonacci number again and again. Hence it takes a lot time to calculate the
answers. Whereas in iterative method we do not rework like recursion. Here we store the
previous two numbers(first and second in above code) and calculate the next one. So its a
efficient way to calculate the number. The time complexity in this will be O(n).
Thanks a lot. Please feel free to ask doubts if you have any. God bless you.
Solution
/******************* using recursive method **********************/
/**
* 1.2.1 Java program to calculate the Fibonacci Number using recursion
*/
import java.util.Scanner;
public class Fibo {
public static void main(String args[]){
Scanner scan= new Scanner(System.in); //Scanner object to read from the user
int n;
System.out.println(\"Enter the value of n\");
n=scan.nextInt();
System.out.println(\"The \"+n+\". Fibonacci number is \"+calFibonacci(n+1));// Calling the
recursive method
scan.close();
}
//method to calculate Fibonacci number recursively
public static long calFibonacci(int index){
if (index == 1)
return 0;
if (index == 2)
return 1;
return calFibonacci(index - 1) + calFibonacci(index - 2);
}
}
/** Outputs
* .
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This document contains notes on Group 17 (halogen) elements from a Class 12 chemistry class. It discusses various properties and reactions of halogens like fluorine and chlorine. Key points include: halogens are strong oxidizing agents due to their high electronegativity and ability to readily accept electrons; fluorine forms only one oxoacid (HOF) due to the absence of d-orbitals; oxygen can form hydrogen bonds due to its smaller size compared to chlorine; and fluorine is a stronger oxidizing agent than chlorine despite having a less negative electron gain enthalpy, due to its lower bond dissociation energy and higher hydration enthalpy.
G 17
This document contains notes on Group 17 (halogen) elements from a Class 12 chemistry class. It discusses various properties and reactions of halogens like fluorine and chlorine. Key points include: halogens are strong oxidizing agents due to their high electronegativity and ability to readily accept electrons; fluorine forms only one oxoacid (HOF) due to the absence of d-orbitals; oxygen can form hydrogen bonds due to its smaller size compared to chlorine; and fluorine is a stronger oxidizing agent than chlorine despite having a less negative electron gain enthalpy, due to its lower bond dissociation energy and higher hydration enthalpy.
Ionic equilibrium
Ionic Equilibrium
Powerpoint presentation for the topics covered under Ionic Equilibrium in class 11.
Hydrogen
This document discusses isotopes of hydrogen, including their properties and natural occurrences. It specifically covers the three main isotopes:
1) Protium (1H1) is the most common isotope of hydrogen, making up 99.98% of naturally occurring hydrogen. It has one proton and no neutrons.
2) Deuterium (1H2) is a stable, non-radioactive isotope that has one proton and one neutron. It occurs naturally at 0.0156%.
3) Tritium (1H3) is a radioactive isotope that has one proton and two neutrons. It occurs naturally at only 4.0 x 10-15%.
The document also discusses the
Hydrogen Class 11th Chemistry NCERT
Hydrogen is the first element in the periodic table. It can form positive ions like alkali metals and negative ions like halogens. It exists as diatomic molecules (H2). There are three isotopes of hydrogen: protium, deuterium, and tritium. Dihydrogen is produced commercially through electrolysis and from the reaction of steam with hydrocarbons. Dihydrogen reacts with many elements and is used to produce ammonia and other important compounds. It forms ionic hydrides with electropositive metals, covalent hydrides with nonmetals, and metallic hydrides with transition metals.
Ch 17sec3
This document summarizes key concepts about aqueous solutions, including:
- Solutions contain a solvent and one or more solutes
- Ionic compounds generally dissolve well in water due to ion separation through solvation
- "Like dissolves like" - polar substances dissolve in water and nonpolar substances do not
- Electrolytes conduct electricity in solution while nonelectrolytes do not
- Hydrates contain water within their crystalline structure.
basics-of-ion-exchange(3)
This document discusses ion exchange and how it is used to produce deionized water. Ion exchange resins remove ions from water by exchanging them for hydrogen or hydroxyl ions. Cation resin has a negative charge and removes positively charged ions by exchanging them for hydrogen ions. Anion resin has a positive charge and removes negatively charged ions by exchanging them for hydroxyl ions. Together this produces pure water without ions. The resins are regenerated using acids and bases to restore the exchange sites.
Bases & alkalis
This document defines bases and alkalis and discusses their properties. Bases are substances that accept hydrogen ions, while alkalis are soluble bases that release hydroxide ions in water and have a pH greater than 7. Alkalis have a bitter taste, turn litmus paper blue, are caustic, and conduct electricity in solution. They can undergo neutralization reactions with acids and precipitation reactions with metal salts to form insoluble metal hydroxides.
ACID BASES AND SALTS.ppt
This document discusses acids, bases, and salts. It explains that acids contain hydrogen ions and bases contain hydroxide ions. When an acid and base react, they form water and a salt through a neutralization reaction. Salts are formed when hydrogen ions in an acid are replaced by metal ions. There are different types of salts including normal salts, acid salts, and basic salts depending on the reaction that forms them. Salts have various uses such as in fertilizers, drying agents, medicines, and gunpowder.
Alkali
The document discusses the definitions and properties of bases and alkalis. It provides examples of metal oxides and metal hydroxides that are bases, including copper(II) oxide, magnesium oxide, sodium hydroxide, and potassium hydroxide. It defines an alkali as a base that is soluble in water, and explains that alkalis ionize in water to produce hydroxide ions. When testing an alkali with red litmus paper, the paper will turn blue. Alkalis also have a bitter taste and slippery feel, and a pH value greater than 7.
10 s-Block Elements.ppt
This document provides information about the s-block elements lithium (Li) through francium (Fr) and the alkaline earth metals beryllium (Be) through radium (Ra). It discusses their electronic configurations, atomic and ionic radii, ionization energies, hydration enthalpies, physical properties, and important compounds such as oxides, hydroxides, halides, and salts. It notes the similarities and differences between lithium and other alkali metals, as well as the similarities between lithium and magnesium. The biological importance of sodium and potassium is also mentioned.
ACID BASES AND SALTS.ppt
The document discusses acids, bases, and salts. It explains that acids contain hydrogen ions and bases contain hydroxide ions. When an acid and base react, they form water and a salt. Salts are formed when the hydrogen ions in an acid are replaced by metal ions from a base. There are different types of salts including normal salts, acid salts, and basic salts depending on the reaction. Common salts have various uses including in fertilizers, batteries, drying agents, and soap making.
ACID BASES AND SALTS.ppt
The document discusses acids, bases, and salts. It explains that acids contain hydrogen ions and bases contain hydroxide ions. When an acid and base react, they form water and a salt through a neutralization reaction. Salts are formed when hydrogen ions in an acid are replaced by metal ions. There are different types of salts including normal salts, acid salts, and basic salts depending on the reaction that forms them. Common salts have various uses including in fertilizers, drying agents, medicines, and other applications.
Chapter 19
The document discusses acids, bases, and the ion product constant of water. It defines acids as substances that produce hydrogen ions (H+) in water and bases as those that produce hydroxide ions (OH-). Water undergoes self-ionization in solution, producing a small concentration of H+ and OH- ions such that their product is 1 x 10-14 at 25°C. Solutions with higher [H+] than [OH-] are acidic, while those with higher [OH-] are basic.
Electrolyte
Electrolytes are substances that can conduct electricity when molten or dissolved in water, as they dissociate into ions that are free to move. Non-electrolytes do not contain ions and cannot conduct electricity either when molten or dissolved. While solid ionic compounds cannot conduct due to their tightly bound ions, they can conduct in molten or aqueous states as their ions gain freedom of movement.
9 Aqueous Solutions
- Water is a polar solvent due to its molecular structure, with oxygen having a partial negative charge and hydrogen having partial positive charges. This allows it to dissolve ionic compounds by interacting with and separating the ions.
- Ionic compounds dissolve to varying degrees in water depending on how strongly the ions are attracted to each other versus water molecules. Solubility can be measured in g/L.
- Acids donate H+ ions in water and are classified as strong or weak based on how completely they ionize. Their strength affects pH calculations. Bases accept H+ and similarly ionize more or less completely.
Similar to a. formic acid (it has more polar components than cyclohexane, inclu.pdf (20)
More from vichu19891
1. Copper is above silver in the activity series. Thus Cu metal will.pdf
1. Copper is above silver in the activity series. Thus Cu metal will displace Ag+ ions from
solution, forming Cu2+ ions and metallic Ag.
Observation: The reddish brown copper metal dissolves and silver metal precipitates out. The
colorless Ag+ solution turns blue due to Cu2+ ions formed.
2. Molecular equation:
Cu(s) + 2 AgNO3(aq) => 2 Ag(s) + Cu(NO3)2(aq)
3. Complete ionic equation (obtained by expanding soluble ions in molecular equation):
Cu(s) + 2 Ag+(aq) + 2 NO3-(aq) => 2 Ag(s) + Cu2+(aq) + 2 NO3-(aq)
4. Net ionic equation (obtained by cancelling common ions in complete ionic equation):
Cu(s) + 2 Ag+(aq) => 2 Ag(s) + Cu2+(aq)
Solution
1. Copper is above silver in the activity series. Thus Cu metal will displace Ag+ ions from
solution, forming Cu2+ ions and metallic Ag.
Observation: The reddish brown copper metal dissolves and silver metal precipitates out. The
colorless Ag+ solution turns blue due to Cu2+ ions formed.
2. Molecular equation:
Cu(s) + 2 AgNO3(aq) => 2 Ag(s) + Cu(NO3)2(aq)
3. Complete ionic equation (obtained by expanding soluble ions in molecular equation):
Cu(s) + 2 Ag+(aq) + 2 NO3-(aq) => 2 Ag(s) + Cu2+(aq) + 2 NO3-(aq)
4. Net ionic equation (obtained by cancelling common ions in complete ionic equation):
Cu(s) + 2 Ag+(aq) => 2 Ag(s) + Cu2+(aq).
using recursive method .pdf
/******************* using recursive method **********************/
/**
* 1.2.1 Java program to calculate the Fibonacci Number using recursion
*/
import java.util.Scanner;
public class Fibo {
public static void main(String args[]){
Scanner scan= new Scanner(System.in); //Scanner object to read from the user
int n;
System.out.println(\"Enter the value of n\");
n=scan.nextInt();
System.out.println(\"The \"+n+\". Fibonacci number is \"+calFibonacci(n+1));// Calling the
recursive method
scan.close();
}
//method to calculate Fibonacci number recursively
public static long calFibonacci(int index){
if (index == 1)
return 0;
if (index == 2)
return 1;
return calFibonacci(index - 1) + calFibonacci(index - 2);
}
}
/** Outputs
* Enter the value of n 25
The 25. Fibonacci number is 75025
*/
/******************************using iterative
method************************************/
/**
* 1.2.2 Java program to calculate the fibonacci number using iterative method
*/
import java.util.Scanner;
public class Fibo2 {
public static void main(String args[]){
Scanner scan= new Scanner(System.in); //Scanner object to read from the user
int n;
System.out.println(\"Enter the value of n\");
n=scan.nextInt();
System.out.println(\"The \"+n+\". Fibonacci number is \"+calFibonacci(n+1)); // Calling the
iterative method to calculate fibonacci number
scan.close();
}
public static long calFibonacci(int index){
int first=0;
int second=1;
for(int i=2;i<=index;i++){
int temp=first;
first=first+second;
second=temp;
}
return first;
}
}
/** Output
*
* Enter the value of n 25
The 25. Fibonacci number is 75025
*
*/
/**************************************************/
The efficient method is the iterative way of calculating the number
Explaination
In recursion method we calculate the the numbers from current index till 0. So for each index we
do calculate the fibonacci number again and again. Hence it takes a lot time to calculate the
answers. Whereas in iterative method we do not rework like recursion. Here we store the
previous two numbers(first and second in above code) and calculate the next one. So its a
efficient way to calculate the number. The time complexity in this will be O(n).
Thanks a lot. Please feel free to ask doubts if you have any. God bless you.
Solution
/******************* using recursive method **********************/
/**
* 1.2.1 Java program to calculate the Fibonacci Number using recursion
*/
import java.util.Scanner;
public class Fibo {
public static void main(String args[]){
Scanner scan= new Scanner(System.in); //Scanner object to read from the user
int n;
System.out.println(\"Enter the value of n\");
n=scan.nextInt();
System.out.println(\"The \"+n+\". Fibonacci number is \"+calFibonacci(n+1));// Calling the
recursive method
scan.close();
}
//method to calculate Fibonacci number recursively
public static long calFibonacci(int index){
if (index == 1)
return 0;
if (index == 2)
return 1;
return calFibonacci(index - 1) + calFibonacci(index - 2);
}
}
/** Outputs
* .
a. Population - families in the state of Florida b. Variable .pdf
a. Population - families in the state of Florida
b. Variable measured - number of children per family
c. Level of measurement - ratio
d. 1.Simple Random Sample
Solution
a. Population - families in the state of Florida
b. Variable measured - number of children per family
c. Level of measurement - ratio
d. 1.Simple Random Sample.
there is no reaction between HNO3 and KCl. S.pdf
There is no reaction observed between HNO3 and KCl in solution. The experiment showed that nitric acid and potassium chloride do not undergo a chemical reaction when combined.
The thickness of non-saturated zone and physico-c.pdf
The thickness of non-saturated zone and physico-chemical conditions are important
parameters to assess the impact of infiltration ponds on water resources with respect to heavy
metals transfer. As changes in physico-chemical parameters of solutions have a strong impact on
the mobility of colloidal phases in sediments and soils, the colloidal facilitated transfer of heavy
metals has to be investigated. Therefore, this study focuses on the characterization of runoff,
surface and interstitial waters in a retention/infiltration pond collecting runoff waters of a bridge
near Nantes. Physico-chemical parameters and chemical analyses were performed on the waters
during about one year. The separation of dissolved and colloidal fractions was carried out by
filtration and ultrafiltration for one sample of surface and interstitial waters. Until now, the
runoff waters were only filtered at 0.45 microm. The comparison of physico-chemical data
shows that the minor variations of runoff water parameters are mitigated in basin and in soils but
strong variations impact the composition of interstitial waters. High concentrations of zinc,
copper and still of lead are measured in runoff. Lead and cadmium seem to be associated to
colloidal and particulate fractions while zinc, copper, nickel and chromium are distributed in all
fractions.
Solution
The thickness of non-saturated zone and physico-chemical conditions are important
parameters to assess the impact of infiltration ponds on water resources with respect to heavy
metals transfer. As changes in physico-chemical parameters of solutions have a strong impact on
the mobility of colloidal phases in sediments and soils, the colloidal facilitated transfer of heavy
metals has to be investigated. Therefore, this study focuses on the characterization of runoff,
surface and interstitial waters in a retention/infiltration pond collecting runoff waters of a bridge
near Nantes. Physico-chemical parameters and chemical analyses were performed on the waters
during about one year. The separation of dissolved and colloidal fractions was carried out by
filtration and ultrafiltration for one sample of surface and interstitial waters. Until now, the
runoff waters were only filtered at 0.45 microm. The comparison of physico-chemical data
shows that the minor variations of runoff water parameters are mitigated in basin and in soils but
strong variations impact the composition of interstitial waters. High concentrations of zinc,
copper and still of lead are measured in runoff. Lead and cadmium seem to be associated to
colloidal and particulate fractions while zinc, copper, nickel and chromium are distributed in all
fractions..
The folding process of proteins is hierarchical, .pdf
The folding process of proteins is hierarchical, with secondary structures forming
before tertiary structures
Solution
The folding process of proteins is hierarchical, with secondary structures forming
before tertiary structures.
Step1 ppt of PbCl2 are soluble in hot water. Ste.pdf
Step1 ppt of PbCl2 are soluble in hot water. Step2 Pb +2 is present in S1 and P1
Step3 Hg and Ag+1 are absent. Step4 Pb+2(aq)+ 2Cl-(aq)---->PbCl2(s)
Solution
Step1 ppt of PbCl2 are soluble in hot water. Step2 Pb +2 is present in S1 and P1
Step3 Hg and Ag+1 are absent. Step4 Pb+2(aq)+ 2Cl-(aq)---->PbCl2(s).
Should take off H from SH group, forming RS-Na+ s.pdf
Should take off H from SH group, forming RS-Na+ salt. note: both S and O are
group 16 elements. However, S is in the third row and O is in the second row. THe atomic size
of S is bigger than that of O, thus S-H bond is weaker than O-H bond. The H in SH is more
acidic than that in OH. the acidity of H: SH > OH > NH > CH
Solution
Should take off H from SH group, forming RS-Na+ salt. note: both S and O are
group 16 elements. However, S is in the third row and O is in the second row. THe atomic size
of S is bigger than that of O, thus S-H bond is weaker than O-H bond. The H in SH is more
acidic than that in OH. the acidity of H: SH > OH > NH > CH.
Rest are okay but B) I think should be vanderwall.pdf
Rest are okay but B) I think should be vanderwall\'s interaction rather than dipole-
dipole. pls ponder over.
Solution
Rest are okay but B) I think should be vanderwall\'s interaction rather than dipole-
dipole. pls ponder over..
pH =4.217 pH = - log(concentration of H+) = -log .pdf
pH =4.217 pH = - log(concentration of H+) = -log (6.06E-05) = 4.217
Solution
pH =4.217 pH = - log(concentration of H+) = -log (6.06E-05) = 4.217.
Liquids may change to a vapor at temperatures bel.pdf
Liquids may change to a vapor at temperatures below their boiling points through
the process of evaporation. Evaporation is a surface phenomenon in which molecules located
near the liquid\'s edge, not contained by enough liquid pressure on that side, escape into the
surroundings as vapor. On the other hand, boiling is a process in which molecules anywhere in
the liquid escape, resulting in the formation of vapor bubbles within the liquid.
Solution
Liquids may change to a vapor at temperatures below their boiling points through
the process of evaporation. Evaporation is a surface phenomenon in which molecules located
near the liquid\'s edge, not contained by enough liquid pressure on that side, escape into the
surroundings as vapor. On the other hand, boiling is a process in which molecules anywhere in
the liquid escape, resulting in the formation of vapor bubbles within the liquid..
intra extra equilibrium potential mEqL mEqL 1.pdf
intra extra equilibrium potential mEq/L mEq/L 12 142 61.77 135 4 -87.97 0.5 *
10^-4 0.75 * 10^-4 120.19 4.5 112 80.36 equlibrium potential = RT/ZF ln(Xo/Xi) =25 mV
ln(Xo/Xi) mEq/L=(mMoles/L) / valency
Solution
intra extra equilibrium potential mEq/L mEq/L 12 142 61.77 135 4 -87.97 0.5 *
10^-4 0.75 * 10^-4 120.19 4.5 112 80.36 equlibrium potential = RT/ZF ln(Xo/Xi) =25 mV
ln(Xo/Xi) mEq/L=(mMoles/L) / valency.
Which of the following forms of DES is considered the most vulnerabl.pdf
Which of the following forms of DES is considered the most vulnerable to attack?
A. CBC
B. ECB
C. CFB
D. OFB
Solution
Which of the following forms of DES is considered the most vulnerable to attack?
A. CBC
B. ECB
C. CFB
D. OFB.
This is actually pretty simple, so Ill help explain. Take a gi.pdf
This is actually pretty simple, so I\'ll help explain.
Take a given hydrogen atom, but we\'ll use an ionic form that has 5 electrons (as you\'ve stated).
Now, for the quantum numbers.
The first one, N, is the electron shell in question. Shells work as providing spaces for electron
density. The first shell, in which N=1, holds two electrons because it only contains one orbital.
N=2 holds two orbitals, which holds four electrons. Combined with N=1, an ion in N=2 has
anywhere between 3 (the minimum of N=1, +1 electron) and 8 electrons (the maximum number
of electrons before entering the third shell. Because the ion in question has 5 electrons, its N
value is 2.
A simple way to calculate the maximum number of electrons in a given atom that occupies a
specific shell is to take that shell number, N, square it and multiply the result by two. That is why
the maximum quantum number N for an atom whose electrons occupy the third shell is 18
(which is (3^2)2).
Onto the next quantum number, which is l. If you understand the order of atomic shells and their
orbitals, this is a simple one.
The first shell is l=0, which is the \"s\" shell of an atom. s shells can only hold one atomic
orbital, which is two electrons. We denote this as \"1s2\". The next shell is 2s, and because we
can fill it, it become 2s2.Because we have 5 electrons, we need to keep going up. The next shell
is l=1, which is the \"p\" shell. That fits our last electron, so your value for the l quantum number
is \"1.\"
Now we\'re at the ml quantum number. These are the numbers we assign to the specific orbitals
in which electrons reside in, within a given shell. We\'re currently in the p shell, which holds 3
orbitals whose names are -1, 0, and +1. Unless you have a picture of the electron shells, we can
only guess that the 5th electron is in any one of these, so the safe bet is to put it into the -1 orbital
(though it can really be in any of them).
Lastly, the ms quantum number. This denotes the spin on the electron. Up spin is +1/2, whilst
down spin is -1/2. Unless you have a picture, I can\'t tell you which way it is spinning, but it\'s a
safe guess to say that it\'s up.
In sum, your quantum numbers as: 2, 1, 0, (1/2)
I hope that helps. (: Feel free to ask any questions.
Solution
This is actually pretty simple, so I\'ll help explain.
Take a given hydrogen atom, but we\'ll use an ionic form that has 5 electrons (as you\'ve stated).
Now, for the quantum numbers.
The first one, N, is the electron shell in question. Shells work as providing spaces for electron
density. The first shell, in which N=1, holds two electrons because it only contains one orbital.
N=2 holds two orbitals, which holds four electrons. Combined with N=1, an ion in N=2 has
anywhere between 3 (the minimum of N=1, +1 electron) and 8 electrons (the maximum number
of electrons before entering the third shell. Because the ion in question has 5 electrons, its N
value is 2.
A simple way to calculate the maximum number of electrons.
There are many test IPv6 networks deployed across the world. For act.pdf
There are many test IPv6 networks deployed across the world. For actual deployment, however,
all the companies need to ensure that the vendors who support companys network have the
requisite IPv6 enhancements.
There are two categories of IPv6 enhancements. The first is the set that supports the packet
forwarding (more commonly referred to as routing) process and the other set comprises
enhancements that support the computing or host infrastructure.
IPv6 enhancements of the first category include larger address formats (the ones that affect the
routing table size and structure), better routing protocols such as Open Shortest First Protocol
(OSPF) and Routing Information Protocol (RIP), and good support for optional extension
headers (which streamline the packet forwarding process) such as the Routing Header. And, the
second category of enhancements comprises enhancements to the Domain Name System (DNS),
the Stateless Auto-configuration (plug and play) process, upgraded Security, and updates to the
Application Programming Interfaces (APIs).
Keeping these requisite enhancements in mind, let us now discuss what kind of support ten of
the premier networking vendors are equipped to provide:
Solution
There are many test IPv6 networks deployed across the world. For actual deployment, however,
all the companies need to ensure that the vendors who support companys network have the
requisite IPv6 enhancements.
There are two categories of IPv6 enhancements. The first is the set that supports the packet
forwarding (more commonly referred to as routing) process and the other set comprises
enhancements that support the computing or host infrastructure.
IPv6 enhancements of the first category include larger address formats (the ones that affect the
routing table size and structure), better routing protocols such as Open Shortest First Protocol
(OSPF) and Routing Information Protocol (RIP), and good support for optional extension
headers (which streamline the packet forwarding process) such as the Routing Header. And, the
second category of enhancements comprises enhancements to the Domain Name System (DNS),
the Stateless Auto-configuration (plug and play) process, upgraded Security, and updates to the
Application Programming Interfaces (APIs).
Keeping these requisite enhancements in mind, let us now discuss what kind of support ten of
the premier networking vendors are equipped to provide:.
The three ways of presenting the changes in the balance of the Compr.pdf
The three ways of presenting the changes in the balance of the Comprehensive Income
(Cumulative) are :
(i) Adding the value to the Stockholders\' Equity Balance.
(ii) Showing the credit balance as a separate item in the Reserves and Surplus section of the
Liabilities side.
(iii) Showing a debit balance in the comprehensive income as an item in the Assets Side of the
Balance Sheet with the heading \"Debit Balance in Profit and Loss Account\".
Solution
The three ways of presenting the changes in the balance of the Comprehensive Income
(Cumulative) are :
(i) Adding the value to the Stockholders\' Equity Balance.
(ii) Showing the credit balance as a separate item in the Reserves and Surplus section of the
Liabilities side.
(iii) Showing a debit balance in the comprehensive income as an item in the Assets Side of the
Balance Sheet with the heading \"Debit Balance in Profit and Loss Account\"..
The RASopathies are a group of genetic syndromes caused by germline .pdf
The RASopathies are a group of genetic syndromes caused by germline mutations in genes that
alter the Ras sub family and Mitogen activated protein kinases that control signal transduction.
Some of these syndromes are neurofibromatosis type 1, Noonan syndrome, Costello syndrome,
cardio-facio-cutaneous syndrome, LEOPARD syndrome and Legius syndrome. RAS opathies
and X-chromosomes are not linked.
Solution
The RASopathies are a group of genetic syndromes caused by germline mutations in genes that
alter the Ras sub family and Mitogen activated protein kinases that control signal transduction.
Some of these syndromes are neurofibromatosis type 1, Noonan syndrome, Costello syndrome,
cardio-facio-cutaneous syndrome, LEOPARD syndrome and Legius syndrome. RAS opathies
and X-chromosomes are not linked..
ThanksSolutionThanks.pdf
This very short document simply contains the word "Thanks" repeated three times. It does not provide much substantive information that can be summarized in 3 sentences or less. The document appears to be acknowledging a solution or response to something but without any other context or details, the intent and meaning cannot be clearly determined from the text alone.
Step1 In O2 ; we have 2 unpaired electrons which occupy pi 2p Anti.pdf
Step1 In O2 ; we have 2 unpaired electrons which occupy pi 2p * Anti bonding orbital. Hence
O2 molecule is paramagnetic.
Step2 C2^- has 1 unpaired electron; is also paramagnetic.
Solution
Step1 In O2 ; we have 2 unpaired electrons which occupy pi 2p * Anti bonding orbital. Hence
O2 molecule is paramagnetic.
Step2 C2^- has 1 unpaired electron; is also paramagnetic..
Quartzite is sandstone that has been converted to a solid quartz roc.pdf
Quartzite is sandstone that has been converted to a solid quartz rock.
It formed when silica is subjected to great heat and pressure. The individual silica particles melt
and glue together to form quartizite.
Solution
Quartzite is sandstone that has been converted to a solid quartz rock.
It formed when silica is subjected to great heat and pressure. The individual silica particles melt
and glue together to form quartizite..
More from vichu19891 (20)
1. Copper is above silver in the activity series. Thus Cu metal will.pdf
1. Copper is above silver in the activity series. Thus Cu metal will.pdf
a. Population - families in the state of Florida b. Variable .pdf
a. Population - families in the state of Florida b. Variable .pdf
The thickness of non-saturated zone and physico-c.pdf
The thickness of non-saturated zone and physico-c.pdf
The folding process of proteins is hierarchical, .pdf
The folding process of proteins is hierarchical, .pdf
Step1 ppt of PbCl2 are soluble in hot water. Ste.pdf
Step1 ppt of PbCl2 are soluble in hot water. Ste.pdf
Should take off H from SH group, forming RS-Na+ s.pdf
Should take off H from SH group, forming RS-Na+ s.pdf
Rest are okay but B) I think should be vanderwall.pdf
Rest are okay but B) I think should be vanderwall.pdf
pH =4.217 pH = - log(concentration of H+) = -log .pdf
pH =4.217 pH = - log(concentration of H+) = -log .pdf
Liquids may change to a vapor at temperatures bel.pdf
Liquids may change to a vapor at temperatures bel.pdf
Which of the following forms of DES is considered the most vulnerabl.pdf
Which of the following forms of DES is considered the most vulnerabl.pdf
This is actually pretty simple, so Ill help explain. Take a gi.pdf
This is actually pretty simple, so Ill help explain. Take a gi.pdf
There are many test IPv6 networks deployed across the world. For act.pdf
There are many test IPv6 networks deployed across the world. For act.pdf
The three ways of presenting the changes in the balance of the Compr.pdf
The three ways of presenting the changes in the balance of the Compr.pdf
The RASopathies are a group of genetic syndromes caused by germline .pdf
The RASopathies are a group of genetic syndromes caused by germline .pdf
Step1 In O2 ; we have 2 unpaired electrons which occupy pi 2p Anti.pdf
Step1 In O2 ; we have 2 unpaired electrons which occupy pi 2p Anti.pdf
Quartzite is sandstone that has been converted to a solid quartz roc.pdf
Quartzite is sandstone that has been converted to a solid quartz roc.pdf
Recently uploaded
LAND USE LAND COVER AND NDVI OF MIRZAPUR DISTRICT, UP
This Dissertation explores the particular circumstances of Mirzapur, a region located in the
core of India. Mirzapur, with its varied terrains and abundant biodiversity, offers an optimal
environment for investigating the changes in vegetation cover dynamics. Our study utilizes
advanced technologies such as GIS (Geographic Information Systems) and Remote sensing to
analyze the transformations that have taken place over the course of a decade.
The complex relationship between human activities and the environment has been the focus
of extensive research and worry. As the global community grapples with swift urbanization,
population expansion, and economic progress, the effects on natural ecosystems are becoming
more evident. A crucial element of this impact is the alteration of vegetation cover, which plays a
significant role in maintaining the ecological equilibrium of our planet.Land serves as the foundation for all human activities and provides the necessary materials for
these activities. As the most crucial natural resource, its utilization by humans results in different
'Land uses,' which are determined by both human activities and the physical characteristics of the
land.
The utilization of land is impacted by human needs and environmental factors. In countries
like India, rapid population growth and the emphasis on extensive resource exploitation can lead
to significant land degradation, adversely affecting the region's land cover.
Therefore, human intervention has significantly influenced land use patterns over many
centuries, evolving its structure over time and space. In the present era, these changes have
accelerated due to factors such as agriculture and urbanization. Information regarding land use and
cover is essential for various planning and management tasks related to the Earth's surface,
providing crucial environmental data for scientific, resource management, policy purposes, and
diverse human activities.
Accurate understanding of land use and cover is imperative for the development planning
of any area. Consequently, a wide range of professionals, including earth system scientists, land
and water managers, and urban planners, are interested in obtaining data on land use and cover
changes, conversion trends, and other related patterns. The spatial dimensions of land use and
cover support policymakers and scientists in making well-informed decisions, as alterations in
these patterns indicate shifts in economic and social conditions. Monitoring such changes with the
help of Advanced technologies like Remote Sensing and Geographic Information Systems is
crucial for coordinated efforts across different administrative levels. Advanced technologies like
Remote Sensing and Geographic Information Systems
9
Changes in vegetation cover refer to variations in the distribution, composition, and overall
structure of plant communities across different temporal and spatial scales. These changes can
occur natural.
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Chapter wise All Notes of First year Basic Civil Engineering.pptx
Chapter wise All Notes of First year Basic Civil Engineering
Syllabus
Chapter-1
Introduction to objective, scope and outcome the subject
Chapter 2
Introduction: Scope and Specialization of Civil Engineering, Role of civil Engineer in Society, Impact of infrastructural development on economy of country.
Chapter 3
Surveying: Object Principles & Types of Surveying; Site Plans, Plans & Maps; Scales & Unit of different Measurements.
Linear Measurements: Instruments used. Linear Measurement by Tape, Ranging out Survey Lines and overcoming Obstructions; Measurements on sloping ground; Tape corrections, conventional symbols. Angular Measurements: Instruments used; Introduction to Compass Surveying, Bearings and Longitude & Latitude of a Line, Introduction to total station.
Levelling: Instrument used Object of levelling, Methods of levelling in brief, and Contour maps.
Chapter 4
Buildings: Selection of site for Buildings, Layout of Building Plan, Types of buildings, Plinth area, carpet area, floor space index, Introduction to building byelaws, concept of sun light & ventilation. Components of Buildings & their functions, Basic concept of R.C.C., Introduction to types of foundation
Chapter 5
Transportation: Introduction to Transportation Engineering; Traffic and Road Safety: Types and Characteristics of Various Modes of Transportation; Various Road Traffic Signs, Causes of Accidents and Road Safety Measures.
Chapter 6
Environmental Engineering: Environmental Pollution, Environmental Acts and Regulations, Functional Concepts of Ecology, Basics of Species, Biodiversity, Ecosystem, Hydrological Cycle; Chemical Cycles: Carbon, Nitrogen & Phosphorus; Energy Flow in Ecosystems.
Water Pollution: Water Quality standards, Introduction to Treatment & Disposal of Waste Water. Reuse and Saving of Water, Rain Water Harvesting. Solid Waste Management: Classification of Solid Waste, Collection, Transportation and Disposal of Solid. Recycling of Solid Waste: Energy Recovery, Sanitary Landfill, On-Site Sanitation. Air & Noise Pollution: Primary and Secondary air pollutants, Harmful effects of Air Pollution, Control of Air Pollution. . Noise Pollution Harmful Effects of noise pollution, control of noise pollution, Global warming & Climate Change, Ozone depletion, Greenhouse effect
Text Books:
1. Palancharmy, Basic Civil Engineering, McGraw Hill publishers.
2. Satheesh Gopi, Basic Civil Engineering, Pearson Publishers.
3. Ketki Rangwala Dalal, Essentials of Civil Engineering, Charotar Publishing House.
4. BCP, Surveying volume 1
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