Solution1.cpp#include iostreamheader for input output function.pdf
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Solution 1.cpp #include //header for input output function using namespace std;//it tells the compiler to link std name space int main() {//main function int n=1,m=-1; if(n<-m)//if else statement { cout<.
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O = O bond is shortest Oxygen electronegativity d.pdf
O = O bond is shortest Oxygen electronegativity difference is higher and also it is
double bond
Solution
O = O bond is shortest Oxygen electronegativity difference is higher and also it is
double bond.
NaCl Na is the symbol for Sodium which bonds to .pdf
NaCl Na is the symbol for Sodium which bonds to Cl which is the symbol for
Chlorine making Sodium Chloride.
Solution
NaCl Na is the symbol for Sodium which bonds to Cl which is the symbol for
Chlorine making Sodium Chloride..
NiCl2 because it not dissociate .pdf
NiCl2 because it not dissociate
Solution
NiCl2 because it not dissociate.
iodide, I-, has the -1 oxidation state. iodine (I.pdf
iodide, I-, has the -1 oxidation state. iodine (I) has 0 oxidation state. I- - e ---> I
Thus, this change is an oxidation reaction.
Solution
iodide, I-, has the -1 oxidation state. iodine (I) has 0 oxidation state. I- - e ---> I
Thus, this change is an oxidation reaction..
Lithium is in group 1 on the peridic table. This .pdf
Lithium is in group 1 on the peridic table. This means that it has only 1 valence
electron. Lithium will tend to lose that electron when it ionizes and become an ion. Remeber, an
ion is any atom, or sometimes molecule, with a charge. When Li loses the electron, it loses one
of its negative charges so the atom becomes an ion with a +1 charge because it now has 3
positively charged protons and only 2 negatively charged electrons.
Solution
Lithium is in group 1 on the peridic table. This means that it has only 1 valence
electron. Lithium will tend to lose that electron when it ionizes and become an ion. Remeber, an
ion is any atom, or sometimes molecule, with a charge. When Li loses the electron, it loses one
of its negative charges so the atom becomes an ion with a +1 charge because it now has 3
positively charged protons and only 2 negatively charged electrons..
Systematic positionKingdom PlantaeSubkingdom Tracheobionta (Vas.pdf
Systematic position:
Kingdom Plantae
Subkingdom Tracheobionta (Vascular plants)
Superdivision Spermatophyta (Seed plants)
Division Gnetophyta (Mormon tea and other gnetophytes)
Class Gnetopsida
Order Ephedrales
Family Ephedraceae (tea family)
Genus Ephedra L.
Common name Jointfir
There are 35 to 40 species in the world. These are wide spread in both in eastern and western
hemispheres
Morphology:
A. Root : Branched root, tap root are present these are characteristic xerophytes, no report of
mycorrhize.
B. Stem: Woody green branched distinctly joined apparently nodes and inter nodes, branches are
photosynthetic. Inter nodes are elongated by the activity of basal meristem.
C. Leaves: Small scaly leaves are present, scale leaves are joined to form sheath, Brown leaves
are non photosynthetic leaves.
Anatomy:
A. Root : It shows usually Gymnosperm structure. Cortex, endoderm, inner periderm, vascular
camblum, primary and secondary xylem was present.
B. Stem:
1. Primary structure: Transverse structure shows ridges and grooves. cortex clorenchyma
elongated, endoderm is present in single layer.
2. Secondary structure: Parenchymalous cells present between the vascular bundles to form a
ring of cambium, ray initials and fusiform initial are present.
C. Leaves: Transverse structure of leaves shows epidermis, cuticle, xylem, phloem, air space,
spongy parenchyma tissues. The vascular bundles are collateral closed and primary phloem and
primary xylem are also present.
Adaptations:
Mostly xerophytes scrubs or woody climbers Ephedra Pedunculata. Most species of ephedra are
branched scrubs, commonly climbing over vegetation many species spread by rhizomes Ephedra
Trifura. Under favorable condition it grows tree like habitat.
Ephedra Foliata - scrambling scrub, Ephedra Compacta - Branching type, Ephedra
Camphylophoda - Spendulus branches, Ephedra Intermedia - Shoot, Ephedra Saxatilis - scrub.
Values and functions:
In pharmaceutical industries it works as to access Ephedra drug materials.
Traditionally it is used by indigenous people for medicinal treatments of asthma, hay fever and
cold.
Alkaloids obtained from the species of Ephedra is used in herbal medicine.
Solution
Systematic position:
Kingdom Plantae
Subkingdom Tracheobionta (Vascular plants)
Superdivision Spermatophyta (Seed plants)
Division Gnetophyta (Mormon tea and other gnetophytes)
Class Gnetopsida
Order Ephedrales
Family Ephedraceae (tea family)
Genus Ephedra L.
Common name Jointfir
There are 35 to 40 species in the world. These are wide spread in both in eastern and western
hemispheres
Morphology:
A. Root : Branched root, tap root are present these are characteristic xerophytes, no report of
mycorrhize.
B. Stem: Woody green branched distinctly joined apparently nodes and inter nodes, branches are
photosynthetic. Inter nodes are elongated by the activity of basal meristem.
C. Leaves: Small scaly leaves are present, scale leaves are joined to form sheath, Brown leaves
are non photosynthetic leaves.
A.
Recommended
O = O bond is shortest Oxygen electronegativity d.pdf
O = O bond is shortest Oxygen electronegativity difference is higher and also it is
double bond
Solution
O = O bond is shortest Oxygen electronegativity difference is higher and also it is
double bond.
NaCl Na is the symbol for Sodium which bonds to .pdf
NaCl Na is the symbol for Sodium which bonds to Cl which is the symbol for
Chlorine making Sodium Chloride.
Solution
NaCl Na is the symbol for Sodium which bonds to Cl which is the symbol for
Chlorine making Sodium Chloride..
NiCl2 because it not dissociate .pdf
NiCl2 because it not dissociate
Solution
NiCl2 because it not dissociate.
iodide, I-, has the -1 oxidation state. iodine (I.pdf
iodide, I-, has the -1 oxidation state. iodine (I) has 0 oxidation state. I- - e ---> I
Thus, this change is an oxidation reaction.
Solution
iodide, I-, has the -1 oxidation state. iodine (I) has 0 oxidation state. I- - e ---> I
Thus, this change is an oxidation reaction..
Lithium is in group 1 on the peridic table. This .pdf
Lithium is in group 1 on the peridic table. This means that it has only 1 valence
electron. Lithium will tend to lose that electron when it ionizes and become an ion. Remeber, an
ion is any atom, or sometimes molecule, with a charge. When Li loses the electron, it loses one
of its negative charges so the atom becomes an ion with a +1 charge because it now has 3
positively charged protons and only 2 negatively charged electrons.
Solution
Lithium is in group 1 on the peridic table. This means that it has only 1 valence
electron. Lithium will tend to lose that electron when it ionizes and become an ion. Remeber, an
ion is any atom, or sometimes molecule, with a charge. When Li loses the electron, it loses one
of its negative charges so the atom becomes an ion with a +1 charge because it now has 3
positively charged protons and only 2 negatively charged electrons..
Systematic positionKingdom PlantaeSubkingdom Tracheobionta (Vas.pdf
Systematic position:
Kingdom Plantae
Subkingdom Tracheobionta (Vascular plants)
Superdivision Spermatophyta (Seed plants)
Division Gnetophyta (Mormon tea and other gnetophytes)
Class Gnetopsida
Order Ephedrales
Family Ephedraceae (tea family)
Genus Ephedra L.
Common name Jointfir
There are 35 to 40 species in the world. These are wide spread in both in eastern and western
hemispheres
Morphology:
A. Root : Branched root, tap root are present these are characteristic xerophytes, no report of
mycorrhize.
B. Stem: Woody green branched distinctly joined apparently nodes and inter nodes, branches are
photosynthetic. Inter nodes are elongated by the activity of basal meristem.
C. Leaves: Small scaly leaves are present, scale leaves are joined to form sheath, Brown leaves
are non photosynthetic leaves.
Anatomy:
A. Root : It shows usually Gymnosperm structure. Cortex, endoderm, inner periderm, vascular
camblum, primary and secondary xylem was present.
B. Stem:
1. Primary structure: Transverse structure shows ridges and grooves. cortex clorenchyma
elongated, endoderm is present in single layer.
2. Secondary structure: Parenchymalous cells present between the vascular bundles to form a
ring of cambium, ray initials and fusiform initial are present.
C. Leaves: Transverse structure of leaves shows epidermis, cuticle, xylem, phloem, air space,
spongy parenchyma tissues. The vascular bundles are collateral closed and primary phloem and
primary xylem are also present.
Adaptations:
Mostly xerophytes scrubs or woody climbers Ephedra Pedunculata. Most species of ephedra are
branched scrubs, commonly climbing over vegetation many species spread by rhizomes Ephedra
Trifura. Under favorable condition it grows tree like habitat.
Ephedra Foliata - scrambling scrub, Ephedra Compacta - Branching type, Ephedra
Camphylophoda - Spendulus branches, Ephedra Intermedia - Shoot, Ephedra Saxatilis - scrub.
Values and functions:
In pharmaceutical industries it works as to access Ephedra drug materials.
Traditionally it is used by indigenous people for medicinal treatments of asthma, hay fever and
cold.
Alkaloids obtained from the species of Ephedra is used in herbal medicine.
Solution
Systematic position:
Kingdom Plantae
Subkingdom Tracheobionta (Vascular plants)
Superdivision Spermatophyta (Seed plants)
Division Gnetophyta (Mormon tea and other gnetophytes)
Class Gnetopsida
Order Ephedrales
Family Ephedraceae (tea family)
Genus Ephedra L.
Common name Jointfir
There are 35 to 40 species in the world. These are wide spread in both in eastern and western
hemispheres
Morphology:
A. Root : Branched root, tap root are present these are characteristic xerophytes, no report of
mycorrhize.
B. Stem: Woody green branched distinctly joined apparently nodes and inter nodes, branches are
photosynthetic. Inter nodes are elongated by the activity of basal meristem.
C. Leaves: Small scaly leaves are present, scale leaves are joined to form sheath, Brown leaves
are non photosynthetic leaves.
A.
Query to list out all the languages in the table LANGUAGES.SELECT .pdf
Query to list out all the languages in the table LANGUAGES.
SELECT LANGUAGE FROM LANGUAGES
Solution
Query to list out all the languages in the table LANGUAGES.
SELECT LANGUAGE FROM LANGUAGES.
public class LunarLander { double currentFuelFlowRate; d.pdf
public class LunarLander
{
double currentFuelFlowRate;
double verticalSpeed;
double altitude;
double amountOfFuel;
double massOfTheLander;
double maxFuelConsumptionRate;
double maxThrust;
/* constructor to initialize member variales*/
public LunarLander(double al, double fuel, double mass, double rate, double thrust)
{
currentFuelFlowRate=0.0;
verticalSpeed=0.0;
altitude=al;
amountOfFuel=fuel;
massOfTheLander=mass;
maxFuelConsumptionRate=rate;
maxThrust=thrust;
}
/* functions to get the values of instance variables*/
public double getCurrentFuelFlowRate()
{
return currentFuelFlowRate;
}
public double getVerticalSpeed()
{
return verticalSpeed;
}
public double getAltitude()
{
return altitude;
}
public double getAmountOfFuel()
{
return amountOfFuel;
}
public double getMassOfTheLander()
{
return massOfTheLander;
}
public double getMaxFuelConsumptionRate()
{
return maxFuelConsumptionRate;
}
public double getMaxThrust()
{
return maxThrust;
}
/* updates the value of currentFuelFlowRate*/
public void setCurrentFuelFlowRate(double rate)
{
currentFuelFlowRate=rate;
}
/* simulating the passage for a small amount of time t*/
public void setPassage(double t)
{
if(currentFuelFlowRate>0)
{
if(amountOfFuel==0)
{
currentFuelFlowRate=0;
}
}
/* velocity is the verticalSpeed*/
double f=maxThrust*currentFuelFlowRate;
double m=massOfTheLander;
verticalSpeed=t*((f/m)-1.62);
//v is verticalSpeed
altitude=t*verticalSpeed;
//ship is landed
if(altitude<0)
{
altitude=0;
verticalSpeed=0;
}
//r is currentFuelFlowRate
//c is maxFuelConsumptionRate
double changeInRemainFuel=t*currentFuelFlowRate*maxFuelConsumptionRate;
amountOfFuel=amountOfFuel-changeInRemainFuel;
if(amountOfFuel<0)
{
amountOfFuel=0;
}
}
}
////////////////////////////////////////////////////////
public class LunarLanderDemo {
public static void main(String[] args) {
LunarLander L=new LunarLander(1000,1700,900,10,5000);
/*give values between 0.0 , 1.0*/
L.setCurrentFuelFlowRate(0.6);
L.setPassage(0.1);
System.out.println(\"Simulation of Lunar Lander Passage at 0.1 seconds\");
System.out.println(\"Current Fuel Flow Rate: \"+L.getCurrentFuelFlowRate());
System.out.println(\"Vertical Speed: \"+L.getVerticalSpeed());
System.out.println(\"Altitude: \"+L.getAltitude());
System.out.println(\"Amount of Fuel: \"+L.getAmountOfFuel());
L.setCurrentFuelFlowRate(0.3);
L.setPassage(0.09);
System.out.println(\"Simulation of Lunar Lander Passage at 0.09 seconds\");
System.out.println(\"Current Fuel Flow Rate: \"+L.getCurrentFuelFlowRate());
System.out.println(\"Vertical Speed: \"+L.getVerticalSpeed());
System.out.println(\"Altitude: \"+L.getAltitude());
System.out.println(\"Amount of Fuel: \"+L.getAmountOfFuel());
}
}
Solution
public class LunarLander
{
double currentFuelFlowRate;
double verticalSpeed;
double altitude;
double amountOfFuel;
double massOfTheLander;
double maxFuelConsumptionRate;
double maxThrust;
/* constructor to initialize member variales*/
public LunarLander(double al, double fuel, double mass, double rate.
Cooling. The reaction is giving off energy(heat) .pdf
Cooling. The reaction is giving off energy(heat) so you need to control this in some
cases or it can be problematic
Solution
Cooling. The reaction is giving off energy(heat) so you need to control this in some
cases or it can be problematic.
php global $bsize,$playerToken,$myToken,$gameOver,$winArr,$rowAr.pdf
Tic Tac Toe in PHP\';
for($i=0; $i<$size; $i++) {
for($j=0; $j<$size; $j++) {
$count++;
$retVal.=\'\';
}
$retVal.=\'
\';
}
$retVal.=\'\';
echo $retVal;
}
function genBox2($size,$arr) {
global
$bsize,$playerToken,$myToken,$gameOver,$winArr,$rowArr,$colArr,$digArr,$vals,$s1,$s2,$s
3,$s4,$s5,$s6,$s7,$s8,$s9;
$count=0;
$retVal=\'\';
for($i=0; $i<$size; $i++) {
for($j=0; $j<$size; $j++) {
$count++;
$retVal.=\'\';
}
$retVal.=\'
\';
}
$retVal.=\'\';
echo $retVal;
}
function isEmpty($who) {
if($who==\"\")
return 1;
else
return 0;
}
function move($bsize,$arr) {
global
$bsize,$playerToken,$myToken,$gameOver,$winArr,$rowArr,$colArr,$digArr,$vals,$s1,$s2,$s
3,$s4,$s5,$s6,$s7,$s8,$s9;
$count=0;
$maxCount=0;
$pos=0;
$retVal=0;
# Build Row Array
for($i=0; $i<$bsize; $i++) {
$maxCount=0;
$fullCounter=0;
for($j=0; $j<$bsize; $j++) {
$count++;
$who=$arr[$count-1];
if($who==$playerToken) {
$maxCount++;
$fullCounter++;
}
if($who==$myToken)
$fullCounter++;
}
$rowArr[$i]=$maxCount;
if($fullCounter==$bsize)
$rowArr[$i]=-1;
}
# Building Column Array
for($i=0; $i<$bsize; $i++) {
$count=$i+1;
$maxCount=0;
$fullCounter=0;
for($j=0; $j<$bsize; $j++) {
$who=$arr[$count-1];
if($who==$playerToken) {
$maxCount++;
$fullCounter++;
}
if($who==$myToken)
$fullCounter++;
$count+=$bsize;
}
$colArr[$i]=$maxCount;
if($fullCounter==$bsize)
$colArr[$i]=-1;
}
# Building Diagonal Array
for($i=0; $i<2; $i++) {
if($i==0)
$count=$i+1;
else
$count=$bsize;
$maxCount=0;
$fullCounter=0;
for($j=0; $j<$bsize; $j++) {
$who=$arr[$count-1];
if($who==$playerToken) {
$maxCount++;
$fullCounter++;
}
if($who==$myToken)
$fullCounter++;
if($i==0)
$count+=$bsize+1;
else
$count+=$bsize-1;
}
$digArr[$i]=$maxCount;
if($fullCounter==$bsize)
$digArr[$i]=-1;
}
# Finding Max Values
$maxRow=myMax(0,$bsize,\"row\",$rowArr);
$maxCol=myMax(0,$bsize,\"col\",$colArr);
$maxDig=myMax(0,$bsize,\"dig\",$digArr);
$maxArrs[0]=myMax(1,$bsize,\"row\",$rowArr);
$maxArrs[1]=myMax(1,$bsize,\"col\",$colArr);
$maxArrs[2]=myMax(1,$bsize,\"dig\",$digArr);
//$maxArrs=array(max(1,$bsize,\"row\",$rowArr),max(1,$bsize,\"col\",$colArr),max(1,$bsize,\"
dig\",$digArr));
if(myMax(0,$bsize,\"x\",$maxArrs)==0)
$pos=$bsize*($maxRow+1)-$bsize;
if(myMax(0,$bsize,\"x\",$maxArrs)==1)
$pos=$maxCol;
if(myMax(0,$bsize,\"x\",$maxArrs)==2)
if($maxDig==0)
$pos=$maxDig;
else
$pos=$bsize-1;
$retFlag=0;
for($y=0; $y<$bsize; $y++) {
if(!$retFlag) {
if($arr[$pos]==\"\") {
$retVal=$pos;
$retFlag=1;
}
if(myMax(0,$bsize,\"x\",$maxArrs)==0)
$pos++;
if(myMax(0,$bsize,\"x\",$maxArrs)==1)
$pos+=$bsize;
if(myMax(0,$bsize,\"x\",$maxArrs)==2)
if($maxDig==0)
$pos+=$bsize+1;
else
$pos+=$bsize-1;
}
}
return $retVal;
}
function myMax($what,$bsize,$type,$arr) {
global
$bsize,$playerToken,$myToken,$gameOver,$winArr,$rowArr,$colArr,$digArr,$vals,$s1,$s2,$s
3,$s4,$s5,$s6,$s7,$s8,$s9;
$max=-1;
$maxIndex=-1;
if($type!=\"dig\") {
for($i=0; $i<$bsize; $i++) {
if($arr[$i]>$max) {
$max=$arr[$i];
$maxIndex=$i;
}
}
}
if($type==\"dig\") {
for($i=0; $i<2; $i++) {
if($arr[$i]>$max) {
$ma.
c. cytochrome C It is the enzyme involved in the .pdf
c. cytochrome C It is the enzyme involved in the reaction, so cannot exchange net
electrons
Solution
c. cytochrome C It is the enzyme involved in the reaction, so cannot exchange net
electrons.
Ka=-log(Ka)=-log(5.610^-10)=9.25 According to Henderson Hasselb.pdf
Ka=-log(Ka)=-log(5.6*10^-10)=9.25
According to Henderson Hasselbalch Equation
pH = pka + log[salt]/[acid]
pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.2/0.6)
pH=8.77
(2)
NH3 + HCl =======> NH4Cl
I 0.2 0.10 0.6
C 0.2-0.1 -0.1 +0.1
E 0.1 0 0.7
pH = pka + log[salt]/[acid]
pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.1/0.7)
pH=8.40
(3)
NH4Cl + NaOH =============> NH3 + NaCl + H2O
I 0.6 0.20 0.2
C -0.20 -0.20 +0.20
E 0.4 0 0.4
pH = pka + log[salt]/[acid]
pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.4/0.4) = 9.25
Solution
Ka=-log(Ka)=-log(5.6*10^-10)=9.25
According to Henderson Hasselbalch Equation
pH = pka + log[salt]/[acid]
pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.2/0.6)
pH=8.77
(2)
NH3 + HCl =======> NH4Cl
I 0.2 0.10 0.6
C 0.2-0.1 -0.1 +0.1
E 0.1 0 0.7
pH = pka + log[salt]/[acid]
pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.1/0.7)
pH=8.40
(3)
NH4Cl + NaOH =============> NH3 + NaCl + H2O
I 0.6 0.20 0.2
C -0.20 -0.20 +0.20
E 0.4 0 0.4
pH = pka + log[salt]/[acid]
pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.4/0.4) = 9.25.
C is wrong 14 7N contains 7 protons and 7 neutron.pdf
C is wrong 14 7N contains 7 protons and 7 neutrons , N/P = 1 which lies in belt of
stability
Solution
C is wrong 14 7N contains 7 protons and 7 neutrons , N/P = 1 which lies in belt of
stability.
1.The following are the main types of service providers to a mutua.pdf
1.
The following are the main types of service providers to a mutual fund:
Service provider refers to the sponsor of funds. From the public who are ready to invest, the fund
sponsor raises funds. The fund sponsor is one among the following:
The duty of the sponsor is to invest the funds of the investor in stocks, bonds, and shares
according to the preference of the investors.
If the investors demand for 100% safety of the invested funds, the fund sponsor invests in such
shares or bonds. If the investor is ready to meet risks even to 100%, then the fund sponsor will
invest accordingly. For this service, the investor has to pay the fund sponsor up to a certain limit.
2.
Determining the fees of mutual fund advisor:
There are certain restrictions that are imposed by SEBI. An advisor has to follow the same in
order to claim his fees. Mutual fund distributors are generally exempted from being registered as
investment advisors.
For a no-load large-cap fund, the percentage of fees is 0.95 whereas for small cap funds it is
1.20%. Therefore, if the advisor charges more than the specified percentage, the shareholders can
challenge that the fees is excessive.
3.
Differences between mutual funds and closed-end funds:
One of the major differences between the mutual funds and the closed-end funds is in the date of
maturity. A mutual fund can be withdrawn whenever the funds are required. However, it is not
the case in closed-end funds. In closed-end funds, the funds can be taken only at its date of
maturity. Some funds can be taken before the date of maturity. However, the rate of interest will
either be very little or even nil.
The closed-end funds have declined in popularity because of the returns and the maturity date.
Investing in this leads to overtrading and poor returns.
4.
Differences between ETFs and UITs from index mutual funds:
ETF is similar to closed-end funds. They trade closer to NAVs and therefore can be created and
retired whenever required. Mostly ETFs are passively managed with the help of indexes.
However, there are actively managed investments too. They are allowed to sell short and buy on
margin. They are even allowed to purchase a single share.
UITs are sold by investment advisors rather than purchasing from a secondary market. This does
not have a board of directors and is registered with Securities and Exchange Board. This
portfolio has different types of securities too.
Index mutual fund is a type of mutual fund in which a sale and buy index is set. When the rate
falls down to the set index, the funds are sold/bought on its own.
Solution
1.
The following are the main types of service providers to a mutual fund:
Service provider refers to the sponsor of funds. From the public who are ready to invest, the fund
sponsor raises funds. The fund sponsor is one among the following:
The duty of the sponsor is to invest the funds of the investor in stocks, bonds, and shares
according to the preference of the investors.
If the investors demand.
D. planned and directed means to improve the functioning of the clie.pdf
D. planned and directed means to improve the functioning of the client system
Solution
D. planned and directed means to improve the functioning of the client system.
domain - -x-1 0 = x+10 =x-1i.e. (-inf,-1)range-(-in.pdf
domain -
-x-1> 0
=> x+1<0
=>x<-1
i.e. (-inf,-1)
range-
(-inf,+inf)
Solution
domain -
-x-1> 0
=> x+1<0
=>x<-1
i.e. (-inf,-1)
range-
(-inf,+inf).
C. Keystroke loggers are a form of malware that records every key th.pdf
C. Keystroke loggers are a form of malware that records every key the computer user presses to
harvest usernames, passwords, social security numbers, and other confidential information and
then send that data to the person who caused the malware installation.
Solution
C. Keystroke loggers are a form of malware that records every key the computer user presses to
harvest usernames, passwords, social security numbers, and other confidential information and
then send that data to the person who caused the malware installation..
Ans.) Researchers have found much more proof that sexual orientation.pdf
Ans.) Researchers have found much more proof that sexual orientation is to a great extent
controlled by hereditary qualities, not decision. That can undermine a noteworthy contention
against the LBGT people group that claims that these individuals are living \"unnaturally.\"
According to latest research conducted and published in scientific journals the two areas of
human genome - one on the X chromosome and one on chromosome 8.
The relationship amongst science and sexual orientation is a subject of deep research and
analysis. A straightforward and particular determinant for sexual orientation has not been
indisputably illustrated; different studies indicate diverse, notwithstanding clashing positions,
however researchers theorize that a blend of genetics, hormonal and social elements decide
sexual orientation. Biological hypotheses for clarifying the reasons for sexual orientation are
more popular and natural elements may include an intricate interaction of hereditary components
and the early uterine, environment. These variables, which might be identified with the
improvement of a hetero, gay person, promiscuous orientation, incorporate qualities, pre-birth
hormones, and cerebrum structure.
According to a study, every single grown-up twin in Sweden (more than 7,600 twins) found that
same-sex conduct was clarified by both genetic elements and individual-particular natural
sources, (for example, pre-birth environment, involvement with sickness and injury and also peer
bunches and sexual encounters), while impacts of shared-environment factors, for example,
familial environment and social demeanors had a weaker, yet noteworthy impact. Women
demonstrated a factually non-critical pattern to weaker impact of genetic impacts, while men
demonstrated no impact of shared ecological impacts.
Chromosome linkage analysis of sexual orientation has shown the presence of different
contributing hereditary elements all through the genome. In 1993 Dean Hamer and partners
published their finding from a linkage examination of a specimen of 76 homo siblings and their
families. The researcher found that the homo men had more homo male uncles and cousins on
the maternal side of the family than on the fatherly side. Gay siblings who demonstrated this
maternal family were then tried for X chromosome linkage, utilizing twenty-two markers on the
X chromosome to test for comparable alleles. In another discovering, thirty-three of the forty kin
sets tried were found to have comparable alleles in the distal district of Xq28, which was
altogether higher than the normal rates of half for friendly siblings. This was prominently named
the \" homo quality\" in the media, bringing on noteworthy discussion. Sanders et al. in 1998
investigated their comparative study, in which they found that 13% of uncles of homosexual
siblings on the maternal side were homo person, contrasted and 6% on the fatherly side.
Hence, it is concluded that up to some extent homosexuality or sex orient.
Amnion - Protection of embryo (It covers the embryo when first forme.pdf
Amnion - Protection of embryo (It covers the embryo when first formed and filled with amniotic
fluid)
Chorion - Waste removal (The chorion is formed by two layers – trophoblast- outer layer and
mesoderm- inner layer. The mesoderm will be in contact with the amnion and removes waste
produced. The trophoblast provides the nutrients for the fetus during its confinement)
Allantois - gas exchange ( is a sac-like structure below the chorion, which consists of all
embryonic and extra-embryonic tissues)
Yolk - Primary nutrition source.
Albumen - Protein, nutrition.
Solution
Amnion - Protection of embryo (It covers the embryo when first formed and filled with amniotic
fluid)
Chorion - Waste removal (The chorion is formed by two layers – trophoblast- outer layer and
mesoderm- inner layer. The mesoderm will be in contact with the amnion and removes waste
produced. The trophoblast provides the nutrients for the fetus during its confinement)
Allantois - gas exchange ( is a sac-like structure below the chorion, which consists of all
embryonic and extra-embryonic tissues)
Yolk - Primary nutrition source.
Albumen - Protein, nutrition..
A. polar covalent NH3 H2O B. nonpolar covalent CH.pdf
A. polar covalent NH3 H2O B. nonpolar covalent CH4 CO2 C. ionic NaI
Solution
A. polar covalent NH3 H2O B. nonpolar covalent CH4 CO2 C. ionic NaI.
1.) If phi is thrival, kernal will be one to one. And kern = 9.pdf
1.) If phi is thrival, kernal will be one to one.
And |kern | = 9
2.) As phi is one to one, kernal will be trivial.
And |ker| = 9
And for c, kernel is also neither.
Solution
1.) If phi is thrival, kernal will be one to one.
And |kern | = 9
2.) As phi is one to one, kernal will be trivial.
And |ker| = 9
And for c, kernel is also neither..
A Lewis acid is a species that can accept a pair .pdf
A Lewis acid is a species that can accept a pair of electrons from a Lewis base,
which is a donator of electrons. In this reaction, AlCl3 is a strong lewis acid and accepts a lone
pair of electrons from the Lewis base Cl-, thereby completing Aluminum\'s octet and forming
AlCl4-.
Solution
A Lewis acid is a species that can accept a pair of electrons from a Lewis base,
which is a donator of electrons. In this reaction, AlCl3 is a strong lewis acid and accepts a lone
pair of electrons from the Lewis base Cl-, thereby completing Aluminum\'s octet and forming
AlCl4-..
Unit 8 - Information and Communication Technology (Paper I).pdf
This slides describes the basic concepts of ICT, basics of Email, Emerging Technology and Digital Initiatives in Education. This presentations aligns with the UGC Paper I syllabus.
Francesca Gottschalk - How can education support child empowerment.pptx
Francesca Gottschalk from the OECD’s Centre for Educational Research and Innovation presents at the Ask an Expert Webinar: How can education support child empowerment?
Thesis Statement for students diagnonsed withADHD.ppt
Presentation required for the master in Education.
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Query to list out all the languages in the table LANGUAGES.SELECT .pdf
Query to list out all the languages in the table LANGUAGES.
SELECT LANGUAGE FROM LANGUAGES
Solution
Query to list out all the languages in the table LANGUAGES.
SELECT LANGUAGE FROM LANGUAGES.
public class LunarLander { double currentFuelFlowRate; d.pdf
public class LunarLander
{
double currentFuelFlowRate;
double verticalSpeed;
double altitude;
double amountOfFuel;
double massOfTheLander;
double maxFuelConsumptionRate;
double maxThrust;
/* constructor to initialize member variales*/
public LunarLander(double al, double fuel, double mass, double rate, double thrust)
{
currentFuelFlowRate=0.0;
verticalSpeed=0.0;
altitude=al;
amountOfFuel=fuel;
massOfTheLander=mass;
maxFuelConsumptionRate=rate;
maxThrust=thrust;
}
/* functions to get the values of instance variables*/
public double getCurrentFuelFlowRate()
{
return currentFuelFlowRate;
}
public double getVerticalSpeed()
{
return verticalSpeed;
}
public double getAltitude()
{
return altitude;
}
public double getAmountOfFuel()
{
return amountOfFuel;
}
public double getMassOfTheLander()
{
return massOfTheLander;
}
public double getMaxFuelConsumptionRate()
{
return maxFuelConsumptionRate;
}
public double getMaxThrust()
{
return maxThrust;
}
/* updates the value of currentFuelFlowRate*/
public void setCurrentFuelFlowRate(double rate)
{
currentFuelFlowRate=rate;
}
/* simulating the passage for a small amount of time t*/
public void setPassage(double t)
{
if(currentFuelFlowRate>0)
{
if(amountOfFuel==0)
{
currentFuelFlowRate=0;
}
}
/* velocity is the verticalSpeed*/
double f=maxThrust*currentFuelFlowRate;
double m=massOfTheLander;
verticalSpeed=t*((f/m)-1.62);
//v is verticalSpeed
altitude=t*verticalSpeed;
//ship is landed
if(altitude<0)
{
altitude=0;
verticalSpeed=0;
}
//r is currentFuelFlowRate
//c is maxFuelConsumptionRate
double changeInRemainFuel=t*currentFuelFlowRate*maxFuelConsumptionRate;
amountOfFuel=amountOfFuel-changeInRemainFuel;
if(amountOfFuel<0)
{
amountOfFuel=0;
}
}
}
////////////////////////////////////////////////////////
public class LunarLanderDemo {
public static void main(String[] args) {
LunarLander L=new LunarLander(1000,1700,900,10,5000);
/*give values between 0.0 , 1.0*/
L.setCurrentFuelFlowRate(0.6);
L.setPassage(0.1);
System.out.println(\"Simulation of Lunar Lander Passage at 0.1 seconds\");
System.out.println(\"Current Fuel Flow Rate: \"+L.getCurrentFuelFlowRate());
System.out.println(\"Vertical Speed: \"+L.getVerticalSpeed());
System.out.println(\"Altitude: \"+L.getAltitude());
System.out.println(\"Amount of Fuel: \"+L.getAmountOfFuel());
L.setCurrentFuelFlowRate(0.3);
L.setPassage(0.09);
System.out.println(\"Simulation of Lunar Lander Passage at 0.09 seconds\");
System.out.println(\"Current Fuel Flow Rate: \"+L.getCurrentFuelFlowRate());
System.out.println(\"Vertical Speed: \"+L.getVerticalSpeed());
System.out.println(\"Altitude: \"+L.getAltitude());
System.out.println(\"Amount of Fuel: \"+L.getAmountOfFuel());
}
}
Solution
public class LunarLander
{
double currentFuelFlowRate;
double verticalSpeed;
double altitude;
double amountOfFuel;
double massOfTheLander;
double maxFuelConsumptionRate;
double maxThrust;
/* constructor to initialize member variales*/
public LunarLander(double al, double fuel, double mass, double rate.
Cooling. The reaction is giving off energy(heat) .pdf
Cooling. The reaction is giving off energy(heat) so you need to control this in some
cases or it can be problematic
Solution
Cooling. The reaction is giving off energy(heat) so you need to control this in some
cases or it can be problematic.
php global $bsize,$playerToken,$myToken,$gameOver,$winArr,$rowAr.pdf
Tic Tac Toe in PHP\';
for($i=0; $i<$size; $i++) {
for($j=0; $j<$size; $j++) {
$count++;
$retVal.=\'\';
}
$retVal.=\'
\';
}
$retVal.=\'\';
echo $retVal;
}
function genBox2($size,$arr) {
global
$bsize,$playerToken,$myToken,$gameOver,$winArr,$rowArr,$colArr,$digArr,$vals,$s1,$s2,$s
3,$s4,$s5,$s6,$s7,$s8,$s9;
$count=0;
$retVal=\'\';
for($i=0; $i<$size; $i++) {
for($j=0; $j<$size; $j++) {
$count++;
$retVal.=\'\';
}
$retVal.=\'
\';
}
$retVal.=\'\';
echo $retVal;
}
function isEmpty($who) {
if($who==\"\")
return 1;
else
return 0;
}
function move($bsize,$arr) {
global
$bsize,$playerToken,$myToken,$gameOver,$winArr,$rowArr,$colArr,$digArr,$vals,$s1,$s2,$s
3,$s4,$s5,$s6,$s7,$s8,$s9;
$count=0;
$maxCount=0;
$pos=0;
$retVal=0;
# Build Row Array
for($i=0; $i<$bsize; $i++) {
$maxCount=0;
$fullCounter=0;
for($j=0; $j<$bsize; $j++) {
$count++;
$who=$arr[$count-1];
if($who==$playerToken) {
$maxCount++;
$fullCounter++;
}
if($who==$myToken)
$fullCounter++;
}
$rowArr[$i]=$maxCount;
if($fullCounter==$bsize)
$rowArr[$i]=-1;
}
# Building Column Array
for($i=0; $i<$bsize; $i++) {
$count=$i+1;
$maxCount=0;
$fullCounter=0;
for($j=0; $j<$bsize; $j++) {
$who=$arr[$count-1];
if($who==$playerToken) {
$maxCount++;
$fullCounter++;
}
if($who==$myToken)
$fullCounter++;
$count+=$bsize;
}
$colArr[$i]=$maxCount;
if($fullCounter==$bsize)
$colArr[$i]=-1;
}
# Building Diagonal Array
for($i=0; $i<2; $i++) {
if($i==0)
$count=$i+1;
else
$count=$bsize;
$maxCount=0;
$fullCounter=0;
for($j=0; $j<$bsize; $j++) {
$who=$arr[$count-1];
if($who==$playerToken) {
$maxCount++;
$fullCounter++;
}
if($who==$myToken)
$fullCounter++;
if($i==0)
$count+=$bsize+1;
else
$count+=$bsize-1;
}
$digArr[$i]=$maxCount;
if($fullCounter==$bsize)
$digArr[$i]=-1;
}
# Finding Max Values
$maxRow=myMax(0,$bsize,\"row\",$rowArr);
$maxCol=myMax(0,$bsize,\"col\",$colArr);
$maxDig=myMax(0,$bsize,\"dig\",$digArr);
$maxArrs[0]=myMax(1,$bsize,\"row\",$rowArr);
$maxArrs[1]=myMax(1,$bsize,\"col\",$colArr);
$maxArrs[2]=myMax(1,$bsize,\"dig\",$digArr);
//$maxArrs=array(max(1,$bsize,\"row\",$rowArr),max(1,$bsize,\"col\",$colArr),max(1,$bsize,\"
dig\",$digArr));
if(myMax(0,$bsize,\"x\",$maxArrs)==0)
$pos=$bsize*($maxRow+1)-$bsize;
if(myMax(0,$bsize,\"x\",$maxArrs)==1)
$pos=$maxCol;
if(myMax(0,$bsize,\"x\",$maxArrs)==2)
if($maxDig==0)
$pos=$maxDig;
else
$pos=$bsize-1;
$retFlag=0;
for($y=0; $y<$bsize; $y++) {
if(!$retFlag) {
if($arr[$pos]==\"\") {
$retVal=$pos;
$retFlag=1;
}
if(myMax(0,$bsize,\"x\",$maxArrs)==0)
$pos++;
if(myMax(0,$bsize,\"x\",$maxArrs)==1)
$pos+=$bsize;
if(myMax(0,$bsize,\"x\",$maxArrs)==2)
if($maxDig==0)
$pos+=$bsize+1;
else
$pos+=$bsize-1;
}
}
return $retVal;
}
function myMax($what,$bsize,$type,$arr) {
global
$bsize,$playerToken,$myToken,$gameOver,$winArr,$rowArr,$colArr,$digArr,$vals,$s1,$s2,$s
3,$s4,$s5,$s6,$s7,$s8,$s9;
$max=-1;
$maxIndex=-1;
if($type!=\"dig\") {
for($i=0; $i<$bsize; $i++) {
if($arr[$i]>$max) {
$max=$arr[$i];
$maxIndex=$i;
}
}
}
if($type==\"dig\") {
for($i=0; $i<2; $i++) {
if($arr[$i]>$max) {
$ma.
c. cytochrome C It is the enzyme involved in the .pdf
c. cytochrome C It is the enzyme involved in the reaction, so cannot exchange net
electrons
Solution
c. cytochrome C It is the enzyme involved in the reaction, so cannot exchange net
electrons.
Ka=-log(Ka)=-log(5.610^-10)=9.25 According to Henderson Hasselb.pdf
Ka=-log(Ka)=-log(5.6*10^-10)=9.25
According to Henderson Hasselbalch Equation
pH = pka + log[salt]/[acid]
pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.2/0.6)
pH=8.77
(2)
NH3 + HCl =======> NH4Cl
I 0.2 0.10 0.6
C 0.2-0.1 -0.1 +0.1
E 0.1 0 0.7
pH = pka + log[salt]/[acid]
pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.1/0.7)
pH=8.40
(3)
NH4Cl + NaOH =============> NH3 + NaCl + H2O
I 0.6 0.20 0.2
C -0.20 -0.20 +0.20
E 0.4 0 0.4
pH = pka + log[salt]/[acid]
pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.4/0.4) = 9.25
Solution
Ka=-log(Ka)=-log(5.6*10^-10)=9.25
According to Henderson Hasselbalch Equation
pH = pka + log[salt]/[acid]
pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.2/0.6)
pH=8.77
(2)
NH3 + HCl =======> NH4Cl
I 0.2 0.10 0.6
C 0.2-0.1 -0.1 +0.1
E 0.1 0 0.7
pH = pka + log[salt]/[acid]
pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.1/0.7)
pH=8.40
(3)
NH4Cl + NaOH =============> NH3 + NaCl + H2O
I 0.6 0.20 0.2
C -0.20 -0.20 +0.20
E 0.4 0 0.4
pH = pka + log[salt]/[acid]
pH = 9.25 log[Nh3]/[Nh4+] = 9.25+log(0.4/0.4) = 9.25.
C is wrong 14 7N contains 7 protons and 7 neutron.pdf
C is wrong 14 7N contains 7 protons and 7 neutrons , N/P = 1 which lies in belt of
stability
Solution
C is wrong 14 7N contains 7 protons and 7 neutrons , N/P = 1 which lies in belt of
stability.
1.The following are the main types of service providers to a mutua.pdf
1.
The following are the main types of service providers to a mutual fund:
Service provider refers to the sponsor of funds. From the public who are ready to invest, the fund
sponsor raises funds. The fund sponsor is one among the following:
The duty of the sponsor is to invest the funds of the investor in stocks, bonds, and shares
according to the preference of the investors.
If the investors demand for 100% safety of the invested funds, the fund sponsor invests in such
shares or bonds. If the investor is ready to meet risks even to 100%, then the fund sponsor will
invest accordingly. For this service, the investor has to pay the fund sponsor up to a certain limit.
2.
Determining the fees of mutual fund advisor:
There are certain restrictions that are imposed by SEBI. An advisor has to follow the same in
order to claim his fees. Mutual fund distributors are generally exempted from being registered as
investment advisors.
For a no-load large-cap fund, the percentage of fees is 0.95 whereas for small cap funds it is
1.20%. Therefore, if the advisor charges more than the specified percentage, the shareholders can
challenge that the fees is excessive.
3.
Differences between mutual funds and closed-end funds:
One of the major differences between the mutual funds and the closed-end funds is in the date of
maturity. A mutual fund can be withdrawn whenever the funds are required. However, it is not
the case in closed-end funds. In closed-end funds, the funds can be taken only at its date of
maturity. Some funds can be taken before the date of maturity. However, the rate of interest will
either be very little or even nil.
The closed-end funds have declined in popularity because of the returns and the maturity date.
Investing in this leads to overtrading and poor returns.
4.
Differences between ETFs and UITs from index mutual funds:
ETF is similar to closed-end funds. They trade closer to NAVs and therefore can be created and
retired whenever required. Mostly ETFs are passively managed with the help of indexes.
However, there are actively managed investments too. They are allowed to sell short and buy on
margin. They are even allowed to purchase a single share.
UITs are sold by investment advisors rather than purchasing from a secondary market. This does
not have a board of directors and is registered with Securities and Exchange Board. This
portfolio has different types of securities too.
Index mutual fund is a type of mutual fund in which a sale and buy index is set. When the rate
falls down to the set index, the funds are sold/bought on its own.
Solution
1.
The following are the main types of service providers to a mutual fund:
Service provider refers to the sponsor of funds. From the public who are ready to invest, the fund
sponsor raises funds. The fund sponsor is one among the following:
The duty of the sponsor is to invest the funds of the investor in stocks, bonds, and shares
according to the preference of the investors.
If the investors demand.
D. planned and directed means to improve the functioning of the clie.pdf
D. planned and directed means to improve the functioning of the client system
Solution
D. planned and directed means to improve the functioning of the client system.
domain - -x-1 0 = x+10 =x-1i.e. (-inf,-1)range-(-in.pdf
domain -
-x-1> 0
=> x+1<0
=>x<-1
i.e. (-inf,-1)
range-
(-inf,+inf)
Solution
domain -
-x-1> 0
=> x+1<0
=>x<-1
i.e. (-inf,-1)
range-
(-inf,+inf).
C. Keystroke loggers are a form of malware that records every key th.pdf
C. Keystroke loggers are a form of malware that records every key the computer user presses to
harvest usernames, passwords, social security numbers, and other confidential information and
then send that data to the person who caused the malware installation.
Solution
C. Keystroke loggers are a form of malware that records every key the computer user presses to
harvest usernames, passwords, social security numbers, and other confidential information and
then send that data to the person who caused the malware installation..
Ans.) Researchers have found much more proof that sexual orientation.pdf
Ans.) Researchers have found much more proof that sexual orientation is to a great extent
controlled by hereditary qualities, not decision. That can undermine a noteworthy contention
against the LBGT people group that claims that these individuals are living \"unnaturally.\"
According to latest research conducted and published in scientific journals the two areas of
human genome - one on the X chromosome and one on chromosome 8.
The relationship amongst science and sexual orientation is a subject of deep research and
analysis. A straightforward and particular determinant for sexual orientation has not been
indisputably illustrated; different studies indicate diverse, notwithstanding clashing positions,
however researchers theorize that a blend of genetics, hormonal and social elements decide
sexual orientation. Biological hypotheses for clarifying the reasons for sexual orientation are
more popular and natural elements may include an intricate interaction of hereditary components
and the early uterine, environment. These variables, which might be identified with the
improvement of a hetero, gay person, promiscuous orientation, incorporate qualities, pre-birth
hormones, and cerebrum structure.
According to a study, every single grown-up twin in Sweden (more than 7,600 twins) found that
same-sex conduct was clarified by both genetic elements and individual-particular natural
sources, (for example, pre-birth environment, involvement with sickness and injury and also peer
bunches and sexual encounters), while impacts of shared-environment factors, for example,
familial environment and social demeanors had a weaker, yet noteworthy impact. Women
demonstrated a factually non-critical pattern to weaker impact of genetic impacts, while men
demonstrated no impact of shared ecological impacts.
Chromosome linkage analysis of sexual orientation has shown the presence of different
contributing hereditary elements all through the genome. In 1993 Dean Hamer and partners
published their finding from a linkage examination of a specimen of 76 homo siblings and their
families. The researcher found that the homo men had more homo male uncles and cousins on
the maternal side of the family than on the fatherly side. Gay siblings who demonstrated this
maternal family were then tried for X chromosome linkage, utilizing twenty-two markers on the
X chromosome to test for comparable alleles. In another discovering, thirty-three of the forty kin
sets tried were found to have comparable alleles in the distal district of Xq28, which was
altogether higher than the normal rates of half for friendly siblings. This was prominently named
the \" homo quality\" in the media, bringing on noteworthy discussion. Sanders et al. in 1998
investigated their comparative study, in which they found that 13% of uncles of homosexual
siblings on the maternal side were homo person, contrasted and 6% on the fatherly side.
Hence, it is concluded that up to some extent homosexuality or sex orient.
Amnion - Protection of embryo (It covers the embryo when first forme.pdf
Amnion - Protection of embryo (It covers the embryo when first formed and filled with amniotic
fluid)
Chorion - Waste removal (The chorion is formed by two layers – trophoblast- outer layer and
mesoderm- inner layer. The mesoderm will be in contact with the amnion and removes waste
produced. The trophoblast provides the nutrients for the fetus during its confinement)
Allantois - gas exchange ( is a sac-like structure below the chorion, which consists of all
embryonic and extra-embryonic tissues)
Yolk - Primary nutrition source.
Albumen - Protein, nutrition.
Solution
Amnion - Protection of embryo (It covers the embryo when first formed and filled with amniotic
fluid)
Chorion - Waste removal (The chorion is formed by two layers – trophoblast- outer layer and
mesoderm- inner layer. The mesoderm will be in contact with the amnion and removes waste
produced. The trophoblast provides the nutrients for the fetus during its confinement)
Allantois - gas exchange ( is a sac-like structure below the chorion, which consists of all
embryonic and extra-embryonic tissues)
Yolk - Primary nutrition source.
Albumen - Protein, nutrition..
A. polar covalent NH3 H2O B. nonpolar covalent CH.pdf
A. polar covalent NH3 H2O B. nonpolar covalent CH4 CO2 C. ionic NaI
Solution
A. polar covalent NH3 H2O B. nonpolar covalent CH4 CO2 C. ionic NaI.
1.) If phi is thrival, kernal will be one to one. And kern = 9.pdf
1.) If phi is thrival, kernal will be one to one.
And |kern | = 9
2.) As phi is one to one, kernal will be trivial.
And |ker| = 9
And for c, kernel is also neither.
Solution
1.) If phi is thrival, kernal will be one to one.
And |kern | = 9
2.) As phi is one to one, kernal will be trivial.
And |ker| = 9
And for c, kernel is also neither..
A Lewis acid is a species that can accept a pair .pdf
A Lewis acid is a species that can accept a pair of electrons from a Lewis base,
which is a donator of electrons. In this reaction, AlCl3 is a strong lewis acid and accepts a lone
pair of electrons from the Lewis base Cl-, thereby completing Aluminum\'s octet and forming
AlCl4-.
Solution
A Lewis acid is a species that can accept a pair of electrons from a Lewis base,
which is a donator of electrons. In this reaction, AlCl3 is a strong lewis acid and accepts a lone
pair of electrons from the Lewis base Cl-, thereby completing Aluminum\'s octet and forming
AlCl4-..
More from anjalitimecenter11 (18)
Query to list out all the languages in the table LANGUAGES.SELECT .pdf
Query to list out all the languages in the table LANGUAGES.SELECT .pdf
public class LunarLander { double currentFuelFlowRate; d.pdf
public class LunarLander { double currentFuelFlowRate; d.pdf
Cooling. The reaction is giving off energy(heat) .pdf
Cooling. The reaction is giving off energy(heat) .pdf
php global $bsize,$playerToken,$myToken,$gameOver,$winArr,$rowAr.pdf
php global $bsize,$playerToken,$myToken,$gameOver,$winArr,$rowAr.pdf
c. cytochrome C It is the enzyme involved in the .pdf
c. cytochrome C It is the enzyme involved in the .pdf
Ka=-log(Ka)=-log(5.610^-10)=9.25 According to Henderson Hasselb.pdf
Ka=-log(Ka)=-log(5.610^-10)=9.25 According to Henderson Hasselb.pdf
C is wrong 14 7N contains 7 protons and 7 neutron.pdf
C is wrong 14 7N contains 7 protons and 7 neutron.pdf
1.The following are the main types of service providers to a mutua.pdf
1.The following are the main types of service providers to a mutua.pdf
D. planned and directed means to improve the functioning of the clie.pdf
D. planned and directed means to improve the functioning of the clie.pdf
domain - -x-1 0 = x+10 =x-1i.e. (-inf,-1)range-(-in.pdf
domain - -x-1 0 = x+10 =x-1i.e. (-inf,-1)range-(-in.pdf
C. Keystroke loggers are a form of malware that records every key th.pdf
C. Keystroke loggers are a form of malware that records every key th.pdf
Ans.) Researchers have found much more proof that sexual orientation.pdf
Ans.) Researchers have found much more proof that sexual orientation.pdf
Amnion - Protection of embryo (It covers the embryo when first forme.pdf
Amnion - Protection of embryo (It covers the embryo when first forme.pdf
A. polar covalent NH3 H2O B. nonpolar covalent CH.pdf
A. polar covalent NH3 H2O B. nonpolar covalent CH.pdf
1.) If phi is thrival, kernal will be one to one. And kern = 9.pdf
1.) If phi is thrival, kernal will be one to one. And kern = 9.pdf
A Lewis acid is a species that can accept a pair .pdf
A Lewis acid is a species that can accept a pair .pdf
Recently uploaded
Unit 8 - Information and Communication Technology (Paper I).pdf
This slides describes the basic concepts of ICT, basics of Email, Emerging Technology and Digital Initiatives in Education. This presentations aligns with the UGC Paper I syllabus.
Francesca Gottschalk - How can education support child empowerment.pptx
Francesca Gottschalk from the OECD’s Centre for Educational Research and Innovation presents at the Ask an Expert Webinar: How can education support child empowerment?
Thesis Statement for students diagnonsed withADHD.ppt
Presentation required for the master in Education.
Azure Interview Questions and Answers PDF By ScholarHat
Azure Interview Questions and Answers PDF By ScholarHat
Digital Artifact 2 - Investigating Pavilion Designs
Digital Artifact 2 - Pavilions
NGV Architecture Commission Competition
MPavilion Commission Competition
Other Pavilion Designs
Operation Blue Star - Saka Neela Tara
Operation “Blue Star” is the only event in the history of Independent India where the state went into war with its own people. Even after about 40 years it is not clear if it was culmination of states anger over people of the region, a political game of power or start of dictatorial chapter in the democratic setup.
The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
Acetabularia Information For Class 9 .docx
Acetabularia acetabulum is a single-celled green alga that in its vegetative state is morphologically differentiated into a basal rhizoid and an axially elongated stalk, which bears whorls of branching hairs. The single diploid nucleus resides in the rhizoid.
1.4 modern child centered education - mahatma gandhi-2.pptx
Child centred education is an educational approach that priorities the interest, needs and abilities of the child in the learning process.
The Challenger.pdf DNHS Official Publication
Read| The latest issue of The Challenger is here! We are thrilled to announce that our school paper has qualified for the NATIONAL SCHOOLS PRESS CONFERENCE (NSPC) 2024. Thank you for your unwavering support and trust. Dive into the stories that made us stand out!
Synthetic Fiber Construction in lab .pptx
Synthetic fiber production is a fascinating and complex field that blends chemistry, engineering, and environmental science. By understanding these aspects, students can gain a comprehensive view of synthetic fiber production, its impact on society and the environment, and the potential for future innovations. Synthetic fibers play a crucial role in modern society, impacting various aspects of daily life, industry, and the environment. ynthetic fibers are integral to modern life, offering a range of benefits from cost-effectiveness and versatility to innovative applications and performance characteristics. While they pose environmental challenges, ongoing research and development aim to create more sustainable and eco-friendly alternatives. Understanding the importance of synthetic fibers helps in appreciating their role in the economy, industry, and daily life, while also emphasizing the need for sustainable practices and innovation.
Supporting (UKRI) OA monographs at Salford.pptx
How libraries can support authors with open access requirements for UKRI funded books
Wednesday 22 May 2024, 14:00-15:00.
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Event Link:-
https://meetups.mulesoft.com/events/details/mulesoft-mysore-presents-exploring-gemini-ai-and-integration-with-mulesoft/
Agenda
● Java 17 Upgrade Overview
● Why and by when do customers need to upgrade to Java 17?
● Is there any immediate impact to upgrading to Mule Runtime 4.6 and beyond?
● Which MuleSoft products are in scope?
For Upcoming Meetups Join Mysore Meetup Group - https://meetups.mulesoft.com/mysore/
YouTube:- youtube.com/@mulesoftmysore
Mysore WhatsApp group:- https://chat.whatsapp.com/EhqtHtCC75vCAX7gaO842N
Speaker:-
Shubham Chaurasia - https://www.linkedin.com/in/shubhamchaurasia1/
Priya Shaw - https://www.linkedin.com/in/priya-shaw
Organizers:-
Shubham Chaurasia - https://www.linkedin.com/in/shubhamchaurasia1/
Giridhar Meka - https://www.linkedin.com/in/giridharmeka
Priya Shaw - https://www.linkedin.com/in/priya-shaw
Shyam Raj Prasad-
https://www.linkedin.com/in/shyam-raj-prasad/
Biological Screening of Herbal Drugs in detailed.
Biological screening of herbal drugs: Introduction and Need for
Phyto-Pharmacological Screening, New Strategies for evaluating
Natural Products, In vitro evaluation techniques for Antioxidants, Antimicrobial and Anticancer drugs. In vivo evaluation techniques
for Anti-inflammatory, Antiulcer, Anticancer, Wound healing, Antidiabetic, Hepatoprotective, Cardio protective, Diuretics and
Antifertility, Toxicity studies as per OECD guidelines
TESDA TM1 REVIEWER FOR NATIONAL ASSESSMENT WRITTEN AND ORAL QUESTIONS WITH A...
TESDA TM1 REVIEWER FOR NATIONAL ASSESSMENT WRITTEN AND ORAL QUESTIONS WITH ANSWERS.
CACJapan - GROUP Presentation 1- Wk 4.pdf
Macroeconomics- Movie Location
This will be used as part of your Personal Professional Portfolio once graded.
Objective:
Prepare a presentation or a paper using research, basic comparative analysis, data organization and application of economic information. You will make an informed assessment of an economic climate outside of the United States to accomplish an entertainment industry objective.
A Strategic Approach: GenAI in Education
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
The approach at University of Liverpool.pptx
How libraries can support authors with open access requirements for UKRI funded books
Wednesday 22 May 2024, 14:00-15:00.
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Thesis Statement for students diagnonsed withADHD.ppt
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Azure Interview Questions and Answers PDF By ScholarHat
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Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
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TESDA TM1 REVIEWER FOR NATIONAL ASSESSMENT WRITTEN AND ORAL QUESTIONS WITH A...
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