A) The possible genotypes are both dominant allele(SS), one dominant.pdf
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A) The possible genotypes are both dominant allele(SS), one dominant and one recessive (Ss) and both recessive(ss). B) 1/4=0.25 Solution A) The possible genotypes are both dominant allele(SS), one dominant and one recessive (Ss) and both recessive(ss). B) 1/4=0.25.
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The equation you should use is just a variation of the PV formula.pdf
The equation you should use is just a variation of the PV formula:
r = (FV/PV)1/t -1
r = ($3.20/$1.50)1/5 - 1
=
Solution
The equation you should use is just a variation of the PV formula:
r = (FV/PV)1/t -1
r = ($3.20/$1.50)1/5 - 1
=.
True short double stranded mRNA are involved in initiating the inte.pdf
True: short double stranded mRNA are involved in initiating the interference Answer: C: genes
are reshuffled in the mRNA segments Answer B: Enhancers can function in either orientation
whilst the UAS only functions in one. False: eukaryotes utilise enhancers
Solution
True: short double stranded mRNA are involved in initiating the interference Answer: C: genes
are reshuffled in the mRNA segments Answer B: Enhancers can function in either orientation
whilst the UAS only functions in one. False: eukaryotes utilise enhancers.
Yes, it would react similarly, since boric oxide .pdf
Yes, it would react similarly, since boric oxide is similar to carbon dioxide in
dissolving water and will produce boric acid as result
Solution
Yes, it would react similarly, since boric oxide is similar to carbon dioxide in
dissolving water and will produce boric acid as result.
Intermediate species means species that have 2 ch.pdf
Intermediate species means species that have 2 characteristics 1. they are a product
in a previous step 2. they are consumed (they are reactants) in the following step Note that they
must have both of these characteristics to be considered intermediates. If they have one or the
other they are not counted as intermediates. Notice how NO is being produced in step I and being
used up in step II. That is why it is an intermediate.
Solution
Intermediate species means species that have 2 characteristics 1. they are a product
in a previous step 2. they are consumed (they are reactants) in the following step Note that they
must have both of these characteristics to be considered intermediates. If they have one or the
other they are not counted as intermediates. Notice how NO is being produced in step I and being
used up in step II. That is why it is an intermediate..
In primitive cells atoms are present at the corne.pdf
In primitive cells atoms are present at the corners only. Among the given unit cells
unit cell 1 is primitive cell. In remaining two unit cells there are atoms at center also.
Solution
In primitive cells atoms are present at the corners only. Among the given unit cells
unit cell 1 is primitive cell. In remaining two unit cells there are atoms at center also..
Recommended
The equation you should use is just a variation of the PV formula.pdf
The equation you should use is just a variation of the PV formula:
r = (FV/PV)1/t -1
r = ($3.20/$1.50)1/5 - 1
=
Solution
The equation you should use is just a variation of the PV formula:
r = (FV/PV)1/t -1
r = ($3.20/$1.50)1/5 - 1
=.
True short double stranded mRNA are involved in initiating the inte.pdf
True: short double stranded mRNA are involved in initiating the interference Answer: C: genes
are reshuffled in the mRNA segments Answer B: Enhancers can function in either orientation
whilst the UAS only functions in one. False: eukaryotes utilise enhancers
Solution
True: short double stranded mRNA are involved in initiating the interference Answer: C: genes
are reshuffled in the mRNA segments Answer B: Enhancers can function in either orientation
whilst the UAS only functions in one. False: eukaryotes utilise enhancers.
Yes, it would react similarly, since boric oxide .pdf
Yes, it would react similarly, since boric oxide is similar to carbon dioxide in
dissolving water and will produce boric acid as result
Solution
Yes, it would react similarly, since boric oxide is similar to carbon dioxide in
dissolving water and will produce boric acid as result.
Intermediate species means species that have 2 ch.pdf
Intermediate species means species that have 2 characteristics 1. they are a product
in a previous step 2. they are consumed (they are reactants) in the following step Note that they
must have both of these characteristics to be considered intermediates. If they have one or the
other they are not counted as intermediates. Notice how NO is being produced in step I and being
used up in step II. That is why it is an intermediate.
Solution
Intermediate species means species that have 2 characteristics 1. they are a product
in a previous step 2. they are consumed (they are reactants) in the following step Note that they
must have both of these characteristics to be considered intermediates. If they have one or the
other they are not counted as intermediates. Notice how NO is being produced in step I and being
used up in step II. That is why it is an intermediate..
In primitive cells atoms are present at the corne.pdf
In primitive cells atoms are present at the corners only. Among the given unit cells
unit cell 1 is primitive cell. In remaining two unit cells there are atoms at center also.
Solution
In primitive cells atoms are present at the corners only. Among the given unit cells
unit cell 1 is primitive cell. In remaining two unit cells there are atoms at center also..
Hg2+ is more water soluble, methyl mercury HgCH3+.pdf
Hg2+ is more water soluble, methyl mercury HgCH3+ is more lipid soluble.
CH3Hg, polar molecules would combine to form a substance similar to water but would be more
acidic as a result of mixing two non-like particles in a solution
Solution
Hg2+ is more water soluble, methyl mercury HgCH3+ is more lipid soluble.
CH3Hg, polar molecules would combine to form a substance similar to water but would be more
acidic as a result of mixing two non-like particles in a solution.
HCl is a polar-covalent compound and also Acetoni.pdf
HCl is a polar-covalent compound and also Acetonitrile they are soluble dueto
hydrogen - bonding between them. So they have London-dispersion forces between them.
Solution
HCl is a polar-covalent compound and also Acetonitrile they are soluble dueto
hydrogen - bonding between them. So they have London-dispersion forces between them..
for a compound to be soluble in water 1. it shou.pdf
for a compound to be soluble in water 1. it should have a large dipolemoment or 2.
it should be ionic or 3. it should form H bond with water in first pair CH3CH2CH3 is non polar
and is not soluble but CH2OHCHOHCH2OH is soluble in water by making H bonds. in second
pair Though CH3Cl has little dipolemoment HCl has very higher dipolemomet because of charge
separation and is readily soluble in water
Solution
for a compound to be soluble in water 1. it should have a large dipolemoment or 2.
it should be ionic or 3. it should form H bond with water in first pair CH3CH2CH3 is non polar
and is not soluble but CH2OHCHOHCH2OH is soluble in water by making H bonds. in second
pair Though CH3Cl has little dipolemoment HCl has very higher dipolemomet because of charge
separation and is readily soluble in water.
e) II and III have the loest, equal stability .pdf
e) II and III have the loest, equal stability
Solution
e) II and III have the loest, equal stability.
TRUE The Reynolds number of falling 2.5 micrometer flyash particles .pdf
TRUE The Reynolds number of falling 2.5 micrometer flyash particles in air exceed that of same
size of soot particle by more than 50% because the soot particle is a outcome produced due to in
complete combustion of carbon particles and very lighter in density compared to the flyash.
Solution
TRUE The Reynolds number of falling 2.5 micrometer flyash particles in air exceed that of same
size of soot particle by more than 50% because the soot particle is a outcome produced due to in
complete combustion of carbon particles and very lighter in density compared to the flyash..
the electron rich molecule is a central Xe with 4 F singlebonded.pdf
the electron rich molecule is a central Xe with 4 F singlebonded to it, 1 lone pair, and 1 oxygen
molecule double bonded toit (if this is confusing the molecule is called xenonoxytetrafluoride so
you can google it or whatever)
XeF6+H2O XeOF4 +2HF
Solution
the electron rich molecule is a central Xe with 4 F singlebonded to it, 1 lone pair, and 1 oxygen
molecule double bonded toit (if this is confusing the molecule is called xenonoxytetrafluoride so
you can google it or whatever)
XeF6+H2O XeOF4 +2HF.
The given formulae themselves are explicit. Explicit means direct i..pdf
The given formulae themselves are explicit. Explicit means direct i.e., the ones having a clear
expression of one variable in terms of several others which is the case here, as an is expressed as
a direct function of n and several lower order terms. By direct or clear we mean that expression
which represents, the variable (on the LHS of the equation) that is expressed, as a function of
other variables does not involve the expressed variable (in the RHS of the equation).
Solution
The given formulae themselves are explicit. Explicit means direct i.e., the ones having a clear
expression of one variable in terms of several others which is the case here, as an is expressed as
a direct function of n and several lower order terms. By direct or clear we mean that expression
which represents, the variable (on the LHS of the equation) that is expressed, as a function of
other variables does not involve the expressed variable (in the RHS of the equation)..
The answer is Meiosis, It occurs in human body at reporductive organ.pdf
The answer is Meiosis, It occurs in human body at reporductive organs only. For healing and
growth Mitosis plays a crucial role.
Solution
The answer is Meiosis, It occurs in human body at reporductive organs only. For healing and
growth Mitosis plays a crucial role..
Structure of HemoglobinHemoglobin is a chromo protein and is found.pdf
Structure of Hemoglobin
Hemoglobin is a chromo protein and is found in red blood cells, it’s a conjugated protein (heme
as prosthetic group + globin as the protein part apoprotein). Adults have 14.0 to 16.0 gm% of Hb
and 90 mg/kg of Hb will be produced and destroyed in the body on daily basis. Molecular weight
of Hb is 67,000 and 3.4 mg of iron present in each gram of Hb.
The combination of iron with a porphyrin ring produces the heme.
Structure of Heme
Heme is derived from porphyrin and porphyrins are cyclic compounds and are formed when 4
pyrrole rings fuse and are linked by methenyl bridges and the four rings are named as I,II,III, IV
and Alpha, beta, gamma and delta are the bridges. To the side chain of Porphyrins, four pyrrole
rings are attached.
We can find one ferrous atom (Fe++) co-ordinated at the centre of the of protoporphyrin IX tetra
pyrrole ring.
Structure of Globin
We can find the tetramer of globin polypeptide chains and each Hb molecule will have 4 Heme
units and the subunits of hemoglobin are found to be arranged in a tetrahedral array. And this
arrangement will give tight spherical overall appearance (which allows the polar residues being
on the exposed surface and keeps the non-polar interactions internal).
A molecule of hemoglobin is known to transport up to four oxygen molecules and here iron ion
interacts with oxygen molecule to form oxyhemoglobin. Oxyhemoglobin blood is bright red and
interactions between the iron–oxygen are very weak and thus can easily be separated without
disturbing the heme unit/ the oxygen molecule (completely reversible binding).
Deoxyhemoglobin is the hemoglobin molecule without oxygen and dark red..
Primary structure of hemoglobin
141 AA residues will be present in linear sequence of alpha chain contains and non- (, and )
chains will be of 146 amino acids in length (here the beta chain will have valine and histidine as
their first residues and Tyr b145 and His b146 found at C-terminal residues). Only 10 residue
difference between the delta chain and the beta chain.
Secondary structure of hemoglobin
We can find nearly 75 percent of the amino acids in or chains in a helical arrangement and 8
helical areas will be found in the chains.
Tertiary structure of or chains
Sphere type o structure will result during the tertiary folding of each globin chain and this
folding brings the Polar or charged side chains directed towards the outer surface of the subunit
and non-polar structures directed inwards making Hb water soluble, Heme pocket will be created
and is open-toped cleft in the surface and this folding will bring Hb in correct orientation to
allow these bonds to form.
Quaternary structure of hemoglobin
Finally the Hb tetramer will be formed composed of two identical dimers ()1and ()2. These two
polypeptide chains are held together tightly (though hydrophobic and ionic interactions andy
hydrogen bonding). The two dimers can move with respect to each other.
T and R forms of Hb
T form (taut structure.
SolutionSelection Sort after two iterations1 14 8 9 5 16 2 1.pdf
Solution
:
Selection Sort after two iterations:
1 14 8 9 5 16 2
1 2 14 9 8 16 5
program:
class Sorting
{
public static void main(String args[])
{
int a[]=new int[]{8,14,2,9,5,16,1};
int i,j,k;
for(i=0;i<2;i++)
{
for(j=i+1;j<7;j++)
{
if(a[i]>a[j])
{
int temp=a[i];
a[i]=a[j];
a[j]=temp;
}
}
for(k=0;k<7;k++)
{
System.out.print(a[k]+\" \");
}
System.out.println();
}
}
}
INsertion Sort: after 2 iterations
8 14 2 9 5 16 1
2 8 14 9 5 16 1
class Sorting
{
public static void main(String args[])
{
int a[]=new int[]{14,8,2,9,5,16,1};
int i,j,k;
for (j = 1; j < 7; j++) {
int temp = a[j];
i = j-1;
while ( (i > -1) && ( a [i] > temp ) ) {
a [i+1] = a [i];
i--;
}
a[i+1] = temp;
for(k=0;k<7;k++)
{
System.out.print(a[k]+\" \");
}
System.out.println();
}
}
}
Bubble Sort: after 1 Iteration:
8 2 9 5 14 1 16
class Sorting
{
public static void main(String args[])
{
int a[]=new int[]{14,8,2,9,5,16,1};
int i,j,k,temp;
for(i=0; i < 7; i++){
for(j=1; j < (7-i); j++){
if(a[j-1] > a[j]){
//swap the elements!
temp = a[j-1];
a[j-1] = a[j];
a[j] = temp;
}
}
for(k=0;k<7;k++)
{
System.out.print(a[k]+\" \");
}
System.out.println();
}
}
}.
Starch agar consists of starch. The secretion of amylase enzyme by s.pdf
Starch agar consists of starch. The secretion of amylase enzyme by some microbes cause
hydrolysis of starch around the colony. The plate flooded with iodine solution after incubation
period.
The amylase prodding colonies appear unstained and form a color zone because they’ve
hydrolyzed starch. The colonies that don’t produce amylase, still have starch in their surrounding
and appear dark-blue or black (color of starch-iodine complex) by reaction of starch with iodine.
There is no clearance zone around the amylase-negative colonies.
Thus, the media differentiates amylase producers from non-amylase produces and can be treated
as a differential media.
Solution
Starch agar consists of starch. The secretion of amylase enzyme by some microbes cause
hydrolysis of starch around the colony. The plate flooded with iodine solution after incubation
period.
The amylase prodding colonies appear unstained and form a color zone because they’ve
hydrolyzed starch. The colonies that don’t produce amylase, still have starch in their surrounding
and appear dark-blue or black (color of starch-iodine complex) by reaction of starch with iodine.
There is no clearance zone around the amylase-negative colonies.
Thus, the media differentiates amylase producers from non-amylase produces and can be treated
as a differential media..
Solution find the solution below and please give the feedback and p.pdf
Solution
: find the solution below and please give the feedback and please let me know if you have any
query regarding this question.
Time (hrs)
Intensity(in/hr)
Intensity (cm/hr)
F value
Infiltration value(f)
0.25
1.2
3.048
1.137
2.503671944
0.5
2.4
6.096
1.6759
1.907607852
0.75
4.8
12.192
2.1167
1.645712666
1
3.6
9.144
2.5073
1.490595461
1.25
1.4
3.556
2.866
1.385389044
1.5
0.6
1.524
3.2022
1.308180314
Total rainfall
3.2
8.128
2.56028932 (Total Infiltration)
Type of soil
silty loam
Features:
Porosity
0.501
residual porosity
0.015
effective porosity
0.486
suction head(cm)
16.68
Conductivity(k) (cm/hr)
0.65
Degree os saturation
0.6
delta =(effective porosity(1-degree of saturation))
0.1944
si
3.242592
F
(F-si*LN((1+(F/si))-k*t)=0
Time (hrs)
Intensity(in/hr)
Intensity (cm/hr)
F value
Infiltration value(f)
0.25
1.2
3.048
1.137
2.503671944
0.5
2.4
6.096
1.6759
1.907607852
0.75
4.8
12.192
2.1167
1.645712666
1
3.6
9.144
2.5073
1.490595461
1.25
1.4
3.556
2.866
1.385389044
1.5
0.6
1.524
3.2022
1.308180314
Total rainfall
3.2
8.128
2.56028932 (Total Infiltration)
Type of soil
silty loam
Features:
Porosity
0.501
residual porosity
0.015
effective porosity
0.486
suction head(cm)
16.68
Conductivity(k) (cm/hr)
0.65
Degree os saturation
0.6
delta =(effective porosity(1-degree of saturation))
0.1944
si
3.242592
F
(F-si*LN((1+(F/si))-k*t)=0.
public class DoubleArraySeq implements Cloneable { Priva.pdf
public class DoubleArraySeq implements Cloneable
{
// Private Instance Variables
private double[] data;
private int manyItems;
private int currentIndex;
//Constructor Methods
/**
* Initialize an empty sequence with an initial capacity of 10.
**/
public DoubleArraySeq()
{
try{
//Set a default capacity for new DoubleArraySeq\'s.
int INITAL_CAPACITY = 10;
//Set each instance variable to its initial value.
data = new double[INITAL_CAPACITY];
manyItems = 0;
currentIndex = 0;
}//end try
catch (OutOfMemoryError e){
throw new OutOfMemoryError (\"There is not enough memory to create a new
sequence!\");
}//end catch
}//end DoubleArraySeq() method
/**
* Initialize an empty sequence with a specified initial capacity. Note that the addAfter and
addBefore methods work
* efficiently (without needing more memory) until this capacity is reached.
**/
public DoubleArraySeq(int initialCapacity)
{
try{
//Set each instance variable to its initial value.
data = new double[initialCapacity];
currentIndex = 0;
manyItems = 0;
}//end try
catch (OutOfMemoryError e){
throw new OutOfMemoryError (\"There is not enough memory to create a new sequence
of capacity \" + initialCapacity + \"!\");
}//end catch
}//end DoubleArraySeq(int initialCapacity) method
// Accessor Methods
/**
**/
public boolean isCurrent()
{
return (currentIndex < manyItems);
}//end isCurrent() method
/**
* Accessor method to get the current element of this sequence.
**/
public double getCurrent()
{
//Confirm that there is a current element first.
if (isCurrent())
return data[currentIndex];
else
throw new IllegalStateException(\"There is no current element! Please specify a current
element first.\");
}//end getCurrent() method
/**
* Accessor method to get the current capacity of this sequence.
**/
public int getCapacity()
{
//Returns the number of indexes in the array.
return data.length;
}//end getCapacity() method
/**
* Accessor method to get the available capacity (number of empty indexes) of this sequence.
* The available capacity (number of empty indexes) of this sequence.
**/
public int getAvailCapacity()
{
//Returns the number of empty indexes in the array.
return data.length - manyItems;
}//end getAvailCapacity() method
/**
**/
public int size()
{
//Returns the number of elements in the sequence.
return manyItems;
}//end size() method
// Setter Methods
/**
* A method to move forward, so the current element is now the next element in this sequence.
**/
public void advance()
{
if (isCurrent())
currentIndex++;
else
throw new IllegalStateException (\"There is no current element! Advance may not be
called.\");
}//end advance() method
/**
* A method to set the current element at the front of this sequence.
**/
public void start()
{
if (manyItems > 0)
currentIndex = 0;
else
throw new IllegalStateException(\"This sequence is empty!\");
}//end start() method
/**
* A method that makes the last element of the sequence the current element.
**/
public void setCurrentLast()
{
if (manyItems > 0)
currentIndex = ma.
plese post the question clearly not able to understandSolution.pdf
plese post the question clearly not able to understand
Solution
plese post the question clearly not able to understand.
Lymphatic system function is to produce T- helper cells and B- helpe.pdf
Lymphatic system function is to produce T- helper cells and B- helper cells,it is majorly
concerned about productions of Antibodies that fights aganish infections.T he B& T cells
produce memory cells which can recognise the germ if it affects second time .It plays a
important role in immune response that protects body from infections
Both cardiovascular and lymphatic systems are considered about transport of fluids entire body
The difference is cardiovascular system considered about transporting materials like
oxygen,hormones,salts etc....whereas lymphatic system considered about removal of wastes from
the body fluids
Solution
Lymphatic system function is to produce T- helper cells and B- helper cells,it is majorly
concerned about productions of Antibodies that fights aganish infections.T he B& T cells
produce memory cells which can recognise the germ if it affects second time .It plays a
important role in immune response that protects body from infections
Both cardiovascular and lymphatic systems are considered about transport of fluids entire body
The difference is cardiovascular system considered about transporting materials like
oxygen,hormones,salts etc....whereas lymphatic system considered about removal of wastes from
the body fluids.
b) 0.2 M HI HI is a strong acid, which can comple.pdf
b) 0.2 M HI HI is a strong acid, which can completely ionized. LiOH is a base
(ionized as Li+ and OH-) CH3COH is a very weak acid, which is only barely ionized.
Solution
b) 0.2 M HI HI is a strong acid, which can completely ionized. LiOH is a base
(ionized as Li+ and OH-) CH3COH is a very weak acid, which is only barely ionized..
Integer version of the member functions are as belowint operate(in.pdf
Integer version of the member functions are as below
int operate(int &a,int &b);
int myOps is also integer member function that overrides the operate function.
a and b are called as member variables
Solution
Integer version of the member functions are as below
int operate(int &a,int &b);
int myOps is also integer member function that overrides the operate function.
a and b are called as member variables.
I have done it under neatbeans IDE and just added ToggleGroup for gr.pdf
I have done it under neatbeans IDE and just added ToggleGroup for grouping radion buttons
inorder to select one at time.
package tempconverter;
import javafx.application.Application;
import javafx.beans.value.ChangeListener;
import javafx.beans.value.ObservableValue;
import javafx.event.ActionEvent;
import javafx.event.EventHandler;
import javafx.geometry.Pos;
import javafx.scene.Scene;
import javafx.scene.control.Button;
import javafx.scene.control.Label;
import javafx.scene.control.RadioButton;
import javafx.scene.control.ScrollPane;
import javafx.scene.control.TextArea;
import javafx.scene.control.TextField;
import javafx.scene.control.Toggle;
import javafx.scene.control.ToggleGroup;
import javafx.scene.layout.BorderPane;
import javafx.scene.layout.StackPane;
import javafx.scene.layout.VBox;
import javafx.stage.Stage;
public class Tempconverter extends Application {
/**
* @param args the command line arguments executes project
*/
final ToggleGroup group1 = new ToggleGroup();
final ToggleGroup group2 = new ToggleGroup();
public static void main(String[] args) {
launch(args);
}
@Override
public void start(Stage primaryStage) throws Exception {
BorderPane pane = new BorderPane(); //the big picture
BorderPane p1 = new BorderPane();//top
BorderPane p2 = new BorderPane(); //bottom
Label temp = new Label(\"Enter a temperature\"); //enter temp label
Label convertTemp = new Label(\"Converted temperature\"); //converted temp label
Label input = new Label(\"Input Scale\");//input scale label
Label output = new Label(\"Output Scale\");//Output Scale label
TextField tempText = new TextField(); //text field to enter temp
TextField convertText = new TextField(); //output textfield for the converted temp
VBox scaleIn = new VBox(15);
VBox scaleOut = new VBox(15);
RadioButton c = new RadioButton(\"celcius\"); //button for celcius on input side
c.setToggleGroup(group1);
RadioButton f = new RadioButton(\"Fahrenheit\");//button for Fahrenheit on input side
f.setToggleGroup(group1);
RadioButton k = new RadioButton(\"kelvin\");//button for kelvin on inputside
k.setToggleGroup(group1);
RadioButton c1 = new RadioButton(\"celcius\");//button for celcius on outputside
c1.setToggleGroup(group2);
RadioButton f1 = new RadioButton(\"Fahrenheit\"); //button for f on output side
f1.setToggleGroup(group2);
RadioButton k1 = new RadioButton(\"kelvin\"); //button for kelvin on output side
k1.setToggleGroup(group2);
p1.setStyle(\"-fx-border-color:red\"); //set top border pane to red
p1.setRight(tempText);
p1.setLeft(temp);
pane.setTop(p1);
pane.setBottom(p2);
p2.setLeft(convertTemp);
p2.setRight(convertText);
p2.setStyle(\"-fx-border-color:red\");
input.setPrefWidth(100);
output.setPrefWidth(100);
c.setPrefWidth(100);
f.setPrefWidth(100);
k.setPrefWidth(100);
c1.setPrefWidth(100);
f1.setPrefWidth(100);
k1.setPrefWidth(100);
scaleIn.setStyle(\"-fx-border-color:red\");
scaleIn.getChildren().addAll(input, c, f, k);
scaleIn.setAlignment(Pos.CENTER);
pane.setLeft(scaleIn);
TextArea.
Model Attribute Check Company Auto Property
In Odoo, the multi-company feature allows you to manage multiple companies within a single Odoo database instance. Each company can have its own configurations while still sharing common resources such as products, customers, and suppliers.
MASS MEDIA STUDIES-835-CLASS XI Resource Material.pdf
mass media studies
CBSE
Full Book
for class 11th
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Hg2+ is more water soluble, methyl mercury HgCH3+.pdf
Hg2+ is more water soluble, methyl mercury HgCH3+ is more lipid soluble.
CH3Hg, polar molecules would combine to form a substance similar to water but would be more
acidic as a result of mixing two non-like particles in a solution
Solution
Hg2+ is more water soluble, methyl mercury HgCH3+ is more lipid soluble.
CH3Hg, polar molecules would combine to form a substance similar to water but would be more
acidic as a result of mixing two non-like particles in a solution.
HCl is a polar-covalent compound and also Acetoni.pdf
HCl is a polar-covalent compound and also Acetonitrile they are soluble dueto
hydrogen - bonding between them. So they have London-dispersion forces between them.
Solution
HCl is a polar-covalent compound and also Acetonitrile they are soluble dueto
hydrogen - bonding between them. So they have London-dispersion forces between them..
for a compound to be soluble in water 1. it shou.pdf
for a compound to be soluble in water 1. it should have a large dipolemoment or 2.
it should be ionic or 3. it should form H bond with water in first pair CH3CH2CH3 is non polar
and is not soluble but CH2OHCHOHCH2OH is soluble in water by making H bonds. in second
pair Though CH3Cl has little dipolemoment HCl has very higher dipolemomet because of charge
separation and is readily soluble in water
Solution
for a compound to be soluble in water 1. it should have a large dipolemoment or 2.
it should be ionic or 3. it should form H bond with water in first pair CH3CH2CH3 is non polar
and is not soluble but CH2OHCHOHCH2OH is soluble in water by making H bonds. in second
pair Though CH3Cl has little dipolemoment HCl has very higher dipolemomet because of charge
separation and is readily soluble in water.
e) II and III have the loest, equal stability .pdf
e) II and III have the loest, equal stability
Solution
e) II and III have the loest, equal stability.
TRUE The Reynolds number of falling 2.5 micrometer flyash particles .pdf
TRUE The Reynolds number of falling 2.5 micrometer flyash particles in air exceed that of same
size of soot particle by more than 50% because the soot particle is a outcome produced due to in
complete combustion of carbon particles and very lighter in density compared to the flyash.
Solution
TRUE The Reynolds number of falling 2.5 micrometer flyash particles in air exceed that of same
size of soot particle by more than 50% because the soot particle is a outcome produced due to in
complete combustion of carbon particles and very lighter in density compared to the flyash..
the electron rich molecule is a central Xe with 4 F singlebonded.pdf
the electron rich molecule is a central Xe with 4 F singlebonded to it, 1 lone pair, and 1 oxygen
molecule double bonded toit (if this is confusing the molecule is called xenonoxytetrafluoride so
you can google it or whatever)
XeF6+H2O XeOF4 +2HF
Solution
the electron rich molecule is a central Xe with 4 F singlebonded to it, 1 lone pair, and 1 oxygen
molecule double bonded toit (if this is confusing the molecule is called xenonoxytetrafluoride so
you can google it or whatever)
XeF6+H2O XeOF4 +2HF.
The given formulae themselves are explicit. Explicit means direct i..pdf
The given formulae themselves are explicit. Explicit means direct i.e., the ones having a clear
expression of one variable in terms of several others which is the case here, as an is expressed as
a direct function of n and several lower order terms. By direct or clear we mean that expression
which represents, the variable (on the LHS of the equation) that is expressed, as a function of
other variables does not involve the expressed variable (in the RHS of the equation).
Solution
The given formulae themselves are explicit. Explicit means direct i.e., the ones having a clear
expression of one variable in terms of several others which is the case here, as an is expressed as
a direct function of n and several lower order terms. By direct or clear we mean that expression
which represents, the variable (on the LHS of the equation) that is expressed, as a function of
other variables does not involve the expressed variable (in the RHS of the equation)..
The answer is Meiosis, It occurs in human body at reporductive organ.pdf
The answer is Meiosis, It occurs in human body at reporductive organs only. For healing and
growth Mitosis plays a crucial role.
Solution
The answer is Meiosis, It occurs in human body at reporductive organs only. For healing and
growth Mitosis plays a crucial role..
Structure of HemoglobinHemoglobin is a chromo protein and is found.pdf
Structure of Hemoglobin
Hemoglobin is a chromo protein and is found in red blood cells, it’s a conjugated protein (heme
as prosthetic group + globin as the protein part apoprotein). Adults have 14.0 to 16.0 gm% of Hb
and 90 mg/kg of Hb will be produced and destroyed in the body on daily basis. Molecular weight
of Hb is 67,000 and 3.4 mg of iron present in each gram of Hb.
The combination of iron with a porphyrin ring produces the heme.
Structure of Heme
Heme is derived from porphyrin and porphyrins are cyclic compounds and are formed when 4
pyrrole rings fuse and are linked by methenyl bridges and the four rings are named as I,II,III, IV
and Alpha, beta, gamma and delta are the bridges. To the side chain of Porphyrins, four pyrrole
rings are attached.
We can find one ferrous atom (Fe++) co-ordinated at the centre of the of protoporphyrin IX tetra
pyrrole ring.
Structure of Globin
We can find the tetramer of globin polypeptide chains and each Hb molecule will have 4 Heme
units and the subunits of hemoglobin are found to be arranged in a tetrahedral array. And this
arrangement will give tight spherical overall appearance (which allows the polar residues being
on the exposed surface and keeps the non-polar interactions internal).
A molecule of hemoglobin is known to transport up to four oxygen molecules and here iron ion
interacts with oxygen molecule to form oxyhemoglobin. Oxyhemoglobin blood is bright red and
interactions between the iron–oxygen are very weak and thus can easily be separated without
disturbing the heme unit/ the oxygen molecule (completely reversible binding).
Deoxyhemoglobin is the hemoglobin molecule without oxygen and dark red..
Primary structure of hemoglobin
141 AA residues will be present in linear sequence of alpha chain contains and non- (, and )
chains will be of 146 amino acids in length (here the beta chain will have valine and histidine as
their first residues and Tyr b145 and His b146 found at C-terminal residues). Only 10 residue
difference between the delta chain and the beta chain.
Secondary structure of hemoglobin
We can find nearly 75 percent of the amino acids in or chains in a helical arrangement and 8
helical areas will be found in the chains.
Tertiary structure of or chains
Sphere type o structure will result during the tertiary folding of each globin chain and this
folding brings the Polar or charged side chains directed towards the outer surface of the subunit
and non-polar structures directed inwards making Hb water soluble, Heme pocket will be created
and is open-toped cleft in the surface and this folding will bring Hb in correct orientation to
allow these bonds to form.
Quaternary structure of hemoglobin
Finally the Hb tetramer will be formed composed of two identical dimers ()1and ()2. These two
polypeptide chains are held together tightly (though hydrophobic and ionic interactions andy
hydrogen bonding). The two dimers can move with respect to each other.
T and R forms of Hb
T form (taut structure.
SolutionSelection Sort after two iterations1 14 8 9 5 16 2 1.pdf
Solution
:
Selection Sort after two iterations:
1 14 8 9 5 16 2
1 2 14 9 8 16 5
program:
class Sorting
{
public static void main(String args[])
{
int a[]=new int[]{8,14,2,9,5,16,1};
int i,j,k;
for(i=0;i<2;i++)
{
for(j=i+1;j<7;j++)
{
if(a[i]>a[j])
{
int temp=a[i];
a[i]=a[j];
a[j]=temp;
}
}
for(k=0;k<7;k++)
{
System.out.print(a[k]+\" \");
}
System.out.println();
}
}
}
INsertion Sort: after 2 iterations
8 14 2 9 5 16 1
2 8 14 9 5 16 1
class Sorting
{
public static void main(String args[])
{
int a[]=new int[]{14,8,2,9,5,16,1};
int i,j,k;
for (j = 1; j < 7; j++) {
int temp = a[j];
i = j-1;
while ( (i > -1) && ( a [i] > temp ) ) {
a [i+1] = a [i];
i--;
}
a[i+1] = temp;
for(k=0;k<7;k++)
{
System.out.print(a[k]+\" \");
}
System.out.println();
}
}
}
Bubble Sort: after 1 Iteration:
8 2 9 5 14 1 16
class Sorting
{
public static void main(String args[])
{
int a[]=new int[]{14,8,2,9,5,16,1};
int i,j,k,temp;
for(i=0; i < 7; i++){
for(j=1; j < (7-i); j++){
if(a[j-1] > a[j]){
//swap the elements!
temp = a[j-1];
a[j-1] = a[j];
a[j] = temp;
}
}
for(k=0;k<7;k++)
{
System.out.print(a[k]+\" \");
}
System.out.println();
}
}
}.
Starch agar consists of starch. The secretion of amylase enzyme by s.pdf
Starch agar consists of starch. The secretion of amylase enzyme by some microbes cause
hydrolysis of starch around the colony. The plate flooded with iodine solution after incubation
period.
The amylase prodding colonies appear unstained and form a color zone because they’ve
hydrolyzed starch. The colonies that don’t produce amylase, still have starch in their surrounding
and appear dark-blue or black (color of starch-iodine complex) by reaction of starch with iodine.
There is no clearance zone around the amylase-negative colonies.
Thus, the media differentiates amylase producers from non-amylase produces and can be treated
as a differential media.
Solution
Starch agar consists of starch. The secretion of amylase enzyme by some microbes cause
hydrolysis of starch around the colony. The plate flooded with iodine solution after incubation
period.
The amylase prodding colonies appear unstained and form a color zone because they’ve
hydrolyzed starch. The colonies that don’t produce amylase, still have starch in their surrounding
and appear dark-blue or black (color of starch-iodine complex) by reaction of starch with iodine.
There is no clearance zone around the amylase-negative colonies.
Thus, the media differentiates amylase producers from non-amylase produces and can be treated
as a differential media..
Solution find the solution below and please give the feedback and p.pdf
Solution
: find the solution below and please give the feedback and please let me know if you have any
query regarding this question.
Time (hrs)
Intensity(in/hr)
Intensity (cm/hr)
F value
Infiltration value(f)
0.25
1.2
3.048
1.137
2.503671944
0.5
2.4
6.096
1.6759
1.907607852
0.75
4.8
12.192
2.1167
1.645712666
1
3.6
9.144
2.5073
1.490595461
1.25
1.4
3.556
2.866
1.385389044
1.5
0.6
1.524
3.2022
1.308180314
Total rainfall
3.2
8.128
2.56028932 (Total Infiltration)
Type of soil
silty loam
Features:
Porosity
0.501
residual porosity
0.015
effective porosity
0.486
suction head(cm)
16.68
Conductivity(k) (cm/hr)
0.65
Degree os saturation
0.6
delta =(effective porosity(1-degree of saturation))
0.1944
si
3.242592
F
(F-si*LN((1+(F/si))-k*t)=0
Time (hrs)
Intensity(in/hr)
Intensity (cm/hr)
F value
Infiltration value(f)
0.25
1.2
3.048
1.137
2.503671944
0.5
2.4
6.096
1.6759
1.907607852
0.75
4.8
12.192
2.1167
1.645712666
1
3.6
9.144
2.5073
1.490595461
1.25
1.4
3.556
2.866
1.385389044
1.5
0.6
1.524
3.2022
1.308180314
Total rainfall
3.2
8.128
2.56028932 (Total Infiltration)
Type of soil
silty loam
Features:
Porosity
0.501
residual porosity
0.015
effective porosity
0.486
suction head(cm)
16.68
Conductivity(k) (cm/hr)
0.65
Degree os saturation
0.6
delta =(effective porosity(1-degree of saturation))
0.1944
si
3.242592
F
(F-si*LN((1+(F/si))-k*t)=0.
public class DoubleArraySeq implements Cloneable { Priva.pdf
public class DoubleArraySeq implements Cloneable
{
// Private Instance Variables
private double[] data;
private int manyItems;
private int currentIndex;
//Constructor Methods
/**
* Initialize an empty sequence with an initial capacity of 10.
**/
public DoubleArraySeq()
{
try{
//Set a default capacity for new DoubleArraySeq\'s.
int INITAL_CAPACITY = 10;
//Set each instance variable to its initial value.
data = new double[INITAL_CAPACITY];
manyItems = 0;
currentIndex = 0;
}//end try
catch (OutOfMemoryError e){
throw new OutOfMemoryError (\"There is not enough memory to create a new
sequence!\");
}//end catch
}//end DoubleArraySeq() method
/**
* Initialize an empty sequence with a specified initial capacity. Note that the addAfter and
addBefore methods work
* efficiently (without needing more memory) until this capacity is reached.
**/
public DoubleArraySeq(int initialCapacity)
{
try{
//Set each instance variable to its initial value.
data = new double[initialCapacity];
currentIndex = 0;
manyItems = 0;
}//end try
catch (OutOfMemoryError e){
throw new OutOfMemoryError (\"There is not enough memory to create a new sequence
of capacity \" + initialCapacity + \"!\");
}//end catch
}//end DoubleArraySeq(int initialCapacity) method
// Accessor Methods
/**
**/
public boolean isCurrent()
{
return (currentIndex < manyItems);
}//end isCurrent() method
/**
* Accessor method to get the current element of this sequence.
**/
public double getCurrent()
{
//Confirm that there is a current element first.
if (isCurrent())
return data[currentIndex];
else
throw new IllegalStateException(\"There is no current element! Please specify a current
element first.\");
}//end getCurrent() method
/**
* Accessor method to get the current capacity of this sequence.
**/
public int getCapacity()
{
//Returns the number of indexes in the array.
return data.length;
}//end getCapacity() method
/**
* Accessor method to get the available capacity (number of empty indexes) of this sequence.
* The available capacity (number of empty indexes) of this sequence.
**/
public int getAvailCapacity()
{
//Returns the number of empty indexes in the array.
return data.length - manyItems;
}//end getAvailCapacity() method
/**
**/
public int size()
{
//Returns the number of elements in the sequence.
return manyItems;
}//end size() method
// Setter Methods
/**
* A method to move forward, so the current element is now the next element in this sequence.
**/
public void advance()
{
if (isCurrent())
currentIndex++;
else
throw new IllegalStateException (\"There is no current element! Advance may not be
called.\");
}//end advance() method
/**
* A method to set the current element at the front of this sequence.
**/
public void start()
{
if (manyItems > 0)
currentIndex = 0;
else
throw new IllegalStateException(\"This sequence is empty!\");
}//end start() method
/**
* A method that makes the last element of the sequence the current element.
**/
public void setCurrentLast()
{
if (manyItems > 0)
currentIndex = ma.
plese post the question clearly not able to understandSolution.pdf
plese post the question clearly not able to understand
Solution
plese post the question clearly not able to understand.
Lymphatic system function is to produce T- helper cells and B- helpe.pdf
Lymphatic system function is to produce T- helper cells and B- helper cells,it is majorly
concerned about productions of Antibodies that fights aganish infections.T he B& T cells
produce memory cells which can recognise the germ if it affects second time .It plays a
important role in immune response that protects body from infections
Both cardiovascular and lymphatic systems are considered about transport of fluids entire body
The difference is cardiovascular system considered about transporting materials like
oxygen,hormones,salts etc....whereas lymphatic system considered about removal of wastes from
the body fluids
Solution
Lymphatic system function is to produce T- helper cells and B- helper cells,it is majorly
concerned about productions of Antibodies that fights aganish infections.T he B& T cells
produce memory cells which can recognise the germ if it affects second time .It plays a
important role in immune response that protects body from infections
Both cardiovascular and lymphatic systems are considered about transport of fluids entire body
The difference is cardiovascular system considered about transporting materials like
oxygen,hormones,salts etc....whereas lymphatic system considered about removal of wastes from
the body fluids.
b) 0.2 M HI HI is a strong acid, which can comple.pdf
b) 0.2 M HI HI is a strong acid, which can completely ionized. LiOH is a base
(ionized as Li+ and OH-) CH3COH is a very weak acid, which is only barely ionized.
Solution
b) 0.2 M HI HI is a strong acid, which can completely ionized. LiOH is a base
(ionized as Li+ and OH-) CH3COH is a very weak acid, which is only barely ionized..
Integer version of the member functions are as belowint operate(in.pdf
Integer version of the member functions are as below
int operate(int &a,int &b);
int myOps is also integer member function that overrides the operate function.
a and b are called as member variables
Solution
Integer version of the member functions are as below
int operate(int &a,int &b);
int myOps is also integer member function that overrides the operate function.
a and b are called as member variables.
I have done it under neatbeans IDE and just added ToggleGroup for gr.pdf
I have done it under neatbeans IDE and just added ToggleGroup for grouping radion buttons
inorder to select one at time.
package tempconverter;
import javafx.application.Application;
import javafx.beans.value.ChangeListener;
import javafx.beans.value.ObservableValue;
import javafx.event.ActionEvent;
import javafx.event.EventHandler;
import javafx.geometry.Pos;
import javafx.scene.Scene;
import javafx.scene.control.Button;
import javafx.scene.control.Label;
import javafx.scene.control.RadioButton;
import javafx.scene.control.ScrollPane;
import javafx.scene.control.TextArea;
import javafx.scene.control.TextField;
import javafx.scene.control.Toggle;
import javafx.scene.control.ToggleGroup;
import javafx.scene.layout.BorderPane;
import javafx.scene.layout.StackPane;
import javafx.scene.layout.VBox;
import javafx.stage.Stage;
public class Tempconverter extends Application {
/**
* @param args the command line arguments executes project
*/
final ToggleGroup group1 = new ToggleGroup();
final ToggleGroup group2 = new ToggleGroup();
public static void main(String[] args) {
launch(args);
}
@Override
public void start(Stage primaryStage) throws Exception {
BorderPane pane = new BorderPane(); //the big picture
BorderPane p1 = new BorderPane();//top
BorderPane p2 = new BorderPane(); //bottom
Label temp = new Label(\"Enter a temperature\"); //enter temp label
Label convertTemp = new Label(\"Converted temperature\"); //converted temp label
Label input = new Label(\"Input Scale\");//input scale label
Label output = new Label(\"Output Scale\");//Output Scale label
TextField tempText = new TextField(); //text field to enter temp
TextField convertText = new TextField(); //output textfield for the converted temp
VBox scaleIn = new VBox(15);
VBox scaleOut = new VBox(15);
RadioButton c = new RadioButton(\"celcius\"); //button for celcius on input side
c.setToggleGroup(group1);
RadioButton f = new RadioButton(\"Fahrenheit\");//button for Fahrenheit on input side
f.setToggleGroup(group1);
RadioButton k = new RadioButton(\"kelvin\");//button for kelvin on inputside
k.setToggleGroup(group1);
RadioButton c1 = new RadioButton(\"celcius\");//button for celcius on outputside
c1.setToggleGroup(group2);
RadioButton f1 = new RadioButton(\"Fahrenheit\"); //button for f on output side
f1.setToggleGroup(group2);
RadioButton k1 = new RadioButton(\"kelvin\"); //button for kelvin on output side
k1.setToggleGroup(group2);
p1.setStyle(\"-fx-border-color:red\"); //set top border pane to red
p1.setRight(tempText);
p1.setLeft(temp);
pane.setTop(p1);
pane.setBottom(p2);
p2.setLeft(convertTemp);
p2.setRight(convertText);
p2.setStyle(\"-fx-border-color:red\");
input.setPrefWidth(100);
output.setPrefWidth(100);
c.setPrefWidth(100);
f.setPrefWidth(100);
k.setPrefWidth(100);
c1.setPrefWidth(100);
f1.setPrefWidth(100);
k1.setPrefWidth(100);
scaleIn.setStyle(\"-fx-border-color:red\");
scaleIn.getChildren().addAll(input, c, f, k);
scaleIn.setAlignment(Pos.CENTER);
pane.setLeft(scaleIn);
TextArea.
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Hg2+ is more water soluble, methyl mercury HgCH3+.pdf
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HCl is a polar-covalent compound and also Acetoni.pdf
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plese post the question clearly not able to understandSolution.pdf
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This will be used as part of your Personal Professional Portfolio once graded.
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