The enthalpy change of reaction = E(bonds broken) - E(bonds formed).pdf
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The enthalpy change of reaction = E(bonds broken) - E(bonds formed) = [2*BE(CO) + 1*BE(O2)] - [2*BE(CO2)] = 2*1074 + 499 - 2*(2*802) = -561 kJ/mol Solution The enthalpy change of reaction = E(bonds broken) - E(bonds formed) = [2*BE(CO) + 1*BE(O2)] - [2*BE(CO2)] = 2*1074 + 499 - 2*(2*802) = -561 kJ/mol.
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![The enthalpy change of reaction = E(bonds broken) - E(bonds formed)
= [2*BE(CO) + 1*BE(O2)] - [2*BE(CO2)]
= 2*1074 + 499 - 2*(2*802)
= -561 kJ/mol
Solution
The enthalpy change of reaction = E(bonds broken) - E(bonds formed)
= [2*BE(CO) + 1*BE(O2)] - [2*BE(CO2)]
= 2*1074 + 499 - 2*(2*802)
= -561 kJ/mol](https://image.slidesharecdn.com/theenthalpychangeofreactionebondsbroken-ebondsformed-230413081330-2c4d59a1/85/The-enthalpy-change-of-reaction-E-bonds-broken-E-bonds-formed-pdf-1-320.jpg)
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W since less electrons means difficult to melt so.pdf
W since less electrons means difficult to melt so high melting point
Solution
W since less electrons means difficult to melt so high melting point.
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Solution
To the left. The ketone is more stable compared to the geminal diol..
The true statement is D. Concentration affects .pdf
The true statement is : D. Concentration affects rate by influencing the frequency of
collisions between reactant particles
Solution
The true statement is : D. Concentration affects rate by influencing the frequency of
collisions between reactant particles.
Smaller mass gases have greater rates of effusion.pdf
Smaller mass gases have greater rates of effusion. Ar < HCl
Solution
Smaller mass gases have greater rates of effusion. Ar < HCl.
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Solution
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melting point of the compound increases due to th.pdf
melting point of the compound increases due to the presence of any insoluble
impurity such as sodium sulfate
Solution
melting point of the compound increases due to the presence of any insoluble
impurity such as sodium sulfate.
molality = moles of solute kg of solvent In thi.pdf
molality = moles of solute / kg of solvent In this equation, H2SO4 is the solute and
water is the solvent 0.91 mol / 0.011 kg = 83 mol/kg
Solution
molality = moles of solute / kg of solvent In this equation, H2SO4 is the solute and
water is the solvent 0.91 mol / 0.011 kg = 83 mol/kg.
Recommended
W since less electrons means difficult to melt so.pdf
W since less electrons means difficult to melt so high melting point
Solution
W since less electrons means difficult to melt so high melting point.
To the left. The ketone is more stable compared .pdf
To the left. The ketone is more stable compared to the geminal diol.
Solution
To the left. The ketone is more stable compared to the geminal diol..
The true statement is D. Concentration affects .pdf
The true statement is : D. Concentration affects rate by influencing the frequency of
collisions between reactant particles
Solution
The true statement is : D. Concentration affects rate by influencing the frequency of
collisions between reactant particles.
Smaller mass gases have greater rates of effusion.pdf
Smaller mass gases have greater rates of effusion. Ar < HCl
Solution
Smaller mass gases have greater rates of effusion. Ar < HCl.
please send the question details I have gone thr.pdf
please send the question details I have gone through the link send by you but
doesnot get anything.
Solution
please send the question details I have gone through the link send by you but
doesnot get anything..
melting point of the compound increases due to th.pdf
melting point of the compound increases due to the presence of any insoluble
impurity such as sodium sulfate
Solution
melting point of the compound increases due to the presence of any insoluble
impurity such as sodium sulfate.
molality = moles of solute kg of solvent In thi.pdf
molality = moles of solute / kg of solvent In this equation, H2SO4 is the solute and
water is the solvent 0.91 mol / 0.011 kg = 83 mol/kg
Solution
molality = moles of solute / kg of solvent In this equation, H2SO4 is the solute and
water is the solvent 0.91 mol / 0.011 kg = 83 mol/kg.
Ionic Equation is MgO(s) + 2 H+(aq) + 2NO3- (aq.pdf
Ionic Equation is MgO(s) + 2 H+(aq) + 2NO3- (aq) ---> Mg+2(aq) +2(NO3)2(aq)
+ H2O(l) if we look at the equation NO3- has not changed it is also in the same physical state so
it is a spectator ion
Solution
Ionic Equation is MgO(s) + 2 H+(aq) + 2NO3- (aq) ---> Mg+2(aq) +2(NO3)2(aq)
+ H2O(l) if we look at the equation NO3- has not changed it is also in the same physical state so
it is a spectator ion.
FeBrs is a typo Lets suppose it is FeBr2. 2Na .pdf
FeBrs is a typo? Let\'s suppose it is FeBr2. 2Na + FeBr2 = 2NaBr + Fe if it is
FeBr3 3Na + FeBr3 = 3NaBr + Fe
Solution
FeBrs is a typo? Let\'s suppose it is FeBr2. 2Na + FeBr2 = 2NaBr + Fe if it is
FeBr3 3Na + FeBr3 = 3NaBr + Fe.
d) in real gases there are attraction between m.pdf
d) in real gases there are attraction between molecules
Solution
d) in real gases there are attraction between molecules.
SolutionTraversing Binary Trees The preorder standard procedure.pdf
Solution
:
Traversing Binary Trees
The preorder standard procedures to traverse a binary tree are defined recursively as follows:
preorder(T):if T then visit root(T); preorder(left(T)); preorder(right(T))
preorder( ) =
preorder(tree(root, Left, Right)) = cat(root,preorder(Left), cat(preorder(Right)))
where cat concatenates two lists and can be defined by,
cat(,L) =L
cat(h ::t, L) = h :: cat(t, L)..
Shareholder’s equity= Current assets+Net fixed assets-Current lia.pdf
Shareholder’s equity:
= Current assets+Net fixed assets-Current liabilities-Long term debt
=$2,340+$11,000-$1,435-$4,150
= $7,755
Net working capital:
= Current assets-Current liabilities
=$2,340-$1,435
= $905
Solution
Shareholder’s equity:
= Current assets+Net fixed assets-Current liabilities-Long term debt
=$2,340+$11,000-$1,435-$4,150
= $7,755
Net working capital:
= Current assets-Current liabilities
=$2,340-$1,435
= $905.
in Reproductive cloning of mammals the Nucelus ( Genetic material) f.pdf
in Reproductive cloning of mammals the Nucelus ( Genetic material) from Donar egg is
removed, so it is enucleated.
the cell is taken from the organism which is to be cloned and remove the nucleus from that cell.
then these nucleus is introduced into the Donar egg (or) fuse these two cells using electrical
shock.
so cloned organism nucleus is fused with donar egg (enucleated egg).
Solution
in Reproductive cloning of mammals the Nucelus ( Genetic material) from Donar egg is
removed, so it is enucleated.
the cell is taken from the organism which is to be cloned and remove the nucleus from that cell.
then these nucleus is introduced into the Donar egg (or) fuse these two cells using electrical
shock.
so cloned organism nucleus is fused with donar egg (enucleated egg)..
a) OH- (aq) is a Lewis base because it can give electrions to Oth.pdf
a) OH- (aq) is a Lewis base because it can give electrions to Other.
b) Fe Br3 (s) is a lewis acid since it will accept electrons from other species.
c) Zn2+ (aq) is a Lewis acid and NH3 is a Lewis base
d) SO2 (g) is a Lewis acid because S has not containing an octate configuration.
Solution
a) OH- (aq) is a Lewis base because it can give electrions to Other.
b) Fe Br3 (s) is a lewis acid since it will accept electrons from other species.
c) Zn2+ (aq) is a Lewis acid and NH3 is a Lewis base
d) SO2 (g) is a Lewis acid because S has not containing an octate configuration..
package Chapter_20;import ToolKit.PostfixNotation;import javaf.pdf
package Chapter_20;
import ToolKit.PostfixNotation;
import javafx.application.Application;
import javafx.geometry.Insets;
import javafx.geometry.Pos;
import javafx.scene.Scene;
import javafx.scene.control.Button;
import javafx.scene.control.Label;
import javafx.scene.control.TextField;
import javafx.scene.image.Image;
import javafx.scene.image.ImageView;
import javafx.scene.layout.BorderPane;
import javafx.scene.layout.HBox;
import javafx.stage.Stage;
import java.util.ArrayList;
import java.util.Stack;
public class Exercise_13 extends Application {
int[] validNumbers = new int[4];
@Override
public void start(Stage primaryStage) {
// Top pane
Button btRefresh = new Button(\"Shuffle\");
Label lblStatus = new Label(\"\");
HBox topPane = new HBox(lblStatus, btRefresh);
topPane.setAlignment(Pos.BASELINE_RIGHT);
topPane.setSpacing(10);
// Center Pane
HBox centerPane = new HBox();
centerPane.setAlignment(Pos.CENTER);
centerPane.setSpacing(10);
centerPane.setPadding(new Insets(10));
// set first 4 random cards
setRandomCards(centerPane);
// Bottom pane
TextField tfExpression = new TextField();
Label lblExpression = new Label(\"Enter an expression:\");
Button btVerify = new Button(\"Verify\");
HBox bottomPane = new HBox(10, lblExpression, tfExpression, btVerify);
// Container Pane
BorderPane borderPane = new BorderPane();
borderPane.setPadding(new Insets(10));
borderPane.setTop(topPane);
borderPane.setCenter(centerPane);
borderPane.setBottom(bottomPane);
// Listeners
btRefresh.setOnAction(e -> {
lblStatus.setText(\"\");
setRandomCards(centerPane);
});
btVerify.setOnAction(e -> {
String expression = tfExpression.getText();
if (isValid(expression) && isCorrect(expression)) {
lblStatus.setText(\"Good job! \" + expression + \" = 24\");
} else {
lblStatus.setText(\"Invalid Expression\");
}
});
Scene scene = new Scene(borderPane);
primaryStage.setTitle(\"4 Random Cards\");
primaryStage.setScene(scene);
primaryStage.show();
// Debug
}
private void setRandomCards(HBox pane) {
boolean[] usedCards = new boolean[52];
// choose 4 random distinct cards from the deck
int count = 0;
pane.getChildren().clear();
while (count < 4) {
int card = (int) (Math.random() * 52);
if (!usedCards[card]) {
usedCards[card] = true;
pane.getChildren().add(new ImageView(new Image(\"image/card/\" + (++card) +
\".png\")));
int value = card % 13;
validNumbers[count] = (value == 0) ? 13 : value;
count++;
}
}
}
private static boolean isOperator(char ch) {
return (ch == \'(\' ||
ch == \')\' ||
isArithmeticOperator(ch));
}
private static boolean isArithmeticOperator(char ch) {
return (ch == \'/\' ||
ch == \'+\' ||
ch == \'-\' ||
ch == \'*\');
}
private static String[] separateExpression(String s) {
ArrayList tokens = new ArrayList<>(30);
char[] chars = s.toCharArray();
String numBuffer = \"\";
for (char ch : chars) {
if (isOperator(ch)) {
if (numBuffer.length() > 0) {
tokens.add(numBuffer);
numBuffer = \"\";
}
tokens.add(ch + \"\");
} else {
if (ch != \' \')
numBuffer += ch;
}
}
if (numBuffe.
Initial concentration of NH3 = molesvolume = 0.2501.00 = 0.250 M.pdf
Initial concentration of NH3 = moles/volume = 0.250/1.00 = 0.250 M
NH3 + H2O <=> NH4+ + OH-
I 0.25 0 0
C -a +a +a
E 0.25-a a a
Kb = [NH4+][OH-]/[NH3]
= a2/(0.25 - a) = 1.8 x 10-5
a2 + 1.8 x 10-5a - 4.5 x 10-6 = 0
a = 0.002112
[OH-] = a = 0.002112 M
[H+] = Kw/[OH-] where Kw is the ionic product of water
= 10-14/0.002112 = 4.735 x 10-12 M
pH = -log[H+] = 11.3
Solution
Initial concentration of NH3 = moles/volume = 0.250/1.00 = 0.250 M
NH3 + H2O <=> NH4+ + OH-
I 0.25 0 0
C -a +a +a
E 0.25-a a a
Kb = [NH4+][OH-]/[NH3]
= a2/(0.25 - a) = 1.8 x 10-5
a2 + 1.8 x 10-5a - 4.5 x 10-6 = 0
a = 0.002112
[OH-] = a = 0.002112 M
[H+] = Kw/[OH-] where Kw is the ionic product of water
= 10-14/0.002112 = 4.735 x 10-12 M
pH = -log[H+] = 11.3.
There are ten guidelines with a broad coverage, ranging from develop.pdf
There are ten guidelines with a broad coverage, ranging from developing a strategic role for
information to the development of capabilities and the marketing of the information unit. There
is a considerable degree of overlap between these guidelines, since many of them support and
depend on each other. Mostly, they are guidelines for development, not maintaining the status
quo. Therefore it is essential that you should regularly devote at least 10% of your time to these
activities, and certainly more in the early stages.
1. Establishing the Strategic Role of Information
This is a two stage process: (1) a research and investigation phase that gives you the information
you need for (2) articulating your mission and strategy. The first phase requires an assessment of
the attitudes of senior management to information and how much they are willing to pay for it. I
recently asked the head of a market research unit how hard it was to justify their existence and
budget. He commented: \"that has not been a problem, ever since we spend hundreds of millions
of dollars on a new product and then lost market share to the Japanese\". This direct link between
a large strategic investment and the bottom line delivered a sharp lesson to senior management
on the value of competitor intelligence.
Some of the strategic decisions whose successful outcomes depend on the availability of good
information include:
Market selection and targeting
New investments
Location of factories and offices
New product development and launch
Pricing and Promotion
Find out how these decisions are made, what information is used, and from where it is sourced.
You may already have data from feedback on how the information you have supplied has helped
such processes (if not, you should get it!). Has your organisation recently had successes or
failures that could be directly attributed to good or bad information? From these investigations
you can determine areas of high information leverage where you could play a role. Identify the
linkages between information and results. This should then be a cornerstone of your strategy.
Use every opportunity to let people, especially senior managers, know about it.
2. Identify Users Real Needs
This is the first of the marketing guidelines. It is essentially about market research. Therefore use
the methods used by researchers - surveys, interviews, usage analysis. You already have users.
Find out how they use your output and again what results and benefits they achieve. One
particularly useful way of teasing this out (used, by the way, to justify office automation
systems) is to ask what would happen if you did not offer that service.
Getting to senior management users and non-users is an important strand of this activity. You
must also learn about their real needs, not the ones they may initially express. Some of these may
be psychological needs such as \"I want to impress our senior management team with the range
of authoritative information I have .
The way Ive been told to look at the classifications is to look at.pdf
The way I\'ve been told to look at the classifications is to look at their limitations. For example,
Arrhenius acids contain H and release H+ in water; this is the strict limitation of an Arrhenius
acid. Here is a chart to help with this
=> Arrhenius
=>acids => contain H and release H+ in water
=>bases => contain OH and release OH- in water.
=> Example => HF is an Arrhenius acid because it releases, or gives away, it\'s H+ proton.
=> HF(acid) + H2O(base) => (H30+)(acid) + F-(base)
I wrote (acid) or (base) to distinguish which is the acid and which is the base because some
molecules are amphoteric
=> Bronsted-Lowry
=> acids => \"proton donors,\" must contain H
=> bases => \"proton acceptors,\" must contain a lone pair to bind to the H+
=> Example => NH3 is a BL base because it has a lone pair to \"accept\" the H+
=> NH3(base) + H2O(acid) => (NH4+) + OH-(base)
=> Can you see the difference between a BL base (NH3) and an Arrhenius base?? The BL base
doesn\'t have an OH, nor does it release an OH- in water. That\'s why acids/bases are classified
under which restrictions they follow. NH3 is a BL base, but not an Arrhenius base.
=> Lewis => this type of definition allows a lot more molecules to be acids or base, due to its
restrictions
=> acids => electron pair acceptors, must have a vacant orbital in order to accept the elctron pair
from the base
=> bases => electron pair donors, must contain an electron pair to donate
=> Notice how the Lewis acids and bases are \"opposite\" of the BL definitions
=> Example => metal cations are Lewis acids because they have vacant orbitals in their valence
shells (Al3+, Fe3+, Ni2+, Cu2+, Ag+,...)
=>Example => F- is a Lewis base because F has 7 valence electrons with one of them being by
itself; F- has 8 valence elctrons, which creates a lone pair from the lone electron from F
=> So to answer your questions, it\'s all about the restrictions of each defintition. If an Arrhenius
acid has an H+ proton to give away, that acid is also a BL acid.
=> Conclusion
=> The Lewis definition has the widest scope of the three acid-base defintions, while the
Arrhenius definition has the narrowest.
=> When classifying the molecules as Lewis, Arrhenius, or BL acids or bases, you must
remember the restrictions each defintion has
Hope this helps!!!
Solution
The way I\'ve been told to look at the classifications is to look at their limitations. For example,
Arrhenius acids contain H and release H+ in water; this is the strict limitation of an Arrhenius
acid. Here is a chart to help with this
=> Arrhenius
=>acids => contain H and release H+ in water
=>bases => contain OH and release OH- in water.
=> Example => HF is an Arrhenius acid because it releases, or gives away, it\'s H+ proton.
=> HF(acid) + H2O(base) => (H30+)(acid) + F-(base)
I wrote (acid) or (base) to distinguish which is the acid and which is the base because some
molecules are amphoteric
=> Bronsted-Lowry
=> acids => \"proton donors,\" must contain H
=> bases => \"proton a.
tanx =1x = tan^-1(1) = pi4 for 0x 90sinx = 1sqrt2 ; cosx = 1.pdf
tanx =1
x = tan^-1(1) = pi/4 for 0
Solution
tanx =1
x = tan^-1(1) = pi/4 for 0.
The answer is b. decreasesSolids have lower entropy than gases.pdf
The answer is: b. decreases
Solids have lower entropy than gases
Solution
The answer is: b. decreases
Solids have lower entropy than gases.
Ques-1Chlamydia trichomatis has small G shaped intracellular bact.pdf
Ques-1:
Chlamydia trichomatis has small G shaped intracellular bacteria with noninfectious reticulate
form and infectious elementary body. It is most sexually transmitting disease in world.
An endocervical swab was obtained from the patient & used to inoculate McCoy cells, after 48
hours of incubation, the McCoy cells were stained with an imunnofluorescent reagent that
demonstrated the presence of inclusions. All Chlamydia are anaerobically performing their
metabolism using fermentation of various body fluids including vaginal fluids and they perform
biphasic life cycle with both infectious replication & non-replicative phases in McCoy cells.
Chlamydia bacteria are ovoid in shape and they are non-motile. They do not produce spores but
possess infectious elementary bodies to release into the host cells.
Ques-2:
Sonographic bimanual examination in the pelvic examination & nucleic acid amplification test
(NAAT) are useful to methodologies currently exist to identify Chlamydia trachomatis. The
results are compared with tissue culture sensitivity based on the growth level on the medium
Ques-3:
Beta-lactam antibiotics in treating infections caused by this organism:
Beta -lactum antibiotics such as penicillins, sulfonamides contain beta-lactum ring in their
structures, examples include penicillin, cephalosporins, carbapenems and monobactum
antibiotics. They act in inhibiting the cell wall synthesis, therefore they considered as
bactericidal. These antibiotics are generally effective against Gram positive bacteria and some
are also effective against gram negative bacteria. Examples include staphylococcus species,
streptococcus species. Doxycycline is going to inhibit protein synthesis of \"this microbial
species\" because beta-lactum are essential to act \"against drug resistance properties of
microbes\". The other epidemiological studies are needed to perform endocervical swab test to
determine any presence of Neisseria gonorrhoeae that cause PID
These penicillins are active against the gram- positive bacteria and they act by inhibiting the cell
wall synthesis. Transition state analog of penicillin & enzyme: Chemical reaction of penicillin
(beta-lactum ring) in reaction with N-C bond of D-ala-D-ala analog i.e. glycoproteinpeptidase
finally result in formation of \"transition state analog\". The mechanism is mainly mediated by
suicide inhibition of the enzyme irreversibly by reacting with enzyme
Solution
Ques-1:
Chlamydia trichomatis has small G shaped intracellular bacteria with noninfectious reticulate
form and infectious elementary body. It is most sexually transmitting disease in world.
An endocervical swab was obtained from the patient & used to inoculate McCoy cells, after 48
hours of incubation, the McCoy cells were stained with an imunnofluorescent reagent that
demonstrated the presence of inclusions. All Chlamydia are anaerobically performing their
metabolism using fermentation of various body fluids including vaginal fluids and they perform.
package net.codejava.swing.mail;import java.awt.Font;import java.pdf
package net.codejava.swing.mail;
import java.awt.Font;
import java.awt.GridBagConstraints;
import java.awt.GridBagLayout;
import java.awt.Insets;
import java.awt.event.ActionEvent;
import java.awt.event.ActionListener;
import java.io.File;
import java.util.Properties;
import javax.swing.JButton;
import javax.swing.JFrame;
import javax.swing.JLabel;
import javax.swing.JMenu;
import javax.swing.JMenuBar;
import javax.swing.JMenuItem;
import javax.swing.JOptionPane;
import javax.swing.JScrollPane;
import javax.swing.JTextArea;
import javax.swing.JTextField;
import javax.swing.SwingUtilities;
import javax.swing.UIManager;
import net.codejava.swing.JFilePicker;
/**
* A Swing application that allows sending e-mail messages from a SMTP server.
* @author www.codejava.net
*
*/
public class SwingEmailSender extends JFrame {
private ConfigUtility configUtil = new ConfigUtility();
private JMenuBar menuBar = new JMenuBar();
private JMenu menuFile = new JMenu(\"File\");
private JMenuItem menuItemSetting = new JMenuItem(\"Settings..\");
private JLabel labelTo = new JLabel(\"To: \");
private JLabel labelSubject = new JLabel(\"Subject: \");
private JTextField fieldTo = new JTextField(30);
private JTextField fieldSubject = new JTextField(30);
private JButton buttonSend = new JButton(\"SEND\");
private JFilePicker filePicker = new JFilePicker(\"Attached\", \"Attach File...\");
private JTextArea textAreaMessage = new JTextArea(10, 30);
private GridBagConstraints constraints = new GridBagConstraints();
public SwingEmailSender() {
super(\"Swing E-mail Sender Program\");
// set up layout
setLayout(new GridBagLayout());
constraints.anchor = GridBagConstraints.WEST;
constraints.insets = new Insets(5, 5, 5, 5);
setupMenu();
setupForm();
pack();
setLocationRelativeTo(null); // center on screen
setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
}
private void setupMenu() {
menuItemSetting.addActionListener(new ActionListener() {
@Override
public void actionPerformed(ActionEvent event) {
SettingsDialog dialog = new SettingsDialog(SwingEmailSender.this, configUtil);
dialog.setVisible(true);
}
});
menuFile.add(menuItemSetting);
menuBar.add(menuFile);
setJMenuBar(menuBar);
}
private void setupForm() {
constraints.gridx = 0;
constraints.gridy = 0;
add(labelTo, constraints);
constraints.gridx = 1;
constraints.fill = GridBagConstraints.HORIZONTAL;
add(fieldTo, constraints);
constraints.gridx = 0;
constraints.gridy = 1;
add(labelSubject, constraints);
constraints.gridx = 1;
constraints.fill = GridBagConstraints.HORIZONTAL;
add(fieldSubject, constraints);
constraints.gridx = 2;
constraints.gridy = 0;
constraints.gridheight = 2;
constraints.fill = GridBagConstraints.BOTH;
buttonSend.setFont(new Font(\"Arial\", Font.BOLD, 16));
add(buttonSend, constraints);
buttonSend.addActionListener(new ActionListener() {
@Override
public void actionPerformed(ActionEvent event) {
buttonSendActionPerformed(event);
}
});
constraints.gridx = 0;
constraints.gridy = 2;
constraints.gridheight = 1;
const.
No. The Oxygens on both sides of the Iodine cancel out, so the mol.pdf
No. The Oxygen\'s on both sides of the Iodine cancel out, so the molecule is non-polar
Solution
No. The Oxygen\'s on both sides of the Iodine cancel out, so the molecule is non-polar.
NameTest.java import java.util.Scanner;public class NameTest {.pdf
NameTest.java
import java.util.Scanner;
public class NameTest {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
System.out.println(\"Enter your name:\");
String name = scan.nextLine();
String first3Letters = name.substring(0, 3).toUpperCase();
System.out.println(\"First three letters are \"+first3Letters);
}
}
Output:
Enter your name:
Jonathan
First three letters are JON
Solution
NameTest.java
import java.util.Scanner;
public class NameTest {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
System.out.println(\"Enter your name:\");
String name = scan.nextLine();
String first3Letters = name.substring(0, 3).toUpperCase();
System.out.println(\"First three letters are \"+first3Letters);
}
}
Output:
Enter your name:
Jonathan
First three letters are JON.
Maryland’s State Finance & Procurement Article §14-410 defines envir.pdf
Maryland’s State Finance & Procurement Article §14-410 defines environmentally preferable
purchasing as the
procurement or acquisition of goods and services that have a lesser or reduced effect on human
health and the
environment when compared with competing goods or services that serve the same purpose.
This includes
considerations based on:
raw materials
manufacturing
packaging and distribution
use, operation and maintenance
refuse and disposal
Article §14-410 also provides clarity that EPP may not require the acquisition of goods or
services that:
do not perform adequately for the intended use
exclude adequate competition
are not available at a reasonable price in a reasonable period of time
Why is Environmentally Preferable Purchasing Important?
As our State\'s population increases, there is more demand for energy,
water, and other resources, putting strain on our transportation
infrastructure, land use and coastal communities and increasing
pollution, air emissions, and waste. Sound and efficient resource
management through EPP has the potential to yield long term cost
savings while minimizing the environmental impact associated with
manufacturing, use, and disposal of the products we purchase. This is
part of the larger move toward sustainability which aspires to meet
\"current human needs without undermining the capacity of the
environment to provide for those needs over the long term.\"
Environmentally Preferable Purchasing Best Practices Manual
3
Benefits to Human Health, the Environment & Economy
EPP can provide a variety of financial, human health, environmental, and social benefits.
Commonly cited
environmental, human health and social benefits of EPP include:
reduced air, water and soil pollution
materials and energy efficiency and reduced consumption
less waste in landfills
reductions in exposure to hazardous and toxic substances
providing a manufacturing demand for collected recycled material
reducing greenhouse gas emissions
increasing the use of renewable materials
improved wildlife habitats
decreased costs associated with waste management, disposal, and cleanup
Financial costs and benefits are easier to quantify. The purchasing price and frequency of
purchase is weighed against
operating costs, maintenance repair and replacement costs, occupational health costs, and
liabilities.
Economic benefits that may not be factored into the initial purchase price, or “first cost”:
Reusable, refillable, durable, and repairable products are usually more cost-effective over time
than singleuse
or disposable products.
Energy, water, or resource efficient products can result in avoided costs for these resources.
Avoiding hazardous substances and preventing pollution can reduce health and disposal costs
and
regulatory liability.
In many instances, a specific value to the benefits cannot be calculated without extensive study
or would be cost
prohibitive. However, in the absence of scientific consensus that an action is not harmful, the
precau.
Habitat,Feeding,Mobilityfresh water habitatNitrogen fixation.pdf
Habitat,
Feeding,
Mobility
fresh water habitat
Nitrogen fixation
non-motile
Litteroal zone of sea
Autotrophs
Non-motile female & motile male
On rocks, lower intertidal and shallow subtidal
Nonmotile
stagnant water, fresh water, sea water
Motile
fresh water
non-motilePhylum/Genus
Habitat,
Feeding,
MobilityLife cycleDominant GenIndependent GenDependent
Gen.Antheridia?Archegonia?sperm?CyanobacteriaAnabaena
fresh water habitat
Nitrogen fixation
non-motileGametophytePhaeophytaLaminariaSublittoral zone of
oceanGametphyte(n),fertilisation,sporophyte (2n),Immature unilocular sporangia (2n),
meiosis,sorus with unilocualr sporangia,spores (n)IndependentFucus
Litteroal zone of sea
Autotrophs
Non-motile female & motile maleGametophyte, gametes, carposporophyte,
spores,tetrasporophyte,meiosis, sporesDominantAntheridiaArchegoniaSargassumShallow
coastal, tropical and temperate marine waterSexual & vegetative reproduction. stolons &
fragments (2n),sporophyte (2n),male antheridium(2n) & female oogonium (2n),
diplophase/haplophase fertilisation, meiosis, ovum, sperm (n),zygote (2n)DiatomsMarine and
fresh waterInitial cell, normal vegetative cell, vegetative cell division,fertile cell,
meiosis,diploid/haploid, fusion,auxosporeRhodophytachondrus
On rocks, lower intertidal and shallow subtidal
Nonmotilemale & female, zygote, diploid carpospores,tetrasporophyte,haploid
tetrasporesPolysiphoniasubtidal zoneChlorophytachlamydomonas
stagnant water, fresh water, sea water
MotileOedogoniumfresh waterAntheridiaArchegoniaSpirogyra
fresh water
non-motileVegetative (filament,akinete) & sexual reproduction.
Filament,direct/indirect/scalriform conjugation,zygospore,meiosis,new
filament.AntheridiaArchegoniaVolvoxfresh waterAsexual & sexual reproduction,
volvox(n),antheridia(n), oogonia (n),antherozoid, egg(n) fusion,zygote(2n),reduction,
biflagellate swarmer,juvenile colonyAntheridiaArchegonia
Solution
Habitat,
Feeding,
Mobility
fresh water habitat
Nitrogen fixation
non-motile
Litteroal zone of sea
Autotrophs
Non-motile female & motile male
On rocks, lower intertidal and shallow subtidal
Nonmotile
stagnant water, fresh water, sea water
Motile
fresh water
non-motilePhylum/Genus
Habitat,
Feeding,
MobilityLife cycleDominant GenIndependent GenDependent
Gen.Antheridia?Archegonia?sperm?CyanobacteriaAnabaena
fresh water habitat
Nitrogen fixation
non-motileGametophytePhaeophytaLaminariaSublittoral zone of
oceanGametphyte(n),fertilisation,sporophyte (2n),Immature unilocular sporangia (2n),
meiosis,sorus with unilocualr sporangia,spores (n)IndependentFucus
Litteroal zone of sea
Autotrophs
Non-motile female & motile maleGametophyte, gametes, carposporophyte,
spores,tetrasporophyte,meiosis, sporesDominantAntheridiaArchegoniaSargassumShallow
coastal, tropical and temperate marine waterSexual & vegetative reproduction. stolons &
fragments (2n),sporophyte (2n),male antheridium(2n) & female oogonium (2n),
diplophase/haplophase fertilisation, meiosis, ovum, sperm (n),zygote (2n)DiatomsMarine a.
Instructions for Submissions thorugh G- Classroom.pptx
This presentation provides a briefing on how to upload submissions and documents in Google Classroom. It was prepared as part of an orientation for new Sainik School in-service teacher trainees. As a training officer, my goal is to ensure that you are comfortable and proficient with this essential tool for managing assignments and fostering student engagement.
How to Create Map Views in the Odoo 17 ERP
The map views are useful for providing a geographical representation of data. They allow users to visualize and analyze the data in a more intuitive manner.
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Ionic Equation is MgO(s) + 2 H+(aq) + 2NO3- (aq.pdf
Ionic Equation is MgO(s) + 2 H+(aq) + 2NO3- (aq) ---> Mg+2(aq) +2(NO3)2(aq)
+ H2O(l) if we look at the equation NO3- has not changed it is also in the same physical state so
it is a spectator ion
Solution
Ionic Equation is MgO(s) + 2 H+(aq) + 2NO3- (aq) ---> Mg+2(aq) +2(NO3)2(aq)
+ H2O(l) if we look at the equation NO3- has not changed it is also in the same physical state so
it is a spectator ion.
FeBrs is a typo Lets suppose it is FeBr2. 2Na .pdf
FeBrs is a typo? Let\'s suppose it is FeBr2. 2Na + FeBr2 = 2NaBr + Fe if it is
FeBr3 3Na + FeBr3 = 3NaBr + Fe
Solution
FeBrs is a typo? Let\'s suppose it is FeBr2. 2Na + FeBr2 = 2NaBr + Fe if it is
FeBr3 3Na + FeBr3 = 3NaBr + Fe.
d) in real gases there are attraction between m.pdf
d) in real gases there are attraction between molecules
Solution
d) in real gases there are attraction between molecules.
SolutionTraversing Binary Trees The preorder standard procedure.pdf
Solution
:
Traversing Binary Trees
The preorder standard procedures to traverse a binary tree are defined recursively as follows:
preorder(T):if T then visit root(T); preorder(left(T)); preorder(right(T))
preorder( ) =
preorder(tree(root, Left, Right)) = cat(root,preorder(Left), cat(preorder(Right)))
where cat concatenates two lists and can be defined by,
cat(,L) =L
cat(h ::t, L) = h :: cat(t, L)..
Shareholder’s equity= Current assets+Net fixed assets-Current lia.pdf
Shareholder’s equity:
= Current assets+Net fixed assets-Current liabilities-Long term debt
=$2,340+$11,000-$1,435-$4,150
= $7,755
Net working capital:
= Current assets-Current liabilities
=$2,340-$1,435
= $905
Solution
Shareholder’s equity:
= Current assets+Net fixed assets-Current liabilities-Long term debt
=$2,340+$11,000-$1,435-$4,150
= $7,755
Net working capital:
= Current assets-Current liabilities
=$2,340-$1,435
= $905.
in Reproductive cloning of mammals the Nucelus ( Genetic material) f.pdf
in Reproductive cloning of mammals the Nucelus ( Genetic material) from Donar egg is
removed, so it is enucleated.
the cell is taken from the organism which is to be cloned and remove the nucleus from that cell.
then these nucleus is introduced into the Donar egg (or) fuse these two cells using electrical
shock.
so cloned organism nucleus is fused with donar egg (enucleated egg).
Solution
in Reproductive cloning of mammals the Nucelus ( Genetic material) from Donar egg is
removed, so it is enucleated.
the cell is taken from the organism which is to be cloned and remove the nucleus from that cell.
then these nucleus is introduced into the Donar egg (or) fuse these two cells using electrical
shock.
so cloned organism nucleus is fused with donar egg (enucleated egg)..
a) OH- (aq) is a Lewis base because it can give electrions to Oth.pdf
a) OH- (aq) is a Lewis base because it can give electrions to Other.
b) Fe Br3 (s) is a lewis acid since it will accept electrons from other species.
c) Zn2+ (aq) is a Lewis acid and NH3 is a Lewis base
d) SO2 (g) is a Lewis acid because S has not containing an octate configuration.
Solution
a) OH- (aq) is a Lewis base because it can give electrions to Other.
b) Fe Br3 (s) is a lewis acid since it will accept electrons from other species.
c) Zn2+ (aq) is a Lewis acid and NH3 is a Lewis base
d) SO2 (g) is a Lewis acid because S has not containing an octate configuration..
package Chapter_20;import ToolKit.PostfixNotation;import javaf.pdf
package Chapter_20;
import ToolKit.PostfixNotation;
import javafx.application.Application;
import javafx.geometry.Insets;
import javafx.geometry.Pos;
import javafx.scene.Scene;
import javafx.scene.control.Button;
import javafx.scene.control.Label;
import javafx.scene.control.TextField;
import javafx.scene.image.Image;
import javafx.scene.image.ImageView;
import javafx.scene.layout.BorderPane;
import javafx.scene.layout.HBox;
import javafx.stage.Stage;
import java.util.ArrayList;
import java.util.Stack;
public class Exercise_13 extends Application {
int[] validNumbers = new int[4];
@Override
public void start(Stage primaryStage) {
// Top pane
Button btRefresh = new Button(\"Shuffle\");
Label lblStatus = new Label(\"\");
HBox topPane = new HBox(lblStatus, btRefresh);
topPane.setAlignment(Pos.BASELINE_RIGHT);
topPane.setSpacing(10);
// Center Pane
HBox centerPane = new HBox();
centerPane.setAlignment(Pos.CENTER);
centerPane.setSpacing(10);
centerPane.setPadding(new Insets(10));
// set first 4 random cards
setRandomCards(centerPane);
// Bottom pane
TextField tfExpression = new TextField();
Label lblExpression = new Label(\"Enter an expression:\");
Button btVerify = new Button(\"Verify\");
HBox bottomPane = new HBox(10, lblExpression, tfExpression, btVerify);
// Container Pane
BorderPane borderPane = new BorderPane();
borderPane.setPadding(new Insets(10));
borderPane.setTop(topPane);
borderPane.setCenter(centerPane);
borderPane.setBottom(bottomPane);
// Listeners
btRefresh.setOnAction(e -> {
lblStatus.setText(\"\");
setRandomCards(centerPane);
});
btVerify.setOnAction(e -> {
String expression = tfExpression.getText();
if (isValid(expression) && isCorrect(expression)) {
lblStatus.setText(\"Good job! \" + expression + \" = 24\");
} else {
lblStatus.setText(\"Invalid Expression\");
}
});
Scene scene = new Scene(borderPane);
primaryStage.setTitle(\"4 Random Cards\");
primaryStage.setScene(scene);
primaryStage.show();
// Debug
}
private void setRandomCards(HBox pane) {
boolean[] usedCards = new boolean[52];
// choose 4 random distinct cards from the deck
int count = 0;
pane.getChildren().clear();
while (count < 4) {
int card = (int) (Math.random() * 52);
if (!usedCards[card]) {
usedCards[card] = true;
pane.getChildren().add(new ImageView(new Image(\"image/card/\" + (++card) +
\".png\")));
int value = card % 13;
validNumbers[count] = (value == 0) ? 13 : value;
count++;
}
}
}
private static boolean isOperator(char ch) {
return (ch == \'(\' ||
ch == \')\' ||
isArithmeticOperator(ch));
}
private static boolean isArithmeticOperator(char ch) {
return (ch == \'/\' ||
ch == \'+\' ||
ch == \'-\' ||
ch == \'*\');
}
private static String[] separateExpression(String s) {
ArrayList tokens = new ArrayList<>(30);
char[] chars = s.toCharArray();
String numBuffer = \"\";
for (char ch : chars) {
if (isOperator(ch)) {
if (numBuffer.length() > 0) {
tokens.add(numBuffer);
numBuffer = \"\";
}
tokens.add(ch + \"\");
} else {
if (ch != \' \')
numBuffer += ch;
}
}
if (numBuffe.
Initial concentration of NH3 = molesvolume = 0.2501.00 = 0.250 M.pdf
Initial concentration of NH3 = moles/volume = 0.250/1.00 = 0.250 M
NH3 + H2O <=> NH4+ + OH-
I 0.25 0 0
C -a +a +a
E 0.25-a a a
Kb = [NH4+][OH-]/[NH3]
= a2/(0.25 - a) = 1.8 x 10-5
a2 + 1.8 x 10-5a - 4.5 x 10-6 = 0
a = 0.002112
[OH-] = a = 0.002112 M
[H+] = Kw/[OH-] where Kw is the ionic product of water
= 10-14/0.002112 = 4.735 x 10-12 M
pH = -log[H+] = 11.3
Solution
Initial concentration of NH3 = moles/volume = 0.250/1.00 = 0.250 M
NH3 + H2O <=> NH4+ + OH-
I 0.25 0 0
C -a +a +a
E 0.25-a a a
Kb = [NH4+][OH-]/[NH3]
= a2/(0.25 - a) = 1.8 x 10-5
a2 + 1.8 x 10-5a - 4.5 x 10-6 = 0
a = 0.002112
[OH-] = a = 0.002112 M
[H+] = Kw/[OH-] where Kw is the ionic product of water
= 10-14/0.002112 = 4.735 x 10-12 M
pH = -log[H+] = 11.3.
There are ten guidelines with a broad coverage, ranging from develop.pdf
There are ten guidelines with a broad coverage, ranging from developing a strategic role for
information to the development of capabilities and the marketing of the information unit. There
is a considerable degree of overlap between these guidelines, since many of them support and
depend on each other. Mostly, they are guidelines for development, not maintaining the status
quo. Therefore it is essential that you should regularly devote at least 10% of your time to these
activities, and certainly more in the early stages.
1. Establishing the Strategic Role of Information
This is a two stage process: (1) a research and investigation phase that gives you the information
you need for (2) articulating your mission and strategy. The first phase requires an assessment of
the attitudes of senior management to information and how much they are willing to pay for it. I
recently asked the head of a market research unit how hard it was to justify their existence and
budget. He commented: \"that has not been a problem, ever since we spend hundreds of millions
of dollars on a new product and then lost market share to the Japanese\". This direct link between
a large strategic investment and the bottom line delivered a sharp lesson to senior management
on the value of competitor intelligence.
Some of the strategic decisions whose successful outcomes depend on the availability of good
information include:
Market selection and targeting
New investments
Location of factories and offices
New product development and launch
Pricing and Promotion
Find out how these decisions are made, what information is used, and from where it is sourced.
You may already have data from feedback on how the information you have supplied has helped
such processes (if not, you should get it!). Has your organisation recently had successes or
failures that could be directly attributed to good or bad information? From these investigations
you can determine areas of high information leverage where you could play a role. Identify the
linkages between information and results. This should then be a cornerstone of your strategy.
Use every opportunity to let people, especially senior managers, know about it.
2. Identify Users Real Needs
This is the first of the marketing guidelines. It is essentially about market research. Therefore use
the methods used by researchers - surveys, interviews, usage analysis. You already have users.
Find out how they use your output and again what results and benefits they achieve. One
particularly useful way of teasing this out (used, by the way, to justify office automation
systems) is to ask what would happen if you did not offer that service.
Getting to senior management users and non-users is an important strand of this activity. You
must also learn about their real needs, not the ones they may initially express. Some of these may
be psychological needs such as \"I want to impress our senior management team with the range
of authoritative information I have .
The way Ive been told to look at the classifications is to look at.pdf
The way I\'ve been told to look at the classifications is to look at their limitations. For example,
Arrhenius acids contain H and release H+ in water; this is the strict limitation of an Arrhenius
acid. Here is a chart to help with this
=> Arrhenius
=>acids => contain H and release H+ in water
=>bases => contain OH and release OH- in water.
=> Example => HF is an Arrhenius acid because it releases, or gives away, it\'s H+ proton.
=> HF(acid) + H2O(base) => (H30+)(acid) + F-(base)
I wrote (acid) or (base) to distinguish which is the acid and which is the base because some
molecules are amphoteric
=> Bronsted-Lowry
=> acids => \"proton donors,\" must contain H
=> bases => \"proton acceptors,\" must contain a lone pair to bind to the H+
=> Example => NH3 is a BL base because it has a lone pair to \"accept\" the H+
=> NH3(base) + H2O(acid) => (NH4+) + OH-(base)
=> Can you see the difference between a BL base (NH3) and an Arrhenius base?? The BL base
doesn\'t have an OH, nor does it release an OH- in water. That\'s why acids/bases are classified
under which restrictions they follow. NH3 is a BL base, but not an Arrhenius base.
=> Lewis => this type of definition allows a lot more molecules to be acids or base, due to its
restrictions
=> acids => electron pair acceptors, must have a vacant orbital in order to accept the elctron pair
from the base
=> bases => electron pair donors, must contain an electron pair to donate
=> Notice how the Lewis acids and bases are \"opposite\" of the BL definitions
=> Example => metal cations are Lewis acids because they have vacant orbitals in their valence
shells (Al3+, Fe3+, Ni2+, Cu2+, Ag+,...)
=>Example => F- is a Lewis base because F has 7 valence electrons with one of them being by
itself; F- has 8 valence elctrons, which creates a lone pair from the lone electron from F
=> So to answer your questions, it\'s all about the restrictions of each defintition. If an Arrhenius
acid has an H+ proton to give away, that acid is also a BL acid.
=> Conclusion
=> The Lewis definition has the widest scope of the three acid-base defintions, while the
Arrhenius definition has the narrowest.
=> When classifying the molecules as Lewis, Arrhenius, or BL acids or bases, you must
remember the restrictions each defintion has
Hope this helps!!!
Solution
The way I\'ve been told to look at the classifications is to look at their limitations. For example,
Arrhenius acids contain H and release H+ in water; this is the strict limitation of an Arrhenius
acid. Here is a chart to help with this
=> Arrhenius
=>acids => contain H and release H+ in water
=>bases => contain OH and release OH- in water.
=> Example => HF is an Arrhenius acid because it releases, or gives away, it\'s H+ proton.
=> HF(acid) + H2O(base) => (H30+)(acid) + F-(base)
I wrote (acid) or (base) to distinguish which is the acid and which is the base because some
molecules are amphoteric
=> Bronsted-Lowry
=> acids => \"proton donors,\" must contain H
=> bases => \"proton a.
tanx =1x = tan^-1(1) = pi4 for 0x 90sinx = 1sqrt2 ; cosx = 1.pdf
tanx =1
x = tan^-1(1) = pi/4 for 0
Solution
tanx =1
x = tan^-1(1) = pi/4 for 0.
The answer is b. decreasesSolids have lower entropy than gases.pdf
The answer is: b. decreases
Solids have lower entropy than gases
Solution
The answer is: b. decreases
Solids have lower entropy than gases.
Ques-1Chlamydia trichomatis has small G shaped intracellular bact.pdf
Ques-1:
Chlamydia trichomatis has small G shaped intracellular bacteria with noninfectious reticulate
form and infectious elementary body. It is most sexually transmitting disease in world.
An endocervical swab was obtained from the patient & used to inoculate McCoy cells, after 48
hours of incubation, the McCoy cells were stained with an imunnofluorescent reagent that
demonstrated the presence of inclusions. All Chlamydia are anaerobically performing their
metabolism using fermentation of various body fluids including vaginal fluids and they perform
biphasic life cycle with both infectious replication & non-replicative phases in McCoy cells.
Chlamydia bacteria are ovoid in shape and they are non-motile. They do not produce spores but
possess infectious elementary bodies to release into the host cells.
Ques-2:
Sonographic bimanual examination in the pelvic examination & nucleic acid amplification test
(NAAT) are useful to methodologies currently exist to identify Chlamydia trachomatis. The
results are compared with tissue culture sensitivity based on the growth level on the medium
Ques-3:
Beta-lactam antibiotics in treating infections caused by this organism:
Beta -lactum antibiotics such as penicillins, sulfonamides contain beta-lactum ring in their
structures, examples include penicillin, cephalosporins, carbapenems and monobactum
antibiotics. They act in inhibiting the cell wall synthesis, therefore they considered as
bactericidal. These antibiotics are generally effective against Gram positive bacteria and some
are also effective against gram negative bacteria. Examples include staphylococcus species,
streptococcus species. Doxycycline is going to inhibit protein synthesis of \"this microbial
species\" because beta-lactum are essential to act \"against drug resistance properties of
microbes\". The other epidemiological studies are needed to perform endocervical swab test to
determine any presence of Neisseria gonorrhoeae that cause PID
These penicillins are active against the gram- positive bacteria and they act by inhibiting the cell
wall synthesis. Transition state analog of penicillin & enzyme: Chemical reaction of penicillin
(beta-lactum ring) in reaction with N-C bond of D-ala-D-ala analog i.e. glycoproteinpeptidase
finally result in formation of \"transition state analog\". The mechanism is mainly mediated by
suicide inhibition of the enzyme irreversibly by reacting with enzyme
Solution
Ques-1:
Chlamydia trichomatis has small G shaped intracellular bacteria with noninfectious reticulate
form and infectious elementary body. It is most sexually transmitting disease in world.
An endocervical swab was obtained from the patient & used to inoculate McCoy cells, after 48
hours of incubation, the McCoy cells were stained with an imunnofluorescent reagent that
demonstrated the presence of inclusions. All Chlamydia are anaerobically performing their
metabolism using fermentation of various body fluids including vaginal fluids and they perform.
package net.codejava.swing.mail;import java.awt.Font;import java.pdf
package net.codejava.swing.mail;
import java.awt.Font;
import java.awt.GridBagConstraints;
import java.awt.GridBagLayout;
import java.awt.Insets;
import java.awt.event.ActionEvent;
import java.awt.event.ActionListener;
import java.io.File;
import java.util.Properties;
import javax.swing.JButton;
import javax.swing.JFrame;
import javax.swing.JLabel;
import javax.swing.JMenu;
import javax.swing.JMenuBar;
import javax.swing.JMenuItem;
import javax.swing.JOptionPane;
import javax.swing.JScrollPane;
import javax.swing.JTextArea;
import javax.swing.JTextField;
import javax.swing.SwingUtilities;
import javax.swing.UIManager;
import net.codejava.swing.JFilePicker;
/**
* A Swing application that allows sending e-mail messages from a SMTP server.
* @author www.codejava.net
*
*/
public class SwingEmailSender extends JFrame {
private ConfigUtility configUtil = new ConfigUtility();
private JMenuBar menuBar = new JMenuBar();
private JMenu menuFile = new JMenu(\"File\");
private JMenuItem menuItemSetting = new JMenuItem(\"Settings..\");
private JLabel labelTo = new JLabel(\"To: \");
private JLabel labelSubject = new JLabel(\"Subject: \");
private JTextField fieldTo = new JTextField(30);
private JTextField fieldSubject = new JTextField(30);
private JButton buttonSend = new JButton(\"SEND\");
private JFilePicker filePicker = new JFilePicker(\"Attached\", \"Attach File...\");
private JTextArea textAreaMessage = new JTextArea(10, 30);
private GridBagConstraints constraints = new GridBagConstraints();
public SwingEmailSender() {
super(\"Swing E-mail Sender Program\");
// set up layout
setLayout(new GridBagLayout());
constraints.anchor = GridBagConstraints.WEST;
constraints.insets = new Insets(5, 5, 5, 5);
setupMenu();
setupForm();
pack();
setLocationRelativeTo(null); // center on screen
setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
}
private void setupMenu() {
menuItemSetting.addActionListener(new ActionListener() {
@Override
public void actionPerformed(ActionEvent event) {
SettingsDialog dialog = new SettingsDialog(SwingEmailSender.this, configUtil);
dialog.setVisible(true);
}
});
menuFile.add(menuItemSetting);
menuBar.add(menuFile);
setJMenuBar(menuBar);
}
private void setupForm() {
constraints.gridx = 0;
constraints.gridy = 0;
add(labelTo, constraints);
constraints.gridx = 1;
constraints.fill = GridBagConstraints.HORIZONTAL;
add(fieldTo, constraints);
constraints.gridx = 0;
constraints.gridy = 1;
add(labelSubject, constraints);
constraints.gridx = 1;
constraints.fill = GridBagConstraints.HORIZONTAL;
add(fieldSubject, constraints);
constraints.gridx = 2;
constraints.gridy = 0;
constraints.gridheight = 2;
constraints.fill = GridBagConstraints.BOTH;
buttonSend.setFont(new Font(\"Arial\", Font.BOLD, 16));
add(buttonSend, constraints);
buttonSend.addActionListener(new ActionListener() {
@Override
public void actionPerformed(ActionEvent event) {
buttonSendActionPerformed(event);
}
});
constraints.gridx = 0;
constraints.gridy = 2;
constraints.gridheight = 1;
const.
No. The Oxygens on both sides of the Iodine cancel out, so the mol.pdf
No. The Oxygen\'s on both sides of the Iodine cancel out, so the molecule is non-polar
Solution
No. The Oxygen\'s on both sides of the Iodine cancel out, so the molecule is non-polar.
NameTest.java import java.util.Scanner;public class NameTest {.pdf
NameTest.java
import java.util.Scanner;
public class NameTest {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
System.out.println(\"Enter your name:\");
String name = scan.nextLine();
String first3Letters = name.substring(0, 3).toUpperCase();
System.out.println(\"First three letters are \"+first3Letters);
}
}
Output:
Enter your name:
Jonathan
First three letters are JON
Solution
NameTest.java
import java.util.Scanner;
public class NameTest {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner scan = new Scanner(System.in);
System.out.println(\"Enter your name:\");
String name = scan.nextLine();
String first3Letters = name.substring(0, 3).toUpperCase();
System.out.println(\"First three letters are \"+first3Letters);
}
}
Output:
Enter your name:
Jonathan
First three letters are JON.
Maryland’s State Finance & Procurement Article §14-410 defines envir.pdf
Maryland’s State Finance & Procurement Article §14-410 defines environmentally preferable
purchasing as the
procurement or acquisition of goods and services that have a lesser or reduced effect on human
health and the
environment when compared with competing goods or services that serve the same purpose.
This includes
considerations based on:
raw materials
manufacturing
packaging and distribution
use, operation and maintenance
refuse and disposal
Article §14-410 also provides clarity that EPP may not require the acquisition of goods or
services that:
do not perform adequately for the intended use
exclude adequate competition
are not available at a reasonable price in a reasonable period of time
Why is Environmentally Preferable Purchasing Important?
As our State\'s population increases, there is more demand for energy,
water, and other resources, putting strain on our transportation
infrastructure, land use and coastal communities and increasing
pollution, air emissions, and waste. Sound and efficient resource
management through EPP has the potential to yield long term cost
savings while minimizing the environmental impact associated with
manufacturing, use, and disposal of the products we purchase. This is
part of the larger move toward sustainability which aspires to meet
\"current human needs without undermining the capacity of the
environment to provide for those needs over the long term.\"
Environmentally Preferable Purchasing Best Practices Manual
3
Benefits to Human Health, the Environment & Economy
EPP can provide a variety of financial, human health, environmental, and social benefits.
Commonly cited
environmental, human health and social benefits of EPP include:
reduced air, water and soil pollution
materials and energy efficiency and reduced consumption
less waste in landfills
reductions in exposure to hazardous and toxic substances
providing a manufacturing demand for collected recycled material
reducing greenhouse gas emissions
increasing the use of renewable materials
improved wildlife habitats
decreased costs associated with waste management, disposal, and cleanup
Financial costs and benefits are easier to quantify. The purchasing price and frequency of
purchase is weighed against
operating costs, maintenance repair and replacement costs, occupational health costs, and
liabilities.
Economic benefits that may not be factored into the initial purchase price, or “first cost”:
Reusable, refillable, durable, and repairable products are usually more cost-effective over time
than singleuse
or disposable products.
Energy, water, or resource efficient products can result in avoided costs for these resources.
Avoiding hazardous substances and preventing pollution can reduce health and disposal costs
and
regulatory liability.
In many instances, a specific value to the benefits cannot be calculated without extensive study
or would be cost
prohibitive. However, in the absence of scientific consensus that an action is not harmful, the
precau.
Habitat,Feeding,Mobilityfresh water habitatNitrogen fixation.pdf
Habitat,
Feeding,
Mobility
fresh water habitat
Nitrogen fixation
non-motile
Litteroal zone of sea
Autotrophs
Non-motile female & motile male
On rocks, lower intertidal and shallow subtidal
Nonmotile
stagnant water, fresh water, sea water
Motile
fresh water
non-motilePhylum/Genus
Habitat,
Feeding,
MobilityLife cycleDominant GenIndependent GenDependent
Gen.Antheridia?Archegonia?sperm?CyanobacteriaAnabaena
fresh water habitat
Nitrogen fixation
non-motileGametophytePhaeophytaLaminariaSublittoral zone of
oceanGametphyte(n),fertilisation,sporophyte (2n),Immature unilocular sporangia (2n),
meiosis,sorus with unilocualr sporangia,spores (n)IndependentFucus
Litteroal zone of sea
Autotrophs
Non-motile female & motile maleGametophyte, gametes, carposporophyte,
spores,tetrasporophyte,meiosis, sporesDominantAntheridiaArchegoniaSargassumShallow
coastal, tropical and temperate marine waterSexual & vegetative reproduction. stolons &
fragments (2n),sporophyte (2n),male antheridium(2n) & female oogonium (2n),
diplophase/haplophase fertilisation, meiosis, ovum, sperm (n),zygote (2n)DiatomsMarine and
fresh waterInitial cell, normal vegetative cell, vegetative cell division,fertile cell,
meiosis,diploid/haploid, fusion,auxosporeRhodophytachondrus
On rocks, lower intertidal and shallow subtidal
Nonmotilemale & female, zygote, diploid carpospores,tetrasporophyte,haploid
tetrasporesPolysiphoniasubtidal zoneChlorophytachlamydomonas
stagnant water, fresh water, sea water
MotileOedogoniumfresh waterAntheridiaArchegoniaSpirogyra
fresh water
non-motileVegetative (filament,akinete) & sexual reproduction.
Filament,direct/indirect/scalriform conjugation,zygospore,meiosis,new
filament.AntheridiaArchegoniaVolvoxfresh waterAsexual & sexual reproduction,
volvox(n),antheridia(n), oogonia (n),antherozoid, egg(n) fusion,zygote(2n),reduction,
biflagellate swarmer,juvenile colonyAntheridiaArchegonia
Solution
Habitat,
Feeding,
Mobility
fresh water habitat
Nitrogen fixation
non-motile
Litteroal zone of sea
Autotrophs
Non-motile female & motile male
On rocks, lower intertidal and shallow subtidal
Nonmotile
stagnant water, fresh water, sea water
Motile
fresh water
non-motilePhylum/Genus
Habitat,
Feeding,
MobilityLife cycleDominant GenIndependent GenDependent
Gen.Antheridia?Archegonia?sperm?CyanobacteriaAnabaena
fresh water habitat
Nitrogen fixation
non-motileGametophytePhaeophytaLaminariaSublittoral zone of
oceanGametphyte(n),fertilisation,sporophyte (2n),Immature unilocular sporangia (2n),
meiosis,sorus with unilocualr sporangia,spores (n)IndependentFucus
Litteroal zone of sea
Autotrophs
Non-motile female & motile maleGametophyte, gametes, carposporophyte,
spores,tetrasporophyte,meiosis, sporesDominantAntheridiaArchegoniaSargassumShallow
coastal, tropical and temperate marine waterSexual & vegetative reproduction. stolons &
fragments (2n),sporophyte (2n),male antheridium(2n) & female oogonium (2n),
diplophase/haplophase fertilisation, meiosis, ovum, sperm (n),zygote (2n)DiatomsMarine a.
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