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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.
W since less electrons means difficult to melt so.pdf
W since less electrons means difficult to melt so.pdf
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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..
To the left. The ketone is more stable compared .pdf
To the left. The ketone is more stable compared .pdf
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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.
The true statement is D. Concentration affects .pdf
The true statement is D. Concentration affects .pdf
sudhirchourasia86
Smaller mass gases have greater rates of effusion. Ar < HCl Solution Smaller mass gases have greater rates of effusion. Ar < HCl.
Smaller mass gases have greater rates of effusion.pdf
Smaller mass gases have greater rates of effusion.pdf
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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..
please send the question details I have gone thr.pdf
please send the question details I have gone thr.pdf
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O3, SO3, SO2 Solution O3, SO3, SO2.
O3, SO3, SO2 .pdf
O3, SO3, SO2 .pdf
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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.
melting point of the compound increases due to th.pdf
melting point of the compound increases due to th.pdf
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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.
molality = moles of solute kg of solvent In thi.pdf
molality = moles of solute kg of solvent In thi.pdf
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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.
W since less electrons means difficult to melt so.pdf
W since less electrons means difficult to melt so.pdf
sudhirchourasia86
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..
To the left. The ketone is more stable compared .pdf
To the left. The ketone is more stable compared .pdf
sudhirchourasia86
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.
The true statement is D. Concentration affects .pdf
The true statement is D. Concentration affects .pdf
sudhirchourasia86
Smaller mass gases have greater rates of effusion. Ar < HCl Solution Smaller mass gases have greater rates of effusion. Ar < HCl.
Smaller mass gases have greater rates of effusion.pdf
Smaller mass gases have greater rates of effusion.pdf
sudhirchourasia86
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..
please send the question details I have gone thr.pdf
please send the question details I have gone thr.pdf
sudhirchourasia86
O3, SO3, SO2 Solution O3, SO3, SO2.
O3, SO3, SO2 .pdf
O3, SO3, SO2 .pdf
sudhirchourasia86
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.
melting point of the compound increases due to th.pdf
melting point of the compound increases due to th.pdf
sudhirchourasia86
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.
molality = moles of solute kg of solvent In thi.pdf
molality = moles of solute kg of solvent In thi.pdf
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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.
Ionic Equation is MgO(s) + 2 H+(aq) + 2NO3- (aq.pdf
Ionic Equation is MgO(s) + 2 H+(aq) + 2NO3- (aq.pdf
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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.
FeBrs is a typo Lets suppose it is FeBr2. 2Na .pdf
FeBrs is a typo Lets suppose it is FeBr2. 2Na .pdf
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Density increases Solution Density increases.
Density increases .pdf
Density increases .pdf
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d) in real gases there are attraction between molecules Solution d) in real gases there are attraction between molecules.
d) in real gases there are attraction between m.pdf
d) in real gases there are attraction between m.pdf
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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)..
SolutionTraversing Binary Trees The preorder standard procedure.pdf
SolutionTraversing Binary Trees The preorder standard procedure.pdf
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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.
Shareholder’s equity= Current assets+Net fixed assets-Current lia.pdf
Shareholder’s equity= Current assets+Net fixed assets-Current lia.pdf
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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)..
in Reproductive cloning of mammals the Nucelus ( Genetic material) f.pdf
in Reproductive cloning of mammals the Nucelus ( Genetic material) f.pdf
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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..
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 Oth.pdf
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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.
package Chapter_20;import ToolKit.PostfixNotation;import javaf.pdf
package Chapter_20;import ToolKit.PostfixNotation;import javaf.pdf
sudhirchourasia86
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.
Initial concentration of NH3 = molesvolume = 0.2501.00 = 0.250 M.pdf
Initial concentration of NH3 = molesvolume = 0.2501.00 = 0.250 M.pdf
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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 .
There are ten guidelines with a broad coverage, ranging from develop.pdf
There are ten guidelines with a broad coverage, ranging from develop.pdf
sudhirchourasia86
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.
The way Ive been told to look at the classifications is to look at.pdf
The way Ive been told to look at the classifications is to look at.pdf
sudhirchourasia86
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.
The enthalpy change of reaction = E(bonds broken) - E(bonds formed).pdf
The enthalpy change of reaction = E(bonds broken) - E(bonds formed).pdf
sudhirchourasia86
tanx =1 x = tan^-1(1) = pi/4 for 0 Solution tanx =1 x = tan^-1(1) = pi/4 for 0.
tanx =1x = tan^-1(1) = pi4 for 0x 90sinx = 1sqrt2 ; cosx = 1.pdf
tanx =1x = tan^-1(1) = pi4 for 0x 90sinx = 1sqrt2 ; cosx = 1.pdf
sudhirchourasia86
The answer is: b. decreases Solids have lower entropy than gases Solution The answer is: b. decreases Solids have lower entropy than gases.
The answer is b. decreasesSolids have lower entropy than gases.pdf
The answer is b. decreasesSolids have lower entropy than gases.pdf
sudhirchourasia86
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.
Ques-1Chlamydia trichomatis has small G shaped intracellular bact.pdf
Ques-1Chlamydia trichomatis has small G shaped intracellular bact.pdf
sudhirchourasia86
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.
package net.codejava.swing.mail;import java.awt.Font;import java.pdf
package net.codejava.swing.mail;import java.awt.Font;import java.pdf
sudhirchourasia86
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.
No. The Oxygens on both sides of the Iodine cancel out, so the mol.pdf
No. The Oxygens on both sides of the Iodine cancel out, so the mol.pdf
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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.
NameTest.java import java.util.Scanner;public class NameTest {.pdf
NameTest.java import java.util.Scanner;public class NameTest {.pdf
sudhirchourasia86
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.
Maryland’s State Finance & Procurement Article §14-410 defines envir.pdf
Maryland’s State Finance & Procurement Article §14-410 defines envir.pdf
sudhirchourasia86
Descrizione dettagliata della antica civiltà dei sumeri
descrizioni della antica civiltà dei sumeri.pptx
descrizioni della antica civiltà dei sumeri.pptx
tecongo2007
Che cos'è la scrittura seo e se è coerente o differente con la scrittura accessibile (maggio 2024)
Scrittura seo e scrittura accessibile
Scrittura seo e scrittura accessibile
Nicola Rabbi
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More from sudhirchourasia86
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.
Ionic Equation is MgO(s) + 2 H+(aq) + 2NO3- (aq.pdf
Ionic Equation is MgO(s) + 2 H+(aq) + 2NO3- (aq.pdf
sudhirchourasia86
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.
FeBrs is a typo Lets suppose it is FeBr2. 2Na .pdf
FeBrs is a typo Lets suppose it is FeBr2. 2Na .pdf
sudhirchourasia86
Density increases Solution Density increases.
Density increases .pdf
Density increases .pdf
sudhirchourasia86
d) in real gases there are attraction between molecules Solution d) in real gases there are attraction between molecules.
d) in real gases there are attraction between m.pdf
d) in real gases there are attraction between m.pdf
sudhirchourasia86
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)..
SolutionTraversing Binary Trees The preorder standard procedure.pdf
SolutionTraversing Binary Trees The preorder standard procedure.pdf
sudhirchourasia86
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.
Shareholder’s equity= Current assets+Net fixed assets-Current lia.pdf
Shareholder’s equity= Current assets+Net fixed assets-Current lia.pdf
sudhirchourasia86
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)..
in Reproductive cloning of mammals the Nucelus ( Genetic material) f.pdf
in Reproductive cloning of mammals the Nucelus ( Genetic material) f.pdf
sudhirchourasia86
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..
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 Oth.pdf
sudhirchourasia86
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.
package Chapter_20;import ToolKit.PostfixNotation;import javaf.pdf
package Chapter_20;import ToolKit.PostfixNotation;import javaf.pdf
sudhirchourasia86
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.
Initial concentration of NH3 = molesvolume = 0.2501.00 = 0.250 M.pdf
Initial concentration of NH3 = molesvolume = 0.2501.00 = 0.250 M.pdf
sudhirchourasia86
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 .
There are ten guidelines with a broad coverage, ranging from develop.pdf
There are ten guidelines with a broad coverage, ranging from develop.pdf
sudhirchourasia86
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.
The way Ive been told to look at the classifications is to look at.pdf
The way Ive been told to look at the classifications is to look at.pdf
sudhirchourasia86
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.
The enthalpy change of reaction = E(bonds broken) - E(bonds formed).pdf
The enthalpy change of reaction = E(bonds broken) - E(bonds formed).pdf
sudhirchourasia86
tanx =1 x = tan^-1(1) = pi/4 for 0 Solution tanx =1 x = tan^-1(1) = pi/4 for 0.
tanx =1x = tan^-1(1) = pi4 for 0x 90sinx = 1sqrt2 ; cosx = 1.pdf
tanx =1x = tan^-1(1) = pi4 for 0x 90sinx = 1sqrt2 ; cosx = 1.pdf
sudhirchourasia86
The answer is: b. decreases Solids have lower entropy than gases Solution The answer is: b. decreases Solids have lower entropy than gases.
The answer is b. decreasesSolids have lower entropy than gases.pdf
The answer is b. decreasesSolids have lower entropy than gases.pdf
sudhirchourasia86
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.
Ques-1Chlamydia trichomatis has small G shaped intracellular bact.pdf
Ques-1Chlamydia trichomatis has small G shaped intracellular bact.pdf
sudhirchourasia86
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.
package net.codejava.swing.mail;import java.awt.Font;import java.pdf
package net.codejava.swing.mail;import java.awt.Font;import java.pdf
sudhirchourasia86
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.
No. The Oxygens on both sides of the Iodine cancel out, so the mol.pdf
No. The Oxygens on both sides of the Iodine cancel out, so the mol.pdf
sudhirchourasia86
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.
NameTest.java import java.util.Scanner;public class NameTest {.pdf
NameTest.java import java.util.Scanner;public class NameTest {.pdf
sudhirchourasia86
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.
Maryland’s State Finance & Procurement Article §14-410 defines envir.pdf
Maryland’s State Finance & Procurement Article §14-410 defines envir.pdf
sudhirchourasia86
More from sudhirchourasia86
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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.pdf
FeBrs is a typo Lets suppose it is FeBr2. 2Na .pdf
FeBrs is a typo Lets suppose it is FeBr2. 2Na .pdf
Density increases .pdf
Density increases .pdf
d) in real gases there are attraction between m.pdf
d) in real gases there are attraction between m.pdf
SolutionTraversing Binary Trees The preorder standard procedure.pdf
SolutionTraversing Binary Trees The preorder standard procedure.pdf
Shareholder’s equity= Current assets+Net fixed assets-Current lia.pdf
Shareholder’s equity= Current assets+Net fixed assets-Current lia.pdf
in Reproductive cloning of mammals the Nucelus ( Genetic material) f.pdf
in Reproductive cloning of mammals the Nucelus ( Genetic material) f.pdf
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 Oth.pdf
package Chapter_20;import ToolKit.PostfixNotation;import javaf.pdf
package Chapter_20;import ToolKit.PostfixNotation;import javaf.pdf
Initial concentration of NH3 = molesvolume = 0.2501.00 = 0.250 M.pdf
Initial concentration of NH3 = molesvolume = 0.2501.00 = 0.250 M.pdf
There are ten guidelines with a broad coverage, ranging from develop.pdf
There are ten guidelines with a broad coverage, ranging from develop.pdf
The way Ive been told to look at the classifications is to look at.pdf
The way Ive been told to look at the classifications is to look at.pdf
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