a. Mean ResponsesSolution a. Mean Responses.pdf

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a. Mean Responses Solution a. Mean Responses.

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Moles of EDTA added initially to the solution = (25.0 mL* 1 L / 1000 mL) * 0.0452 mol/L =0.00113 mol Excess EDTA present in the solution after reacting withNi+2 and Zn+2 = (12.4 mL * 1 L /1000mL) * 0.0123 mol/L 0.00015 mol Moles of EDTA reacting with Zn+2 andNi+2 = 0.00113 mol - 0.00015 mol = 0.00098 mol Zn+2 and Ni+2 react with EDTA in1:1 ratio Moles of Zn+2 and Ni+2 inthe solution = 0.00098 mol EDTA was displaced from Zn and the liberated EDTA wastitrated with Mg+2 Moles of Mg+2 reacting with EDTA liberatedfrom Zn+2 = (29.2 mL * 1 L / 1000 mL) * 0.0123mol/L = 0.00035916 mol Each mole of EDTA is bound to 1 mole ofZn+2 Moles of Zn+2 in the solution = 0.00035916mol Zn+2 Moles of Ni+2 in the solution = 0.00098 mol -0.00035916 mol =0.00062084 mol Ni+2 [Zn+2] = 0.00035916 molZn+2 / 0.050 L = 0.0072 M Zn+2 [Ni+2] = 0.00062084 mol Ni+2 /0.050 L = 0.0124 M Ni+2 Solution Moles of EDTA added initially to the solution = (25.0 mL* 1 L / 1000 mL) * 0.0452 mol/L =0.00113 mol Excess EDTA present in the solution after reacting withNi+2 and Zn+2 = (12.4 mL * 1 L /1000mL) * 0.0123 mol/L 0.00015 mol Moles of EDTA reacting with Zn+2 andNi+2 = 0.00113 mol - 0.00015 mol = 0.00098 mol Zn+2 and Ni+2 react with EDTA in1:1 ratio Moles of Zn+2 and Ni+2 inthe solution = 0.00098 mol EDTA was displaced from Zn and the liberated EDTA wastitrated with Mg+2 Moles of Mg+2 reacting with EDTA liberatedfrom Zn+2 = (29.2 mL * 1 L / 1000 mL) * 0.0123mol/L = 0.00035916 mol Each mole of EDTA is bound to 1 mole ofZn+2 Moles of Zn+2 in the solution = 0.00035916mol Zn+2 Moles of Ni+2 in the solution = 0.00098 mol -0.00035916 mol =0.00062084 mol Ni+2 [Zn+2] = 0.00035916 molZn+2 / 0.050 L = 0.0072 M Zn+2 [Ni+2] = 0.00062084 mol Ni+2 /0.050 L = 0.0124 M Ni+2.

Moles of EDTA added initially to the solution = (.pdf

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IV Water has a high boiling compared to H2S. .pdf

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it is NO2 addition in meta position giving 1,3 di.pdf

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in saturn weight is 1.064 times that of the earth.pdf

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If I am reading the question right, this would be similar to something like a pond or a creek receiving fertilizer runoff. This is a more biological standpoint rather than chemical, but the ideas should be the same. #1. Influx of nutrients causes a bacterial/algal bloom, depleting water of oxygen. #2. As oxygen levels drop, algae/bacteria begin to die depleting water even more. #3. Animal live begins to die, causing another increase in nitrogen (ammonia etc) in the water #4. Anaerobic bacteria as well as plants begin to remove carbon dioxide, phosphates, nitrogen etc from the water and begin to bring system back into equilbrium. #5. Assuming influx of fertilizer runoff it not constant, the system will either return to its original state, or a new equilibrium, depending on how extreme the case was. Another possibility, considering the \"dynamic surface\", would mean that the extra carbon dioxide would outgas and allow oxygen to reenter the system. This is the general order, and there might be repetition of steps or one or two might switch places. Solution If I am reading the question right, this would be similar to something like a pond or a creek receiving fertilizer runoff. This is a more biological standpoint rather than chemical, but the ideas should be the same. #1. Influx of nutrients causes a bacterial/algal bloom, depleting water of oxygen. #2. As oxygen levels drop, algae/bacteria begin to die depleting water even more. #3. Animal live begins to die, causing another increase in nitrogen (ammonia etc) in the water #4. Anaerobic bacteria as well as plants begin to remove carbon dioxide, phosphates, nitrogen etc from the water and begin to bring system back into equilbrium. #5. Assuming influx of fertilizer runoff it not constant, the system will either return to its original state, or a new equilibrium, depending on how extreme the case was. Another possibility, considering the \"dynamic surface\", would mean that the extra carbon dioxide would outgas and allow oxygen to reenter the system. This is the general order, and there might be repetition of steps or one or two might switch places..

If I am reading the question right, this would be.pdf

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Which of the following are true statements about serial interfaces? Answer D. The default encapsulation is HDLC, but it can be changed Explanation:- Serial interfaces is used to connect to devices outside the network boundary or WAN. They are multilink point-to-point connections. The default encapsulation is HDLC(High-Level Data Link Controller), but it can be changed to Frame Relay, PPP, Synchronous Data Link Control (SDLC), SMDS, Cisco Serial Tunnel (STUN), Cisco Bisync Serial Tunnel (BSTUN), X.25-based encapsulations. Solution Which of the following are true statements about serial interfaces? Answer D. The default encapsulation is HDLC, but it can be changed Explanation:- Serial interfaces is used to connect to devices outside the network boundary or WAN. They are multilink point-to-point connections. The default encapsulation is HDLC(High-Level Data Link Controller), but it can be changed to Frame Relay, PPP, Synchronous Data Link Control (SDLC), SMDS, Cisco Serial Tunnel (STUN), Cisco Bisync Serial Tunnel (BSTUN), X.25-based encapsulations..

Which of the following are true statements about serial interfaces.pdf

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Thermostability is the quality of a substance to resist irreversible change in its chemical or physical structure, often by resisting decomposition or polymerisation, at a high relative temperature. Thermostable materials may be used industrially as fire retardants. A thermostableplastic, an uncommon and unconventional term, is likely to refer to a thermosetting plastic that cannot be reshaped when heated, than to a thermoplastic that can be remelted and recast. Thermostability also commonly refers to a protein resistant to change in its protein structure due to applied heat. Thermostable proteins Most life-forms on Earth live at temperatures of less than 50 °C, commonly from 15 to 50 °C. Above this, thermal energy may cause the unfolding of the protein structure, where the activity of the protein is abolished and a condition understandably deleterious to continuing life- functions. The denaturing of proteins in albumen from a clear, nearly colourless liquid to an opaque white, insoluble gel is a common example of this. Certain thermophilic life-forms exist which can withstand temperatures above this, and have corresponding adaptations to preserve protein function at these temperatures. . These can include altered bulk properties of the cell to stabilize all proteins, and specific changes to individual proteins. Examining homologous proteins present in these thermophiles and other organisms reveal only slight differences in the protein structure. One notable difference is the presence of extra hydrogen bonds in the thermophile\'s proteins—meaning that the protein structure is more resistant to unfolding. The presence of certain types of salt has been observed to alter thermostability in the proteins, indicating that salt bridges likely also play a role in thermostability Other factors of protein thermostability are compactness of protein structure,. oligomerization and strength interaction between subunits. Thermostable enzymes such as Taq polymerase and Pfu DNA polymerase are used in polymerase chain reactions where temperatures of 94 °C or over are used to melt apart DNA strands. Approaches to improve thermostability of proteins Protein engineering can be used to enhance the thermostability of proteins. A number of site- directed and random mutagenesis techniques, in addition to directed evolution .have been used to increase the thermostability of target proteins. Comparative methods have been used to increase the stability of mesophilic proteins based on comparison to thermophilic homologs. Additionally, analysis of the protein unfolding by molecular dynamics can be used to understand the process of unfolding and then design stabilizing mutations. Rational protein engineering for increasing protein thermostability includes mutations which truncate loops, increase salt bridges. or hydrogen bonds, introduced disulfide bonds. In addition, ligand binding can increase the stability of the protein, particularly when purified. Bacillus lichenifor.

Thermostability is the quality of a subst.pdf

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VictimizedSolutionVictimized.pdf

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True,Design activities for high risk system interfaces might come up.pdf

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WAN is a wide area network in which two or more computer or other devices connected to share data and resources. There are two types of medias are used for creating a comouter network i.e. Wired or Guided Media and Wirless or Unguided Media. For the connectivity of two points over a large distance we can choose any of the media on the basis of following charactersitics. 1. Wired Media- There are three types of wired medias are used for connectivity ie. Twisted, Co- axial and Fiber Optics for the long distance of connectivity we choose Fiber Optics due to it provide very high speed of data transfer. but the cost of these types of media is very expensive. If we choose Twisted Pair it has a limitation of transfer rate is maximum of 100mbps which is feasible for LAN not for the WAN. 2. Wireless Media- Wirelesss connectity can be established with the help of Wi-Fi Router, Satelite Communication, Radar etc. No doubt it provide fast transfer of data over the network but mail problem of these types of devices are first it is very expensive, second issue is with the range offered by the Wireless Access point is at limited distance so a large number of Wireless access points are required to establish a long distance based wan. Also backbone is required to be establish with wired media, which again restricted to the speed of the network. Solution WAN is a wide area network in which two or more computer or other devices connected to share data and resources. There are two types of medias are used for creating a comouter network i.e. Wired or Guided Media and Wirless or Unguided Media. For the connectivity of two points over a large distance we can choose any of the media on the basis of following charactersitics. 1. Wired Media- There are three types of wired medias are used for connectivity ie. Twisted, Co- axial and Fiber Optics for the long distance of connectivity we choose Fiber Optics due to it provide very high speed of data transfer. but the cost of these types of media is very expensive. If we choose Twisted Pair it has a limitation of transfer rate is maximum of 100mbps which is feasible for LAN not for the WAN. 2. Wireless Media- Wirelesss connectity can be established with the help of Wi-Fi Router, Satelite Communication, Radar etc. No doubt it provide fast transfer of data over the network but mail problem of these types of devices are first it is very expensive, second issue is with the range offered by the Wireless Access point is at limited distance so a large number of Wireless access points are required to establish a long distance based wan. Also backbone is required to be establish with wired media, which again restricted to the speed of the network..

WAN is a wide area network in which two or more computer or other de.pdf

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The most important characteristics that comprise the software quality are: • Functionality • Reliability • Usability • Efficiency • Maintainability • Portability Functionality: Functionality is the essential characteristic of any product or the service that refers to the specification of the functions of the software and to the correctness of the functions. As the system does not typically function in isolation, it helps the system to interact with the other components. This relates to the unauthorized access to the software functions. Reliability: This characteristic concerns the frequency of failure of the software and has the capacity of software to withstand and recover from components or environment and failure. It is also capable of bringing back the failed system to the full operation, including data and network connections. Usability: It is used to determine the simplicity of the systems function, relate the user models to computer interaction methods and has the ability of software to be easily operated by the user. Efficiency: This characteristic is concerned with the system resources used when providing the required functionality. The amount of disk space, memory, network etc provides a good indication of this characteristic. Maintainability: This characteristic has the ability to identify the root cause of a failure within the software, the amount of effort to change a system and the effort needed to test the system changes. Portability: The ability of the system to change to new specifications or operating environments. To achieve the software of high quality it is necessary to understand the ways how the software system fails. The challenges are: Solution The most important characteristics that comprise the software quality are: • Functionality • Reliability • Usability • Efficiency • Maintainability • Portability Functionality: Functionality is the essential characteristic of any product or the service that refers to the specification of the functions of the software and to the correctness of the functions. As the system does not typically function in isolation, it helps the system to interact with the other components. This relates to the unauthorized access to the software functions. Reliability: This characteristic concerns the frequency of failure of the software and has the capacity of software to withstand and recover from components or environment and failure. It is also capable of bringing back the failed system to the full operation, including data and network connections. Usability: It is used to determine the simplicity of the systems function, relate the user models to computer interaction methods and has the ability of software to be easily operated by the user. Efficiency: This characteristic is concerned with the system resources used when providing the required functionality. The amount of disk space, memory, network etc provides a good indication of this characteristic. Maintainability: This characteristic has .

The most important characteristics that comprise the software qualit.pdf

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The electronic configuration of H is 1s1 It has one valence electron like the ist group and It gets octet configuration [He] when it accepts one electron like the Group 7 elements (Halogens) Hence Hydrogen is like group 1 and group 7. Solution The electronic configuration of H is 1s1 It has one valence electron like the ist group and It gets octet configuration [He] when it accepts one electron like the Group 7 elements (Halogens) Hence Hydrogen is like group 1 and group 7..

The electronic configuration of H is 1s1 It has one valence electr.pdf

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public class FullListException extends Exception { /** * Default constructor */ public FullListException() { super(); } /** * param message */ public FullListException(String message) { super(message); } } public class NonPositivePriceException extends Exception{ /** * Default constructor */ public NonPositivePriceException() { super(); } /** * param message */ public NonPositivePriceException(String message) { super(message); } } public class MenuItem { //Attributes private String name; private String description; private double price; /** * Default constructor */ public MenuItem() { this.name = \"\"; this.description = \"\"; this.price = 0.0; } /** * Parameterized Constructor * param name * param description * param price */ public MenuItem(String name, String description, double price) { this.name = name; this.description = description; this.price = price; } /** * return the name */ public String getName() { return name; } /** * param name the name to set */ public void setName(String name) { this.name = name; } /** * return the description */ public String getDescription() { return description; } /** * param description the description to set */ public void setDescription(String description) { this.description = description; } /** * return the price */ public double getPrice() { return price; } /** * param price the price to set * throws NonPositivePriceException */ public void setPrice(double price) throws NonPositivePriceException { if(price <= 0) throw new NonPositivePriceException(\"Price cannot be set to a non positive value\"); this.price = price; } /** * Checks whether this MenuItem has the same attribute values as obj */ Override public boolean equals(Object obj) { MenuItem item = (MenuItem)obj; if((this.name.equalsIgnoreCase(item.name)) && (this.description.equalsIgnoreCase(item.description)) && (this.price == item.price)) return true; else return false; } Override public String toString() { System.out.printf(\"\ %2s %-25s %-75s %-10s\", \"#\", \"Item Name\", \"Item Description\", \"Item Price\"); System.out.printf(\"\ %50s\", new String(new char[115]).replace(\'\\0\', \'-\')); System.out.printf(\"\ %2s %-25s %-75s %10.2f\", \"1\", this.name, this.description, this.price); return \"\"; } } public class Menu{ static final int MAX_ITEMS = 50; //Attributes private MenuItem[] list; /** * Default Constructor * Construct an instance of the Menu class with no MenuItem objects in it * * Postcondition: This Menu has been initialized to an empty list of MenuItems. */ public Menu() { list = new MenuItem[MAX_ITEMS]; } /** * Generates a copy of this Menu. * return The return value is a copy of this Menu. * Subsequent changes to the copy will not affect the original, nor vice versa. * Note that the return value is typecast to an Menu before it can be used. */ Override public Object clone() { Menu clone = new Menu(); if(this.size() > 0) { for(int i = 0 ; i < MAX_ITEMS ; i++) { if(this.list[i] != null) clone.list[i] = this.list[i]; } } return cl.

public class FullListException extends Exception { Default .pdf

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program for JRiddle.java package riddle; import javax.swing.*; import java.awt.event.*; import java.awt.*; public class Riddle implements ActionListener { private final String LABEL_TEXT = \"What is your longterm goal?\"; JFrame frame; JPanel contentPane; JLabel label, label1; JButton button; JButton button1; private static int i; public Riddle() { /* Create and set up the frame */ frame = new JFrame(LABEL_TEXT); frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); /* Create a content pane with a GridLayout and empty borders */ contentPane = new JPanel(); contentPane.setLayout(new GridLayout(0, 2, 10, 5)); /* Create and add label that is centered and has empty borders */ label = new JLabel(\"Why did the chicken cross the road?\"); label.setAlignmentX(JButton.LEFT_ALIGNMENT); label.setBorder(BorderFactory.createEmptyBorder(20, 50, 20, 50)); contentPane.add(label); label1 = new JLabel(\" \"); label1.setAlignmentX(JButton.LEFT_ALIGNMENT); label1.setBorder(BorderFactory.createEmptyBorder(20, 50, 20, 50)); contentPane.add(label1); /* Create and add button that is centered */ button = new JButton(\"Answer\"); button.setAlignmentX(JButton.RIGHT_ALIGNMENT); button.setActionCommand(\"Show Answer\"); button.addActionListener(this); contentPane.add(button); /* Add content pane to frame */ frame.setContentPane(contentPane); /* Size and then display the frame */ frame.pack(); frame.setVisible(true); } /** Handle button click action event * pre: * post: clicked button shows answer */ public void actionPerformed(ActionEvent event) { String eventName = event.getActionCommand(); if (eventName.equals(\"Show Answer\")) { label1.setText(\"To become architect \"); label1.setFont(new Font(\"Serif\",Font.PLAIN,30)); button.setText(\"Answer\"); button.setActionCommand(\"Answer\"); } } /** * Create and show the GUI */ private static void runGUI() { JFrame.setDefaultLookAndFeelDecorated(true); Riddle greeting = new Riddle(); } public static void main(String[] args) { javax.swing.SwingUtilities.invokeLater(new Runnable() { public void run() { runGUI(); } }); } } output: what is your longterm goal? architectanswer Solution program for JRiddle.java package riddle; import javax.swing.*; import java.awt.event.*; import java.awt.*; public class Riddle implements ActionListener { private final String LABEL_TEXT = \"What is your longterm goal?\"; JFrame frame; JPanel contentPane; JLabel label, label1; JButton button; JButton button1; private static int i; public Riddle() { /* Create and set up the frame */ frame = new JFrame(LABEL_TEXT); frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); /* Create a content pane with a GridLayout and empty borders */ contentPane = new JPanel(); contentPane.setLayout(new GridLayout(0, 2, 10, 5)); /* Create and add label that is centered and has empty borders */ label = new JLabel(\"Why did the chicken cross the road?\"); label.setAlignmentX(JButton.LEFT_ALIGNMENT); label.setBorder(BorderFactory.createEmptyBorder(20, 50, 20, 50)); contentPa.

program for JRiddle.javapackage riddle;import javax.swing.; i.pdf

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Litumus is a weak acid which can be simply written as HLit. The un-ionised litmus is red, whereas the ion is blue. When the solution is basic, hydroxide ions are added. So more of Litumus ions are obtained. Litumus ions absorb orange color. Hence we observe a blue color. Similarly, when the solution is acidic, the equilibrium shifts to unionized litumus which absorbs blue green. Hence we observe a red color. Hope this helps Solution Litumus is a weak acid which can be simply written as HLit. The un-ionised litmus is red, whereas the ion is blue. When the solution is basic, hydroxide ions are added. So more of Litumus ions are obtained. Litumus ions absorb orange color. Hence we observe a blue color. Similarly, when the solution is acidic, the equilibrium shifts to unionized litumus which absorbs blue green. Hence we observe a red color. Hope this helps.

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