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phosphate group and deoxyribose The groups are 1. Phosphate 2. Deoxyribose sugar 3. Nitrogen base The phosphates and deoxyribose sugars make up the sides of the \"ladder\" (alternating one after the other) and nitrogen bases are the \"rungs\" of the ladder. Solution phosphate group and deoxyribose The groups are 1. Phosphate 2. Deoxyribose sugar 3. Nitrogen base The phosphates and deoxyribose sugars make up the sides of the \"ladder\" (alternating one after the other) and nitrogen bases are the \"rungs\" of the ladder..
phosphate group and deoxyribose The groups are .pdf
phosphate group and deoxyribose The groups are .pdf
anwarfoot
Osmosis Solution Osmosis.
Osmosis .pdf
Osmosis .pdf
anwarfoot
in case of solid oxygen the atoms of oxygen are sharing elctrons and have a covalent bond between them so there is a dipole dipole bond Solution in case of solid oxygen the atoms of oxygen are sharing elctrons and have a covalent bond between them so there is a dipole dipole bond.
in case of solid oxygen the atoms of oxygen are s.pdf
in case of solid oxygen the atoms of oxygen are s.pdf
anwarfoot
Yeast is a microorganism and doesn\'t itself rise. Yeast eats carbohydrate and makes Carbon Dioxide (CO2) - the gas in soda water - as a waste product and it is this gas that makes the bread or cake \"rise\". Solution Yeast is a microorganism and doesn\'t itself rise. Yeast eats carbohydrate and makes Carbon Dioxide (CO2) - the gas in soda water - as a waste product and it is this gas that makes the bread or cake \"rise\"..
Yeast is a microorganism and doesnt itself rise.pdf
Yeast is a microorganism and doesnt itself rise.pdf
anwarfoot
First, let\'s analyze what each component of the solution is (strong base/acid, weak base/acid, salt?) HF: This is a weak acid (something you should probably know off the top of your head, but if you look at Ka, it\'s a very small value which indicates that it is weak). NaOH: A strong base (once again, something you should know off the top of your head. Virtually any ionic compound with Na is strong). Since it\'s strong, we can assume that however much NaOH we have will give us an equivalent amount of OH- ions. NaF: This is the salt for HF (as previously mentioned, this is completely soluble in water and adds F- ions, which will affect the equilibrium of HF). Like before, we can assume that however much NaF we have gives us the same amount of F-, since all of it dissolves. Now, let\'s think about the reactions that are happening. First, whenever you have an acid and a base, you have a neutralization reaction (I\'m leaving Na out of all the following reactions since it is a spectator ion): HF(aq) + OH-(aq) ==> F-(aq) + H2O(l) This means that the OH- we\'ve added will neutralize our initial amount of HF into F- and water. Since we have more HF than OH-, all of the OH- will go towards neutralizing the HF. Thus, 2.00M HF and 1.00M NaOH is essentially equivalent to 1.00M HF and 0 NaOH. Now, all we have is HF and F-. Recall the Henderson-Hasselbalch equation, which tells you how to calculate pH based on the ratio of an acid and its salt. In the case of HF, it will be pH = pKa + log([F-]/[HF]) You have [F-] in the form of [NaF], and you have [HF] which we just determined. You can calculate pKa from the given Ka value. Thus, you should be able to calculate pH. Solution First, let\'s analyze what each component of the solution is (strong base/acid, weak base/acid, salt?) HF: This is a weak acid (something you should probably know off the top of your head, but if you look at Ka, it\'s a very small value which indicates that it is weak). NaOH: A strong base (once again, something you should know off the top of your head. Virtually any ionic compound with Na is strong). Since it\'s strong, we can assume that however much NaOH we have will give us an equivalent amount of OH- ions. NaF: This is the salt for HF (as previously mentioned, this is completely soluble in water and adds F- ions, which will affect the equilibrium of HF). Like before, we can assume that however much NaF we have gives us the same amount of F-, since all of it dissolves. Now, let\'s think about the reactions that are happening. First, whenever you have an acid and a base, you have a neutralization reaction (I\'m leaving Na out of all the following reactions since it is a spectator ion): HF(aq) + OH-(aq) ==> F-(aq) + H2O(l) This means that the OH- we\'ve added will neutralize our initial amount of HF into F- and water. Since we have more HF than OH-, all of the OH- will go towards neutralizing the HF. Thus, 2.00M HF and 1.00M NaOH is essentially equivalent to 1.00M HF and 0 NaOH. Now,.
First, lets analyze what each component of the .pdf
First, lets analyze what each component of the .pdf
anwarfoot
when they reach equilibrium te cell potential is zero Solution when they reach equilibrium te cell potential is zero.
when they reach equilibrium te cell potential is .pdf
when they reach equilibrium te cell potential is .pdf
anwarfoot
E° = 0.07 - (0.14) = 0.21 V 2UO22+(aq) + Sn(s) 2UO2+(aq) + Sn2+(aq) [UO22+] = [UO2+] =[Sn2+] = 1.1mM = 0.0011 M Q =[UO2+]2[Sn2+]/[UO22+]2 = 0.0011 E = E° - 0.0592/n logQ = 0.21 - 0.0592/2 log 0.0011 = 0.2976 V Solution E° = 0.07 - (0.14) = 0.21 V 2UO22+(aq) + Sn(s) 2UO2+(aq) + Sn2+(aq) [UO22+] = [UO2+] =[Sn2+] = 1.1mM = 0.0011 M Q =[UO2+]2[Sn2+]/[UO22+]2 = 0.0011 E = E° - 0.0592/n logQ = 0.21 - 0.0592/2 log 0.0011 = 0.2976 V.
E° = 0.07 - (0.14) = 0.21 V 2UO22+(aq) + Sn(s.pdf
E° = 0.07 - (0.14) = 0.21 V 2UO22+(aq) + Sn(s.pdf
anwarfoot
D) Not B) because Cl-Benzene bond develops a double bond character due to resonance Solution D) Not B) because Cl-Benzene bond develops a double bond character due to resonance.
D) Not B) because Cl-Benzene bond develops a doub.pdf
D) Not B) because Cl-Benzene bond develops a doub.pdf
anwarfoot
Recommended
phosphate group and deoxyribose The groups are 1. Phosphate 2. Deoxyribose sugar 3. Nitrogen base The phosphates and deoxyribose sugars make up the sides of the \"ladder\" (alternating one after the other) and nitrogen bases are the \"rungs\" of the ladder. Solution phosphate group and deoxyribose The groups are 1. Phosphate 2. Deoxyribose sugar 3. Nitrogen base The phosphates and deoxyribose sugars make up the sides of the \"ladder\" (alternating one after the other) and nitrogen bases are the \"rungs\" of the ladder..
phosphate group and deoxyribose The groups are .pdf
phosphate group and deoxyribose The groups are .pdf
anwarfoot
Osmosis Solution Osmosis.
Osmosis .pdf
Osmosis .pdf
anwarfoot
in case of solid oxygen the atoms of oxygen are sharing elctrons and have a covalent bond between them so there is a dipole dipole bond Solution in case of solid oxygen the atoms of oxygen are sharing elctrons and have a covalent bond between them so there is a dipole dipole bond.
in case of solid oxygen the atoms of oxygen are s.pdf
in case of solid oxygen the atoms of oxygen are s.pdf
anwarfoot
Yeast is a microorganism and doesn\'t itself rise. Yeast eats carbohydrate and makes Carbon Dioxide (CO2) - the gas in soda water - as a waste product and it is this gas that makes the bread or cake \"rise\". Solution Yeast is a microorganism and doesn\'t itself rise. Yeast eats carbohydrate and makes Carbon Dioxide (CO2) - the gas in soda water - as a waste product and it is this gas that makes the bread or cake \"rise\"..
Yeast is a microorganism and doesnt itself rise.pdf
Yeast is a microorganism and doesnt itself rise.pdf
anwarfoot
First, let\'s analyze what each component of the solution is (strong base/acid, weak base/acid, salt?) HF: This is a weak acid (something you should probably know off the top of your head, but if you look at Ka, it\'s a very small value which indicates that it is weak). NaOH: A strong base (once again, something you should know off the top of your head. Virtually any ionic compound with Na is strong). Since it\'s strong, we can assume that however much NaOH we have will give us an equivalent amount of OH- ions. NaF: This is the salt for HF (as previously mentioned, this is completely soluble in water and adds F- ions, which will affect the equilibrium of HF). Like before, we can assume that however much NaF we have gives us the same amount of F-, since all of it dissolves. Now, let\'s think about the reactions that are happening. First, whenever you have an acid and a base, you have a neutralization reaction (I\'m leaving Na out of all the following reactions since it is a spectator ion): HF(aq) + OH-(aq) ==> F-(aq) + H2O(l) This means that the OH- we\'ve added will neutralize our initial amount of HF into F- and water. Since we have more HF than OH-, all of the OH- will go towards neutralizing the HF. Thus, 2.00M HF and 1.00M NaOH is essentially equivalent to 1.00M HF and 0 NaOH. Now, all we have is HF and F-. Recall the Henderson-Hasselbalch equation, which tells you how to calculate pH based on the ratio of an acid and its salt. In the case of HF, it will be pH = pKa + log([F-]/[HF]) You have [F-] in the form of [NaF], and you have [HF] which we just determined. You can calculate pKa from the given Ka value. Thus, you should be able to calculate pH. Solution First, let\'s analyze what each component of the solution is (strong base/acid, weak base/acid, salt?) HF: This is a weak acid (something you should probably know off the top of your head, but if you look at Ka, it\'s a very small value which indicates that it is weak). NaOH: A strong base (once again, something you should know off the top of your head. Virtually any ionic compound with Na is strong). Since it\'s strong, we can assume that however much NaOH we have will give us an equivalent amount of OH- ions. NaF: This is the salt for HF (as previously mentioned, this is completely soluble in water and adds F- ions, which will affect the equilibrium of HF). Like before, we can assume that however much NaF we have gives us the same amount of F-, since all of it dissolves. Now, let\'s think about the reactions that are happening. First, whenever you have an acid and a base, you have a neutralization reaction (I\'m leaving Na out of all the following reactions since it is a spectator ion): HF(aq) + OH-(aq) ==> F-(aq) + H2O(l) This means that the OH- we\'ve added will neutralize our initial amount of HF into F- and water. Since we have more HF than OH-, all of the OH- will go towards neutralizing the HF. Thus, 2.00M HF and 1.00M NaOH is essentially equivalent to 1.00M HF and 0 NaOH. Now,.
First, lets analyze what each component of the .pdf
First, lets analyze what each component of the .pdf
anwarfoot
when they reach equilibrium te cell potential is zero Solution when they reach equilibrium te cell potential is zero.
when they reach equilibrium te cell potential is .pdf
when they reach equilibrium te cell potential is .pdf
anwarfoot
E° = 0.07 - (0.14) = 0.21 V 2UO22+(aq) + Sn(s) 2UO2+(aq) + Sn2+(aq) [UO22+] = [UO2+] =[Sn2+] = 1.1mM = 0.0011 M Q =[UO2+]2[Sn2+]/[UO22+]2 = 0.0011 E = E° - 0.0592/n logQ = 0.21 - 0.0592/2 log 0.0011 = 0.2976 V Solution E° = 0.07 - (0.14) = 0.21 V 2UO22+(aq) + Sn(s) 2UO2+(aq) + Sn2+(aq) [UO22+] = [UO2+] =[Sn2+] = 1.1mM = 0.0011 M Q =[UO2+]2[Sn2+]/[UO22+]2 = 0.0011 E = E° - 0.0592/n logQ = 0.21 - 0.0592/2 log 0.0011 = 0.2976 V.
E° = 0.07 - (0.14) = 0.21 V 2UO22+(aq) + Sn(s.pdf
E° = 0.07 - (0.14) = 0.21 V 2UO22+(aq) + Sn(s.pdf
anwarfoot
D) Not B) because Cl-Benzene bond develops a double bond character due to resonance Solution D) Not B) because Cl-Benzene bond develops a double bond character due to resonance.
D) Not B) because Cl-Benzene bond develops a doub.pdf
D) Not B) because Cl-Benzene bond develops a doub.pdf
anwarfoot
Yes Because (1,1) is missing, it is not reflexive though (3,3) (2,2) are there Solution Yes Because (1,1) is missing, it is not reflexive though (3,3) (2,2) are there.
YesBecause (1,1) is missing, it is not reflexive though (3,3) (2,2.pdf
YesBecause (1,1) is missing, it is not reflexive though (3,3) (2,2.pdf
anwarfoot
We know that from Nernst Equation , E o cell = E o – ( 0.059 / n ) log ( [ products ] / [reactants ]) Given Eo cell = 0.17 V Eo = 0.24 V [products ] = [Cd 2+ ] = ? [reactants ] = [ Ni 2+ ] = 1.0 M N = no . of electrons transferred = 2 Solution We know that from Nernst Equation , E o cell = E o – ( 0.059 / n ) log ( [ products ] / [reactants ]) Given Eo cell = 0.17 V Eo = 0.24 V [products ] = [Cd 2+ ] = ? [reactants ] = [ Ni 2+ ] = 1.0 M N = no . of electrons transferred = 2.
We know that from Nernst Equation ,E o cell = E o – ( 0.059 n ) .pdf
We know that from Nernst Equation ,E o cell = E o – ( 0.059 n ) .pdf
anwarfoot
lower.. think about it.. take a straw draw a line on it and fill it with water, then look at it from different angles Solution lower.. think about it.. take a straw draw a line on it and fill it with water, then look at it from different angles.
lower.. think about it.. take a straw draw a line.pdf
lower.. think about it.. take a straw draw a line.pdf
anwarfoot
r= 1+ sin theta dr/d = cos = 0 Therefore = /2 Solution r= 1+ sin theta dr/d = cos = 0 Therefore = /2.
r= 1+ sin theta drd = cos = 0Therefore = 2Solutionr=.pdf
r= 1+ sin theta drd = cos = 0Therefore = 2Solutionr=.pdf
anwarfoot
From left to right: iodocyclopropane; 1-bromo-3-methylcyclopentane; 1-iodo-2,2- dimethylpropane; Solution From left to right: iodocyclopropane; 1-bromo-3-methylcyclopentane; 1-iodo-2,2- dimethylpropane;.
From left to right iodocyclopropane; 1-bromo-3-m.pdf
From left to right iodocyclopropane; 1-bromo-3-m.pdf
anwarfoot
For inorganic compounds Chemical properties remain same but colour or physical appearance changes. CuSO4 .5H2O is blue and crystalline but CuSO4 anhydrous is colourless.Colour change can be explained by theories of Co ordination compounds. In organic compounds ; chemical properties also change as dehydration changes the chemical nature; C2H5OH---> C2H4+H2O Solution For inorganic compounds Chemical properties remain same but colour or physical appearance changes. CuSO4 .5H2O is blue and crystalline but CuSO4 anhydrous is colourless.Colour change can be explained by theories of Co ordination compounds. In organic compounds ; chemical properties also change as dehydration changes the chemical nature; C2H5OH---> C2H4+H2O.
For inorganic compounds Chemical properties remai.pdf
For inorganic compounds Chemical properties remai.pdf
anwarfoot
n=4 l=3 we know that L=h/2(l(l+1)) L=3.7*10-34 Solution n=4 l=3 we know that L=h/2(l(l+1)) L=3.7*10-34.
n=4 l=3 we know that L=h2(l(l+1))L=3.710-34Solutionn.pdf
n=4 l=3 we know that L=h2(l(l+1))L=3.710-34Solutionn.pdf
anwarfoot
Over forty interstellar molecules have been found, ranging from simple OH radicals to molecules with over a dozen different atoms, including a few polycyclic. Their significance lies in that planets are formed from such nebulas( area of the cloud), so this tells us what may be found. Solution Over forty interstellar molecules have been found, ranging from simple OH radicals to molecules with over a dozen different atoms, including a few polycyclic. Their significance lies in that planets are formed from such nebulas( area of the cloud), so this tells us what may be found..
Over forty interstellar molecules have been found, ranging from simp.pdf
Over forty interstellar molecules have been found, ranging from simp.pdf
anwarfoot
One of the four prominent symptoms of typhoid is coated tongue. Even before the onset of fever, the tongue appears furred, with centrally coated and a red margin around. There are other illness also which shows coated tongue, but with different appearance. In malaria, the tongue is uniformly coated whereas in candidia thrush, a thick white layer of which patch is formed. In typhoid, the patch is centrally located on the tongue surrounded by a red margin. During the first week of typhoid, the dorsum of tongue is covered with a thin dirty white fur, which occurs due to the toxins released by the pathogen. The coat gradually clears from the tip and edges and completely disappears by the thid week of infection, leaving a glazed, dry or smooth red tongue. By observing the coat pattern, the Doctor would have diagnosed the fever well in advance. Solution One of the four prominent symptoms of typhoid is coated tongue. Even before the onset of fever, the tongue appears furred, with centrally coated and a red margin around. There are other illness also which shows coated tongue, but with different appearance. In malaria, the tongue is uniformly coated whereas in candidia thrush, a thick white layer of which patch is formed. In typhoid, the patch is centrally located on the tongue surrounded by a red margin. During the first week of typhoid, the dorsum of tongue is covered with a thin dirty white fur, which occurs due to the toxins released by the pathogen. The coat gradually clears from the tip and edges and completely disappears by the thid week of infection, leaving a glazed, dry or smooth red tongue. By observing the coat pattern, the Doctor would have diagnosed the fever well in advance..
One of the four prominent symptoms of typhoid is coated tongue. Even.pdf
One of the four prominent symptoms of typhoid is coated tongue. Even.pdf
anwarfoot
Meselson and Stahl in 1957 gave experimental evidence that each DNA strand served as a template for new DNA synthesis, a process called semi-conservative replication. At that time, there were three proposed models for DNA replication put forward by the scientific community after DNA structure had been discovered- Semi-conservative, conservative and dispersive replication. Semi-conservative replication: In this model, the two strands of DNA unwind from each other, and each acts as a template for synthesis of a new, complementary strand. This results in two DNA molecules with one original strand and one new strand. Conservative replication. In this model, DNA replication results in one molecule that consists of both original DNA strands and another molecule that consists of two new strands with exactly the same sequences as the original molecule. Meselson and Stahl used the density gradient sedimentation experiment (done with the E.Coli bacteria) to establish that DNA replicates using the semi-conservative model of replication as follows: Results of analysis: Solution Meselson and Stahl in 1957 gave experimental evidence that each DNA strand served as a template for new DNA synthesis, a process called semi-conservative replication. At that time, there were three proposed models for DNA replication put forward by the scientific community after DNA structure had been discovered- Semi-conservative, conservative and dispersive replication. Semi-conservative replication: In this model, the two strands of DNA unwind from each other, and each acts as a template for synthesis of a new, complementary strand. This results in two DNA molecules with one original strand and one new strand. Conservative replication. In this model, DNA replication results in one molecule that consists of both original DNA strands and another molecule that consists of two new strands with exactly the same sequences as the original molecule. Meselson and Stahl used the density gradient sedimentation experiment (done with the E.Coli bacteria) to establish that DNA replicates using the semi-conservative model of replication as follows: Results of analysis:.
Meselson and Stahl in 1957 gave experimental evidence that each DNA .pdf
Meselson and Stahl in 1957 gave experimental evidence that each DNA .pdf
anwarfoot
Introduction One of the key goals for the Windows Subsystem for Linux is to allow users to work with their files as they would on Linux, while giving full interoperability with files the user already has on their Windows machine. Unlike a virtual machine, where you have to use network shares or other solutions to share files between the host and guest OS, WSL has direct access to all your Windows drives to allow for easy interop. Windows file systems differ substantially from Linux file systems, and this post looks into how WSL bridges those two worlds. File systems on Linux Linux abstracts file systems operations through the Virtual File System (VFS), which provides both an interface for user mode programs to interact with the file system (through system calls such as open, read, chmod, stat, etc.) and an interface that file systems have to implement. This allows multiple file systems to coexist, providing the same operations and semantics, with VFS giving a single namespace view of all these file systems to the user. File systems are mounted on different directories in this namespace. For example, on a typical Linux system your hard drive may be mounted at the root, /, with directories such as /dev, /proc, /sys, and /mnt/cdrom all mounting different file systems which may be on different devices. Examples of file systems used on Linux include ext4, rfs, FAT, and others. VFS implements the various system calls for file system operations by using a number of data structures such as inodes, directory entries and files, and related callbacks that file systems must implement. Inodes The inode is the central data structure used in VFS. It represents a file system object such as a regular file, directory, symbolic link, etc. An inode contains information about the file type, size, permissions, last modified time, and other attributes. For many common Linux disk file systems such as ext4, the on-disk data structures used to represent file metadata directly correspond to the inode structure used by the Linux kernel. While an inode represents a file, it does not represent a file name. A single file may have multiple names, or hard links, but only one inode. File systems provide a lookup callback to VFS which is used to retrieve an inode for a particular file, based on the parent inode and the child name. File systems must implement a number of other inode operations such as chmod, stat, open, etc. Directory entries VFS uses a directory entry cache to represent your file system namespace. Directory entries only exist in memory, and contain a pointer to the inode for the file. For example, if you have a path like /home/user/foo, there is a directory entry for home, user, and foo, each with a pointer to an inode. Directory entries are cached for fast lookup, but if an entry is not yet in the cache, the inode lookup operation is used to retrieve the inode from the file system so a new directory entry can be created. File objects When an inode is opened, .
Introduction One of the key goals for the Windows Subsystem for Li.pdf
Introduction One of the key goals for the Windows Subsystem for Li.pdf
anwarfoot
import java.awt.BorderLayout; import java.awt.Color; import java.awt.event.ActionEvent; import java.awt.event.ActionListener; import javax.swing.JButton; import javax.swing.JColorChooser; import javax.swing.JFrame; import javax.swing.JMenu; import javax.swing.JMenuBar; import javax.swing.JMenuItem; import javax.swing.JToolBar; public class DrawApp extends JFrame { //declaring data fields private Drawable selectedFig = null; //default border color private Color presentBorderColor = Color.BLACK; private Color presentInteriorrColor = Color.WHITE; private Drawable[] theFigKinds; // Constructor private DrawApp(String args[]) { //title super(\"Draw App\"); //frame size setSize(650, 650); setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); loadFigKinds(args); JMenu drawingToolMenu = createDrawingToolMenu(); JToolBar drawingToolToolBar = createDrawingToolToolBar(); selectedFig = theFigKinds[0]; JMenu colorMenu = createColorMenu(); JMenuBar menuBar = new JMenuBar(); menuBar.add(colorMenu); menuBar.add(drawingToolMenu); setJMenuBar(menuBar); getContentPane().add(drawingToolToolBar, BorderLayout.EAST); getContentPane().add(new DrawPanel(this), BorderLayout.CENTER); } // creating drawing tool menu private JMenu createDrawingToolMenu() { JMenu drawingToolMenu = new JMenu(\"Drawing Tool\"); for (int i = 0; i < theFigKinds.length; i++) { // create a menu item for this figure kind JMenuItem item = new JMenuItem(theFigKinds[i].getName()); // set the action listener item.addActionListener(new ToolSelector(theFigKinds[i])); // add the item to the menu drawingToolMenu.add(item); } return drawingToolMenu; } // creating menu tool bar JToolBar createDrawingToolToolBar() { JToolBar drawingToolToolBar = new JToolBar(JToolBar.VERTICAL); for (int i = 0; i < theFigKinds.length; i++) { JButton aButton = new JButton(theFigKinds[i].getIcon(16)); aButton.addActionListener(new ToolSelector(theFigKinds[i])); drawingToolToolBar.add(aButton); } return drawingToolToolBar; } //creating color menu private JMenu createColorMenu() { JMenu colorMenu = new JMenu(\"Colors\"); JMenuItem setInteriorColor = new JMenuItem(\"Set Interior Color\"); setInteriorColor.addActionListener(new SetInteriorColor()); colorMenu.add(setInteriorColor); JMenuItem setBorderColor = new JMenuItem(\"Set Border Color\"); setBorderColor.addActionListener(new SetBorderColor()); colorMenu.add(setBorderColor); return colorMenu; } public Drawable getSelectedFig() { return selectedFig; } public static void main(String args[]) { DrawApp drawApp = new DrawApp(args); drawApp.setVisible(true); } // Inner Classes private class ToolSelector implements ActionListener { // Data Fields private Drawable desiredFig; // Constructor public ToolSelector(Drawable desiredFig) { this.desiredFig = desiredFig; } // Methods @Override public void actionPerformed(ActionEvent e) { selectedFig = desiredFig; } } private class SetInteriorColor implements ActionListener { @Override public void actionPerformed(ActionEvent e) { presentInteriorrColo.
import java.awt.BorderLayout; import java.awt.Color; import java.pdf
import java.awt.BorderLayout; import java.awt.Color; import java.pdf
anwarfoot
Higher lattice energies imply stronger bonding between the ions. This means that they will hold on to each other more strongly so that the polar water molecules will have a harder time separating them and thus dissolving them I would expect that a cation with a high charge density (smaller size for the same charge) would attract more polar water molecules and be more strongly hydrated. Solution Higher lattice energies imply stronger bonding between the ions. This means that they will hold on to each other more strongly so that the polar water molecules will have a harder time separating them and thus dissolving them I would expect that a cation with a high charge density (smaller size for the same charge) would attract more polar water molecules and be more strongly hydrated..
Higher lattice energies imply stronger bonding between the ions. Thi.pdf
Higher lattice energies imply stronger bonding between the ions. Thi.pdf
anwarfoot
Golden poison frog ( Phyllobates terribilis ) Kingdom: Animalia Phylum :Chordata Class: Amphibia The golden poison frog also known as the golden dart frog, most poisonous frog in the world.This species is originated from the Pacific coast of Colombia.This species is native to the lowland tropical rainforests. P. terribilis occurs in many colours such as yellow, golden yellow, mint green,metallic green,pale green,and orange.Adult females are typically larger than males.The golden poison frog likes to eat ants,flies and crickets and hunts insects using its long tongue,to strike and pull the prey to its mouth.This species is diurnal,meaning it is active during the day. P. terribilis is considered to be the most intelligent anurans,they are also extremely successful tongue hunters. The golden poison frogs are strictly terrestrial.The toxins they produce can be harmful for the heart of the victims and it can cause the breakdown of nervous system.The golden poison frog stores its poison in skin glands.The golden poison frog is polygynandrous.Males attract females by using high pitched calls.Courtship behaviour can last for several hours.Poison frogs in general can live for five years. Solution Golden poison frog ( Phyllobates terribilis ) Kingdom: Animalia Phylum :Chordata Class: Amphibia The golden poison frog also known as the golden dart frog, most poisonous frog in the world.This species is originated from the Pacific coast of Colombia.This species is native to the lowland tropical rainforests. P. terribilis occurs in many colours such as yellow, golden yellow, mint green,metallic green,pale green,and orange.Adult females are typically larger than males.The golden poison frog likes to eat ants,flies and crickets and hunts insects using its long tongue,to strike and pull the prey to its mouth.This species is diurnal,meaning it is active during the day. P. terribilis is considered to be the most intelligent anurans,they are also extremely successful tongue hunters. The golden poison frogs are strictly terrestrial.The toxins they produce can be harmful for the heart of the victims and it can cause the breakdown of nervous system.The golden poison frog stores its poison in skin glands.The golden poison frog is polygynandrous.Males attract females by using high pitched calls.Courtship behaviour can last for several hours.Poison frogs in general can live for five years..
Golden poison frog ( Phyllobates terribilis )Kingdom Animalia Phy.pdf
Golden poison frog ( Phyllobates terribilis )Kingdom Animalia Phy.pdf
anwarfoot
e) Constitutive transcription. The abberated Lac repressor does not bind to the promoter region of the operon. Thus leads to binding of the RNApolymerase and Constitutive expression of the operon Solution e) Constitutive transcription. The abberated Lac repressor does not bind to the promoter region of the operon. Thus leads to binding of the RNApolymerase and Constitutive expression of the operon.
e) Constitutive transcription.The abberated Lac repressor does not.pdf
e) Constitutive transcription.The abberated Lac repressor does not.pdf
anwarfoot
Comments added... import java.io.File; import java.io.FileNotFoundException; import java.util.Scanner; /** * * @author prmsh */ public class SaveMatrix { public static void main(String args[]) { //declaring matrix.. String[][] data = new String[4][4]; int row = 0; // opening and reading file File file = new File(\"matrix.txt\"); try { Scanner sc = new Scanner(file); while (sc.hasNextLine()) { String line = sc.nextLine(); //splitting each row by spaces String[] lineArray = line.split(\" \"); //storing into array for(int i=0; i Solution Comments added... import java.io.File; import java.io.FileNotFoundException; import java.util.Scanner; /** * * @author prmsh */ public class SaveMatrix { public static void main(String args[]) { //declaring matrix.. String[][] data = new String[4][4]; int row = 0; // opening and reading file File file = new File(\"matrix.txt\"); try { Scanner sc = new Scanner(file); while (sc.hasNextLine()) { String line = sc.nextLine(); //splitting each row by spaces String[] lineArray = line.split(\" \"); //storing into array for(int i=0; i.
Comments added... import java.io.File; import java.io.FileNotF.pdf
Comments added... import java.io.File; import java.io.FileNotF.pdf
anwarfoot
Connection may be defined as connectivity between many devices which can create a network, where we can send or receive data or any other informative signals that can used for some matter. There are many devcices used for creating connections in a network such as router, switches, hubs, bridges and etc etc. Here available port are being connected then that port remains in forwarding state mainly when we talk about switches. These ports become main port of that particular switch and they are kept in forwarding state and these ports send and receive signals from other ports. basically switches sends frames thats layer-2 data. Talking about available ports we have TCP and UPD ports as well, where UDP stands user datagram protocol and TCP stands for transmission control protocol, both these protocols are in Transport Layer on OSI model. where TCP is connection oriented protocol and UDP is connection less prototocol. if u have made connectivity where TCP is enabled then that connectivity mkes it sure that data should arrive anyhow by receiving acknowledgement from end devices where as in UDP it is not necessary that ack should be recieved or not then accordingly connections are made. Solution Connection may be defined as connectivity between many devices which can create a network, where we can send or receive data or any other informative signals that can used for some matter. There are many devcices used for creating connections in a network such as router, switches, hubs, bridges and etc etc. Here available port are being connected then that port remains in forwarding state mainly when we talk about switches. These ports become main port of that particular switch and they are kept in forwarding state and these ports send and receive signals from other ports. basically switches sends frames thats layer-2 data. Talking about available ports we have TCP and UPD ports as well, where UDP stands user datagram protocol and TCP stands for transmission control protocol, both these protocols are in Transport Layer on OSI model. where TCP is connection oriented protocol and UDP is connection less prototocol. if u have made connectivity where TCP is enabled then that connectivity mkes it sure that data should arrive anyhow by receiving acknowledgement from end devices where as in UDP it is not necessary that ack should be recieved or not then accordingly connections are made..
Connection may be defined as connectivity between many devices which.pdf
Connection may be defined as connectivity between many devices which.pdf
anwarfoot
a. CH3CH2CH2OH is more soluble in water since alcohol can form hydrogen bonds with water. e. C5H5CH3 (toluene) is more soluble in hexane since it has very low polarity. Solution a. CH3CH2CH2OH is more soluble in water since alcohol can form hydrogen bonds with water. e. C5H5CH3 (toluene) is more soluble in hexane since it has very low polarity..
a. CH3CH2CH2OH is more soluble in water since alc.pdf
a. CH3CH2CH2OH is more soluble in water since alc.pdf
anwarfoot
Ans) D) B and C The fraudulent practice of using another person\'s name and personal information in order to obtain credit, loan ext is known as identity theft Solution Ans) D) B and C The fraudulent practice of using another person\'s name and personal information in order to obtain credit, loan ext is known as identity theft.
Ans) D) B and CThe fraudulent practice of using another persons .pdf
Ans) D) B and CThe fraudulent practice of using another persons .pdf
anwarfoot
Allopatric speciation occurs by seperation of organisms by geographic isolation. Allopatric speciation may occur by migration. Since, the gene flow between organisms is inhibited, the geographically isolated organisms form into two separate species due to adaptations to new geographical environment. Allopatric speciation is most common type of speciation observed in various organisms. Speciation may occur in population inhabited in same geographical area but exhibit reproductive barriers. Such speciation is known sympatric speciation and it is rare and it may result in polyploidy. Solution Allopatric speciation occurs by seperation of organisms by geographic isolation. Allopatric speciation may occur by migration. Since, the gene flow between organisms is inhibited, the geographically isolated organisms form into two separate species due to adaptations to new geographical environment. Allopatric speciation is most common type of speciation observed in various organisms. Speciation may occur in population inhabited in same geographical area but exhibit reproductive barriers. Such speciation is known sympatric speciation and it is rare and it may result in polyploidy..
Allopatric speciation occurs by seperation of organisms by geographi.pdf
Allopatric speciation occurs by seperation of organisms by geographi.pdf
anwarfoot
EADTU-EU Summit 2024
e-Sealing at EADTU by Kamakshi Rajagopal
e-Sealing at EADTU by Kamakshi Rajagopal
EADTU
https://app.box.com/s/71kthbth9ww0fyjrppmh1p2gasinqj5z
ĐỀ THAM KHẢO KÌ THI TUYỂN SINH VÀO LỚP 10 MÔN TIẾNG ANH FORM 50 CÂU TRẮC NGHI...
ĐỀ THAM KHẢO KÌ THI TUYỂN SINH VÀO LỚP 10 MÔN TIẾNG ANH FORM 50 CÂU TRẮC NGHI...
Nguyen Thanh Tu Collection
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More from anwarfoot
Yes Because (1,1) is missing, it is not reflexive though (3,3) (2,2) are there Solution Yes Because (1,1) is missing, it is not reflexive though (3,3) (2,2) are there.
YesBecause (1,1) is missing, it is not reflexive though (3,3) (2,2.pdf
YesBecause (1,1) is missing, it is not reflexive though (3,3) (2,2.pdf
anwarfoot
We know that from Nernst Equation , E o cell = E o – ( 0.059 / n ) log ( [ products ] / [reactants ]) Given Eo cell = 0.17 V Eo = 0.24 V [products ] = [Cd 2+ ] = ? [reactants ] = [ Ni 2+ ] = 1.0 M N = no . of electrons transferred = 2 Solution We know that from Nernst Equation , E o cell = E o – ( 0.059 / n ) log ( [ products ] / [reactants ]) Given Eo cell = 0.17 V Eo = 0.24 V [products ] = [Cd 2+ ] = ? [reactants ] = [ Ni 2+ ] = 1.0 M N = no . of electrons transferred = 2.
We know that from Nernst Equation ,E o cell = E o – ( 0.059 n ) .pdf
We know that from Nernst Equation ,E o cell = E o – ( 0.059 n ) .pdf
anwarfoot
lower.. think about it.. take a straw draw a line on it and fill it with water, then look at it from different angles Solution lower.. think about it.. take a straw draw a line on it and fill it with water, then look at it from different angles.
lower.. think about it.. take a straw draw a line.pdf
lower.. think about it.. take a straw draw a line.pdf
anwarfoot
r= 1+ sin theta dr/d = cos = 0 Therefore = /2 Solution r= 1+ sin theta dr/d = cos = 0 Therefore = /2.
r= 1+ sin theta drd = cos = 0Therefore = 2Solutionr=.pdf
r= 1+ sin theta drd = cos = 0Therefore = 2Solutionr=.pdf
anwarfoot
From left to right: iodocyclopropane; 1-bromo-3-methylcyclopentane; 1-iodo-2,2- dimethylpropane; Solution From left to right: iodocyclopropane; 1-bromo-3-methylcyclopentane; 1-iodo-2,2- dimethylpropane;.
From left to right iodocyclopropane; 1-bromo-3-m.pdf
From left to right iodocyclopropane; 1-bromo-3-m.pdf
anwarfoot
For inorganic compounds Chemical properties remain same but colour or physical appearance changes. CuSO4 .5H2O is blue and crystalline but CuSO4 anhydrous is colourless.Colour change can be explained by theories of Co ordination compounds. In organic compounds ; chemical properties also change as dehydration changes the chemical nature; C2H5OH---> C2H4+H2O Solution For inorganic compounds Chemical properties remain same but colour or physical appearance changes. CuSO4 .5H2O is blue and crystalline but CuSO4 anhydrous is colourless.Colour change can be explained by theories of Co ordination compounds. In organic compounds ; chemical properties also change as dehydration changes the chemical nature; C2H5OH---> C2H4+H2O.
For inorganic compounds Chemical properties remai.pdf
For inorganic compounds Chemical properties remai.pdf
anwarfoot
n=4 l=3 we know that L=h/2(l(l+1)) L=3.7*10-34 Solution n=4 l=3 we know that L=h/2(l(l+1)) L=3.7*10-34.
n=4 l=3 we know that L=h2(l(l+1))L=3.710-34Solutionn.pdf
n=4 l=3 we know that L=h2(l(l+1))L=3.710-34Solutionn.pdf
anwarfoot
Over forty interstellar molecules have been found, ranging from simple OH radicals to molecules with over a dozen different atoms, including a few polycyclic. Their significance lies in that planets are formed from such nebulas( area of the cloud), so this tells us what may be found. Solution Over forty interstellar molecules have been found, ranging from simple OH radicals to molecules with over a dozen different atoms, including a few polycyclic. Their significance lies in that planets are formed from such nebulas( area of the cloud), so this tells us what may be found..
Over forty interstellar molecules have been found, ranging from simp.pdf
Over forty interstellar molecules have been found, ranging from simp.pdf
anwarfoot
One of the four prominent symptoms of typhoid is coated tongue. Even before the onset of fever, the tongue appears furred, with centrally coated and a red margin around. There are other illness also which shows coated tongue, but with different appearance. In malaria, the tongue is uniformly coated whereas in candidia thrush, a thick white layer of which patch is formed. In typhoid, the patch is centrally located on the tongue surrounded by a red margin. During the first week of typhoid, the dorsum of tongue is covered with a thin dirty white fur, which occurs due to the toxins released by the pathogen. The coat gradually clears from the tip and edges and completely disappears by the thid week of infection, leaving a glazed, dry or smooth red tongue. By observing the coat pattern, the Doctor would have diagnosed the fever well in advance. Solution One of the four prominent symptoms of typhoid is coated tongue. Even before the onset of fever, the tongue appears furred, with centrally coated and a red margin around. There are other illness also which shows coated tongue, but with different appearance. In malaria, the tongue is uniformly coated whereas in candidia thrush, a thick white layer of which patch is formed. In typhoid, the patch is centrally located on the tongue surrounded by a red margin. During the first week of typhoid, the dorsum of tongue is covered with a thin dirty white fur, which occurs due to the toxins released by the pathogen. The coat gradually clears from the tip and edges and completely disappears by the thid week of infection, leaving a glazed, dry or smooth red tongue. By observing the coat pattern, the Doctor would have diagnosed the fever well in advance..
One of the four prominent symptoms of typhoid is coated tongue. Even.pdf
One of the four prominent symptoms of typhoid is coated tongue. Even.pdf
anwarfoot
Meselson and Stahl in 1957 gave experimental evidence that each DNA strand served as a template for new DNA synthesis, a process called semi-conservative replication. At that time, there were three proposed models for DNA replication put forward by the scientific community after DNA structure had been discovered- Semi-conservative, conservative and dispersive replication. Semi-conservative replication: In this model, the two strands of DNA unwind from each other, and each acts as a template for synthesis of a new, complementary strand. This results in two DNA molecules with one original strand and one new strand. Conservative replication. In this model, DNA replication results in one molecule that consists of both original DNA strands and another molecule that consists of two new strands with exactly the same sequences as the original molecule. Meselson and Stahl used the density gradient sedimentation experiment (done with the E.Coli bacteria) to establish that DNA replicates using the semi-conservative model of replication as follows: Results of analysis: Solution Meselson and Stahl in 1957 gave experimental evidence that each DNA strand served as a template for new DNA synthesis, a process called semi-conservative replication. At that time, there were three proposed models for DNA replication put forward by the scientific community after DNA structure had been discovered- Semi-conservative, conservative and dispersive replication. Semi-conservative replication: In this model, the two strands of DNA unwind from each other, and each acts as a template for synthesis of a new, complementary strand. This results in two DNA molecules with one original strand and one new strand. Conservative replication. In this model, DNA replication results in one molecule that consists of both original DNA strands and another molecule that consists of two new strands with exactly the same sequences as the original molecule. Meselson and Stahl used the density gradient sedimentation experiment (done with the E.Coli bacteria) to establish that DNA replicates using the semi-conservative model of replication as follows: Results of analysis:.
Meselson and Stahl in 1957 gave experimental evidence that each DNA .pdf
Meselson and Stahl in 1957 gave experimental evidence that each DNA .pdf
anwarfoot
Introduction One of the key goals for the Windows Subsystem for Linux is to allow users to work with their files as they would on Linux, while giving full interoperability with files the user already has on their Windows machine. Unlike a virtual machine, where you have to use network shares or other solutions to share files between the host and guest OS, WSL has direct access to all your Windows drives to allow for easy interop. Windows file systems differ substantially from Linux file systems, and this post looks into how WSL bridges those two worlds. File systems on Linux Linux abstracts file systems operations through the Virtual File System (VFS), which provides both an interface for user mode programs to interact with the file system (through system calls such as open, read, chmod, stat, etc.) and an interface that file systems have to implement. This allows multiple file systems to coexist, providing the same operations and semantics, with VFS giving a single namespace view of all these file systems to the user. File systems are mounted on different directories in this namespace. For example, on a typical Linux system your hard drive may be mounted at the root, /, with directories such as /dev, /proc, /sys, and /mnt/cdrom all mounting different file systems which may be on different devices. Examples of file systems used on Linux include ext4, rfs, FAT, and others. VFS implements the various system calls for file system operations by using a number of data structures such as inodes, directory entries and files, and related callbacks that file systems must implement. Inodes The inode is the central data structure used in VFS. It represents a file system object such as a regular file, directory, symbolic link, etc. An inode contains information about the file type, size, permissions, last modified time, and other attributes. For many common Linux disk file systems such as ext4, the on-disk data structures used to represent file metadata directly correspond to the inode structure used by the Linux kernel. While an inode represents a file, it does not represent a file name. A single file may have multiple names, or hard links, but only one inode. File systems provide a lookup callback to VFS which is used to retrieve an inode for a particular file, based on the parent inode and the child name. File systems must implement a number of other inode operations such as chmod, stat, open, etc. Directory entries VFS uses a directory entry cache to represent your file system namespace. Directory entries only exist in memory, and contain a pointer to the inode for the file. For example, if you have a path like /home/user/foo, there is a directory entry for home, user, and foo, each with a pointer to an inode. Directory entries are cached for fast lookup, but if an entry is not yet in the cache, the inode lookup operation is used to retrieve the inode from the file system so a new directory entry can be created. File objects When an inode is opened, .
Introduction One of the key goals for the Windows Subsystem for Li.pdf
Introduction One of the key goals for the Windows Subsystem for Li.pdf
anwarfoot
import java.awt.BorderLayout; import java.awt.Color; import java.awt.event.ActionEvent; import java.awt.event.ActionListener; import javax.swing.JButton; import javax.swing.JColorChooser; import javax.swing.JFrame; import javax.swing.JMenu; import javax.swing.JMenuBar; import javax.swing.JMenuItem; import javax.swing.JToolBar; public class DrawApp extends JFrame { //declaring data fields private Drawable selectedFig = null; //default border color private Color presentBorderColor = Color.BLACK; private Color presentInteriorrColor = Color.WHITE; private Drawable[] theFigKinds; // Constructor private DrawApp(String args[]) { //title super(\"Draw App\"); //frame size setSize(650, 650); setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); loadFigKinds(args); JMenu drawingToolMenu = createDrawingToolMenu(); JToolBar drawingToolToolBar = createDrawingToolToolBar(); selectedFig = theFigKinds[0]; JMenu colorMenu = createColorMenu(); JMenuBar menuBar = new JMenuBar(); menuBar.add(colorMenu); menuBar.add(drawingToolMenu); setJMenuBar(menuBar); getContentPane().add(drawingToolToolBar, BorderLayout.EAST); getContentPane().add(new DrawPanel(this), BorderLayout.CENTER); } // creating drawing tool menu private JMenu createDrawingToolMenu() { JMenu drawingToolMenu = new JMenu(\"Drawing Tool\"); for (int i = 0; i < theFigKinds.length; i++) { // create a menu item for this figure kind JMenuItem item = new JMenuItem(theFigKinds[i].getName()); // set the action listener item.addActionListener(new ToolSelector(theFigKinds[i])); // add the item to the menu drawingToolMenu.add(item); } return drawingToolMenu; } // creating menu tool bar JToolBar createDrawingToolToolBar() { JToolBar drawingToolToolBar = new JToolBar(JToolBar.VERTICAL); for (int i = 0; i < theFigKinds.length; i++) { JButton aButton = new JButton(theFigKinds[i].getIcon(16)); aButton.addActionListener(new ToolSelector(theFigKinds[i])); drawingToolToolBar.add(aButton); } return drawingToolToolBar; } //creating color menu private JMenu createColorMenu() { JMenu colorMenu = new JMenu(\"Colors\"); JMenuItem setInteriorColor = new JMenuItem(\"Set Interior Color\"); setInteriorColor.addActionListener(new SetInteriorColor()); colorMenu.add(setInteriorColor); JMenuItem setBorderColor = new JMenuItem(\"Set Border Color\"); setBorderColor.addActionListener(new SetBorderColor()); colorMenu.add(setBorderColor); return colorMenu; } public Drawable getSelectedFig() { return selectedFig; } public static void main(String args[]) { DrawApp drawApp = new DrawApp(args); drawApp.setVisible(true); } // Inner Classes private class ToolSelector implements ActionListener { // Data Fields private Drawable desiredFig; // Constructor public ToolSelector(Drawable desiredFig) { this.desiredFig = desiredFig; } // Methods @Override public void actionPerformed(ActionEvent e) { selectedFig = desiredFig; } } private class SetInteriorColor implements ActionListener { @Override public void actionPerformed(ActionEvent e) { presentInteriorrColo.
import java.awt.BorderLayout; import java.awt.Color; import java.pdf
import java.awt.BorderLayout; import java.awt.Color; import java.pdf
anwarfoot
Higher lattice energies imply stronger bonding between the ions. This means that they will hold on to each other more strongly so that the polar water molecules will have a harder time separating them and thus dissolving them I would expect that a cation with a high charge density (smaller size for the same charge) would attract more polar water molecules and be more strongly hydrated. Solution Higher lattice energies imply stronger bonding between the ions. This means that they will hold on to each other more strongly so that the polar water molecules will have a harder time separating them and thus dissolving them I would expect that a cation with a high charge density (smaller size for the same charge) would attract more polar water molecules and be more strongly hydrated..
Higher lattice energies imply stronger bonding between the ions. Thi.pdf
Higher lattice energies imply stronger bonding between the ions. Thi.pdf
anwarfoot
Golden poison frog ( Phyllobates terribilis ) Kingdom: Animalia Phylum :Chordata Class: Amphibia The golden poison frog also known as the golden dart frog, most poisonous frog in the world.This species is originated from the Pacific coast of Colombia.This species is native to the lowland tropical rainforests. P. terribilis occurs in many colours such as yellow, golden yellow, mint green,metallic green,pale green,and orange.Adult females are typically larger than males.The golden poison frog likes to eat ants,flies and crickets and hunts insects using its long tongue,to strike and pull the prey to its mouth.This species is diurnal,meaning it is active during the day. P. terribilis is considered to be the most intelligent anurans,they are also extremely successful tongue hunters. The golden poison frogs are strictly terrestrial.The toxins they produce can be harmful for the heart of the victims and it can cause the breakdown of nervous system.The golden poison frog stores its poison in skin glands.The golden poison frog is polygynandrous.Males attract females by using high pitched calls.Courtship behaviour can last for several hours.Poison frogs in general can live for five years. Solution Golden poison frog ( Phyllobates terribilis ) Kingdom: Animalia Phylum :Chordata Class: Amphibia The golden poison frog also known as the golden dart frog, most poisonous frog in the world.This species is originated from the Pacific coast of Colombia.This species is native to the lowland tropical rainforests. P. terribilis occurs in many colours such as yellow, golden yellow, mint green,metallic green,pale green,and orange.Adult females are typically larger than males.The golden poison frog likes to eat ants,flies and crickets and hunts insects using its long tongue,to strike and pull the prey to its mouth.This species is diurnal,meaning it is active during the day. P. terribilis is considered to be the most intelligent anurans,they are also extremely successful tongue hunters. The golden poison frogs are strictly terrestrial.The toxins they produce can be harmful for the heart of the victims and it can cause the breakdown of nervous system.The golden poison frog stores its poison in skin glands.The golden poison frog is polygynandrous.Males attract females by using high pitched calls.Courtship behaviour can last for several hours.Poison frogs in general can live for five years..
Golden poison frog ( Phyllobates terribilis )Kingdom Animalia Phy.pdf
Golden poison frog ( Phyllobates terribilis )Kingdom Animalia Phy.pdf
anwarfoot
e) Constitutive transcription. The abberated Lac repressor does not bind to the promoter region of the operon. Thus leads to binding of the RNApolymerase and Constitutive expression of the operon Solution e) Constitutive transcription. The abberated Lac repressor does not bind to the promoter region of the operon. Thus leads to binding of the RNApolymerase and Constitutive expression of the operon.
e) Constitutive transcription.The abberated Lac repressor does not.pdf
e) Constitutive transcription.The abberated Lac repressor does not.pdf
anwarfoot
Comments added... import java.io.File; import java.io.FileNotFoundException; import java.util.Scanner; /** * * @author prmsh */ public class SaveMatrix { public static void main(String args[]) { //declaring matrix.. String[][] data = new String[4][4]; int row = 0; // opening and reading file File file = new File(\"matrix.txt\"); try { Scanner sc = new Scanner(file); while (sc.hasNextLine()) { String line = sc.nextLine(); //splitting each row by spaces String[] lineArray = line.split(\" \"); //storing into array for(int i=0; i Solution Comments added... import java.io.File; import java.io.FileNotFoundException; import java.util.Scanner; /** * * @author prmsh */ public class SaveMatrix { public static void main(String args[]) { //declaring matrix.. String[][] data = new String[4][4]; int row = 0; // opening and reading file File file = new File(\"matrix.txt\"); try { Scanner sc = new Scanner(file); while (sc.hasNextLine()) { String line = sc.nextLine(); //splitting each row by spaces String[] lineArray = line.split(\" \"); //storing into array for(int i=0; i.
Comments added... import java.io.File; import java.io.FileNotF.pdf
Comments added... import java.io.File; import java.io.FileNotF.pdf
anwarfoot
Connection may be defined as connectivity between many devices which can create a network, where we can send or receive data or any other informative signals that can used for some matter. There are many devcices used for creating connections in a network such as router, switches, hubs, bridges and etc etc. Here available port are being connected then that port remains in forwarding state mainly when we talk about switches. These ports become main port of that particular switch and they are kept in forwarding state and these ports send and receive signals from other ports. basically switches sends frames thats layer-2 data. Talking about available ports we have TCP and UPD ports as well, where UDP stands user datagram protocol and TCP stands for transmission control protocol, both these protocols are in Transport Layer on OSI model. where TCP is connection oriented protocol and UDP is connection less prototocol. if u have made connectivity where TCP is enabled then that connectivity mkes it sure that data should arrive anyhow by receiving acknowledgement from end devices where as in UDP it is not necessary that ack should be recieved or not then accordingly connections are made. Solution Connection may be defined as connectivity between many devices which can create a network, where we can send or receive data or any other informative signals that can used for some matter. There are many devcices used for creating connections in a network such as router, switches, hubs, bridges and etc etc. Here available port are being connected then that port remains in forwarding state mainly when we talk about switches. These ports become main port of that particular switch and they are kept in forwarding state and these ports send and receive signals from other ports. basically switches sends frames thats layer-2 data. Talking about available ports we have TCP and UPD ports as well, where UDP stands user datagram protocol and TCP stands for transmission control protocol, both these protocols are in Transport Layer on OSI model. where TCP is connection oriented protocol and UDP is connection less prototocol. if u have made connectivity where TCP is enabled then that connectivity mkes it sure that data should arrive anyhow by receiving acknowledgement from end devices where as in UDP it is not necessary that ack should be recieved or not then accordingly connections are made..
Connection may be defined as connectivity between many devices which.pdf
Connection may be defined as connectivity between many devices which.pdf
anwarfoot
a. CH3CH2CH2OH is more soluble in water since alcohol can form hydrogen bonds with water. e. C5H5CH3 (toluene) is more soluble in hexane since it has very low polarity. Solution a. CH3CH2CH2OH is more soluble in water since alcohol can form hydrogen bonds with water. e. C5H5CH3 (toluene) is more soluble in hexane since it has very low polarity..
a. CH3CH2CH2OH is more soluble in water since alc.pdf
a. CH3CH2CH2OH is more soluble in water since alc.pdf
anwarfoot
Ans) D) B and C The fraudulent practice of using another person\'s name and personal information in order to obtain credit, loan ext is known as identity theft Solution Ans) D) B and C The fraudulent practice of using another person\'s name and personal information in order to obtain credit, loan ext is known as identity theft.
Ans) D) B and CThe fraudulent practice of using another persons .pdf
Ans) D) B and CThe fraudulent practice of using another persons .pdf
anwarfoot
Allopatric speciation occurs by seperation of organisms by geographic isolation. Allopatric speciation may occur by migration. Since, the gene flow between organisms is inhibited, the geographically isolated organisms form into two separate species due to adaptations to new geographical environment. Allopatric speciation is most common type of speciation observed in various organisms. Speciation may occur in population inhabited in same geographical area but exhibit reproductive barriers. Such speciation is known sympatric speciation and it is rare and it may result in polyploidy. Solution Allopatric speciation occurs by seperation of organisms by geographic isolation. Allopatric speciation may occur by migration. Since, the gene flow between organisms is inhibited, the geographically isolated organisms form into two separate species due to adaptations to new geographical environment. Allopatric speciation is most common type of speciation observed in various organisms. Speciation may occur in population inhabited in same geographical area but exhibit reproductive barriers. Such speciation is known sympatric speciation and it is rare and it may result in polyploidy..
Allopatric speciation occurs by seperation of organisms by geographi.pdf
Allopatric speciation occurs by seperation of organisms by geographi.pdf
anwarfoot
More from anwarfoot
(20)
YesBecause (1,1) is missing, it is not reflexive though (3,3) (2,2.pdf
YesBecause (1,1) is missing, it is not reflexive though (3,3) (2,2.pdf
We know that from Nernst Equation ,E o cell = E o – ( 0.059 n ) .pdf
We know that from Nernst Equation ,E o cell = E o – ( 0.059 n ) .pdf
lower.. think about it.. take a straw draw a line.pdf
lower.. think about it.. take a straw draw a line.pdf
r= 1+ sin theta drd = cos = 0Therefore = 2Solutionr=.pdf
r= 1+ sin theta drd = cos = 0Therefore = 2Solutionr=.pdf
From left to right iodocyclopropane; 1-bromo-3-m.pdf
From left to right iodocyclopropane; 1-bromo-3-m.pdf
For inorganic compounds Chemical properties remai.pdf
For inorganic compounds Chemical properties remai.pdf
n=4 l=3 we know that L=h2(l(l+1))L=3.710-34Solutionn.pdf
n=4 l=3 we know that L=h2(l(l+1))L=3.710-34Solutionn.pdf
Over forty interstellar molecules have been found, ranging from simp.pdf
Over forty interstellar molecules have been found, ranging from simp.pdf
One of the four prominent symptoms of typhoid is coated tongue. Even.pdf
One of the four prominent symptoms of typhoid is coated tongue. Even.pdf
Meselson and Stahl in 1957 gave experimental evidence that each DNA .pdf
Meselson and Stahl in 1957 gave experimental evidence that each DNA .pdf
Introduction One of the key goals for the Windows Subsystem for Li.pdf
Introduction One of the key goals for the Windows Subsystem for Li.pdf
import java.awt.BorderLayout; import java.awt.Color; import java.pdf
import java.awt.BorderLayout; import java.awt.Color; import java.pdf
Higher lattice energies imply stronger bonding between the ions. Thi.pdf
Higher lattice energies imply stronger bonding between the ions. Thi.pdf
Golden poison frog ( Phyllobates terribilis )Kingdom Animalia Phy.pdf
Golden poison frog ( Phyllobates terribilis )Kingdom Animalia Phy.pdf
e) Constitutive transcription.The abberated Lac repressor does not.pdf
e) Constitutive transcription.The abberated Lac repressor does not.pdf
Comments added... import java.io.File; import java.io.FileNotF.pdf
Comments added... import java.io.File; import java.io.FileNotF.pdf
Connection may be defined as connectivity between many devices which.pdf
Connection may be defined as connectivity between many devices which.pdf
a. CH3CH2CH2OH is more soluble in water since alc.pdf
a. CH3CH2CH2OH is more soluble in water since alc.pdf
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