Element# of protons# of neutrons# of electronsCarbon (atomic.pdf
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Element # of protons # of neutrons # of electrons Carbon (atomic # 6) Neutral atom: Ion: Isotope: Neutral atom: Ion: Isotope: Neutral atom: Ion: Isotope: Oxygen (atomic # 8) Neutral atom: Ion: Isotope: Neutral atom: Ion: Isotope: Neutral atom: Ion: Isotope: Hydrogen (atomic # 1) Neutral atom: Ion: Isotope: Neutral atom: Ion: Isotope: Neutral atom: Ion: Isotope: Nitrogen (atomic # 7) Neutral atom: Ion: Isotope: Neutral atom: Ion: Isotope: Neutral atom: Ion: Isotope: How does the number of electrons affect how many bonds are formed by these atoms? Define the following (see boxes on pp 31, 33) Polar covalent bond Non-polar covalent bond Ionic bond Hydrogen bond Hydrophobic Hydrophilic Amphipathic Element # of protons # of neutrons # of electrons Carbon (atomic # 6) Neutral atom: Ion: Isotope: Neutral atom: Ion: Isotope: Neutral atom: Ion: Isotope: Oxygen (atomic # 8) Neutral atom: Ion: Isotope: Neutral atom: Ion: Isotope: Neutral atom: Ion: Isotope: Hydrogen (atomic # 1) Neutral atom: Ion: Isotope: Neutral atom: Ion: Isotope: Neutral atom: Ion: Isotope: Nitrogen (atomic # 7) Neutral atom: Ion: Isotope: Neutral atom: Ion: Isotope: Neutral atom: Ion: Isotope: Solution Element # of protons # of electrons # of neutrons Ion Isotopes Carbon 6 6 6 C+4 or C-4 Carbon-12 which has 6 protons and 6 neutrons and 6 electrons. Carbon-14 which has 6 protons, 8 neutrons and 6 electrons. Oxygen 8 8 8 Oxygen generally forms O-2. It has 6 electrons in the outermost orbit. Three stable isotopes of oxygen are found- O-16,O17 and O-18. Hydrogen 1 1 1 Hydrogen has a charge of +1. Hydrogen has 3 stable isotopes – protium (1 proton and 1 electron), deuterium (1 proton, 1 electron and 1 neutron) and tritium (1 proton, 1 electron and 2 neutrons) Nitrogen 7 7 7 Usually has the charge -3. N-14 has odd number of protons and neutrons (7 each). N-15 has protons and 8 neutrons. Carbon does not usually form an ion by loss or gain of 4 electrons. It shares its 4 electrons with 4 others to complete its orbit and forms a covalent bond. Oxygen has 6 valence lectrons in its outermost shell, therefore it can form 2 bonds Hydrogen atom has one valence electron in its outermost orbit. The maximum capacity of its orbital is 2 electrons. Nonpolar covalent bond Non-polar covalent bond is formed when the atoms (with similar electron affinity) share electrons equally. The electronegativity difference between the two bonded atoms is usually less than 0.5. Examples – diatomic molecules H2 and N2. Polar covalent bond Polar covalent bond is formed when the electrons are not equally shared i.e. the atom with higher electronegativity (affinity) pulls the electrons towards itself. The compound which is formed has electrostatic potential and has the ability to form weak bonds with other polar molecules. The electronegativity difference between bonded atoms is 0.5 to 1.9. Examples include water, ozone. Ionic bond – the valence electrons are completely transferred from one atom to another. The metal lo.
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This document discusses chemical bonding and molecular structure. It begins by introducing atoms, molecules, and the forces that hold atoms together in molecules. It then defines chemical bonding and describes the main types of bonds: ionic, covalent, hydrogen, and polar bonds. The remainder of the document discusses these bond types in more detail, including how to represent bonds using Lewis structures, the characteristics of ionic compounds, factors that influence ionic bond formation, and more. It also introduces concepts like formal charge, valence shell electron pair repulsion theory (VSEPR) for predicting molecular geometry, and hybridization.
The Structure of Matter.ppt Scienceeee 9
This document provides an overview of the structure of matter and different types of bonds between atoms. It defines key terms like elements, compounds, and chemical bonds. It describes the three main types of bonds - covalent, ionic, and metallic - and explains how they form. Covalent bonds form when atoms share electrons. Ionic bonds form when electrons are transferred between atoms to form ions. Metallic bonds form due to attraction between positively charged metal ions and delocalized electrons. The document also discusses naming conventions for compounds and polyatomic ions.
Covalent bond
A covalent bond involves the sharing of electron pairs between atoms. There are two types of covalent bonding: non-polar, with equal sharing, and polar, with unequal sharing. A non-polar bond forms between like atoms that share electrons equally, while a polar bond forms between different atoms where one atom attracts the electrons more than the other due to differing electronegativity. Examples of non-polar molecules are H2, Cl2, and O2, where the atoms share electrons equally. Examples of polar molecules are HCl and H2O, where the more electronegative atom (Cl or O) attracts the electrons slightly more, giving it a partial negative charge and leaving the other atom with
Chemical bonding sk0023
Lewis symbols and the octet rule are used to represent valence electrons and chemical bonding. Lewis symbols show valence electrons as dots around the element symbol. The octet rule states that atoms bond to gain, lose, or share electrons to achieve an octet of 8 valence electrons. Exceptions include odd total electron molecules, molecules where atoms have less than an octet, and hypervalent molecules where the central atom has more than 8 electrons.
Basic Chemistry Review.pptx
This document provides information on key concepts in earth and space science including:
- Atoms are made up of protons, neutrons, and electrons. Protons and neutrons are in the nucleus, while electrons orbit the nucleus.
- Isotopes are atoms of the same element that differ in the number of neutrons. Chemical bonds such as covalent and ionic bonds form compounds with different properties than their constituent elements.
- Acids and bases are defined by their release of hydrogen and hydroxide ions in water, and indicators are used to determine if a solution is acidic or basic. Water has unique properties due to its polarity that allow it to dissolve many substances and transport materials in living things.
Chapter 8, section 8.4 (1)
This document discusses bond polarity and molecular polarity. It explains that polar covalent bonds form when electrons are shared unequally between atoms of different electronegativity. The more electronegative atom attracts the electrons more strongly, acquiring a partial negative charge, while the other atom acquires a partial positive charge. Whether a bond is nonpolar, moderately polar, or ionic depends on the electronegativity difference between the atoms. Polar bonds can result in polar molecules if the bond polarities do not cancel out. Intermolecular forces like dipole interactions and dispersion forces are responsible for molecular phase changes and arise from attractions between slightly positively and negatively charged regions on different molecules. Hydrogen bonds are especially strong dipole interactions that
Similar to Element# of protons# of neutrons# of electronsCarbon (atomic.pdf (20)
Introduction to Biology Lecture Chapter 2 Study Guide.pdf
Introduction to Biology Lecture Chapter 2 Study Guide.pdf
The electrons which are involved in bond formatio.pdf
The electrons which are involved in bond formatio.pdf
Class 11 Chemistry Revision Notes Chemical Bonding and Molecular Structure.pdf
Class 11 Chemistry Revision Notes Chemical Bonding and Molecular Structure.pdf
More from arjuncorner565
Write about Participatory Design Issues and Concerns (use your own.pdf
Write about Participatory Design: Issues and Concerns? (use your own word about 500 words
not copy paste from another website pl.)
Solution
Participatory design is also called as co-operative design or co-design.It can be said as an
approach to design which invloves active participation of all the stakeholders in the design
process to ensure the final result meets its requirement.It is rather a process and procedure of
design than a design style.According to market study and surveys designers do a better job in a
co-design environment than they do with there own ideas.
We discuss three main issues addressed by Participatory Design researchers -
1)politics of design : This is one of the main concern for many design researchers and
practitioners.For example, the introduction of new frameworks such as system design or
platforms related to the introduction of computer-based systems and power dynamics that
emerge within the workspace. The automation process of system platforms has created a huge
concern within the workers as it is threatening their ownership over the work and involvement.
2)nature of participation : One of the major international organization Project for Public Spaces
create opportunities for continuous participation in the design and creation of place which is
believed to be the essential ingredient for successful environments. Project for Public Spaces
creates a new platform for the community to participate and reflect their initimate knowledge.A
method called Place Performance Evaluation or Place Game is used where groups from the
community are taken on the site of proposed development.They use their knowledge to design
strategies from which society will be benefited.
3)method, tools and techniques for participation : Distributed participatory design is a design
approach that involves direct participation of users and other stakeholders in system analysis and
design work.At present design teams most often are distributed leads to a need for support and
knowledge gathered from design of distributed systems. Distributed Participatory design focuses
on developing an understanding between different stakeholders in distributed design teams by
giving each the opportunity to engage in hands-on activities.
Why do some vertebrates, like a hellbender salamander, exchange oxyg.pdf
Why do some vertebrates, like a hellbender salamander, exchange oxygen as well as carbon
dioxide through their skin but bony fishes, like a goldfish, only exchange oxygen? (Choose One)
A. Goldfish do not produce CO2 since they live completely underwater
B. Goldfish have gills that diffuse CO2 while hellbender salamanders do not
C. Only young goldfish diffuse just O2 through their skin but adult goldfish diffuse both gases
D. Hellbender salamanders have hair-like projections on their skin, enabling them to diffuse both
gases
Solution
Answer - Option D - Hellbender salamanders have hair like projections on their skin, enabling
them to diffuse both gases.
Because they have cutaneous respiration..
Which of the following is the most direct source of energy for cotran.pdf
Which of the following is the most direct source of energy for cotransport? The movement of
one of the transported substances up its concentration gradient The movement of one of the
transported substances down its concentration gradient ATP hydrolysis ATP formation
Cotransport requires no energy
Solution
ATP hydrolysis
The direct source of energy for Cotransport is ATP hydrolysis to ADP means it can be achieved
by active transport..
Which minerals commonly show pleochroisma. Isometric minerals.pdf
Which minerals commonly show pleochroism?
a. Isometric minerals
b. Triclinic minerals
c. Framework silicates
d. Mafic minerals
e. Quartz
Solution
A substance which appears to be different colors when observed at different angles, with
polarized light is called Pleochroism. it is an optical phoenomenon.
Triclinic minerals show pleochroism.
Answer is (b)..
Two cards are drawn from a standard deck of cards. Determine the pro.pdf
Two cards are drawn from a standard deck of cards. Determine the probability that the first card
is a Heart and the second card is an Ace.
Solution
Here,
P(heart then ace) = P(heart) P(ace|first is heart of ace) + P(heart) P(ace|first is not heart of ace)
Thus,
P(heart then ace) = (1/4)(3/51) + (1/4)(4/51)
= 7/204 or 0.034313725 [ANSWER].
The probability that a student will find a (desirable) job before gr.pdf
The probability that a student will find a (desirable) job before graduation is .47 and the
probability that a student will get married before graduation in .38. The probability that a student
will get a job and get married before graduation is .17. What is the probability that a student will
either find a job or get married before graduation?
Solution.
The Northern Pacific Railroad actively lobbied for which of the foll.pdf
The Northern Pacific Railroad actively lobbied for which of the following?
Solution
The Northern Pacific Railroad did eventually lobby for the idea of a national park when in 1871
when Judge William D. Kelley sent a letter by way of Northern Pacific Railroad executive A. B.
Nettleton to Ferdinand Hayden to lobby in Congress for a national park..
The difference between vertical and horizontal transmission of plant .pdf
The difference between vertical and horizontal transmission of plant viruses is that vertical
transmission is transmission of a virus from a parent plant to its progeny, and horizontal
transmission is one plant spreading the virus to another plant the spread of viruses from trees
and tall plants to bushes and other smaller plants, and horizontal transmission is the spread of
viruses among plants of similar size the spread of viruses from upper leaves to lower leaves of
the plant, and horizontal transmission is the spread of a virus among leaves at the same general
level the transfer of DNA from a plant of one species to a plant of a different species, and
horizontal transmission is the spread of viruses among plants of the same species What is the
main structural difference between enveloped and nonenveloped viruses? Enveloped viruses
have their genetic material enclosed by a layer made only of protein. Both types of viruses have
a capsid and phospholipid membrane; but in the nonenveloped virus the genetic material is
between these two membranes, while in the enveloped virus the genetic material is inside both
membranes. Enveloped viruses have a phospholipid membrane outside their capsid, whereas
nonenveloped viruses do not have a phospholipid membrane. Nonenveloped viruses have only a
phospholipid membrane, while enveloped viruses have two membranes, the other one being a
protein capsid. If all prokaryotes on Earth suddenly vanished, which of the following would be
the most likely and most direct result? There would be no more pathogens on Earth. Human
populations would thrive in the absence of disease. The recycling of nutrients would be greatly
reduced, at least initially. Bacteriophage numbers would dramatically increase. Viruses are
referred to as obligate parasites because they cannot reproduce outside of a host cell viral DNA
always inserts itself into host DNA. they invariably kill any cell they infect. they can
iocorporate nucleic acids from other viruses they must use enzymes encoded by the virus itself
Bacteriophage DNA that have become integrated into the host cell chromosome are called
intemperate bacteriophages. traosposons. prophages. T-even phages. plasmids. Virulent
pahges undergo a(n) life cycles, whereas temperate phages are capable of undergoing a(n)
cycle. infective; retroviral lysogenic; lytic lytic; lysogenic retroviral; infective Infective;
benign
Solution
36.Answer a. In Horizontal transmission spread of virus takes place from one plant to other
while vertical transmission involves transfer of virus from plant to offspring
37.Answer c. Phospholipid enclosed virus they are enveloped virus while other lack
phospholipid membrane
38.Answer c.microbes help in recycling of material.
39.Answer a. Reproduction takes place inside host
40. Answer d. Known as phage
41.Answer c : Lyric and lysogenic phage.
select all that apply to ground substanceSolutionLets us see w.pdf
select all that apply to ground substance
Solution
Lets us see what is ground substance.
It is an amorphous gel-like or a transparent flud substance surrounding the cells,present in the
connective tissue .
The ground substance and fibers between cells in a connective tissue (most abundant tissues in
the body-functions : binding and supporting; consists of relatively few cells in a generous matrix
) are surrounded and supported by matrix called as extracellular matrix.
The Ground substance includes all the other components of the extracellular matrix excluding
fibers (Eg: collagen and elastic fibers).
Depending upon the tissue, the components of the ground substances vary .
Ground substance is primarily composed of water, proteoglycans,glycosaminoglycans (most
notably hyaluronan-hyaluronic acid), calcium and glycoproteins.
In cytology, it may also refers to cytosol or protoplasm.
The stroma is the tissue that forms the ground substance, base or the foundation, or the
framework of an organ, as opposed to its functional parts (parenchyma)..
Relying on the above pathway for biosynthesis of tryptophan, fill i.pdf
Relying on the above pathway for biosynthesis of tryptophan, fill in the table below to predict
growth (+) or failure to grow (-) when mutants defective in each of the trp genes are placed on
media supplemented with the compounds listed.
Solution
Answer:
A mutant for a particular gene in the tryptophan pathway will only grow when it is supplemented
with the required nutrient so that the biosynthetic pathway for the tryptophan synthesis is
resumed. The mutant strain (auxotroph) will proliferate only when the medium is supplemented
with the required substance leading to the completion of the tryptophan biosynthetic pathway.
+ indicates growth when a particular supplement is given. Mutants which have dysfunctional
trpB, trpC, trpD or trpE can still grow if supplemented with any of Indole, tryptophan or
Indole+tryptophan because in all of these cases the strain will get the required tryptophan
necessary for its survival.SupplementMutanttrpAtrpBtrpCtrpDtrpENone-----Anthranilic acid----
+Indole-++++Tryptophan+++++Typtophan and indole+++++.
Question 7 (2 points)A researcher conducts a study on the realtion.pdf
Question 7 (2 points)
A researcher conducts a study on the realtionship between caffeine consumption and
performance on a math test. What kind of study is this?
Question 7 options:
There is not enough information here to tell.
Correlational
Quasi-experimental
Experimental
Save
Question 8 (2 points)
Which researcher is most likely to engage in content analysis?
Question 8 options:
Dr. Williams, who studies transcripts of police interviews with suspects.
Dr. McKenzie, who studies relationships between various personality traits.
Dr. Perryman, who studies the social skills of indigenous people.
Dr. Ramirez, who studies mathematical competence in adolescents.
Save
Question 9 (2 points)
Which of the following would be more likely to be found in an empirical research article on
bullying than in a popular magazine article on the same topic?
Question 9 options:
Opinions about why people bully from the point of view of the author of the article.
Excerpts from actual interviews with bullies and/or victims.
Summaries of the methods and results found by previous researchers on bullying accompanied
by reference citations.
Humerous anecdotes about how victims fantasize about getting back at their bullies.
Save
Question 10 (2 points)
Dr. Simmons is interested in studying attachment behavior in infants. She trained her research
assistants to recognize classic attachment behaviors such as fearful reactions to strangers and
proximity seeking with the attachment figure. The research assistants then learned to code these
behaivors while watching video tapes of infants and caregivers interacting while they waited in
the doctor\'s waiting room. What type of research is going on here?
Question 10 options:
Quasi-experimental research
Naturalistic observation
Archival data collection
Interrupted time-series designa)
There is not enough information here to tell.b)
Correlationalc)
Quasi-experimentald)
Experimental
Solution
b)
Correlational
a)
Dr. Williams, who studies transcripts of police interviews with suspects.
c)
Summaries of the methods and results found by previous researchers on bullying accompanied
by reference citations
c)
Archival data collection
b)
Correlational.
Primary difference between the income sections for form 1120 and .pdf
Primary difference between the income sections for form 1120 and 1120s
Primary difference between the income sections for form 1120 and 1120s
Solution
Answer :- The primary difference between form 1120 and 1120s is that the form 1120 is filed by
corporations on annual basis under internal revenue service (IRS) code whereas the form 1120s
is filed by S corporation (corporation who have opted for \"S\" status) on annual basis under
internal revenue service (IRS) code..
Once its readable i will be okay.SolutionI will be okay in.pdf
Once it\'s readable i will be okay.
Solution
\"I will be okay in the end. If it\'s not okay, it\'s not the end.\" What are some incidences when
this John Lennon quote was proved true in real life?
Many inspiring stories of people\'s lives show it...
Consider Abraham Lincoln. He met failure after failure throughout his life before he became the
President of USA at 52.
Thomas Edison... 10,000 failures before inventing the light bulb.
Henry Ford... was broke at the age of 40.
Conolel Sanders... broke at 65... knocked at 1000 doors before he could make his first sale of a
chicken dish...founder of KFC....
Each and everytime in life ,
On all the small tasks and bigger challenges...its proven true.
It may be preparing a cup of tea or coffee or
Working on a home . The tea or coffee after two boils are perfectly okay for me but you may
boil it more with extra milk.
I work on decor with few frames of modern art and exclusive vases but you may go for plain
walls.
A waiter takes the order and serves and is okay with it...but another one wears a smile and is
quite alert and won\'t let the glass of water to be empty because its his level of \"okay\".
So each of us working and completing our daily task work on this principle.
We try hard and reach our okay\'s.
There\'s a variation in \"okay\" of all of us...that\'s it..
Need the mutiple choice answered uslde .An)_can omunicate with a sys.pdf
Need the mutiple choice answered uslde .An)_can omunicate with a system inside a firewall
system to be used to filter out undesirable traffic and prevent employees from accessing
potentially hostile web sites Multiple-Choice Quiz e three types of event logs generated What are
the three t t bw Windows Which of the following is not a capability of 6. Windows 2003 and
Vista by A. Event, Process, and Security B. Application, User, and Security c. User, Event, and
Security network-based IDS A. Can detect denial-of-service attacks B. Can decrypt and read
encrypted traffic C. Can decode UDP and TCP packets D. Can be tuned to a particular network
Application, System, and Security What are the two main types of intrusion detection systems?
A. Network-based and host-based B. Signature-based and event-based C. Active and reactive D.
Intelligent and passive environment An active IDS can: A. Respond to attacks with TCP resets
B. Monitor for malicious activity C. A and B D. None of the above Honeypots are used to: , 7. s.
What was the first commercial, network-based 8. IDS product? A. Stalker B. NetRanger C. IDES
D. RealSecure What are the two main types of IDS signatures? 9. Egress filtering is used to
detect spam that is: A. Network-based and file-based B. Context-based and content-based C
Active and reactive A. Attract attackers by simulating systems with open network services B.
Monitor network usage by employees C. Process alarms from other IDSs D. Attract customers to
e-commerce sites A. Coming into an organization B. Sent from known spammers outside your
organization C. Leaving an organization D. Sent to mailing lists in your organization
Preventative intrusion detection systems: A. Are cheaper B. D. None of the above . Which of the
following describes a passive, host based IDS? 10. A. Runs on the local system Are designed to
stop malicious activity from occurring B. Does not C. Can l D. All of the above not interact with
the traffic around it C. Can only monitor activity D. Were the first types of IDS ook at system
event and error logs r 13:Indusion Detection Systems Detection Systems and Network Security
Solution
1. What are the three types of event logs generated by Windows 2003 and Vista systems?
D. Application, System, and Security
2. What are the two main types of intrusion detection systems?
A. Network-based and host-based
3. What was the first commercial, network-based IDS product?
B. NetRanger
4. What are the two main types of IDS signatures?
B. Context-based and content-based
5. Which of the following describes a passive, host-based IDS?
D. All of the above
6. Which of the following is not a capability of network-based IDS?
B. Can decrypt and read encrypted traffic
7. An active IDS can:
C. A and B
8. Honeypots are used to:
A. Attract attackers by simulating systems with open network services
9. Egress filtering is used to detect spam that is:
C. Leaving an organization
10. Preventative intrusion detection systems:
B. Are designed to stop m.
Need a strong biologistgeneticist to answer this P element question.pdf
Need a strong biologist/geneticist to answer this P element question for Drosophila... PLEASE
What do you know about P-elements in Drosophila? How might P-elements have entered
Drosophila melanogaster? What do you know about P-element structure? How big are the native
elements? How do P-elements move around the genome? Connect your ideas to the special P-
elements used in this lab. What is the connection between transposition of mobile DNA elements
and DNA replication? Why are P-elements important for biologists and Drosophila researchers
in particular? What do you know about the Gene Disruption Project (GDP) in Drosophila?
Solution
Ans.) P element is a type of transposon found particularly in Drosophila and is used for
mutagenesis and for the production of genetically modified flies that is extensively used in
various genetic and medical researches. It has ability to initiate a phenotype known as hybrid
dysgenesis. It encodes the protein P transposase and is flanked by terminal inverted repeats
which are important for its mobility.
The P element enters in Drosophila by the horizontal gene transfer method. It has a sequence of 4
exons with 3 introns. The absolute splicing of the intron produces the transposase enzyme,
whereas the alternative partial splicing of intron 1 and 2 produce the P element repressor.
The P element is a class II transposon and it is moved by DNA-based \"cut and paste\"
mechanism. The P element encodes the transposase enzyme which recognizes the 31 bp terminal
inverted repeats of the P element and catalyzes P element excision and re-insertion.
P element is important for biologist and genetics because nowadays it is widely used to solve
many puzzles of genetics through its use in fly transformation, insertional mutagenesis as well as
reverse genetics.
The Gene Disruption Project (GDP) is a public collection of mutant that is created by applying a
non-targeted transposon mutagenesis strategy. Its main objective is to acquire a transposable
element insertion in each gene and this collection is available online and new mutants may be
requested while they are being balanced. These balanced stocks are available at the Bloomington
Drosophila Stock Center..
Modify HuffmanTree.java and HuffmanNode.java to allow the user to se.pdf
Modify HuffmanTree.java and HuffmanNode.java to allow the user to select the value of n 9 to
construct an n ary Huffman Tree. Once finished, verify the accuracy of your tree by running it
for the English Alphabet for n = 3, 4, 5, 6, 7, 8, 9. For each value of n, compute the average
codeword length. Write these in a table (n vs. ACW L(Cn)).
/ HuffmanTree.java
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.util.ArrayList;
import java.util.Scanner;
public class HuffmanTree {
public static void main(String[] args){
Scanner in = new Scanner(System.in);
System.out.print(\"Enter the file name: \");
String fileName = in.nextLine();
ArrayList nodes = new ArrayList<>();
ArrayList temp = new ArrayList<>();
try {
Scanner fin = new Scanner(new FileReader(fileName));
String symbol;
double prob;
while(fin.hasNext()){
symbol = fin.next();
prob = fin.nextDouble();
nodes.add(findInsertIndex(nodes, prob), new HuffmanNode(symbol, prob));
}
fin.close();
temp.addAll(nodes);
HuffmanNode root = createHuffmanTree(temp);
setCodeWords(root);
double avg = 0;
for (HuffmanNode node : nodes) {
System.out.println(node.getS() + \": \" + node.getC() + \": \" + node.getL());
avg += node.getL();
}
if(avg != 0)
avg /= nodes.size();
System.out.println(\"Average code length is: \" + avg);
} catch (FileNotFoundException e) {
e.printStackTrace();
}
}
private static HuffmanNode createHuffmanTree(ArrayList nodes){
HuffmanNode root = null;
if(nodes.size() > 0) {
HuffmanNode node0, node1, node;
while (nodes.size() > 1) {
node0 = nodes.get(0);
node1 = nodes.get(1);
node = new HuffmanNode(null, node0.getP() + node1.getP());
node.setLeft(node0);
node.setRight(node1);
nodes.remove(0);
nodes.remove(0);
nodes.add(findInsertIndex(nodes, node.getP()), node);
}
root = nodes.get(0);
}
return root;
}
private static void setCodeWords(HuffmanNode root){
if(root != null){
setCodeWords(root, \"\");
}
}
private static void setCodeWords(HuffmanNode root, String codeWord){
if(root.getS() != null){
root.setC(codeWord);
root.setL(codeWord.length());
}
else{
setCodeWords(root.getLeft(), codeWord + \"0\");
setCodeWords(root.getRight(), codeWord + \"1\");
}
}
private static int findInsertIndex(ArrayList nodes, double prob){
for(int i = 0; i < nodes.size(); ++i){
if(prob < nodes.get(i).getP()){
return i;
}
}
return nodes.size();
}
}
// HuffmanNode.java
public class HuffmanNode {
private String S;
private double P;
private String C;
private int L;
private HuffmanNode left, right;
public HuffmanNode(String s, double p) {
S = s;
P = p;
}
public String getS() {
return S;
}
public void setS(String s) {
S = s;
}
public double getP() {
return P;
}
public void setP(double p) {
P = p;
}
public String getC() {
return C;
}
public void setC(String c) {
C = c;
}
public int getL() {
return L;
}
public void setL(int l) {
L = l;
}
public HuffmanNode getLeft() {
return left;
}
public void setLeft(HuffmanNode left) {
this.left = left;
}
public HuffmanNode getRight() {
return right;
}
public void se.
Listed are 2 of the core theories of evolution. For each, give a def.pdf
Listed are 2 of the core theories of evolution. For each, give a definition or brief explanation of
the theory and 2 examples of the type of evidence we have that supports the theory.
The theory of common descent –
2 examples of supporting evidence:
The theory of natural selection –
2 examples of supporting evidence:
Solution
Evolution:
Evolution is a process of change through time. It is the process by which modern organisms have
descended from ancient organisms.
In genetic terms, the change in the frequency of alleles in populations from one generation to
another generation. (change over time).
A. The theory of common descent:
According to the principle of common descent, all species (living and extinct) are descended
from ancient common ancestors.
In common descent theory darwin explains, that organisms are descended from common
ancestors that gradually change over time into entirely new organism.
To create new species, many generations must pass on traits with slight modifications.
Examples:
1. Archaeopteryx is between reptiles and birds.
2. Eustheopteron is an amphibious fish.
3. Seymouria is a reptile-like amphibian.
4. Therapsids were mammal-like reptiles.
B. Darwin\'s Natural selection:
According to the Darwin, It is the process that results in adaptation of an organism to its
environment which means of selectivity reproducing changes in the genotype.
In the process of evolution the natural selection is a basic mechanism, it is mainly happens due to
the mutations, migrations and genetic drift.
Examples:
1. The beaks of Darwin finches. The finches of the Galapagos islands were adopted to the island
in different ways, but are believed to share a common ancestor.
2. Peppered moths and industrial pollution: Natural selection favored dark colored moths in areas
of heavy pollution, while light colored moths are survived better in less polluted area.
3.The soapberry bugs have must mutated their beaks length genes, when they ran out of food.
Beak size variation:
In South Florida soapberry bugs feed on balloon vine tree with large fruits, longer beak present
in them.
In North Florida soapberry bugs feed on golden rain tree with small fruits, they have shorter
beaks..
Light is confined within the core of a simple clad optical fiber by.pdf
Light is confined within the core of a simple clad optical fiber by?
Solution
An optical fiber is a cylindrical dielec wave guide (nonconducting waveguide) that transmits
light along its axis, by the process of total internal reflection. To confine the optical signal in the
core, therefractive index of the core must be greater than that of the cladding..
Let L R^n rightarrow R^m be a linear transformation. Show that if n .pdf
Let L: R^n rightarrow R^m be a linear transformation. Show that if n > m, then L must have a
nontrivial kernel, i.e., dim(ker(L)) > 0. If n lessthanorequalto m does L have to be one-to-one?
Solution
For second proof,
If T is one-one, it can\'t map a non-zero vector into 0, since it already maps 0 into 0. This means
that its kernel can contain only 0.
Suppose the kernel contains just the zero vector. To check if T is one-one, check if it can map
two different vectors u and v in Rn into the same vector in Rm. If T(u) = T(v), then T(u – v) =
T(u) – T(v) = 0. This means that the non-zero vectoru – v is in the kernel of T, which is
impossible since we assumed the kernel contained just 0. Then T cannot map different vectors
into the same vector, and must be one-one..
IS path vector routing algorithm closer to the distance vector rou.pdf
: IS path vector routing algorithm closer to the distance vector routing algorithm or to the link
state routing algorithm? Explain.
Solution
The path-vector routing algorithm is truly distance-vector routing using the finest path instead of
the direct distance as the metric. Each node first create a forwarding table. Assuming it can only
arrive at instant neighbors . The forwarding table is slowly better as path vectors turn up from the
instant neighbors..
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Element# of protons# of neutrons# of electronsCarbon (atomic.pdf
- 1. Element # of protons # of neutrons # of electrons Carbon (atomic # 6) Neutral atom: Ion: Isotope: Neutral atom: Ion: Isotope: Neutral atom: Ion: Isotope: Oxygen (atomic # 8) Neutral atom: Ion: Isotope: Neutral atom: Ion: Isotope: Neutral atom: Ion: Isotope: Hydrogen (atomic # 1) Neutral atom: Ion: Isotope: Neutral atom: Ion: Isotope: Neutral atom: Ion: Isotope: Nitrogen (atomic # 7)
- 2. Neutral atom: Ion: Isotope: Neutral atom: Ion: Isotope: Neutral atom: Ion: Isotope: How does the number of electrons affect how many bonds are formed by these atoms? Define the following (see boxes on pp 31, 33) Polar covalent bond Non-polar covalent bond Ionic bond Hydrogen bond Hydrophobic Hydrophilic Amphipathic Element # of protons # of neutrons # of electrons Carbon (atomic # 6) Neutral atom: Ion: Isotope: Neutral atom: Ion: Isotope:
- 3. Neutral atom: Ion: Isotope: Oxygen (atomic # 8) Neutral atom: Ion: Isotope: Neutral atom: Ion: Isotope: Neutral atom: Ion: Isotope: Hydrogen (atomic # 1) Neutral atom: Ion: Isotope: Neutral atom: Ion: Isotope: Neutral atom: Ion: Isotope: Nitrogen (atomic # 7) Neutral atom: Ion: Isotope: Neutral atom: Ion: Isotope: Neutral atom: Ion: Isotope: Solution
- 4. Element # of protons # of electrons # of neutrons Ion Isotopes Carbon 6 6 6 C+4 or C-4 Carbon-12 which has 6 protons and 6 neutrons and 6 electrons. Carbon-14 which has 6 protons, 8 neutrons and 6 electrons. Oxygen 8 8 8 Oxygen generally forms O-2. It has 6 electrons in the outermost orbit. Three stable isotopes of oxygen are found- O-16,O17 and O-18. Hydrogen 1 1 1 Hydrogen has a charge of +1. Hydrogen has 3 stable isotopes – protium (1 proton and 1 electron), deuterium (1 proton, 1 electron and 1 neutron) and tritium (1 proton, 1 electron and 2 neutrons) Nitrogen 7 7 7 Usually has the charge -3. N-14 has odd number of protons and neutrons (7 each). N-15 has protons and 8 neutrons. Carbon does not usually form an ion by loss or gain of 4 electrons. It shares its 4 electrons with 4 others to complete its orbit and forms a covalent bond. Oxygen has 6 valence lectrons in its outermost shell, therefore it can form 2 bonds Hydrogen atom has one valence electron in its outermost orbit. The maximum capacity of its
- 5. orbital is 2 electrons. Nonpolar covalent bond Non-polar covalent bond is formed when the atoms (with similar electron affinity) share electrons equally. The electronegativity difference between the two bonded atoms is usually less than 0.5. Examples – diatomic molecules H2 and N2. Polar covalent bond Polar covalent bond is formed when the electrons are not equally shared i.e. the atom with higher electronegativity (affinity) pulls the electrons towards itself. The compound which is formed has electrostatic potential and has the ability to form weak bonds with other polar molecules. The electronegativity difference between bonded atoms is 0.5 to 1.9. Examples include water, ozone. Ionic bond – the valence electrons are completely transferred from one atom to another. The metal loses electrons to become positively charged cation while non-metals accept electrons to form a negatively charged ion. The net charge of the compound formed via ionic bonding has a net charge of 0. Example – NaCl, CaCl2. Hydrogen bonding is a special dipole dipole weak attractive foece which forms between hydrogen atom of one electronegative atom and an electronegative atom of different molecule. The aforementioned electronegative atom must has one or more unshared electron pairs and a partial negative charge. The hydrogen atom with a partial positive charge searches for another atom of oxygen or nitrogen with excess electrons and gets attracted to its partial negative charge. In other words, hydrogen bonding occurs between hydrogen and fluorine/oxygen/nitrogen. Moelcules that exhibit hydrogen bonding include ammonia, chloroform and nylon. Hydrophobicity refers to repulsion from water. The tendancy of nonpolar molecules to aggregate in aqueous medium by excluding water molecules is known as hydrophobicity. Hydrophobicity plays an important role in protein folding, insertion of membrane proteins into non-polar lipid environment, formation of cell membrane. Hydrophilic refers to thermodynamically favourable interactions between molecules and water or polar molecules. Hydrophilicity basically means affinity towards water. The molecules which are hydrophilic usually carry a charge and therefore are polar in nature. Amphipathic are molecules which have both hydrophilic (polar) and hydrophobic (non-polar) parts. Examples include detergents, phospholipids of biological membranes. Element # of protons # of electrons # of neutrons Ion Isotopes
- 6. Carbon 6 6 6 C+4 or C-4 Carbon-12 which has 6 protons and 6 neutrons and 6 electrons. Carbon-14 which has 6 protons, 8 neutrons and 6 electrons. Oxygen 8 8 8 Oxygen generally forms O-2. It has 6 electrons in the outermost orbit. Three stable isotopes of oxygen are found- O-16,O17 and O-18. Hydrogen 1 1 1 Hydrogen has a charge of +1. Hydrogen has 3 stable isotopes – protium (1 proton and 1 electron), deuterium (1 proton, 1 electron and 1 neutron) and tritium (1 proton, 1 electron and 2 neutrons) Nitrogen 7 7 7 Usually has the charge -3. N-14 has odd number of protons and neutrons (7 each). N-15 has protons and 8 neutrons.