The correct statements are When two atoms of app.pdf

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The correct statements are When two atoms of approach each other, the two nuclei repel each other. When two atoms of approach each other, each electron is attracted by the other nucleus Solution The correct statements are When two atoms of approach each other, the two nuclei repel each other. When two atoms of approach each other, each electron is attracted by the other nucleus.

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One P2O5 contains five oxygen 3.45 x 5 = 17.3 mol.pdf

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Na2O is basic NO2 is acidic SO2 is acidic CO2 is .pdf

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moles of CO2 = moles of NaHCO3 = 284 =0.0238 .pdf

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Eukaryotic Chromosome Structure The length of DNA in the nucleus is far greater than the size of the compartment in which it is contained. To fit into this compartment the DNA has to be condensed in some manner. The degree to which DNA is condensed is expressed as its packing ratio. Packing ratio - the length of DNA divided by the length into which it is packaged For example, the shortest human chromosome contains 4.6 x 107 bp of DNA (about 10 times the genome size of E. coli). This is equivalent to 14,000 µm of extended DNA. In its most condensed state during mitosis, the chromosome is about 2 µm long. This gives a packing ratio of 7000 (14,000/2). To achieve the overall packing ratio, DNA is not packaged directly into final structure of chromatin. Instead, it contains several hierarchies of organization. The first level of packing is achieved by the winding of DNA around a protein core to produce a \"bead-like\" structure called a nucleosome. This gives a packing ratio of about 6. This structure is invariant in both the euchromatin and heterochromatin of all chromosomes. The second level of packing is the coiling of beads in a helical structure called the 30 nm fiber that is found in both interphase chromatin and mitotic chromosomes. This structure increases the packing ratio to about 40. The final packaging occurs when the fiber is organized in loops, scaffolds and domains that give a final packing ratio of about 1000 in interphase chromosomes and about 10,000 in mitotic chromosomes. Eukaryotic chromosomes consist of a DNA-protein complex that is organized in a compact manner which permits the large amount of DNA to be stored in the nucleus of the cell. The subunit designation of the chromosome is chromatin. The fundamental unit of chromatin is the nucleosome. Chromatin - the unit of analysis of the chromosome; chromatin reflects the general structure of the chromosome but is not unique to any particular chromosome Nucleosome - simplest packaging structure of DNA that is found in all eukaryotic chromosomes; DNA is wrapped around an octamer of small basic proteins called histones; 146 bp is wrapped around the core and the remaining bases link to the next nucleosome; this structure causes negative supercoiling The nucleosome consists of about 200 bp wrapped around a histone octamer that contains two copies of histone proteins H2A, H2B, H3 and H4. These are known as the core histones. Histones are basic proteins that have an affinity for DNA and are the most abundant proteins associated with DNA. The amino acid sequence of these four histones is conserved suggesting a similar function for all. The length of DNA that is associated with the nucleosome unit varies between species. But regardless of the size, two DNA components are involved. Core DNA is the DNA that is actually associated with the histone octamer. This value is invariant and is 146 base pairs. The core DNA forms two loops around the octamer, and this permits two regions that are .

Eukaryotic Chromosome Structure The length of DN.pdf

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Fe + S ----> FeS moles of Fe = 9.42/56 = 0.1682 moles of S = 8.5/32 = 0.265625 here Fe is limiting so, all the Fe will react to form same amount of FeS as Fe. S0, amount of FeS = 0.1682moles mass of FeS = 0.1682*88 = 14.8g Solution Fe + S ----> FeS moles of Fe = 9.42/56 = 0.1682 moles of S = 8.5/32 = 0.265625 here Fe is limiting so, all the Fe will react to form same amount of FeS as Fe. S0, amount of FeS = 0.1682moles mass of FeS = 0.1682*88 = 14.8g.

Fe + S ---- FeS moles of Fe = 9.4256 = 0.1682 m.pdf

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D. is correct. Solution .pdf

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The various ways to finance a project are: 1)Debt financing-One can take out a loan and finance the project costs Pros: Ownership is maintained Lower interest rate Lower cost of capital Cons: Repayment needed even if business goes bankrupt High rates for old debt even if macroeconomic conditions change Collateral needed 2) Equity financing is when project financing provides share in a firm in lieu of financing Pros- Easily availaible than debt Not to be repaid in case of bankrupcy Periodic payment as in case of debt is not required. Dividends can be paid as when the owner feels so Cons- have to share profit High return required by investor May lose control of company Private equity funding is when a large investor invests in company and his share cannot be publicly tarded: Pros- Capital is easily availaible and all capital can be raised at once The private equity investor provides his insights No return requirement and no cash flow has to be paid out as in case of equity Cons- More interference in day to day matters by investor investor may want to recoup his investment when he feels so but company may still have better profits Solution The various ways to finance a project are: 1)Debt financing-One can take out a loan and finance the project costs Pros: Ownership is maintained Lower interest rate Lower cost of capital Cons: Repayment needed even if business goes bankrupt High rates for old debt even if macroeconomic conditions change Collateral needed 2) Equity financing is when project financing provides share in a firm in lieu of financing Pros- Easily availaible than debt Not to be repaid in case of bankrupcy Periodic payment as in case of debt is not required. Dividends can be paid as when the owner feels so Cons- have to share profit High return required by investor May lose control of company Private equity funding is when a large investor invests in company and his share cannot be publicly tarded: Pros- Capital is easily availaible and all capital can be raised at once The private equity investor provides his insights No return requirement and no cash flow has to be paid out as in case of equity Cons- More interference in day to day matters by investor investor may want to recoup his investment when he feels so but company may still have better profits.

The various ways to finance a project are 1)Debt financing-One ca.pdf

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boiling point: 1-propanol > propanone note: there exists intermolecular hydrogen bonding interaction between 1-propanol (i.e., H-O...H bond), whereas no such interaction between propanone molecules. Thus, bp of 1-propanol is higher than propanone. Solution boiling point: 1-propanol > propanone note: there exists intermolecular hydrogen bonding interaction between 1-propanol (i.e., H-O...H bond), whereas no such interaction between propanone molecules. Thus, bp of 1-propanol is higher than propanone..

boiling point 1-propanol propanone note there.pdf

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The organs of urinary system includesPaired kidneys, paired ureter.pdf

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The incorrect statement ise. Glucose is a stereoisomer of ribose..pdf

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The answer is: a. The intermolecular forces are much greater in water than in hydrogen sulfide and hydrogen selenide. Boiling point increases with strength of intermolecular forces as these intermolecular bonds need to be broken for the molecules to escape into the gas phase. Water has strong intermolecular hydrogen bonds not present in hydrogen sulfide and hydrogen selenide. Thus water has a much higher boiling point. Solution The answer is: a. The intermolecular forces are much greater in water than in hydrogen sulfide and hydrogen selenide. Boiling point increases with strength of intermolecular forces as these intermolecular bonds need to be broken for the molecules to escape into the gas phase. Water has strong intermolecular hydrogen bonds not present in hydrogen sulfide and hydrogen selenide. Thus water has a much higher boiling point..

The answer is a. The intermolecular forces are much greater in wate.pdf

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potassium and sulfur - K2S Ionic Bondcalcium and chlorine - CaCl2 .pdf

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About elements, atoms and compounds definition an.pdf

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Part A: Jean Baptiste de Lamarck is best known for his Theory of Inheritance of Acquired Characteristics, first presented in 1801. Here are some of the basic theories of Lamarckism: The theory of evolution by natural selection, first formulated in Darwin\'s book \"On the Origin of Species\" in 1859, is the process by which organisms change over time as a result of changes in heritable physical or behavioral traits. The main points regarding Darwinism are as follows: Darwin\'s theory has been supported by a lot of evidence. Hence, it was accepted by the scientific community during his time. Lamarck\'s Theory of Inheritance of Acquired Characteristics has been disproved. This was done in two major ways. The first is by experiment. We have seen through many real examples and observations that changes that occur in an animal during life are not passed on to the animal\'s offspring. If a dog\'s ears are cropped short, its puppies are still born with long ears. The other way that Lamarck\'s theory has been proven wrong is the study of genetics. Darwin knew that traits are passed on, but he never understood how they are passed on. During the time when Darwin\'s first book first came out, Gregor Mendel, who discovered genetics, was just starting his experiments. These and other examples show that Lamarck\'s theory does not explain how life formed and became the way it is. Part B: (i) Alfred Russel Wallace: He discovered the concept of evolution by natural selection. Although he is now rarely mentioned as the discoverer (Darwin, who discovered the theory independently, is usually cited) Wallace enjoyed a very high reputation in his lifetime and was awarded many of science’s most prestigious awards.Alfred Russel Wallace did not exactly influence Charles Darwin, but rather was his contemporary and collaborated with Darwin on solidifying his Theory of Evolution through Natural Selection.The two pooled their data to present the idea jointly to the Linnaean Society of London. (ii)Stanley Miller: He was an American chemist who designed the first experiment to produce organic molecules from some of the inorganic components of the Earth’s prebiotic atmosphere. Miller’s procedure (which was co-designed by Harold Urey and is known as the Miller-Urey experiment) contained three key elements: a heated pool of water meant to simulate the primitive Earth’s ocean; an atmosphere of water vapour, methane, ammonia, and molecular hydrogen; and storms of “lightning” in the form of continuous electric discharges. After one week, 10–15% of the system’s carbon was found in organic molecules such as amino acids, the building blocks of proteins. Miller’s experiment was not only a groundbreaking moment for research into the origin of life on Earth but also a breakthrough that captured the popular imagination and gave rise to the term prebiotic soup. Miller\'s experiment was important because it showed that organic compounds such as amino acids, which are essential to cellular .

Part AJean Baptiste de Lamarck is best known for his Theory of In.pdf

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a) The solubility of the ionic compound is reduce.pdf

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Hyperventilation is “clinical reduction in PCO2 caused by increasing the alveolar minute volume of ventilation”. Caused by rapid/deep breathing. Tidal volume (TV) – air that moves into and out of the lungs with each breath (approximately 500 ml) Alveolar ventilation rate (AVR) – measures the flow of fresh gases into and out of the alveoli during a particular time When alveolar ventilation is excessive, more carbon dioxide will be removed from the blood stream than the body can produce. This causes the concentration of carbon dioxide in the blood stream to fall and produces a state known as hypocapnia (reduced Carbon dioxide). The body normally attempts to compensate for this metabolically. Thus there is decrease in respiration rate and tidal volume as metabolic measure to avoid side effects from reduction of carbon dioxide by naturally trying to bring the breathing pattern back to normal. Solution Hyperventilation is “clinical reduction in PCO2 caused by increasing the alveolar minute volume of ventilation”. Caused by rapid/deep breathing. Tidal volume (TV) – air that moves into and out of the lungs with each breath (approximately 500 ml) Alveolar ventilation rate (AVR) – measures the flow of fresh gases into and out of the alveoli during a particular time When alveolar ventilation is excessive, more carbon dioxide will be removed from the blood stream than the body can produce. This causes the concentration of carbon dioxide in the blood stream to fall and produces a state known as hypocapnia (reduced Carbon dioxide). The body normally attempts to compensate for this metabolically. Thus there is decrease in respiration rate and tidal volume as metabolic measure to avoid side effects from reduction of carbon dioxide by naturally trying to bring the breathing pattern back to normal..

Hyperventilation is “clinical reduction in PCO2 caused by increasing.pdf

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import java.util.Scanner; import java.math.BigInteger; public class Money{ private int dollars; private int cents; public Money(){ dollars=0; cents=0; } public Money(int d,int c){ dollars=d+c/100; cents=c%100; } public Money(int c){ dollars=c/100; cents=c%100; } public void add(int c){ cents+=c; dollars+=cents/100; cents=cents%100; } public void add(int d,int c){ cents+=c; dollars=dollars+d+cents/100; cents=cents%100; } public void remove(int c){ cents=dollars*100+cents; if(c>cents){ cents=0; } else{ cents-=c; } dollars=cents/100; cents=cents%100; } public String getMoney(){ return dollars+\" dollars and \"+cents+\" cents\"; } public static void main(String []args){ Money h=new Money(1,75); h.add(190); System.out.println(h.getMoney()); } } Solution import java.util.Scanner; import java.math.BigInteger; public class Money{ private int dollars; private int cents; public Money(){ dollars=0; cents=0; } public Money(int d,int c){ dollars=d+c/100; cents=c%100; } public Money(int c){ dollars=c/100; cents=c%100; } public void add(int c){ cents+=c; dollars+=cents/100; cents=cents%100; } public void add(int d,int c){ cents+=c; dollars=dollars+d+cents/100; cents=cents%100; } public void remove(int c){ cents=dollars*100+cents; if(c>cents){ cents=0; } else{ cents-=c; } dollars=cents/100; cents=cents%100; } public String getMoney(){ return dollars+\" dollars and \"+cents+\" cents\"; } public static void main(String []args){ Money h=new Money(1,75); h.add(190); System.out.println(h.getMoney()); } }.

import java.util.Scanner; import java.math.BigInteger; public cl.pdf

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#include #include\"Data.h\" #include\"Person.h\" #include\"Report.h\" #include #include #include \"Menu.h\" using namespace std; /* Programmers: Javier A. Molina, Eduardo Bonnet Students #:79249 */ int main() { int cnt; int idx; string fileName; Report * telefonia; Menu menu(\"Reports\"); bool fileNameExists(string fileName); Report * addReport(Report * reports, int & totalReports); Report * delReport(Report * reports, int & totalReports, int idx); int displayReports(Report * reports, int totalReports); menu.addItem(\"New report\"); menu.addItem(\"Delete a report\"); menu.addItem(\"Display a report\"); cnt = 0; telefonia = NULL; do { cout << menu; cin >> menu; //cls(); switch (menu.getOption()) { case 1: cout << \"Enter report name (TE?????): \"; cin >> fileName; if (fileNameExists(fileName + \".tel\")) { telefonia = addReport(telefonia, cnt); idx = cnt - 1; telefonia[idx].setFileName(fileName); // telefonia[idx].loadReport(); } else { cls(); cerr << \"Error: Report name \'\" << fileName << \"\' does not exists!\ \ \ \"; //pause(); } break; case 2: if (telefonia != NULL) { do { cls(); cout << \"Delete a Report:\ \ \"; idx = displayReports(telefonia, cnt); if (between(idx, 0, cnt - 1)) { cls(); telefonia = delReport(telefonia, cnt, idx); idx = 0; // pause(); } } while (idx != cnt); } else { cerr << \"No reports available for deleting!\ \ \ \"; //pause(); } break; case 3: char ans; if (telefonia != NULL) { do { // cls(); cout << \"Display/Filed a Report:\ \ \"; idx = displayReports(telefonia, cnt); if (between(idx, 0, cnt - 1)) { // cls(); cout << \"Would you like the report to be display otherwise save into a file? \"; cin >> ans; cls(); cin.ignore(); if (upper(ans) == \'Y\') { cout << telefonia[idx] << endl; } else { //telefonia[idx].saveReport(); cout << \"Report \'\" << telefonia[idx].getFileName() << \"\' saved as \'\" << telefonia[idx].getFileName() << \".rep\'.\ \ \"; } //pause(); } } while (idx != cnt); } else { cerr << \"No reports available for displaying!\ \ \ \"; // pause(); } break; case 4: cout << \"Thanks for using Telefonia Reports System!!!\ \ \ \"; break; } } while (menu.getLast() != menu.getOption()); delete[] telefonia; return 0; } //Funcion que verifica si existe un archivo en disco bool fileNameExists(string fileName) { ifstream fin(fileName.c_str()); bool exists = !(fin.fail()); if (exists) fin.close(); return(exists); } //Anade dinamicamente objetos tipo Report al arreglo reports Report * addReport(Report * reports, int & totalReports) { if (totalReports == 0) { reports = new Report[1]; } else { Report * tmp = new Report[totalReports]; for (int i = 0; i < totalReports; ++i) { tmp[i] = reports[i]; } delete[] reports; reports = new Report[totalReports + 1]; for (int i = 0; i < totalReports; ++i) { reports[i] = tmp[i]; } delete[] tmp; } totalReports++; return(reports); } //Borra objetos del arreglo reports segun el idx enviado Report * delReport(Report * reports, int & totalReports, int idx) { Report * tmp = new Report[tota.

#include iostream #includeData.h #includePerson.h#in.pdf

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GDP is defined as \"an aggregate measure of production equal to the sum of the gross values added of all resident, institutional units engaged in production (plus any taxes, and minus any subsidies, on products not included in the value of their outputs)\" by the OECD. It can be measured through any of the following three ways: In general the various components of GDP includes the following: Y = C + I + G + (X M) where Y = GDP C = Consumption I = INvestment G = Govt. Spending X-M = Net Export When talking about US\'s GDP, the major contribution is made by Consumption. Nearly 70% of what the United States produces is for personal consumption. In 2014, $11.929 trillion of the the total $17.418 trillion produced in the U.S, where goods contributed almost 1/4 of total GDP and service nearly half (45.7%) of GDP. Because of post recession period, investment is not that much contributing. Although fixed investment (specially real estate development) has a better role to play than private inventory change. Government spending added $3.176 trillion to the economy in 2014, 18.2% of total GDP. In the end, the United States imports more than it exports, creating a trade deficit hence having negative effect on GDP. In 2014, imports were $2.875 trillion, while exports were $2.335 trillion. As a result, international tradesubtracted $540 billion from GDP. All data are sourced from U.S. Bureau of Economic Analysis, National Income and Product Accounts Tables. Differences between real GDP and nominal GDP are given in the following table: Real GDP = pbqt where b denotes the base year. Nominal GDP = ptqt where p refers to price, q is quantity, and t indicates the year in question (usually the current year). Business Cycle refers to economic contractions, troughs, expansions and peaks that are unpredictable phases of economic activity. As the economy goes through business cycle changes, these positively or negatively affect the GDP. During a contraction, economic output slows, usually due to decreased demand for products and services, an increase in the cost of raw materials or both. An economic trough occurs after a contraction. A historically high national unemployment rate and low economic output usually mark this trough, which often signals that the economy is already in or heading toward a recession. Unlike a contractionary phase in which the GDP decreases but is still positive, during a trough the GDP is negative.After the \"rock bottom\" of an economic trough, expansion is its recovery. If the economy grows for two or three consecutive calendar quarters, it indicates that it is beginning its recovery and GDP begins to increase. An economic peak is like a mountain summit. Once the economy reaches this peak, it must come down. This economic peak is the highest point of economic growth and output, resulting in an increase in the GDP.BasisReal GDPNominal GDPMeaning/MeasureIt measures the value of output economy, adjusted for price changes.It measures the mark.

GDP is defined as an aggregate measure of production equal to the .pdf

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code: ----- public void add(int key, T value) { Link previous = null; // start at first Link current = first; // until end of list, while( current != null && key > current.getKey() ) { // or current > key, previous = current; current = current.next; // go to next item } if(previous==null) // if beginning of list, first = theLink; // first --> new link else // not at beginning, previous.next = theLink; // prev --> new link theLink.next = current; // new link --> current } Pseudo code: ------------ 1. start from first node Link current=first; 2. Traverse the hash table till the end of the list and the key value becomes greater than the current key value, and locate a position,called previous. 3. If the obtained position is null, make the new node as first node. Place the value in that node. 4. if the position is not null,insert the new node as next to previous node. Solution code: ----- public void add(int key, T value) { Link previous = null; // start at first Link current = first; // until end of list, while( current != null && key > current.getKey() ) { // or current > key, previous = current; current = current.next; // go to next item } if(previous==null) // if beginning of list, first = theLink; // first --> new link else // not at beginning, previous.next = theLink; // prev --> new link theLink.next = current; // new link --> current } Pseudo code: ------------ 1. start from first node Link current=first; 2. Traverse the hash table till the end of the list and the key value becomes greater than the current key value, and locate a position,called previous. 3. If the obtained position is null, make the new node as first node. Place the value in that node. 4. if the position is not null,insert the new node as next to previous node..

code ----- public void add(int key, T value) { Link previou.pdf

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