15.0mL = 0.015L phenylacetic acid 0.50molL 0.01.pdf

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15.0mL = 0.015L phenylacetic acid 0.50mol/L *0.015= 0.0075 mol phenylacetic acid 20.0mL= 0.020L sodium phenylacetate 0.40mol/L * 0.020L=.008 mol sodium phenylacetate due to the fact that this acts as a buffer Using henderson hasselblach pH= pKa + log(base/acid) and finding that the pKa of phenylaceticacid is 4.28 pH= 4.28 + log (.008/.0075) pH= 4.28 + 0.093 pH= 4.373 Solution 15.0mL = 0.015L phenylacetic acid 0.50mol/L *0.015= 0.0075 mol phenylacetic acid 20.0mL= 0.020L sodium phenylacetate 0.40mol/L * 0.020L=.008 mol sodium phenylacetate due to the fact that this acts as a buffer Using henderson hasselblach pH= pKa + log(base/acid) and finding that the pKa of phenylaceticacid is 4.28 pH= 4.28 + log (.008/.0075) pH= 4.28 + 0.093 pH= 4.373.

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We ca use HashSet data structure to remove duplicats from an array in O(n) time. HashSet does not allow duplicate elements. So add all elements from an array to HashSet. While adding first check whether current element is already in Set or not. If it is already there then remove from an array. PSUDO CODE: void removeDuplicateFromArray(int []array, int size): 1. create a HashSet: HashSet set = new HashSet(); 2. Itereate give array and add element in Set one by one: int k = 0; for i=0 to size-1: if(!set.contains(array[i]) // if current element is not in Set, then add it in set and include in array also array[k++] = array[i]; set.add(array[i]) Runnning TIme: O(n) Solution We ca use HashSet data structure to remove duplicats from an array in O(n) time. HashSet does not allow duplicate elements. So add all elements from an array to HashSet. While adding first check whether current element is already in Set or not. If it is already there then remove from an array. PSUDO CODE: void removeDuplicateFromArray(int []array, int size): 1. create a HashSet: HashSet set = new HashSet(); 2. Itereate give array and add element in Set one by one: int k = 0; for i=0 to size-1: if(!set.contains(array[i]) // if current element is not in Set, then add it in set and include in array also array[k++] = array[i]; set.add(array[i]) Runnning TIme: O(n).

We ca use HashSet data structure to remove duplicats from an array i.pdf

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this eample is of t-test for comparing two means here mean1 = mean of FIFO and mean 2 : Mean of LIFO so here we use student t-test for comparision of 2 means to evaluate both methods in this procedure ans: Student t-test for comparision of means Solution this eample is of t-test for comparing two means here mean1 = mean of FIFO and mean 2 : Mean of LIFO so here we use student t-test for comparision of 2 means to evaluate both methods in this procedure ans: Student t-test for comparision of means.

this eample is of t-test for comparing two meanshere mean1 = m.pdf

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The middle ear bones in mammals are derived from bones in the dentary bone (mandible) of Therapsids. In the Therapsid dentary, there were three bones used in the jaw hinge: the angular, articular and quadrate. The tympanic, malleus, and incus are homologous to the angular, articular, and quadrate Tympanic = Angular Malleus = Articular Incus = Quadrate In early Therapsids, the quadrate, angular and articular were directly connected to the dentary bone in the jaw hinge. These bones were essential for jaw movement. These bones were also utilized for crude hearing capabilities. The next stage of evolution came with the separation of these bones and the dentary. After the bones separated from the jaw, they formed the postdentary rod. It is believed that these the mammalian middle ear bones evolved at least twice. It is believed that monotreme and therian ancestors evolved very similar middle ear bones, but in two separate paths. In monotreme ancestors, there is a mandibular trough, that is believed to have accommodated the postdentary rod. This rod is believed to have been connected to the trough by ligaments. This rod was used for hearing and the quadrate, angular, and articular were no longer needed in the jaw hinge. In therians, the mandibular trough no longer held the quadrate, angular, and articular. These bones were held to the jaw by ossified Meckel’s cartilage. This bridge of cartilage and bone extended away from the dentary and approached the area that would make up the modern ear cavity. The evolution of the mammalian middle ear and jaw joint were pivotal steps in the evolution of mammals. It is also a great example of how classical comparative anatomy, paleontology and developmental biology have come together to piece together how this remarkable transformation of jaw joint to ear ossicles was able to come about. The homologies of the malleus, incus and stapes to the articular, quadrate and columella, and tympanic ring and gonial to the angular and prearticular suggested by comparative anatomy 175 years ago have been recently confirmed by molecular and developmental biology. The recent discovery of new mammali form fossils has allowed careful documentation of the shift from primary to secondary jaw articulation, creating an opportunity to follow the transformation of the post-dentary skeletal elements. This fossil data has been complemented by the study of marsupial development, providing insight into the changing role of the malleus and incus, and the relationship of the primary and secondary jaw joints. We now have a number of unanswered questions. What are the signalling molecules involved in the interactions between the condylar process and the glenoid fossa that create the novel mammalian jaw joint? What are the mechanisms that lead to transformation of Meckel’s cartilage into a ligament allowing isolation of the ear from the jaw? What controls the distribution of secondary cartilages? What controls the timing of differentiation an.

The middle ear bones in mammals are derived from bones in the dentar.pdf

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There is an old saying “Do not put all eggs in one basket” Why? Because risk is high.By keeping eggs in different baskets, we reduce risk We try to do same thing through diversification. By investing our capital in different assets , we try to reduce risk. Diversified folios reduce the risk and also the ratio of Risk to reward. If w1, w2 , w3 …wn are weight in the portfolio for assets 1, 2,3 ….n Then,w1+w2+w3+……………………+wn=1 R1, R2,R3,…….Rn are the return of the assets 1, 2 , 3 ….n S1, S2, S3……Sn are the standard deviation of the assets 1, 2, 3 …n Portfolio Return=w1R1+w2R2+w3R3+…….+wnRn Portfolio Variance=(w1^2)*(S1^2)+(w2^2)(S2^2)+………….(wn^2)*(Sn^2)+2w1w2*Cov(1,2)+2w1w3* Cov(1,3)+………+w(n-1)wn*Cov(n,(n-1) Cov(1,2)=Covariance of returns of asset1 and asset2 Portfolio Standard Deviation =Square root of Portfolio variance Risk of a stock is measured by standard deviation. Hence reduction of standard deviation through diversification means reduction of risk. We can take a simple example of two assets 1 and 2 Return of asset1=R1=15% Return of asset2=R2=12% Standard deviation of asset 1=S1=10% Standard deviation of asset 2=S2=8% Correlation of asset 1 and 2=Corr(1,2)=0.1 Covariance(1,2)=Corr(1,2)*S1*S2=0.1*10*8=8 Assume for simplicity, equal amount is invested in asset 1 and asset 2 Hence, w1=w2=0.5 Portfolio Return; 0.5*15+0.5*12=13.5% Portfolio Variance=(0.5^2)*(10^2)+(0.5^2)*(8^2)+2*0.5*0.5*8=45 Portfolio Standard Deviation=Square root of Variance=(45^0.5)= 6.708204 We can see, the risk of portfolio as measured by Standard Deviation has reduced significantly to 6,7 whereas the assets in the portfolio had standard deviation of 10 and 8 Risk / Return ratio of the portfolio=6.7/13.5=0.496 Risk/Return ratio of asset1=10/15= 0.666667 Risk/Return ratio of asset2=8/12= 0.666667 Risk return ratio of the portfolio is lower WHY INVEST IN DIFFERENT INDUSTRIES TO DIVERSIFY? Return of same types of assets are highly correlated. Suppose, if you invest in 10 different Technology Companies, you may not be able to reduce risk, because all the companies will have similar return and highly correlated. We have seen in the equation of Portfolio Variance, one factor=2 w1*w2*Cov(1,2) If there is high correlation between return of asset1 and 2, the Covariance(1,2) will be high . Hence this factor 2w1w2Cov(1,2) will be high. As a result portfolio standard deviation will be high and diversification will not serve any purpose. Hence, in order to get benefit of diversification, we need to invest in different type of industries so that the correlation between different assets are low. Consequently, the portfolio risk (measured through standard deviation) is low. Solution There is an old saying “Do not put all eggs in one basket” Why? Because risk is high.By keeping eggs in different baskets, we reduce risk We try to do same thing through diversification. By investing our capital in different assets , we try to reduce risk. Diversified folios reduce the risk and also the rati.

There is an old saying “Do not put all eggs in one basket”WhyBe.pdf

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The ER Model is focussed to be a description of real-world entities. Although its development/construction is done in such a way as to allow fair and easy translation to the relational schema model, this is not fully trivial process. The ER diagram represents the conceptual level of database design while the relational schema represents the logical level for the database design. Following simple rules are followed: ADDRESS( cityAddress, city, state ) candidate key:cityAddress, city , state PERSON( pid,name ) candidate key: pid OBSERVATION (oid, date, temperature, humidity, precip. ,city), candidate key: oid foreign key: city OBSERVER ( pid ), candidate key: pid METEREOLOGIST( pid ), candidate key: pid CONSUMER( pid ), candidate key: pid WARNINGS( wwid,description,date,city) candidate key: wwid FORECAST( fid,city,date,low,high) candidate key: fid WEATHERPROGRAM( wid,name ) candidatekey; wid LIVESAT(city, streerAddress, state,pid) RECORDS(oid,pid,method) ISSUES(wwid.pid) PRESENTS(fid,pid) VIEWS(pid,fid) MAKES(wid,fid,oid) 15 tables will be created in all along with relationship tables and only 9 tables if relationships are shown as foreign keys Solution The ER Model is focussed to be a description of real-world entities. Although its development/construction is done in such a way as to allow fair and easy translation to the relational schema model, this is not fully trivial process. The ER diagram represents the conceptual level of database design while the relational schema represents the logical level for the database design. Following simple rules are followed: ADDRESS( cityAddress, city, state ) candidate key:cityAddress, city , state PERSON( pid,name ) candidate key: pid OBSERVATION (oid, date, temperature, humidity, precip. ,city), candidate key: oid foreign key: city OBSERVER ( pid ), candidate key: pid METEREOLOGIST( pid ), candidate key: pid CONSUMER( pid ), candidate key: pid WARNINGS( wwid,description,date,city) candidate key: wwid FORECAST( fid,city,date,low,high) candidate key: fid WEATHERPROGRAM( wid,name ) candidatekey; wid LIVESAT(city, streerAddress, state,pid) RECORDS(oid,pid,method) ISSUES(wwid.pid) PRESENTS(fid,pid) VIEWS(pid,fid) MAKES(wid,fid,oid) 15 tables will be created in all along with relationship tables and only 9 tables if relationships are shown as foreign keys.

The ER Model is focussed to be a description of real-world entities..pdf

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Please find the answers below: Answer 16: Option A (IEP). (IEP or individual educational plan is an official document which carries the information depicting the personal/specific educatinal goals of children with special needs. It is not only a document, but refers to a protocol/programme which clearly demarcates the specific needs of a student. It has some features such as specilization, personalized, validity, usability as a shared document, information about requirement of a specific teacher/mentor or subject expert etc.) Answer 17: The statement is true. Technically, based upon the physical, physiological and psychological development of children, the age group of 6-11 has been globally accepted as the time of adolescence for growing children. This age group is associated with discrete phsycial growth (puberty), psycholgoical growth (feelings, emotions etc) and physiological growth (activation of reproductive organs). Solution Please find the answers below: Answer 16: Option A (IEP). (IEP or individual educational plan is an official document which carries the information depicting the personal/specific educatinal goals of children with special needs. It is not only a document, but refers to a protocol/programme which clearly demarcates the specific needs of a student. It has some features such as specilization, personalized, validity, usability as a shared document, information about requirement of a specific teacher/mentor or subject expert etc.) Answer 17: The statement is true. Technically, based upon the physical, physiological and psychological development of children, the age group of 6-11 has been globally accepted as the time of adolescence for growing children. This age group is associated with discrete phsycial growth (puberty), psycholgoical growth (feelings, emotions etc) and physiological growth (activation of reproductive organs)..

Please find the answers belowAnswer 16 Option A (IEP). (IEP or i.pdf

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Question 1 answer is A Global trade allows wealthy countries to use their resources - whether labor, technology or capital - more efficiently. Because countries are endowed with different assets and natural resources (land, labor, capital and technology), some countries may produce the same good more efficiently and therefore sell it more cheaply than other countries. If a country cannot efficiently produce an item, it can obtain the item by trading with another country that can. This is known as specialization in international trade. Question 2 answer is C All options are correct Question 3 answer is D Revenue tariffs are levied to raise money. Question 4 answer is D.(cannot promote third world development, unless governments offer some incentive for importing from third world economies). Question 5 answer is A Question 6 answer is C Solution Question 1 answer is A Global trade allows wealthy countries to use their resources - whether labor, technology or capital - more efficiently. Because countries are endowed with different assets and natural resources (land, labor, capital and technology), some countries may produce the same good more efficiently and therefore sell it more cheaply than other countries. If a country cannot efficiently produce an item, it can obtain the item by trading with another country that can. This is known as specialization in international trade. Question 2 answer is C All options are correct Question 3 answer is D Revenue tariffs are levied to raise money. Question 4 answer is D.(cannot promote third world development, unless governments offer some incentive for importing from third world economies). Question 5 answer is A Question 6 answer is C.

Question 1 answer is A Global trade allows wealthy countries to use .pdf

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Precipitation hardening, or age hardening, provides one of the most widely used mechanisms for the strengthening of metal alloys. The fundamental understanding and basis for this technique was established in early work at the U. S. Bureau of Standards on Duralumin. The importance of theoretical suggestion for the development of new alloys is clear from the historical record. At the end of the 19th century, cast iron was the only important commercial alloy not already known to western technology at the time of the Romans. When age hardening of aluminum was discovered accidentally by Wilm, during the years 1903 -1911, it quickly became an important commercial alloy under the trade name Duralumin. The strength and hardness of some metal alloys may be enhanced by the formation of extremely small uniformly dispersed second-phase particles within the original phase matrix in a process known as precipitation or age hardening. The precipitate particles act as obstacles to dislocation movement and thereby strengthen the heat-treated alloys. Many aluminum based alloys, copper- tin, certain steels, nickel based super-alloys and titanium alloys can be strengthened by age hardening processes. In order for an alloy system to be able to be precipitation-strengthened, there must be a terminal solid solution that has a decreasing solid solubility as the temperature decreases. The Al-Cu (Duralumin is an aluminum alloy of 2XXX group) phase diagram shown in Figure 1 shows this type of decrease along the solvus between the and + regions. Consider a 96wt%Al – 4wt%Cu alloy which is chosen since there is a large degrease in the solid solubility of solid solution in decreasing the temperature from 550°C to 75°C. Solution Precipitation hardening, or age hardening, provides one of the most widely used mechanisms for the strengthening of metal alloys. The fundamental understanding and basis for this technique was established in early work at the U. S. Bureau of Standards on Duralumin. The importance of theoretical suggestion for the development of new alloys is clear from the historical record. At the end of the 19th century, cast iron was the only important commercial alloy not already known to western technology at the time of the Romans. When age hardening of aluminum was discovered accidentally by Wilm, during the years 1903 -1911, it quickly became an important commercial alloy under the trade name Duralumin. The strength and hardness of some metal alloys may be enhanced by the formation of extremely small uniformly dispersed second-phase particles within the original phase matrix in a process known as precipitation or age hardening. The precipitate particles act as obstacles to dislocation movement and thereby strengthen the heat-treated alloys. Many aluminum based alloys, copper- tin, certain steels, nickel based super-alloys and titanium alloys can be strengthened by age hardening processes. In order for an alloy system to be able to be precipitation-strengthene.

Precipitation hardening, or age hardening, provides one of the most .pdf

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Protists are beneficial to the ecosystem in generating oxygen as they are photosynthetic & single celled in aquatic system result in higher dissolved oxygen finally other trophic levels are in equilibrium Some protists are natural scavangers & profoundly promote \"decomposition of natural organic matter & dead bodies\" therefore nitrogen balance in the ecosystem Some protist are harmful to the ecosystem and they cause disease to various animal and humans result in ecological imbalance in trophic levels of food web Economic importance of protists: Bio-medical importance of protists: Ecological importance of protists: Ques-4: Green algae: These are the algae, which were originated predominantly from the endosymbiotic cyanobacteria, and these algae possess both plastids and chloroplasts for photosynthesis to perform their nutrients synthesis. Volvox is multicellular green algae and possesses cell differentiation as an evolutionary reproductive adaptation. The cells are associated with flagella outside for motility and vacuoles & chloroplasts inside. They mediate cell –to- cell communication through sensory perception obtained from “eyespot” finally neighboring cells join through “cytoplasmic strands” that promote cell –to-cell communication. Recent research has illustrated that cellular adhesion molecules (Algal-CAMs) are found in multicellular Volvox. However, the bulk transport has not observed in the Volvox, a primitive feature still yet to develop in them. Brown algae are the algae, which possess secondary chloroplasts as their functional units to perform their photosynthesis and these algae are commonly originated from the endosymbiotic red algae. Examples are diatoms. Phylogenetic relationship with land plants: Red algae do not possess property of photosynthesis & as they do not contain green pigments Ques-3: Outer coverings: Sphingolipids Pellicle Phospholipids & glycolipids Solution Protists are beneficial to the ecosystem in generating oxygen as they are photosynthetic & single celled in aquatic system result in higher dissolved oxygen finally other trophic levels are in equilibrium Some protists are natural scavangers & profoundly promote \"decomposition of natural organic matter & dead bodies\" therefore nitrogen balance in the ecosystem Some protist are harmful to the ecosystem and they cause disease to various animal and humans result in ecological imbalance in trophic levels of food web Economic importance of protists: Bio-medical importance of protists: Ecological importance of protists: Ques-4: Green algae: These are the algae, which were originated predominantly from the endosymbiotic cyanobacteria, and these algae possess both plastids and chloroplasts for photosynthesis to perform their nutrients synthesis. Volvox is multicellular green algae and possesses cell differentiation as an evolutionary reproductive adaptation. The cells are associated with flagella outside for motility and vacuoles & chloroplasts inside. They mediate.

Protists are beneficial to the ecosystem in generating oxygen as the.pdf

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Now a normal O2 molecule has 12 electrons in the valence shell so, O2+ has 11 electrons in its valence shell. The first two go in the s2s and the next two go in the s*2s. Now two go in the s2p. Now thats 6 of the electrons. 4 electrons go in the p2p level and the remaining one goes in the p*2p level. Now count that up and the bond order is 1/2 the number of bonding electrons minus antibonding electrons: Order = 1/2 (8 - 3) = 5/2 Solution Now a normal O2 molecule has 12 electrons in the valence shell so, O2+ has 11 electrons in its valence shell. The first two go in the s2s and the next two go in the s*2s. Now two go in the s2p. Now thats 6 of the electrons. 4 electrons go in the p2p level and the remaining one goes in the p*2p level. Now count that up and the bond order is 1/2 the number of bonding electrons minus antibonding electrons: Order = 1/2 (8 - 3) = 5/2.

Now a normal O2 molecule has 12 electrons in the valence shell so, O.pdf

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P1.javaimport java.util.Scanner;keyboard inputting package pub.pdf

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//operating system linux,ubuntu,Mac #include #include #include #include // for Time using namespace std; struct Node { int number; struct Node* nextNode; }; //append Method Implementation void append(struct Node** head, int num) { if (*head == NULL) { // checking if first node is NULL or not (*head) = (struct Node*)malloc(sizeof(struct Node)); (*head)->number = num; (*head)->nextNode = NULL; } else { struct Node* temp = *head; while (temp->nextNode != NULL) { temp = temp->nextNode; } struct Node* newNode = (struct Node*)malloc(sizeof(struct Node)); //dynamically allocation of memory newNode->number = num; newNode->nextNode = NULL; temp->nextNode = newNode; } } /* Getting Node with value from List*/ Node* search(struct Node** head, int value) { clock_t begin = clock(); std::cout<number == value) { clock_t end = clock(); std::cout<nextNode; } return NULL; } /*Display List Method Implementation*/ void printList(struct Node* head) { while (head != NULL) { std::cout << head->number << \" \"; head = head->nextNode; } std::cout << endl; } int main(int argc, char *argv[]) { struct Node* head = NULL,*temp; srand(time(0)); if ( argc != 3 ) // argc should be 2 for correct execution // We print error message std::cout<<\"Please Enter 3 argument\"< Solution //operating system linux,ubuntu,Mac #include #include #include #include // for Time using namespace std; struct Node { int number; struct Node* nextNode; }; //append Method Implementation void append(struct Node** head, int num) { if (*head == NULL) { // checking if first node is NULL or not (*head) = (struct Node*)malloc(sizeof(struct Node)); (*head)->number = num; (*head)->nextNode = NULL; } else { struct Node* temp = *head; while (temp->nextNode != NULL) { temp = temp->nextNode; } struct Node* newNode = (struct Node*)malloc(sizeof(struct Node)); //dynamically allocation of memory newNode->number = num; newNode->nextNode = NULL; temp->nextNode = newNode; } } /* Getting Node with value from List*/ Node* search(struct Node** head, int value) { clock_t begin = clock(); std::cout<number == value) { clock_t end = clock(); std::cout<nextNode; } return NULL; } /*Display List Method Implementation*/ void printList(struct Node* head) { while (head != NULL) { std::cout << head->number << \" \"; head = head->nextNode; } std::cout << endl; } int main(int argc, char *argv[]) { struct Node* head = NULL,*temp; srand(time(0)); if ( argc != 3 ) // argc should be 2 for correct execution // We print error message std::cout<<\"Please Enter 3 argument\"<.

operating system linux,ubuntu,Mac#include iostream #include .pdf

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Network Protocol: 1) A Protocol is used for the communication between two or more machines. 2) We can even define the network protocol as, it is for the communication between two network devices. 3) Network protocols include some mechanisms such as to identify the devices and make connections with each other. 4) After the connection is obtained, data is packaged into messages are sent and received. 5) Some protocols even support message acknowledgement and data compression is used for high-performance. What is a network service: 1) A network service is an application. 2) This runs at application layer and above. 3) This is used in computer networking. 4) Network service provides data Storage, manipulation ,presentation , communication 5) This is implemented using client-server architecture or peer-to-peer architecture. Service Interface: 1) This interface is used to implement a Service-Oriented Architecture(SOA). 2) This is implemented to achieve interoperability among the applications among various technologies. Solution Network Protocol: 1) A Protocol is used for the communication between two or more machines. 2) We can even define the network protocol as, it is for the communication between two network devices. 3) Network protocols include some mechanisms such as to identify the devices and make connections with each other. 4) After the connection is obtained, data is packaged into messages are sent and received. 5) Some protocols even support message acknowledgement and data compression is used for high-performance. What is a network service: 1) A network service is an application. 2) This runs at application layer and above. 3) This is used in computer networking. 4) Network service provides data Storage, manipulation ,presentation , communication 5) This is implemented using client-server architecture or peer-to-peer architecture. Service Interface: 1) This interface is used to implement a Service-Oriented Architecture(SOA). 2) This is implemented to achieve interoperability among the applications among various technologies..

Network Protocol1) A Protocol is used for the communication betwe.pdf

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Lipids are madeup with an alcahol and fatty acids that are linked by ester linkages (Some times Ether link). If the alcahol part is Glycerol, then it is called Glycero Phospho Lipids. Among the Glycero Phospho Lipids, Phospho tidyl ethanol amine is called cephalin CH2---coo--(CH2)n---CH3 | CH----Coo--(CH2)n----CH3 | CH2---o-PoH-o-o-CH2--CH2---NH2 Phosphoidyl Serne is another exampe for Gyceophospho lipids. In which it contain serine at ethanol amine place (Underlined portion of above structure) NH2-----CHCH2OH--COOH Solution Lipids are madeup with an alcahol and fatty acids that are linked by ester linkages (Some times Ether link). If the alcahol part is Glycerol, then it is called Glycero Phospho Lipids. Among the Glycero Phospho Lipids, Phospho tidyl ethanol amine is called cephalin CH2---coo--(CH2)n---CH3 | CH----Coo--(CH2)n----CH3 | CH2---o-PoH-o-o-CH2--CH2---NH2 Phosphoidyl Serne is another exampe for Gyceophospho lipids. In which it contain serine at ethanol amine place (Underlined portion of above structure) NH2-----CHCH2OH--COOH.

Lipids are madeup with an alcahol and fatty acids that are linked by.pdf

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Introns are nucleotide sequences in gene which are removed during RN.pdf

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import java.util.Random; //ASSIGNMENT #2: MATRIX ARITHMETIC //Class Matrix. File: Matrix.java public class Matrix { public static final int MAX = 20; //Attributes private int size; private int[][] table; /** * Default constructor * Initializes the table with size MAX */ public Matrix() { this.table = new int[MAX][MAX]; } /** * Parameterized constructor * Initializes the table with size s */ public Matrix(int s) { this.size = s; this.table = new int[this.size][this.size]; } /** * Returns the size of the matrix * @return */ public int getSize() { return this.size; } /** * The the number at position (r, c) * @param r * @param c * @return */ public int getElement(int r, int c) { return this.table[r][c]; } /** * Set a number at position (r, c) * @param r * @param c * @param value */ public void setElement(int r, int c, int value) { this.table[r][c] = value; } /** * Initializes the matrix with random numbers * @param low * @param up */ public void init(int low, int up) { Random rand = new Random(); for (int r = 0 ; r < this.size ; r++) { for (int c = 0 ; c < this.size ; c++) { setElement(r, c, rand.nextInt(up) + low); } } } /** * Prints the matrix */ public void print() { for (int r = 0 ; r < this.size ; r++) { for (int c = 0 ; c < this.size ; c++) { System.out.printf(\" %2s\", getElement(r, c)); } System.out.printf(\"\ \"); } } /** * Adds this matrix with Matrix a * @param a * @return */ public Matrix add(Matrix a) { Matrix result = new Matrix(this.size); for (int r = 0 ; r < this.size ; r++) { for (int c = 0 ; c < this.size ; c++) { int value = this.getElement(r, c) + a.getElement(r, c); result.setElement(r, c, value); } } return result; } /** * Subtracts this matrix with Matrix a * @param a * @return */ public Matrix subtract(Matrix a) { Matrix result = new Matrix(this.size); for (int r = 0 ; r < this.size ; r++) { for (int c = 0 ; c < this.size ; c++) { int value = this.getElement(r, c) - a.getElement(r, c); result.setElement(r, c, value); } } return result; } /** * Multiplies this matrix with Matrix a * @param a * @return */ public Matrix multiply(Matrix a) { Matrix result = new Matrix(this.size); for (int r = 0 ; r < this.size ; r++) { for (int c = 0 ; c < this.size ; c++) { int value = 0; for (int k = 0 ; k < this.size ; k++) { value += this.getElement(r, k) * a.getElement(k, c); } result.setElement(r, c, value); } } return result; } /** * Multiplies this matrix with a constant whatConst * @param a * @return */ public Matrix multiplyConst(int whatConst) { Matrix result = new Matrix(this.size); for (int r = 0 ; r < this.size ; r++) { for (int c = 0 ; c < this.size ; c++) { result.setElement(r, c, getElement(r, c) * whatConst); } } return result; } /** * Transposes this matrix * @param a * @return */ public Matrix transpose() { Matrix result = new Matrix(this.size); for (int r = 0 ; r < this.size ; r++) { for (int c = 0 ; c < this.size ; c++) { result.setElement(r, c, this.getElement(c, r)); } } return result; } /** * * @param a * @return */ p.

import java.util.Random;ASSIGNMENT #2 MATRIX ARITHMETIC Cla.pdf

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HELIX3 is a live CD-based digital forensic suite generated to be used in occurrence response. It comes with a lot of open source digital forensics tools including hex editors, data carving and password cracking tools. This tool can collect data from physical memory, network connections, user accounts, executing processes and services, scheduled jobs, Windows Fegistry, chat logs, screen captures, SAM files, applications, drivers, environment variables and Internet history. Advantages: Disadvantages: Solution HELIX3 is a live CD-based digital forensic suite generated to be used in occurrence response. It comes with a lot of open source digital forensics tools including hex editors, data carving and password cracking tools. This tool can collect data from physical memory, network connections, user accounts, executing processes and services, scheduled jobs, Windows Fegistry, chat logs, screen captures, SAM files, applications, drivers, environment variables and Internet history. Advantages: Disadvantages:.

HELIX3 is a live CD-based digital forensic suite generated to be use.pdf

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IT risk management is that the application of risk management ways to info technology so as to manage IT risk, i.e.: The business risk related to the employment, ownership, operation, involvement, influence ANd adoption of IT at intervals an enterprise or organization IT risk management may be thought-about a part of a wider enterprise risk management system. The institution, maintenance ANd continuous update of an ISMS offer a robust indication that an Different methodologies are projected to manage IT risks, every of them divided in processes and steps. Solution IT risk management is that the application of risk management ways to info technology so as to manage IT risk, i.e.: The business risk related to the employment, ownership, operation, involvement, influence ANd adoption of IT at intervals an enterprise or organization IT risk management may be thought-about a part of a wider enterprise risk management system. The institution, maintenance ANd continuous update of an ISMS offer a robust indication that an Different methodologies are projected to manage IT risks, every of them divided in processes and steps..

IT risk management is that the application of risk management ways t.pdf

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First, let\'s find out what we need. The question asks for the empirical formula of some compound. To do this, we are going to need to find the amount of moles of each substance. To find the moles of Sb, all we will need to do is divide the mass of Sb by the molar mass of Sb. To find the moles of S will be a bit more complicated. Since we know that Sulfur is left over after the reaction, we know that sulfur is in excess (meaning Sb is the only thing that will limit our reaction). Assuming this is an ideal gas, we can take the original amount of moles of Sulfur and subtract it from the moles we get later (after it is transferred to the other container). This will give us the amount of moles of Sulfur that reacted with the moles of Sb. So this is what we need: moles Sb = ? moles S (reacted with Sb) = ? Calculate moles Sb. moles Sb = (mass Sb) / (molar mass Sb) = (0.350g) / (121.76g/mol) = 2.87 x 10^-3 moles Calculate moles S. For this, we are going to have to use the PV = nRT equation two times. One for the S that is placed with the Sb, and one for the S that is placed in the container after the reaction. Moles of S BEFORE reaction: PV = nRT (10.72 atm)(0.100L) = n(0.082)(373.15 K) n = (10.72 atm)(0.100L) / (0.082)(373.15 K) = 0.035 moles S Moles of S AFTER reaction: PV = nRT (14.8 atm)(0.050L) = n(0.082)(323.15 K) n = (14.8 atm)(0.050L) / (0.082)(323.15 K) = 0.028 moles S Calculate moles of S that reacted with the Sb: Since the moles of S went down after the reaction, we know that the moles of S lost must have reacted with the Sb. So, we can simply find the change in moles of S to find the amount of S that reacted with the Sb. Moles of S before reaction - Moles of S after reaction = Moles of S reacted 0.035 moles S - 0.028 moles S = 7.07 x 10^-3 moles S Now that we have our moles of each element that reacted, we can find the empirical formula. To do this, we need to divide through by the element with the least amount of moles. So, in this case, we must divide through by the moles of Sb. (2.87 x 10^-3 moles Sb) / (2.87 x 10^-3 moles Sb) = 1 (7.07 x 10^-3 moles S) / (2.87 x 10^-3 moles Sb) = 2.46 = 2.5 We need subscripts that are integers in empirical formulas. To make 2.5 an integer, we can simply multiply it by 2 to make it 5. When we do this, we also need to multiply 1 by 2. 2.5 x 2 = 5 1 x 2 = 2 These numbers represent the subscripts for each of our elements in our compound. So the empirical formula would be: Sb2S5 Solution First, let\'s find out what we need. The question asks for the empirical formula of some compound. To do this, we are going to need to find the amount of moles of each substance. To find the moles of Sb, all we will need to do is divide the mass of Sb by the molar mass of Sb. To find the moles of S will be a bit more complicated. Since we know that Sulfur is left over after the reaction, we know that sulfur is in excess (meaning Sb is the only thing that will limit our reaction). Assuming this is an ideal gas, we can take.

First, lets find out what we need.The question asks for the empi.pdf

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due to repulsions between lone pair and bond pair of electrons on Te.pdf

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