Molarity = moles volume in L Molarity = 0.0342.pdf

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Molarity = moles / volume in L Molarity = 0.0342/8.39 Molarity = 0.004076 M solution Now use the Ka equation to calculate the [H+] of the solution: Ka for CH3COOH = 1.8*10^-5 Ka = [H+] / [CH3COOH] 1.8*10^-5 = [H+]²/ 0.004076 [H+]² = (1.8910^-5) * ( 0.004076 [H+]² = 7.34*10^-8 [H+] = 2.71*10^-4 pH = -log [H+] pH = -log 2.71*10^-4 pH = 3.57 Solution Molarity = moles / volume in L Molarity = 0.0342/8.39 Molarity = 0.004076 M solution Now use the Ka equation to calculate the [H+] of the solution: Ka for CH3COOH = 1.8*10^-5 Ka = [H+] / [CH3COOH] 1.8*10^-5 = [H+]²/ 0.004076 [H+]² = (1.8910^-5) * ( 0.004076 [H+]² = 7.34*10^-8 [H+] = 2.71*10^-4 pH = -log [H+] pH = -log 2.71*10^-4 pH = 3.57.

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Rio de Janeiro is a city located on Brazil\'s south-east coast. It is one of Brazil\'s largest settlements with a population of approximately 11.7 million people. The population of Rio de Janeiro has grown for a number of reasons. Natural Increase is one reason for its growth (this is when the birth rate is higher than the death rate). The population has also grown as the result of urbanisation. The has been caused by rural to urban migration. Millions of people have migrated from Brazil\'s rural areas to Rio de Janeiro. 65% of urban growth is a result of migration. This is caused by a variety of push and pull factors. The rapid growth of Rio de Janeiro\'s population has led to a severe shortage of housing. Millions of people have been forced to construct their own homes from scrap materials such as wood, corrugated iron and metals. These areas of temporary accommodation are known as favelas in Brazil. The conditions associated with favelas are very poor. Often families have to share one tap, there is no sewerage provision, disease is common and many people are unemployed. Favelas are located on the edge of most major Brazilian cities. They are located here for a number of reasons. Firstly, this is the only available land to build on within the city limits. Secondly, industry is located on the edge of the cities. Many people need jobs therefore they locate close to factories. Some of these settlements may be 40 or 50 km from the city centre (on the edge of the city), along main roads and up very steep hillsides. Causes of urban growth; Although the process of urbanisation happens in both MEDCs(more economically developed countries)and LEDCs,(less economically developed countries) the fastest-growing cities in the world are in LEDCs. The reasons for the growth of urban areas include: There are many problems associated with the rapid growth. These include unplanned housing (squatter settlements/shanty towns), dealing with urban waste, pollution and stress on the infrastructure and the city\'s services. Shanty towns/sqatter settlements; The fact that cities in LEDCs are growing rapidly means that conditions can be poor. There are often great inequalities within LEDC urban areas and they are even more pronounced in LEDCs. Some of the worst conditions are found in the shanty towns on the edge of the city, near the CBD(convention on biological diversity) or along main transport routes. They tend to be unplanned and are often illegal. Houses are self-built using basic materials and shanty towns have few services. Shanty town residents face many problems on a daily basis. Problems in shanty towns/squatter settlements this is how the squatter settlements effects the urban geography of brazil Solution Rio de Janeiro is a city located on Brazil\'s south-east coast. It is one of Brazil\'s largest settlements with a population of approximately 11.7 million people. The population of Rio de Janeiro has grown for a number of reasons. Natural Increase is one .

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Na2SeO3 ---> 2 Na+ + SeO32- 0.08 0.16 0.08 SeO32- + H2O <---> HSeO3- + OH- 0.08 0 0 -x x x 0.08-x x x Kb2 = Kw/Ka2 = 1E-14/(4.8E-9) Kb2 = 0.21E-5 0.21E-5 = x2/(0.08-x) x = 4.1E-4 HSeO3- + H2O <---> H2SeO3 + OH- 4.1E-4 0 0 -y y y 4.1E-4 - y y y Kb1 = Kw/Ka1 Kb1 = 1E-14/(2.4E-3) = 0.42E-11 The second hydrolysis constant is much smaller than the first, so we can the assume that the first step dominates. [H+]*[OH-]=1E-14 [H+]*4.1E-4=1E-14 [H+] = 0.24E-10 pH = 10.6 Ok? Solution Na2SeO3 ---> 2 Na+ + SeO32- 0.08 0.16 0.08 SeO32- + H2O <---> HSeO3- + OH- 0.08 0 0 -x x x 0.08-x x x Kb2 = Kw/Ka2 = 1E-14/(4.8E-9) Kb2 = 0.21E-5 0.21E-5 = x2/(0.08-x) x = 4.1E-4 HSeO3- + H2O <---> H2SeO3 + OH- 4.1E-4 0 0 -y y y 4.1E-4 - y y y Kb1 = Kw/Ka1 Kb1 = 1E-14/(2.4E-3) = 0.42E-11 The second hydrolysis constant is much smaller than the first, so we can the assume that the first step dominates. [H+]*[OH-]=1E-14 [H+]*4.1E-4=1E-14 [H+] = 0.24E-10 pH = 10.6 Ok?.

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Let the normal gene be represented by P, and the two mutations be represented by p# and p*. Given that the plant is p#p*, and that the two mutations seggregate independently. Therefore, the gametes p# and p* will be obtained in equal frequencies (1:1). Upon self-cross, the genotypes obtained will be: p# p* p# p#p# (1) p#p* (1) p* p#p* (1) p*p* (1) In the table given above, it can be seen that p#p* appears twice (1 + 1 = 2), whereas p#p# and p*p* appear only once (1). Since, both the mutations are recessive, three phenotypes would be obtained in the ratio: mutant 1 : normal : mutant 2 :: 1:2:1 Three genotypes will be obtained in the ratio: p#p# : p#p* : p*p* :: 1:2:1 p# p* p# p#p# (1) p#p* (1) p* p#p* (1) p*p* (1) Solution Let the normal gene be represented by P, and the two mutations be represented by p# and p*. Given that the plant is p#p*, and that the two mutations seggregate independently. Therefore, the gametes p# and p* will be obtained in equal frequencies (1:1). Upon self-cross, the genotypes obtained will be: p# p* p# p#p# (1) p#p* (1) p* p#p* (1) p*p* (1) In the table given above, it can be seen that p#p* appears twice (1 + 1 = 2), whereas p#p# and p*p* appear only once (1). Since, both the mutations are recessive, three phenotypes would be obtained in the ratio: mutant 1 : normal : mutant 2 :: 1:2:1 Three genotypes will be obtained in the ratio: p#p# : p#p* : p*p* :: 1:2:1 p# p* p# p#p# (1) p#p* (1) p* p#p* (1) p*p* (1).

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Imagine you are an Information Systems Security Officer for a medium-sized financial services firm that has operations in four (4) states (Virginia, Florida, Arizona, and California). Due to the highly sensitive data created, stored, and transported by your organization, the CIO is concerned with implementing proper security controls for the LAN-to-WAN domain. Specifically, the CIO is concerned with the following areas: Protecting data privacy across the WAN Filtering undesirable network traffic from the Internet Filtering the traffic to the Internet that does not adhere to the organizational acceptable use policy (AUP) for the Web Having a zone that allows access for anonymous users but aggressively controls information exchange with internal resources Having an area designed to trap attackers in order to monitor attacker activities Allowing a means to monitor network traffic in real time as a means to identify and block unusual activity Hiding internal IP addresses Allowing operating system and application patch management The CIO has tasked you with proposing a series of hardware and software controls designed to provide security for the LAN-to-WAN domain. The CIO anticipates receiving both a written report and diagram(s) to support your recommendations. Write a three to five (3-5) page paper in which you: Use MS Visio or an open source equivalent to graphically depict a solution for the provided scenario that will: filter undesirable network traffic from the Internet filter Web traffic to the Internet that does not adhere to the organizational AUP for the Web allow for a zone for anonymous users but aggressively controls information exchange with internal resources allow for an area designed to trap attackers in order to monitor attacker activities offer a means to monitor network traffic in real time as a means to identify and block unusual activity hide internal IP addresses Identify the fundamentals of public key infrastructure (PKI). Describe the manner in which your solution will protect the privacy of data transmitted across the WAN. Protecting data privacy across the WAN Filtering undesirable network traffic from the Internet Filtering the traffic to the Internet that does not adhere to the organizational acceptable use policy (AUP) for the Web Having a zone that allows access for anonymous users but aggressively controls information exchange with internal resources Having an area designed to trap attackers in order to monitor attacker activities Allowing a means to monitor network traffic in real time as a means to identify and block unusual activity Hiding internal IP addresses Allowing operating system and application patch management Solution Imagine you are an Information Systems Security Officer for a medium-sized financial services firm that has operations in four (4) states (Virginia, Florida, Arizona, and California). Due to the highly sensitive data created, stored, and transported by your organization, the CIO is concern.

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Knowing the electronegativities of each atom involved in the bonds will be helpful in doing this problem. The greater the ionic character of a bond, the more ionic it is. When doing a problem like this, you have to keep in mind of the trends for electronegativity (EN): (1) EN increases from left to right in a period. (2) EN increases from bottom to top in a group. Na-F K-O Li-Li S-O K-Br Therefore, looking at the bonds above, you can list the bonds in the following order: Li-Li < S-O < K-Br < K-O < Na-F. Logically think this problem out using the trends first before checking with the electronegativites (change in EN will tell you how polar the bond is; the larger the change in electronegativity, the more polar/ionic the bond). This problem requires that you know the relative locations of the elements on the periodic table. If this helps you, then please rate lifesaver. Thanks. Solution Knowing the electronegativities of each atom involved in the bonds will be helpful in doing this problem. The greater the ionic character of a bond, the more ionic it is. When doing a problem like this, you have to keep in mind of the trends for electronegativity (EN): (1) EN increases from left to right in a period. (2) EN increases from bottom to top in a group. Na-F K-O Li-Li S-O K-Br Therefore, looking at the bonds above, you can list the bonds in the following order: Li-Li < S-O < K-Br < K-O < Na-F. Logically think this problem out using the trends first before checking with the electronegativites (change in EN will tell you how polar the bond is; the larger the change in electronegativity, the more polar/ionic the bond). This problem requires that you know the relative locations of the elements on the periodic table. If this helps you, then please rate lifesaver. Thanks..

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Hi I have added a loop for adding values to list. Highlighted the code changes below. public static ArrayList doArrayListAdd(int numItems) { System.out.print(\"doArrayListAdd: \"); ArrayList list = new ArrayList<>(); long startTime = getTimestamp(); // TODO Write code that adds 0 through (numitems - 1) // to list inside a loop for(int i=0; i Solution Hi I have added a loop for adding values to list. Highlighted the code changes below. public static ArrayList doArrayListAdd(int numItems) { System.out.print(\"doArrayListAdd: \"); ArrayList list = new ArrayList<>(); long startTime = getTimestamp(); // TODO Write code that adds 0 through (numitems - 1) // to list inside a loop for(int i=0; i.

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How a network connection is created ? A network connection is initiated by a client program when it creates a socket for the communication with the server. To create the socket in Java, the client calls the Socket constructor and passes the server address and the the specific server port number to it. At this stage the server must be started on the machine having the specified address and listening for connections on its specific port number. The server uses a specific port dedicated only to listening for connection requests from clients. It can not use this specific port for data communication with the clients because the server must be able to accept the client connection at any instant. So, its specific port is dedicated only to listening for new connection requests. The server side socket associated with specific port is called server socket. When a connection request arrives on this socket from the client side, the client and the server establish a connection. This connection is established as follows: The java.net package in the Java development environment provides the class Socket that implements the client side and the class serverSocket class that implements the server side sockets. The client and the server must agree on a protocol. They must agree on the language of the information transferred back and forth through the socket. There are two communication protocols : The stream communication protocol is known as TCP (transfer control protocol). TCP is a connection-oriented protocol. It works as described in this document. In order to communicate over the TCP protocol, a connection must first be established between two sockets. While one of the sockets listens for a connection request (server), the other asks for a connection (client). Once the two sockets are connected, they can be used to transmit and/or to receive data. When we say \"two sockets are connected\" we mean the fact that the server accepted a connection. As it was explained above the server creates a new local socket for the new connection. The process of the new local socket creation, however, is transparent for the client. The datagram communication protocol, known as UDP (user datagram protocol), is a connectionless protocol. No connection is established before sending the data. The data are sent in a packet called datagram. The datagram is sent like a request for establishing a connection. However, the datagram contains not only the addresses, it contains the user data also. Once it arrives to the destination the user data are read by the remote application and no connection is established. This protocol requires that each time a datagram is sent, the local socket and the remote socket addresses must also be sent in the datagram. These addresses are sent in each datagram. The java.net package in the Java development environment provides the class DatagramSocket for programming datagram communications. UDP is an unreliable protocol. There is no guarantee that the .

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DNA is the callde as molucule which gives genetic information. It gives or stores instruction to give large molucules called proteins. These instructions are there in side of every cells, distributed by 46 long structures called chromosomes. These are made up of thusands of DNA, called genes. Each gene stores the directions for making protein fragments, whole proteins, or multiple specific proteins. DNA is well-suited to perform this biological function because of its molecular structure, and because of the development of a series of high performance enzymes that are fine-tuned to interact with this molecular structure in specific ways. The match between DNA structure and the activities of these enzymes is so effective and well-refined that DNA has become, over evolutionary time, the universal information-storage molecule for all forms of life. Nature has yet to find a better solution than DNA for storing, expressing, and passing along instructions for making proteins. Ribonuclic acid functions as genetic messengers and builders of the cellular world. 3 types of RNA mainly r-RNA (ribosomal RNA) , t-RNA (transfer RNA) and m-RNA (messenger RNA) one of the main function of RNA is protein synthesis Differences between RNA and DNA Genetic code is stored in DNA as linear/ non-overlaping form in the nitrogenous base of adenine, guanine, cytosine and thymineDNARNALong polymer with deoxyriboses and phosphate backboneRNA is a polymer with a ribose and phosphate backboneHaving nitrogenous bases: adenine, guanine, cytosine and thymineHaving nitrogenous bases: adenine, guanine, cytosine, and uracilIt usually accours in side the nucleus and some cell organellesVery little present in side nucleus most of it accours in cytoplasmDauble standard exception of some virusesSingle standard exception of some virusesFunction is to transfer gentitic information to one generation to other generation.Main function is protein synthesisLong life spanshart life span Solution DNA is the callde as molucule which gives genetic information. It gives or stores instruction to give large molucules called proteins. These instructions are there in side of every cells, distributed by 46 long structures called chromosomes. These are made up of thusands of DNA, called genes. Each gene stores the directions for making protein fragments, whole proteins, or multiple specific proteins. DNA is well-suited to perform this biological function because of its molecular structure, and because of the development of a series of high performance enzymes that are fine-tuned to interact with this molecular structure in specific ways. The match between DNA structure and the activities of these enzymes is so effective and well-refined that DNA has become, over evolutionary time, the universal information-storage molecule for all forms of life. Nature has yet to find a better solution than DNA for storing, expressing, and passing along instructions for making proteins. Ribonuclic acid functions as gen.

DNA is the callde as molucule which gives genetic information. It gi.pdf

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false, Because Conservatsim is define as a plitical policy that promote the conservation of traditional ,cultural and social stability. establising institutions, and preferring gradual development to abrupt change. Solution false, Because Conservatsim is define as a plitical policy that promote the conservation of traditional ,cultural and social stability. establising institutions, and preferring gradual development to abrupt change..

false,Because Conservatsim is define as a plitical policy that pro.pdf

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d. Fungiform, foliate, and circumvallate papillae Fungiform papillae are mushroom shaped, present at the dorsal surface of the tongue Foliate papillae are ridges and grooves present at the posterior edge of the tongue circumvallate papillae are only few in number, and are located posterior to the oral part of the tongue. ** Filiform papillae are not associated with taste buds. Solution d. Fungiform, foliate, and circumvallate papillae Fungiform papillae are mushroom shaped, present at the dorsal surface of the tongue Foliate papillae are ridges and grooves present at the posterior edge of the tongue circumvallate papillae are only few in number, and are located posterior to the oral part of the tongue. ** Filiform papillae are not associated with taste buds..

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Carrier sense multiple access with collision detection (CSMA/CD) is a media access method used by twisted-pair Ethernet networks. CSMA/CA is used in wireless networks . Token passing is used in token ring and ARCNET. OFDMA is used in WiMAX. Solution Carrier sense multiple access with collision detection (CSMA/CD) is a media access method used by twisted-pair Ethernet networks. CSMA/CA is used in wireless networks . Token passing is used in token ring and ARCNET. OFDMA is used in WiMAX..

Carrier sense multiple access with collision detection (CSMACD) is .pdf

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c) Why are financial markets essential for a healthy economy and economic growth? Financial markets are a common place where the funds are transferred from people who have surplus funds to those who have shortage of funds. It leads to exchange of funds between the borrowers and the lenders. There are two forms of exchange that can take place in an economy. They are direct exchange and indirect exchange. Direct exchange means that the borrowers and lenders meet each other and exchange their securities like stock, bonds etc. Direct exchange takes place through financial markets only. For an economy to function properly, the financial markets have to function efficiently. The health of an economy depends on the transfer of funds. Thus, financial markets are important for economic growth and healthy economy. d) What are derivatives? How can derivatives be used to reduce risk? Can derivatives be used to increase risk? Explain. Derivatives are financial instruments that are exchanged between two parties and the instrument is based on an underlying asset. The value of the derivative is derived from the value of the underlying asset. Derivatives can be used to reduce risk as well as to increase risk. Some companies use derivatives to reduce their exposure to risk while some companies use derivatives to speculate. The companies that use derivatives to speculate are actually trying to increase their income based on increase in risk. e) Briefly describe each of the following financial institutions: investment banks, commercial banks, financial services corporations, pension funds, mutual funds, exchange traded funds, hedge funds, and private equity companies. Invetsment banks: These are financial institutions that help businesses to launch their Initial Public Offerings and start trading their shares in the market. It helps the business to raise funds through equity. Commercial banks: These are financial institutions that help the individual borrowers to borrow money from the bank and other individuals to save their money in the banks through savings accounts. They accept deposits and provide loans as well as offer investment products for individual investors. Financial services corporations: These are conglomerates that provide financial services to individual investors. A group of such financial service providers are grouped together as a financial service corporation. For example: Citi group. Pension funds: Pension funds are plans or schemes that provide retirement benefits or retirement income to the employees of a company. These are invested in by the individual investors or by the employers or both. Mutual funds: These are financial instruments that consist of a combination of a variety of investment options. It can be a combination of stocks, bonds, gold etc. It attracts the investment from individual investors as well as companies. The mutual funds are managed by professional managers. The aim here is to diversify the risk over a variety of inve.

c) Why are financial markets essential for a healthy economy and eco.pdf

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