The answer is CH2Since moles = massmolar massMoles of C H = .pdf

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The answer is: CH2 Since moles = mass/molar mass Moles of C : H = 85.5/12 : 14.5/1 = 7.125 : 14.5 = 1 : 2 Thus the empircal formula is CH2 Solution The answer is: CH2 Since moles = mass/molar mass Moles of C : H = 85.5/12 : 14.5/1 = 7.125 : 14.5 = 1 : 2 Thus the empircal formula is CH2.

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Mg(0H)2 as there is ionic between Mg & OH and co.pdf

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It depends on how many unique protons are adjacent to the proton(s) that you are looking at. First, if you have an isolated proton that is not adjacent to any other protons, it will be a singlet. A good example is a tert-butyl group. The three CH3 appear equivalent and none of them has directly adjacent protons. The signal appears as a huge singlet that integrates to nine protons. Doublets arise when there is a single proton next to the proton you are observing. My favorite example of this is 2-methyl cyclohexanone. If you were to look at the CH3 group, you would see a doublet because there is one proton on the neighboring carbon. Now for some of the more complicated stuff: Suppose you have an ethyl ether group in your molecule. If you look at the signal for the CH3 at the end of the chain, you have a signal which would be split by the two protons on the next carbon over. The rule is, you get n + 1 peaks, where n is the number of protons that are causing the splitting. Since in the case of Et there are two equivalent protons on the CH2, you end up with 2 + 1 = 3 peaks, a triplet. The peaks should appear in a 1:2:1 height ratio. Now suppose we look at the CH2 protons, which have a different chemical shift. In this case, the protons are adjacent to three equivalent protons (on the methyl group). As a result you get 3 + 1 = 4 peaks, a quartet. In this case, the peaks will show up in a 1:3:3:1 height ratio. Solution It depends on how many unique protons are adjacent to the proton(s) that you are looking at. First, if you have an isolated proton that is not adjacent to any other protons, it will be a singlet. A good example is a tert-butyl group. The three CH3 appear equivalent and none of them has directly adjacent protons. The signal appears as a huge singlet that integrates to nine protons. Doublets arise when there is a single proton next to the proton you are observing. My favorite example of this is 2-methyl cyclohexanone. If you were to look at the CH3 group, you would see a doublet because there is one proton on the neighboring carbon. Now for some of the more complicated stuff: Suppose you have an ethyl ether group in your molecule. If you look at the signal for the CH3 at the end of the chain, you have a signal which would be split by the two protons on the next carbon over. The rule is, you get n + 1 peaks, where n is the number of protons that are causing the splitting. Since in the case of Et there are two equivalent protons on the CH2, you end up with 2 + 1 = 3 peaks, a triplet. The peaks should appear in a 1:2:1 height ratio. Now suppose we look at the CH2 protons, which have a different chemical shift. In this case, the protons are adjacent to three equivalent protons (on the methyl group). As a result you get 3 + 1 = 4 peaks, a quartet. In this case, the peaks will show up in a 1:3:3:1 height ratio..

It depends on how many unique protons are adjacen.pdf

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x = 1087 n = 1430 p =x/n = 0.76 standard error, SE = sqrt(p(1-p)/n) SE = sqrt(0.76(1-0.76) / 1430) = 0.0113 a. alpha,a = 1-0.90 = 0.10 Za/2 = Z0.05= 1.645 confidence interval: p +/- [Za/2 * SE] 0.76 +/- 0.0186 ( 0.7416 , 0.7787 ) b. alpha,a = 1-0.95 = 0.05 Za/2 = Z0.025= 1.96 confidence interval: p +/- [Za/2 * SE] 0.76 +/- 0.0221 ( 0.7380 , 0.7823 ) c. as the confidence level increases, the confidence interval get wider Solution x = 1087 n = 1430 p =x/n = 0.76 standard error, SE = sqrt(p(1-p)/n) SE = sqrt(0.76(1-0.76) / 1430) = 0.0113 a. alpha,a = 1-0.90 = 0.10 Za/2 = Z0.05= 1.645 confidence interval: p +/- [Za/2 * SE] 0.76 +/- 0.0186 ( 0.7416 , 0.7787 ) b. alpha,a = 1-0.95 = 0.05 Za/2 = Z0.025= 1.96 confidence interval: p +/- [Za/2 * SE] 0.76 +/- 0.0221 ( 0.7380 , 0.7823 ) c. as the confidence level increases, the confidence interval get wider.

x = 1087n = 1430p =xn = 0.76standard error, SE = sqrt(p(1.pdf

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When two genes are located on the same chromosome they are called as linked. That means they are inherited at the same time together thereby violating independent law of assortment. They can be separated during meiosis by recombination process. Solution When two genes are located on the same chromosome they are called as linked. That means they are inherited at the same time together thereby violating independent law of assortment. They can be separated during meiosis by recombination process..

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We have layered model used in network design since these layers group functions according to the task that needs to be performed. Every function in this model is targeted to help a specific layer perform its job. 5 layers in network model : 1) Application layer: It governs how to application work with each other.It provides applications the ability to access the services of the other layers and defines the protocols that applications use to exchange data. Addressing : NA Application layer Protocols: HTTP, FTP, SMTP, Telnet 2) Transport layer: It establishes the connection between applications running on different hosts.It keeps track of the processes running in the applications above it by assigning port numbers to them and uses the Network layer to access the TCP/IP network. Addressing : Port address Transport layer Protocols: TCP (Connection-oriented) , UDP (Connection-less) 3) Network layer: It governs the transmission of packets across an internet by sending them through several routers along the route. It uses IP addresses to identify the packet’s source and destination. Addressing : IP Address Network layer Protocols : IP, ARP, ICMP, IGMP 4) Data link layer: It governs transmission of frames across a single network.These frames encapsulate the packets and use MAC addresses to identify the source and destination. Addressing : MAC (Physical) address Data link layer Protocols : WLAN protocols, Token rings 5) Physical layer: It governs transmission between adjacent devices connected by a transmission medium. Addressing : NA Solution We have layered model used in network design since these layers group functions according to the task that needs to be performed. Every function in this model is targeted to help a specific layer perform its job. 5 layers in network model : 1) Application layer: It governs how to application work with each other.It provides applications the ability to access the services of the other layers and defines the protocols that applications use to exchange data. Addressing : NA Application layer Protocols: HTTP, FTP, SMTP, Telnet 2) Transport layer: It establishes the connection between applications running on different hosts.It keeps track of the processes running in the applications above it by assigning port numbers to them and uses the Network layer to access the TCP/IP network. Addressing : Port address Transport layer Protocols: TCP (Connection-oriented) , UDP (Connection-less) 3) Network layer: It governs the transmission of packets across an internet by sending them through several routers along the route. It uses IP addresses to identify the packet’s source and destination. Addressing : IP Address Network layer Protocols : IP, ARP, ICMP, IGMP 4) Data link layer: It governs transmission of frames across a single network.These frames encapsulate the packets and use MAC addresses to identify the source and destination. Addressing : MAC (Physical) address Data link layer Protocols : WLAN protocols, Token rings 5) .

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The half-reactions are: Anode: Fe => [Fe]2+ + 2 e- Fe(s) releases electrons and is oxidized to Fe2+(aq) at the anode. Cathode: [Ag]+ + e- => Ag Ag+(aq) accepts electrons and is reduced to Ag(s) at the cathode Solution The half-reactions are: Anode: Fe => [Fe]2+ + 2 e- Fe(s) releases electrons and is oxidized to Fe2+(aq) at the anode. Cathode: [Ag]+ + e- => Ag Ag+(aq) accepts electrons and is reduced to Ag(s) at the cathode.

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Since HCL is a strong acid we can assume that it will fulyy ionise HCl -----> H+ + Cl- 0.2 0 0 0 0.2 0.2 H2S ------- > H+ + HS- Initial 0.10 0.2 0 Final 0.10-x 0.2+x x I am taking pKa = 6.9 => Ka = 1.259 *10-7 = x*(0.2+x)/(0.10-x) => x2 + 0.2x - 1.259*10-8 = 0 => x = 6.295* 10-8 => Concentration of HS- = [HS-] = 6 * 10-8 M Solution Since HCL is a strong acid we can assume that it will fulyy ionise HCl -----> H+ + Cl- 0.2 0 0 0 0.2 0.2 H2S ------- > H+ + HS- Initial 0.10 0.2 0 Final 0.10-x 0.2+x x I am taking pKa = 6.9 => Ka = 1.259 *10-7 = x*(0.2+x)/(0.10-x) => x2 + 0.2x - 1.259*10-8 = 0 => x = 6.295* 10-8 => Concentration of HS- = [HS-] = 6 * 10-8 M.

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d) Peptide Note that forming a peptide bond results in elimination of a molecule of water. since isoleucine, leucine and valine, these are hydrophobic and tend to orient towards the interior of the folded protein molecule. Hence Peptide bonds are the ideal to form..... Hope u understand.... :) Solution d) Peptide Note that forming a peptide bond results in elimination of a molecule of water. since isoleucine, leucine and valine, these are hydrophobic and tend to orient towards the interior of the folded protein molecule. Hence Peptide bonds are the ideal to form..... Hope u understand.... :).

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Introduction •Super Computer developed by IBM Research •Named for IMB’s founder: Thomas J. Watson •Initially created for Jeopardy! Game Show Dr. David Ferrucci leads the Watson project Development •Search engines deliver thousands of results that match keywords •University’s have worked on a consistent question answering software for years •Programmed by 25 IBM scientists Not connected to the internet Jeopardy! Challenge •Set out to answer complex Jeopardy! Questions •Language is hard for computers because of “intended meaning” •During trials, it won 70% of practice games •February 2011: First computer to compete against humans in Jeopardy! •Defeated shows greatest two champions Ken Jennings and Brad Rutter Purpose of Watson •Idea formed during IBM top executives brainstorming publicity stunts •DeepBlue– chess supercomputer defeated Garry Kasparov •IBM’s previous most advanced machine was slow and inaccurate •Overall goal: “to create a new generation of technology that can find answers from data more effectively than current search engines” Technology •Question-answering Technology •Deep understanding of natural language. –Process and answer complex questions that have puns, irony, and/or riddles •Computer running Software called Deep QA •Runs on cluster of Power 750 computers –ten racks holding 90 servers, for a total of 2880 processor cores running DeepQA software and storage –Holds approximately one million books worth of information Managerial Purposes •Financial Assistant: Working with Citi Bank –Help analyze customer needs –Process financial, economic, product, and client data –Help financial professionals make better decision •Could IBM Watson rival complex derivatives on the trade floor? Other Possibilities? •Travel •Retail •Healthcare Classroom Advantages •Provides Services that revolve around the new, digital world •Gives immediate answers instead of search results •Healthcare uses –Diagnosis –Information Warehouse •Efficiency and Organization Disadvantages •Cannot read PET and CT scans to identify tumors •Questions asked must be in text •Less human effort –Too reliant upon technology –Less personal interaction between doctors and patients •Not cognitive –Only manipulates symbols •Limits understanding and reasoning behind decisions Competition •Thus far there are no other computers that are near the performance level of Watson •Microsoft and GE announced plans to create something similar to use in the healthcare industry –Aim to use analytics, high performance software technologies to deliver patient outcomes and clinical applications Applications to SIT •New technology for a business’ Decision Making Processes –Organizational structure may shift based on allocation of decision making •Potentially eliminates the needs/advantages of Virtual Teams (especially in healthcare) •Changes Knowledge Management Processes Solution Introduction •Super Computer developed by IBM Research •Named for IMB’s founder: Thomas J. Watson •Initially creat.

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includes cilia, mucous membranes and dendritic cells-Innate immunity immunobiological response bought by production of antibodies-Humaral immunity immunological response that kills infected host cells-cellular immunity use(s) BCRs to recognize etitope. First step in clonal selection-immature B cells Phagocytes that engulf anything foreign. Eventually display etitope to helper T cells using MHC I or II-dendritic cells lymphocytes that activate B cells and CTLs-Th cells produce and secrete antibodies-plasma cells Kill(s) infected host cells-cytotoxic T cells Solution includes cilia, mucous membranes and dendritic cells-Innate immunity immunobiological response bought by production of antibodies-Humaral immunity immunological response that kills infected host cells-cellular immunity use(s) BCRs to recognize etitope. First step in clonal selection-immature B cells Phagocytes that engulf anything foreign. Eventually display etitope to helper T cells using MHC I or II-dendritic cells lymphocytes that activate B cells and CTLs-Th cells produce and secrete antibodies-plasma cells Kill(s) infected host cells-cytotoxic T cells.

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In internetworking, Multiplexing is a process in which multiple data channels are combined into a single data or physical channel at the source. Multiplexing can be implemented at any of the OSI layers. Conversely, demultiplexing is the process of separating multiplexed data channels at the destination. In this way student data & class assignment data can be combined & separated. Types of Multiplexing There are two basic forms of multiplexing used: Time Division Multiplexing Time Division Multiplexing works by the multiplexor collecting and storing the incoming transmissions from all of the slow lines connected to it and allocating a time slice on the fast link to each in turn. The messages are sent down the high speed link one after the other. Each transmission when received can be separated according to the time slice allocated. Theoretically, the available speed of the fast link should at least be equal to the total of all of the slow speeds coming into the multiplexor so that its maximum capacity is not exceeded. Two ways of implementing TDM are: Synchronous TDM Synchronous TDM works by the muliplexor giving exactly the same amount of time to each device connected to it. This time slice is allocated even if a device has nothing to transmit. This is wasteful in that there will be many times when allocated time slots are not being used. Therefore, the use of Synchronous TDM does not guarantee maximum line usage and efficiency. Synchronous TDM is used in T1 and E1 connections. Asynchronous TDM Asynchronous TDM is a more flexible method of TDM. With Asynchronous TDM the length of time allocated is not fixed for each device but time is given to devices that have data to transmit. This version of TDM works by tagging each frame with an identification number to note which device it belongs to. This may require more processing by the multiplexor and take longer, however, the time saved by efficient and effective bandwidth utilization makes it worthwhile. Asynchronous TDM allows more devices than there is physical bandwidth for. This type of TDM is used in Asynchronous Transfer Mode (ATM) networks. Frequency Division Multiplexing Frequency Division Multiplexing (FDM) works by transmitting all of the signals along the same high speed link simultaneously with each signal set at a different frequency. For FDM to work properly frequency overlap must be avoided. Therefore, the link must have sufficient bandwidth to be able to carry the wide range of frequencies required. The demultiplexor at the receiving end works by dividing the signals by tuning into the appropriate frequency. FDM operates in a similar way to radio broadcasting where a number of different stations will broadcast simultaneously but on different frequencies. Listeners can then \"tune\" their radio so that it captures the frequency or station they want. FDM gives a total bandwidth greater than the combined bandwidth of the signals to be transmitted. In order to prevent signal overlap t.

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