c) For each of the three products that are observed in Spectrum C- the.docx

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c) For each of the three products that are observed in Spectrum C, there is a diagnostic proton that can be used to accurately calculate the product ratios. (Remember that a good diagnostic proton has a \"clean\" signal that has minimal overlap with other signals- see \"Handout 05: Calculating Compound Ratios\" on CourseWeb.) What diagnostic proton can be used to calculate product ratios for each isomer? If the isomer was not formed, write \"N/A.\" (4 points) p-bromoaniline o-bromoaniline o,p- dibromoaniline O,o,p tribromoaniline d) Using the diagnostic protons you listed in part c, calculate the percent of each isomer. Show calculations for each isomer. If an isomer was not formed, write \"0%\" (8 points) %p-bromoaniline % o-bromoaniline % o,o,p- tribromoaniline 960,p. dibromoaniline Solution Solution: The spectrum interpretation are as shown below, Part (C) p - bromoaniline = 7.529 o - bromoaniline = 7.230 o,p - dibromoaniline = 6.651 o,o,p - tribromoaniline = 6.566 Part (D) p - bromoaniline = (2*1):(2*0.83) = 2/(2+1.66)*100% = 54.65 % o - bromoaniline = 0 o,p - dibromoaniline = 1.15 /(5.43+1.15+0.61)*100% = 15.99 = 16% o,o,p - tribromoaniline = 5.95/(0.80+1.35+5.95) *100% = 73.46 = 74% .

c) For each of the three products that are observed in Spectrum C, there is a diagnostic proton
that can be used to accurately calculate the product ratios. (Remember that a good diagnostic
proton has a "clean" signal that has minimal overlap with other signals- see "Handout 05:
Calculating Compound Ratios" on CourseWeb.) What diagnostic proton can be used to calculate
product ratios for each isomer? If the isomer was not formed, write "N/A." (4 points) p-
bromoaniline o-bromoaniline o,p- dibromoaniline O,o,p tribromoaniline d) Using the diagnostic
protons you listed in part c, calculate the percent of each isomer. Show calculations for each
isomer. If an isomer was not formed, write "0%" (8 points) %p-bromoaniline % o-bromoaniline
% o,o,p- tribromoaniline 960,p. dibromoaniline
Solution
Solution:
The spectrum interpretation are as shown below,
Part (C)
p - bromoaniline = 7.529
o - bromoaniline = 7.230
o,p - dibromoaniline = 6.651
o,o,p - tribromoaniline = 6.566
Part (D)
p - bromoaniline = (2*1):(2*0.83) = 2/(2+1.66)*100% = 54.65 %
o - bromoaniline = 0
o,p - dibromoaniline = 1.15 /(5.43+1.15+0.61)*100% = 15.99 = 16%
o,o,p - tribromoaniline = 5.95/(0.80+1.35+5.95) *100% = 73.46 = 74%

c) For each of the three products that are observed in Spectrum C- the.docx

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Briefly explain the purposes of adjustments Briefly explain the purposes of adjustments Solution The accounting process involves the preparation of adjusting entries. Adjusting journal entries are made to update the accounts and bring them to their correct balances. The preparation of adjusting entries is an application of the accrual concept of accounting and the matching principle . The accrual concept states that income is recognized when earned regardless of when collected and expenses are recognized when incurred regardless of when paid. Also, the matching principle aims to align expenses with revenues. Expenses should be recognized in the period when the revenues earned with them are recognized. Why Are Adjusting Entries Necessary The purpose of adjusting entries is to update the accounts to conform with the accrual concept . At the end of the accounting period, some income and expenses may have not been recorded, taken up or updated; hence, there is a need to update the accounts. If adjusting entries are not prepared, some income, expense, asset, and liability accounts may not reflect their true values when reported in the financial statements. For this reason, adjusting entries are necessary. BELOW are the types of journal entires. Generally, there are 4 types of adjusting entries. Adjusting entries are prepared for the following: .

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Budget Solution Answer: Caroline must recognise the following incomes: 1. Long term capital gain of $50,000 as she sold the stock one year later at a price of $300 per share , while she exercised the stoock at a price of $250 per share. [(1000 shares*300)-(1000shares*250)] 2. And Oridinary Income of $100,000 as she exercised the option at the rate of $250 per share while the stok was granted at the rate of $150.[(1000shares*250)-(1000 shares*150)] .

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Calculate fost of goods sold and ending inventory and analyze effect of each method on financial statements. Jefferson company had the following sales and purchases during 2011, its first year of business: January 5 Purchased 40 units at $100 each February 15 Sold 15 units at $150 each April 10 Sold 10 units at $150 each June 30 Purchased 30 units at $105 each August 15 Sold 25 units at $150 each November 28 Purchased 30 untis at $110 each Solution Part A: Record the effect of the following transactions on the accounting equation assuming LIFO perpetual: Transactions in January, 2011: Assets Liabilities Shareholders .

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Calculate fost of goods sold and ending inventory and analyze effect of each method on financial statements. Jefferson company had the following sales and purchases during 2011, its first year of business: January 5 Purchased 40 units at $100 each February 15 Sold 15 units at $150 each April 10 Sold 10 units at $150 each June 30 Purchased 30 units at $105 each August 15 Sold 25 units at $150 each November 28 Purchased 30 untis at $110 each Solution Part A: Record the effect of the following transactions on the accounting equation assuming LIFO perpetual: Transactions in January, 2011: Assets Liabilities Shareholders .

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Ca(OH)2 is insoluble. What is the total concentration of ions if 2M of Ca(OH)2 dissolves in water. Ca(OH)2 is insoluble. What is the total concentration of ions if 2M of Ca(OH)2 dissolves in water. Solution Ca(OH)2 is partially soluble in water and we can find solubility based on solubility product Ca(OH)2 (s) <-----> Ca2+ (aq) + 2OH- (aq) at equilibrium [Ca2+] = S , [OH-] = 2S Ksp = [Ca2+] [OH-]^2 ( where Ksp is solubility product of Ca(OH)2 in water) 5.5 x 10^ -6 = (S) (2S)^2 S = 0.0111 M Thus [Ca2+] = 0.0111 , [OH-] = 2 x 0.0111 = 0.0222 M Total concetrations of ions = [Ca2+] + 2 [OH-] = ( 0.0111) + 2 ( 0.0222) = 0.0555 M .

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c. Since BEEF hooks to Metasploit, is there any safe way to use the web today? Explain. Solution a.) The safe way of using web is by implementing firewalls,secure virtual private networks and antivirus. b.) Implementing the firewalls is an imperative solution for all organizations to protect their sensitive data fro getting lost.Firewall keeps everything secure from the possible attacks.This is done by analysing the data that gets transmitted back and forth in the network. c.) Implementation is possible through the following steps - defining the security needs to finf out the level of security necessary. - researching soutions to match the specified needs - and finally installing,configuring and maintaining the firewalls. d.) Virtual private network as the name itself suggest extends a private network across the public network of internet.I tprovides functionality,network management and security benefits to the user. e.) The antivirus helps in preventing,detecting and removing the malicious software.I tprovides protection from various computer threats phishing attacks,online identity attacks,malicious URL\'s and etc. .

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9. Fill in the following table with names or formulas, as appropriate: Names silver sulfate hydrocyanic acid gold(III) chlorite perbromic acid calcium phosphide mercury(I) nitrate sulfurous acid aluminum chromate tin(IV) carbonate oxalic acid Formulas Fe(HC03)2 HNO2(aq (NH4)3BO3 Li2Cr20, Hl(aq) Solution silver sulfate = Ag2So4 hydrocyanic acid = HCN gold(III) chlorite = Au(Clo2)3 perbromic acid = Hbro4 calcium phosphide = ca3p2 mercury(I)nitrate = hg2(no3)2 sulfurous acid = H2So3 aluminum chromate = Al2(cro4)3 tin(iv)carbonate = sn(co3)2 oxalic acid = (cooh)2 iron(ii)hydrogen carbonate = Fe(hco3)2 aqueous nitrous acid =HNo2(aq) ammonium borate = (NH4)3Bo3 lithium dichromate = Li2Cr2o7 aqueous hydro iodic acid = HI(aq) the above names based on the type of elements or substituents attached to the compounds .

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9- Fill in the following table with names or formulas- as appropriate-.docx

9. A combination of sand, salt, and water is an example of a A) homogeneous mixture B) heterogeneous mixture C) compound D) pure substance E) solid 10. Forthe following reaction Chldh +Oz ? CO2 + 11,0 what is the coefficient of oxygen? a. 2 b. 4 c. 6 d. 8 e. None of the above Solution 9) Salt, sand and water mixture contains two different phases. liquid phase and solid phase. and moreover they dont mix and form separate phases. So, this is example of heterogeneous mixture. (B) 10) Balance the reaction : C 3 H 4 O2 + O 2 -----------> CO 2 + H 2 O balance the reaction for Carbon , C 3 H 4 O2 + O 2 -----------> 3 CO 2 + H 2 O balance the reaction for hydrogen, C 3 H 4 O2 + O 2 -----------> 3 CO 2 + 2 H 2 O now balance the reaction for oxygen, C 3 H 4 O2 + 3 O 2 -----------> 3 CO 2 + 2 H 2 O So, the answer is 3 and option is none of the above. .

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8.6. Which of the highlighted elements in Figure P8.6 has the greatest electronegativity? 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 FIGURE P8.6 Solution Electronegativity is a chemical property of an atom that describes how well the atom withdraws shared pair of electrons towards it. It is measured in terms of Pauling Scale from Â Â 0 to 4. The higher the value, the more the electronegativity of the atom. The highlighted element 17 (Fluorine, F) has the highest electronegativity value is 3.98 . The highlighted element 1 (Lithium, Li) has the highest electronegativity value is 0.98. The highlighted element 2 (Beryllium, Be) has the highest electronegativity value is 1.57. The highlighted element 13 (Boron, B) has the highest electronegativity value is 2.04. The highlighted element 14 (Carbon, C) has the highest electronegativity value is 2.55. The highlighted element 15 (Nitrogen, N) has the highest electronegativity value is 3.04. The highlighted element 16 (Oxygen, O) has the highest electronegativity value is 3.44. The highlighted element 18 (Neon, Ne) has the highest electronegativity value is 0. .

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870 The plates of a capacitor are not quite parallel, the distance between them at the bottom being (D a) and at the top (Dta), where a Solution let length of each plate be L and width be W. then area A=L*W at a distance x from the bottom end of the plate, distance between the plates as a function of x is given as d=D-a+(2*a/L)*x consider a small segment of length dx at a distance of x from the bottom end area of the segment=W*dx distance between the two plates=d=(D-a+(2*a/L)*x) then capacitance=dC=epsilon*area/distance between plates =epsilon*W*dx/(D-a+(2*a/L)*x) let D-a+(2*a/L)*x=y ==>(2*a/L)*dx=dy ==>dx=L*dy/(2*a) substituting, dC=epsilon*W*L*dy/(2*a*y) integrating both sides, C=(epsilon*W*L/(2*a))*ln(y) as y varies from D-a to D+a C=(epsilon*W*L/(2*a))*ln((D+a)/(D-a)) as A=W*L and ln((D+a)/(D-a)) =ln((1+(a/D))/(1-(a/D))) =2*(a/D)*(1+(a^2/(3*D^2))) then C=(epslion*A/(2*a))*(2*a/D)*(1+(a^2/(3*D^2))) =(epsilon*A/D)*(1+(a^2/(3*D^2))) .

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8. Calculate the distance from the donor (D) to the acceptor (A) in the protein represented. Assume that there is a single D-A distance, and that diffusion does not occur during the donor excited-state lifetime. Given: A.The F DA /F D ratio is 0.50 (that is, the ratio between the fluorescence intensity of the donor in the presence of the acceptor (F DA ) and the fluorescence of the respective control molecule which lacks the acceptor (F D )). B.The spectral overlap integralÂ Â ( J (l)) for this donor-acceptor pair is 4.5 x 10 13 M -1 cm -1 nm 4 . C.The refractive index of the solution is 1.4. D.The fluorescence quantum yield of the donor in the absence of acceptor is 0.3. Solution Fret E = 1- FDA/FD E = 1-0.50 = 0.50 Rn = R0 (n(1-E)/E)1/6 Rn is the distance between A and D E = 0.5 R0 = 4.5 X 1013 unit Hence Rn = 4.5 X 1013 X 1.4 (1-0.5/0.5)1/6 =4.5 X 1013 X (1.4)1/6 = 4.5 X 1013 X 8.4 = 38291.4 angstrom N = 1.4 .

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7. You want to create an SSIS package that copies data from a source to a destination. You want to run the package after you create it. Which is the simplest method to accomplish that? Solution SQL Server Integration Services ( SSIS ) is a component of the Microsoft SQL Server database software that can be used to perform a broad range of data migration tasks. SSIS is a platform for data integration and workflow applications. It features a fast and flexible data warehousing tool used for dataextraction, transformation, and loading (ETL). The tool may also be used to automate maintenance of SQL Server databases and updates to multidimensional cube data. First released with Microsoft SQL Server 2005, SSIS replaced Data Transformation Services, which had been a feature of SQL Server since Version 7.0. Unlike DTS, which was included in all versions, SSIS is only available in the \"Standard\", \"Business Intelligence\" and \"Enterprise\" editions. .

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9. List four interface design issues present in the development of most user interfaces? Solution The interface design issues are: There are four interface design issues they are 1) Access and short cut keys 2) Bit maps and Icons 3) UI controls 4) Messages. Access and shortcut keys: Short cut keys doesnot contains characters from the double bite character. Bitmaps and Icons: These functionality dont use in text when working with bit maps and icons consider of following rules. Bit maps which are not international standards avoid using them. And we need to check whether the Access key and short cut key are there are not. Limitation of this functionality is ude function keys and numbers instead of letters in short cut keys. UI controls: There are so many UI controls like buttons, List boxes, Combo boxes etc.,These are the building blocks of dailogue boxes. - Hiding a control behind another control is not the correct way to locate it. It will happen due to language expansion. - By using controls as part os scentences the sentences suffer the same problem as concatenation strings. Messages: Text strings are usually larger than english text strings in other languages. The growth rate which is acute in short strings become less of a problem as string length increases. .

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1. c) For each of the three products that are observed in Spectrum C, there is a diagnostic proton that can be used to accurately calculate the product ratios. (Remember that a good diagnostic proton has a "clean" signal that has minimal overlap with other signals- see "Handout 05: Calculating Compound Ratios" on CourseWeb.) What diagnostic proton can be used to calculate product ratios for each isomer? If the isomer was not formed, write "N/A." (4 points) p- bromoaniline o-bromoaniline o,p- dibromoaniline O,o,p tribromoaniline d) Using the diagnostic protons you listed in part c, calculate the percent of each isomer. Show calculations for each isomer. If an isomer was not formed, write "0%" (8 points) %p-bromoaniline % o-bromoaniline % o,o,p- tribromoaniline 960,p. dibromoaniline Solution Solution: The spectrum interpretation are as shown below, Part (C) p - bromoaniline = 7.529 o - bromoaniline = 7.230 o,p - dibromoaniline = 6.651 o,o,p - tribromoaniline = 6.566 Part (D) p - bromoaniline = (2*1):(2*0.83) = 2/(2+1.66)*100% = 54.65 % o - bromoaniline = 0 o,p - dibromoaniline = 1.15 /(5.43+1.15+0.61)*100% = 15.99 = 16% o,o,p - tribromoaniline = 5.95/(0.80+1.35+5.95) *100% = 73.46 = 74%