The name of a binary acid a) has no prefix b) begins with the prefix b.docx

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The name of a binary acid a) has no prefix b) begins with the prefix bi- c) ends with the suffix -ous d) begins with the prefix hydro- The name of a binary acid a) has no prefix b) begins with the prefix bi- c) ends with the suffix -ous d) begins with the prefix hydro- a) has no prefix b) begins with the prefix bi- c) ends with the suffix -ous d) begins with the prefix hydro- Solution Binary acids are the compounds in which hydrogen is combined with some non metallic component suh as HF,HCl,HBr and name of such acids are hyochloric acid,hydrobromic acid,hydrofluoric acid eetc. so binary acids begins with prefix Hydro option d is correct .

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Please show ALL work. If written please show with neat handwriting. I will rate! Thank You! Sketch a realistic titration curve for the following situation. For last name starting: A-L Titrate 20.00 mL of 0.300 M morpholine with 0.200 M HCl M-O Titrate 10.00 mL of 0.300 M sodium 4-nitrophenolate with 0.100 M HCI P-Z Titrate 30.00 mL of 0.100 M sodium cyanide with 0.200 M HOI You should show proof of some specific points in the titration curve. i) initial pH ii) pH at 9.00 mL addition i) pH at the equiv. point iv) pH at 2.00 mL past the equiv. pt. Solution Titration Morpholine with HCl i) initial pH [morpholine] = 0.3 M morpholine + H2O <==> morpholineH+ + OH- let x amount hydrolyzed Kb = [morpholineH+][OH-]/[morpholine] 2.5 x 10^-6 = x^2/0.3 x = [OH-] = 8.7 x 10^-4 M [H+] = 1 x 10^-14/8.7 x 10^-4 = 1.15 x 10^-11 M pH = -log[H+] = 10.94 ii) 9.0 ml HCl added moles morpholine = 0.3 M x 20 ml = 6 mmol moles HCl = 0.2 M x 9 ml = 1.8 mmol morpholineH+ formed = 1.8 mmol morpholine remained = 6-1.8 = 4.2 mmol pH = pKa + log(morpholine/morpholneH+) = 8.4 + log(4.2/1.8) = 8.77 iii) pH at equivalence point volume HCl added = 6 mmol/0.2 M = 30 ml [morpholineH+] formed = 6 mmo/50 ml = 0.12 M morpholineH+ + H2O <==> morpholine + H3O+ let x amount hydrolyzed Ka = [morpholine][H3O+]/[morpholineH+] 4 x 10^-9 = x^2/0.12 x = [H3O+] = 2.2 x 10^-5 M pH = -log[H3O+] = 4.66 iv) pH at 2 ml past equivalence point excess [HCl] = [H+] = 0.2 M x 2 ml/52 ml = 0.008 M pH = -log(0.008) = 2.11 v) pH at half-equivalence point [morpholineH+] formed = [morpholine] remained pH = pKa = 8.4 Plot, volume HCl on x-axis and pH on y-axis .

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Please only answer if you know the answer. I have posted this twice and both times received incorrect answers that didn\'t even match up to the journal format. The account tile options are: salaries expense, Terrell Capital, Donley Capital, Account receivable, distributions, accounts payable, Pinkston capital, Lamar Capital and Cash. At April 30, partners\' capital balances in PDL Company are: G. Donley $51,400, C. Lamar $50,100, and J. Pinkston $15,200. The income sharing ratios are 5 4: 1, respectively. On May 1, the PDLT Company is formed by admitting Â·Terrell to the firm as a partner Journalize the admission of Terrell under each of the following independent assumptions. (Credit account titles are automatically indented when amount is entered. Do not indent manually.) (1) (2) (3) (4) Terrell purchases 50% of Pinkston\'s ownership interest by paying Pinkston $16,140 in cash. Terrell purchases 331/3% of Lamar\'s ownership interest by paying Lamar $15,450 in cash. Terrell invests $63,100 for a 30% ownership interest, and bonuses are given to the old partners Terrell invests $43,500 for a 30% ownership interest, which includes a bonus to the new partner No. Account Titles and Explanation Debit Credit 1. 2. 3. Lamar\'s capital balance is $31,570 after admitting Terrell to the partnership by investment. If Lamar\'s ownership interest is 20% of total partnership capital, what were Terrell\'s cash investment and the bonus to the new partner? Terrell\'s cash investment Bonus to new partner Solution I have already answered this question once. .

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Please make comments of the following discussion in 150-200 words.If you agree or disagree.why or why not? EEO laws apply to recruiting activities because a firm is required by the laws to utilize a sufficiently diverse workforce without violating the privacy of others and without and kind of bias based on sex, race, religon, disabilities, and other such things. Without these laws, recruiting activities could become invasive, obtrusive, unfair, and biased in a multitude of ways. The five main things i would do to help promote diversity in the workforce: 1.) Allow for flex-hours and as much remote working as possible for single parents and the elderly or disabled 2.) Hire on all mediums and through many agencies to try and reach the most demographically diverse audience possible 3.) Offer full-time, part-time, and temporary jobs in a way that is dynamic around the employees available hours 4.) Go through the Office of Disability Employement to provide work for the disabled 5.) Offer counseling and training for uneducated or low skill employees Solution Flexible working hours would offer an immense advantage to the employees as it would help to maintain a balance between their personal life and work obligations. Providing the facility of work from home or flexible work hours would help to employ single parents, elderly people and disabled. Thus, it would create diversity in the workforce. Different agencies including the Office of Disability Employment would help to acquire diverse pool of employee. Full-time, temporary and part-time jobs would help to attract students, other part-time workers, housewives etc. Hiring uneducated or low-skills employees after providing them some training would also help to increase the diversity in workforce. Above techniques are efficient to maintain a diverse work environment at the workplace without any kind of bias including sex, race, religion, or disabilities. .

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Please help!! I need to make the instructions for these by manipulating the registers such as CPP, SP, TOS, H The Level-2 instruction formats are: Q1) ROT // rotate: Pop two words from stack as // RotationCount: m[SP] // ValueToBeRotated: m[SP-1] // and push the rotated value on the stack m[SP-1] Q2) ADDD // integer double word addition: // Pop two double words (top four values) from the stack as // B_Low = m[SP], B_High = m[SP - 1], // A_Low = m[SP - 2], A_High = m[SP - 3] // Perform addition on A_Low and B_Low, and propagate the carry // to the addition result of A_High and B_High // Push the result as // m[SP - 2] = Result_Low // m[SP - 3] = Result_High where m stands for a 1-d array these will be uploaded as a .ijvm to a mic-1 simulator Solution Ans; .

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PLEASE HELP! 1. The integrating power of globalization forces managers and their firms to be knowledgeable about the nations in which they operate. (Points : 5) Â Â Â Â Â Â True False True False Solution 1. True 2. denationalization; internationalization 3. all of the above 4. all of the above 5. True 6. False 7. False 8. True 9. False 10. False 11. True 12. True 13. False 14. True 15.Â Â the tangible expression of an idea; the idea itself 16. True 17. value-added taxes 18. Civil law 19. True 20. True .

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please help with answer and explain homo and lumo in terms of acid/base. thanks! 10) A molecule with a high HOMO and a low LUMO can act as: (A) a Lewis base only (B) a Lewis acid only (C) a Lewis base and Lewis acid (D) neither a Lewis base, nor a Lewis acid (E) an inert molecule 11) A molecule with a low HOMO and a high LUMO can act as: (A) a Lewis base only (B) a Lewis acid only (C) a Lewis base and Lewis acid (D) an inert molecule (E) none of the above Solution 10) a molecule with a high HOMO and low LUMO can act as a Lewis Base or Lewis Acid. 11) a molecule with low HOMO and high LUMO can act as an inert molecule. According to lewis acid base concept, the Homo and Lumo are the pair of orbitals that lie closest. When having a high HOMO, this is lewis base. When having a low LUMO , this is lewis acid. When the HOMO and LUMO are far they have a larger energy gap, and the HOMO-LUMO interactions do not happen. And when the HOMO and LUMO of a molecule are closer they interact with each other. .

please help with answer and explain homo and lumo in terms of acid-bas.docx

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Please help me with this crossword! Solution Across: 4.Constitutional Isomers Cis and trans isomers are formed when the substituents are on opposite sides. These are constitutional Isomers. 5. Levorotatory When rotation of plane polarized light is towards the left, it is called levorotatory, and if to the right then dextrorotatory. 6. Stereoisomers Isomeric compounds with same connections but different spatial arrangements. 8. Stereocenters 11.Chiral Have non-superimposible mirror images 13. Constitutional Isomers Have different connectivities. 14. Raesmic Mixture Have no optical activity due to external compensation by as equal amounts of + and - enantiomers cancel each other. 15. Skeletol These are skeletol isomers. Down: 1. Diastereomers They are stereoisomers which are not mirror images of each other or are not enantiomers. 2. Achiral 3. Dextrorotatory 7. Meso Compound Show no optical rotation of plane polarized light due to internal compensation. 9. Functional These are functional group isomers. 10. Enantiomers 12. Positional Different position of functional groups will make them positional isomers. .

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Please Type Answer! Compare and contrast the properties of acids and bases. Solution Acids are species which have the ability to accept a pair of electrons. Acids are sour in taste, corrosive in nature, and turn blue litmus paper to red. Reaction of acids with a base results into loss of acidity (acidic nature) of acid. while the bases have the ability to donate a pair of electron. And, bases turn red litmum paper to blue. Bases lose their basic nature when they react with an acid. The reaction between an acid and a base is called acid-base reaction. Two examples are discussed below #1 HCl + NaOH ----> NaCl + H 2 O (H + + OH - -----> H 2 O) #2 BH 3 + NH 3 ----> BH 3 NH 3 In reaction 1, as H + ion of HCl can accept a pair of electron (from OH - of NaOH), thus HCl is an acid and NaOH is a base. Going by the same argument BH 3 in reaction 2, can accept a pair of elctron from NH 3 , thus BH3 is an acid, while NH3 is a base. Acids and bases contrast in their ability to change the color of litmum. Acids turn blue litmum paper to red, while bases changes the red litmum to blue. .

Please Type Answer! Compare and contrast the properties of acids and b.docx

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