Based on differences in electronegativity- how would you characterize.docx

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Based on differences in electronegativity, how would you characterize the bonding in sulfur dioxide, SO2? Essay answers are limited to about 500 words (3800 characters maximum, including spaces). 3800 Character(s) remaining Solution The atom with less electronegative is central atom and the atoms with more electronegtive are terminal atoms. So, Sulphur with less electronegative is central atom and Oxygens are terminal atoms. S ( Z = 16 ) 1st excited state electronic configuration is, 1s2 s2 2p6 3s2 3p3 3d1 S in its first exicted state undergoes sp2 hybirdisation due ot which it has 1 paired sp2 hybrid orbital and 2 unpaired sp2 hybrid orbitals. And also S has one p and one d orbital with unpaired electrons. O (Z = 8) E.C is 1s2 2s2 2p4 Each O has two unpaired electrons in its valence shell. The each sp2 unpaired hybrid orbital of S overlaps with one of the unpaired p - orbital ofeach oxygen along the internuclear axis to form single S - O sigma bond with each Oxygen. The reamining unpaired pure p-orbitals of S overlaps with unpaired p - orbital of each oxygen perpendicular to the internuclear axis to form one pi bonds with each oxygen. Still S ia remained with one lone pair of electons. Due to this SO 2 is in bent shape and has the bond angle < 120 0 .

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Balance the redox reaction by inserting the appropriate coefficients. redox reaction: | Fe3+ + NO + H2O- Fe2+ + H+ + No; Solution Fe in Fe+3 has oxidation state of +3 Fe in Fe+2 has oxidation state of +2 So, Fe in Fe+3 is reduced to Fe+2 N in NO2- has oxidation state of +3 N in NO3- has oxidation state of +5 So, N in NO2- is oxidised to NO3- Reduction half cell: Fe+3 + 1e- --> Fe+2 Oxidation half cell: NO2- --> NO3- + 2e- Balance number of electrons to be same in both half reactions Reduction half cell: 2 Fe+3 + 2e- --> 2 Fe+2 Oxidation half cell: NO2- --> NO3- + 2e- Lets combine both the reactions. 2 Fe+3 + NO2- --> 2 Fe+2 + NO3- Balance Oxygen by adding water 2 Fe+3 + NO2- + H2O --> 2 Fe+2 + NO3- Balance Hydrogen by adding H+ 2 Fe+3 + NO2- + H2O --> 2 Fe+2 + NO3- + 2 H+ This is balanced chemical equation in acidic medium Answer: 2 Fe 3+ + NO 2 - + H 2 O â€”> 2 Fe 2+ + 2 H + + NO 3 - .

Balance the redox reaction by inserting the appropriate coefficients-.docx

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Auditing and Accounting Cases Investigating Issues of Fraud and Professional Ethics 4th Edition Case 2.6 Page 78 1. Refer to the fundamental principles governing an audit. Under the respon-sibilities principle, auditors are required to exercise due care and maintain professional skepticism throughout the audit. Based on the case information, discuss the ways in which the BLMIS auditor, David Friehling, disregarded his responsibility to uphold the fundamental principles governing an audit Solution Based on the facts of the case, due care was not exercised by the auditors. Auditors failed to conduct independent verification of BLMIS assets, review material sources of BLMIS revenue, examine bank account through which client funds flowed, verify liabilities related to client accounts or verify the purchase and custody of securities by BLMIS. This all shows a lack of intent on part of the auditors to gather and evaluate evidence to provide an opinion. Hence, they did not fulfill the Responsibilities principle. The auditor and/ or his relatives had accounts with BLMIS showing lack of professional independence as well. .

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b. Impact testing was used to assess which property of the materials? For metals, what crystal structure was more susceptible to the ductile-to-brittle transition temperature? Why was this the case? Solution Answer:- The Imapct testing was used to to asses the required energy to fracture the material and it can be measure the toughness of the material. we can also known the yield strength of material by using impact testing but we dont have any analytical formual for. we can also calcualte the strain rate for its effect on fracture of material. There is BCC and FCC type of crystal which is more susceptible to the ductile to brittle transiotion temperature. the low carbon steel BCC metals, which become brittle at low temperature or at very high strain rate. incase of FCC metals its remain ductiule at low temperature. In metals there is plastic deformation at room temperature occurs due to dislocation motion. this dislocation motion required a stress to move and its depend on the atomic bonding which is a crustal structure. In this case brittle fracture occurs at low temperature and ductile fracture occurs at high temperature. ductile-brittle transformation occured in between low and high temperature. .

b- Impact testing was used to assess which property of the materials-.docx

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B. lonic Bonds: Some atoms have such a strong attraction for electrons (called electronegativity) that they will steal electrons that are loosely held in place from other atoms. The specific structure of an atom will determine whether it gains or loses electrons to form ions. Ion are charged atoms, and atoms electrons have a negative charge (one negative for each extra electron), and atoms which have a positive charge are atoms which have lost electrons Positively charged ions are called cations. Negatively charged ions are anions. Like charges repel, while unlike charges attract (Opposites attract). Cations are then attracted to anions, and together, they will from ionic compounds. 1. Using the periodic table of elements, analyze the following element, ions and compounds. You will be asked to determine the total number of electrons ro protons in the atom or molecule as written. a. Protons in Na b Protons in Na* c. Protons in Cl d. Protons in Cr e. Protons in Ca f. Protons in Ca2 g. Protons in O h. Protons in O2 i. Protons in CaO k. Electrons in NacI m. Electrons in Na* j. Electrons in Cao I. Electrons in Na n. Electrons in Cl o. Electrons in C p. Electrons in Ca q. Electrons in Ca r. Electrons in O . Electrons in O2 Solution a. protons in Na = 11 b. protons in Na + = 11 c. protons in Cl = 17 d. protons in Cl - = 17 e. protons in Ca = 20 f. protons in Ca +2 = 20 g. protons in O = 8 h. protons in O -2 = 8 i. protons in CaO = 28 j. electrons in CaO = 28 K. protons in NaCl = 28 L. electrons in Na = 11 m. electrons in Na + = 10 n. electrons in Cl = 17 o. electrons in Cl - = 16 p. electrons in Ca = 20 q. electrons in Ca +2 = 18 r. electrons in O = 8 s. electrons in O -2 = 10 .

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Fill out the following table for four different IT systems- Note two.docx

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