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A robust whistleblowing regime is now an integral part of governance best practice for all companies, regardless of their size or location. If you have not already done so, you should think about introducing a whistleblowing policy or code of conduct. But simply having a whistleblowing policy and hotline is not enough to create a culture in which employees are genuinely encouraged to make disclosures. So what can you do to embed an open culture and make sure your whistleblowing regime is effective? Top-level commitment The CEO and board should clearly support and sponsor any whistleblowing regime. Board members or other senior managers must be seen to respect the policy. If anyone discourages whistleblowing this will significantly undermine other efforts. Senior accountability A senior member of management must have overall responsibility for embedding the culture of internal disclosure throughout the company – particularly within management.This person should also announce the policy to all employees, manage and review it, and feedback on it to the board. What’s more, he or she must have enough resources to be able to do this well. Communication and training Employees must know the regime is in place and understand when and how to use it. Some businesses ensure this with regular emails, videos and presentations from the CEO, responsible manager or general counsel. Comprehensive training must be compulsory for all, run regularly and records of employee attendance should be kept. It may be sensible to require employees to sign an annual declaration that they have read and understood the policy and received training. If there is a danger of training fatigue, make whistleblowing training part of the company’s general compliance training and reserve more intensive training for employees with managerial responsibilities. Regular review and audit Regularly review any whistleblowing policies and prepare reports on the number and types of disclosures received in any given year. Asking the questions below will help your board to assess how the policy can be applied most effectively. • Are disclosures widespread across the company or very limited? • Are they concentrated in a particular business area or scattered across all areas? • Are the numbers of disclosures going up or down? • Do employees feel able to make disclosures? • Are the types of issues being disclosed through the policy appropriate and sensible? • How have disclosures been investigated and followed up? You could also give employees updates on a more general level. This lets them see that people are making disclosures and that those disclosures are being encouraged and dealt with appropriately. Bring the information to life with anonymised examples of disclosures and how they were handled Proper investigation and action It is essential that the whistleblowing policy is enforced. Any disclosure must be investigated promptly and properly. Take appropriate action at the right level. The pol.
A robust whistleblowing regime is now an integral part of governance.pdf
A robust whistleblowing regime is now an integral part of governance.pdf
anjanaarts2014
1. In higher parts of the reef, where the wave energy is high, the coral forms tend to assume either encrusting or massive, ex:brain corals. Going further deep down, massive corals dominate and eventually give way to branching coral forms (because as depth increases, wave energy decerases). At the base of the reef front, there are plate-like coral forms. They are deep down (diminished light) and hence orient their flat surface to gain maximum amount of sunlight. Further deep, limestone boulders, coral branches, soft corals sponges and algae are found. Thus, the answer for this question is, brain coral forms dominate where there is shallow water and high wave energy. 2. Menhadens spawn in the inshores water of the Atlantic coast. The eggs hatch in the ocean and drift to estuaries along with ocean currents. These young ones stay around for certain period of time before, they return to the ocean. their early stage is known as \"peanut bunker\". The larval fish moves into low salinity waters in estuaries and the juveniles fish remain the bay untill the fall. Thus, the answer for this question is menhaden fish spend their larval and juvenile (option d) ontogenic stage at the Chesapeake Bay . 3. Sea urchins are the biggest kelp grazers. A massive density of sea urchins is suffiecient enough to eliminate kelp forest. But at times, when densities of kelps dominate, they secrete secondary compounds to defend themselves from the sea urchin. Thus it basically depends upon the densities of each species and their dominance over an are. However, sea urchins are the greatest grazers of kelps that play a role in their abundance. 4. Recruitment in marine biology means, surving of new young one. Recruitment limitation is the limit to ho much recruitment is possible. In this case, greater the recruitment of species, the greater the diversity of niche and hence greater is the diversity of high reef fish.The recruitment limitation model suggest that high reef fish diversity is due to niche diversity. 5. Reef building corals are temperature sensitive and that is why confined to tropical and semi tropical waters. Their optimal temperature range is 23o- 29o celsius. They cannot tolerate temperature below 18o celsius, and very few can tolerate temperatures as high as 40o celsius. Thus, coral reefs forms in areas where seawater temperatures averages no less than 18o celsius. Solution 1. In higher parts of the reef, where the wave energy is high, the coral forms tend to assume either encrusting or massive, ex:brain corals. Going further deep down, massive corals dominate and eventually give way to branching coral forms (because as depth increases, wave energy decerases). At the base of the reef front, there are plate-like coral forms. They are deep down (diminished light) and hence orient their flat surface to gain maximum amount of sunlight. Further deep, limestone boulders, coral branches, soft corals sponges and algae are found. Thus, the answer for this question is.
1. In higher parts of the reef, where the wave energy is high, the c.pdf
1. In higher parts of the reef, where the wave energy is high, the c.pdf
anjanaarts2014
a) dy/dx + 2y/x = x I.F. = x^2 solution i y * x^2 = int [ x^2 * x]dx + C = [x^4]/4 + C Solution a) dy/dx + 2y/x = x I.F. = x^2 solution i y * x^2 = int [ x^2 * x]dx + C = [x^4]/4 + C.
a)dydx + 2yx = xI.F. = x^2solution iy .pdf
a)dydx + 2yx = xI.F. = x^2solution iy .pdf
anjanaarts2014
23.C. 1 in 4 24. B. Homozygous 25. B. 50% 26.All females will be hetrozygous for the trait. 27. B. Ee. 28 D. All of the above. 29.A. Only once and then devided between the resulting four gamates 30.1/36 31.Both mom and dad are homozygous for blood type B There is more than one possible type of blood type for the offspring. Solution 23.C. 1 in 4 24. B. Homozygous 25. B. 50% 26.All females will be hetrozygous for the trait. 27. B. Ee. 28 D. All of the above. 29.A. Only once and then devided between the resulting four gamates 30.1/36 31.Both mom and dad are homozygous for blood type B There is more than one possible type of blood type for the offspring..
23.C. 1 in 424. B. Homozygous25. B. 5026.All females will be .pdf
23.C. 1 in 424. B. Homozygous25. B. 5026.All females will be .pdf
anjanaarts2014
A is correct. The ntp server 10.0.0.5 command is entered in Global Configuration mode. B, C, and D are incorrect. B and D are incorrect because they are not real NAT commands. C is incorrect because the command was entered in Interface Configuration mode, not Global Configuration mode. 12. C is correct. The show ntp associations command is entered in Privileged (enable) mode. A, B, and D are incorrect. All are the right command, just in the wrong configuration mode Solution A is correct. The ntp server 10.0.0.5 command is entered in Global Configuration mode. B, C, and D are incorrect. B and D are incorrect because they are not real NAT commands. C is incorrect because the command was entered in Interface Configuration mode, not Global Configuration mode. 12. C is correct. The show ntp associations command is entered in Privileged (enable) mode. A, B, and D are incorrect. All are the right command, just in the wrong configuration mode.
A is correct. The ntp server 10.0.0.5 command is entered in Global C.pdf
A is correct. The ntp server 10.0.0.5 command is entered in Global C.pdf
anjanaarts2014
A. Gene expression. B. Transcription factor. C. Gene expression. D.Operon.,Promoter E.Lactose F.Negative regulator. G. Lactose., Promoter H. Negative regulation., Operator. I. Inducible operon., Inducer. J. .Transcription factors, lac operon, CRP-cAMP. K. Lactase,Lactose intolerent. Solution A. Gene expression. B. Transcription factor. C. Gene expression. D.Operon.,Promoter E.Lactose F.Negative regulator. G. Lactose., Promoter H. Negative regulation., Operator. I. Inducible operon., Inducer. J. .Transcription factors, lac operon, CRP-cAMP. K. Lactase,Lactose intolerent..
A. Gene expression.B. Transcription factor.C. Gene expression..pdf
A. Gene expression.B. Transcription factor.C. Gene expression..pdf
anjanaarts2014
All are capable to grow in lactose medium . In presence of lactose the repressor gets inactivated and transcription continues. Here all mero diploids have functional B- galactosidase (Z) and permease gene (Y) Solution All are capable to grow in lactose medium . In presence of lactose the repressor gets inactivated and transcription continues. Here all mero diploids have functional B- galactosidase (Z) and permease gene (Y).
All are capable to grow in lactose medium . In presence of lactose t.pdf
All are capable to grow in lactose medium . In presence of lactose t.pdf
anjanaarts2014
1)The viruse which is the most dangerous threat.In DTI survey,72% of all the companies received infected e-mails last year.the internet experienceed three worms in 12 day which causes 1.8bn damages,according to Symantec\'s internet Security Threat Report. 2)Social network attacks which are cyber security threat due to social networks like facebook,twitter.We can access them by even phone. While subscribing to these we provide some information regarding us like phone number to the web thus the hackers can easily get our information through internet.They use one Technic called \"Intelligence Information gathering\" which is more advance than google.They can see our posts also so try to avoid posting sensitive and personal information. 3)By gathering the Celebrity information the hackers sent malware(.pif,.scr,.exe,.com,.vbs,.bat,.cmd or .hta have extenstion ) to their targets. When the user run that program,hackers trigger a Throjan horse or Remote Administration Tool(RAT) by this hackers can remotely acces users machine. SSL protocal is used for internet banking.By using Man-in-middle software which employs the ARP posioning technique to can sniff the password. So be alert while using ATM,Credit card and check the statements carefully. 4)Insider threats or organised crime more than 50 percent cyber security threats are caused by internal figures means computer crime may be carried out by one person by hacking prepaid phone card system.That is \"Organised crime\" china,Brazil,Russia cooperating with internal employees in criminal model CaaS. 5)These internal attacks are most dangerous threat to data and systems.Rogue employees,who are members of the IT team with knowledge of and access to networks,data centers and admin accounts, can cause serious damage. 6)Misunderstanding GRC:So many managements dont\'t understand the concept of GRC(Governance,Risk Management,Compliance) for IT governance,information security governance and corporate governance.To deal with training in IT governance,corporate governance,information security governance are best practice. ITIL and CobiT helps to deal with this problem. Solution 1)The viruse which is the most dangerous threat.In DTI survey,72% of all the companies received infected e-mails last year.the internet experienceed three worms in 12 day which causes 1.8bn damages,according to Symantec\'s internet Security Threat Report. 2)Social network attacks which are cyber security threat due to social networks like facebook,twitter.We can access them by even phone. While subscribing to these we provide some information regarding us like phone number to the web thus the hackers can easily get our information through internet.They use one Technic called \"Intelligence Information gathering\" which is more advance than google.They can see our posts also so try to avoid posting sensitive and personal information. 3)By gathering the Celebrity information the hackers sent malware(.pif,.scr,.exe,.com,.vbs,.bat,.cmd or .h.
1)The viruse which is the most dangerous threat.In DTI survey,72 of.pdf
1)The viruse which is the most dangerous threat.In DTI survey,72 of.pdf
anjanaarts2014
Recommended
A robust whistleblowing regime is now an integral part of governance best practice for all companies, regardless of their size or location. If you have not already done so, you should think about introducing a whistleblowing policy or code of conduct. But simply having a whistleblowing policy and hotline is not enough to create a culture in which employees are genuinely encouraged to make disclosures. So what can you do to embed an open culture and make sure your whistleblowing regime is effective? Top-level commitment The CEO and board should clearly support and sponsor any whistleblowing regime. Board members or other senior managers must be seen to respect the policy. If anyone discourages whistleblowing this will significantly undermine other efforts. Senior accountability A senior member of management must have overall responsibility for embedding the culture of internal disclosure throughout the company – particularly within management.This person should also announce the policy to all employees, manage and review it, and feedback on it to the board. What’s more, he or she must have enough resources to be able to do this well. Communication and training Employees must know the regime is in place and understand when and how to use it. Some businesses ensure this with regular emails, videos and presentations from the CEO, responsible manager or general counsel. Comprehensive training must be compulsory for all, run regularly and records of employee attendance should be kept. It may be sensible to require employees to sign an annual declaration that they have read and understood the policy and received training. If there is a danger of training fatigue, make whistleblowing training part of the company’s general compliance training and reserve more intensive training for employees with managerial responsibilities. Regular review and audit Regularly review any whistleblowing policies and prepare reports on the number and types of disclosures received in any given year. Asking the questions below will help your board to assess how the policy can be applied most effectively. • Are disclosures widespread across the company or very limited? • Are they concentrated in a particular business area or scattered across all areas? • Are the numbers of disclosures going up or down? • Do employees feel able to make disclosures? • Are the types of issues being disclosed through the policy appropriate and sensible? • How have disclosures been investigated and followed up? You could also give employees updates on a more general level. This lets them see that people are making disclosures and that those disclosures are being encouraged and dealt with appropriately. Bring the information to life with anonymised examples of disclosures and how they were handled Proper investigation and action It is essential that the whistleblowing policy is enforced. Any disclosure must be investigated promptly and properly. Take appropriate action at the right level. The pol.
A robust whistleblowing regime is now an integral part of governance.pdf
A robust whistleblowing regime is now an integral part of governance.pdf
anjanaarts2014
1. In higher parts of the reef, where the wave energy is high, the coral forms tend to assume either encrusting or massive, ex:brain corals. Going further deep down, massive corals dominate and eventually give way to branching coral forms (because as depth increases, wave energy decerases). At the base of the reef front, there are plate-like coral forms. They are deep down (diminished light) and hence orient their flat surface to gain maximum amount of sunlight. Further deep, limestone boulders, coral branches, soft corals sponges and algae are found. Thus, the answer for this question is, brain coral forms dominate where there is shallow water and high wave energy. 2. Menhadens spawn in the inshores water of the Atlantic coast. The eggs hatch in the ocean and drift to estuaries along with ocean currents. These young ones stay around for certain period of time before, they return to the ocean. their early stage is known as \"peanut bunker\". The larval fish moves into low salinity waters in estuaries and the juveniles fish remain the bay untill the fall. Thus, the answer for this question is menhaden fish spend their larval and juvenile (option d) ontogenic stage at the Chesapeake Bay . 3. Sea urchins are the biggest kelp grazers. A massive density of sea urchins is suffiecient enough to eliminate kelp forest. But at times, when densities of kelps dominate, they secrete secondary compounds to defend themselves from the sea urchin. Thus it basically depends upon the densities of each species and their dominance over an are. However, sea urchins are the greatest grazers of kelps that play a role in their abundance. 4. Recruitment in marine biology means, surving of new young one. Recruitment limitation is the limit to ho much recruitment is possible. In this case, greater the recruitment of species, the greater the diversity of niche and hence greater is the diversity of high reef fish.The recruitment limitation model suggest that high reef fish diversity is due to niche diversity. 5. Reef building corals are temperature sensitive and that is why confined to tropical and semi tropical waters. Their optimal temperature range is 23o- 29o celsius. They cannot tolerate temperature below 18o celsius, and very few can tolerate temperatures as high as 40o celsius. Thus, coral reefs forms in areas where seawater temperatures averages no less than 18o celsius. Solution 1. In higher parts of the reef, where the wave energy is high, the coral forms tend to assume either encrusting or massive, ex:brain corals. Going further deep down, massive corals dominate and eventually give way to branching coral forms (because as depth increases, wave energy decerases). At the base of the reef front, there are plate-like coral forms. They are deep down (diminished light) and hence orient their flat surface to gain maximum amount of sunlight. Further deep, limestone boulders, coral branches, soft corals sponges and algae are found. Thus, the answer for this question is.
1. In higher parts of the reef, where the wave energy is high, the c.pdf
1. In higher parts of the reef, where the wave energy is high, the c.pdf
anjanaarts2014
a) dy/dx + 2y/x = x I.F. = x^2 solution i y * x^2 = int [ x^2 * x]dx + C = [x^4]/4 + C Solution a) dy/dx + 2y/x = x I.F. = x^2 solution i y * x^2 = int [ x^2 * x]dx + C = [x^4]/4 + C.
a)dydx + 2yx = xI.F. = x^2solution iy .pdf
a)dydx + 2yx = xI.F. = x^2solution iy .pdf
anjanaarts2014
23.C. 1 in 4 24. B. Homozygous 25. B. 50% 26.All females will be hetrozygous for the trait. 27. B. Ee. 28 D. All of the above. 29.A. Only once and then devided between the resulting four gamates 30.1/36 31.Both mom and dad are homozygous for blood type B There is more than one possible type of blood type for the offspring. Solution 23.C. 1 in 4 24. B. Homozygous 25. B. 50% 26.All females will be hetrozygous for the trait. 27. B. Ee. 28 D. All of the above. 29.A. Only once and then devided between the resulting four gamates 30.1/36 31.Both mom and dad are homozygous for blood type B There is more than one possible type of blood type for the offspring..
23.C. 1 in 424. B. Homozygous25. B. 5026.All females will be .pdf
23.C. 1 in 424. B. Homozygous25. B. 5026.All females will be .pdf
anjanaarts2014
A is correct. The ntp server 10.0.0.5 command is entered in Global Configuration mode. B, C, and D are incorrect. B and D are incorrect because they are not real NAT commands. C is incorrect because the command was entered in Interface Configuration mode, not Global Configuration mode. 12. C is correct. The show ntp associations command is entered in Privileged (enable) mode. A, B, and D are incorrect. All are the right command, just in the wrong configuration mode Solution A is correct. The ntp server 10.0.0.5 command is entered in Global Configuration mode. B, C, and D are incorrect. B and D are incorrect because they are not real NAT commands. C is incorrect because the command was entered in Interface Configuration mode, not Global Configuration mode. 12. C is correct. The show ntp associations command is entered in Privileged (enable) mode. A, B, and D are incorrect. All are the right command, just in the wrong configuration mode.
A is correct. The ntp server 10.0.0.5 command is entered in Global C.pdf
A is correct. The ntp server 10.0.0.5 command is entered in Global C.pdf
anjanaarts2014
A. Gene expression. B. Transcription factor. C. Gene expression. D.Operon.,Promoter E.Lactose F.Negative regulator. G. Lactose., Promoter H. Negative regulation., Operator. I. Inducible operon., Inducer. J. .Transcription factors, lac operon, CRP-cAMP. K. Lactase,Lactose intolerent. Solution A. Gene expression. B. Transcription factor. C. Gene expression. D.Operon.,Promoter E.Lactose F.Negative regulator. G. Lactose., Promoter H. Negative regulation., Operator. I. Inducible operon., Inducer. J. .Transcription factors, lac operon, CRP-cAMP. K. Lactase,Lactose intolerent..
A. Gene expression.B. Transcription factor.C. Gene expression..pdf
A. Gene expression.B. Transcription factor.C. Gene expression..pdf
anjanaarts2014
All are capable to grow in lactose medium . In presence of lactose the repressor gets inactivated and transcription continues. Here all mero diploids have functional B- galactosidase (Z) and permease gene (Y) Solution All are capable to grow in lactose medium . In presence of lactose the repressor gets inactivated and transcription continues. Here all mero diploids have functional B- galactosidase (Z) and permease gene (Y).
All are capable to grow in lactose medium . In presence of lactose t.pdf
All are capable to grow in lactose medium . In presence of lactose t.pdf
anjanaarts2014
1)The viruse which is the most dangerous threat.In DTI survey,72% of all the companies received infected e-mails last year.the internet experienceed three worms in 12 day which causes 1.8bn damages,according to Symantec\'s internet Security Threat Report. 2)Social network attacks which are cyber security threat due to social networks like facebook,twitter.We can access them by even phone. While subscribing to these we provide some information regarding us like phone number to the web thus the hackers can easily get our information through internet.They use one Technic called \"Intelligence Information gathering\" which is more advance than google.They can see our posts also so try to avoid posting sensitive and personal information. 3)By gathering the Celebrity information the hackers sent malware(.pif,.scr,.exe,.com,.vbs,.bat,.cmd or .hta have extenstion ) to their targets. When the user run that program,hackers trigger a Throjan horse or Remote Administration Tool(RAT) by this hackers can remotely acces users machine. SSL protocal is used for internet banking.By using Man-in-middle software which employs the ARP posioning technique to can sniff the password. So be alert while using ATM,Credit card and check the statements carefully. 4)Insider threats or organised crime more than 50 percent cyber security threats are caused by internal figures means computer crime may be carried out by one person by hacking prepaid phone card system.That is \"Organised crime\" china,Brazil,Russia cooperating with internal employees in criminal model CaaS. 5)These internal attacks are most dangerous threat to data and systems.Rogue employees,who are members of the IT team with knowledge of and access to networks,data centers and admin accounts, can cause serious damage. 6)Misunderstanding GRC:So many managements dont\'t understand the concept of GRC(Governance,Risk Management,Compliance) for IT governance,information security governance and corporate governance.To deal with training in IT governance,corporate governance,information security governance are best practice. ITIL and CobiT helps to deal with this problem. Solution 1)The viruse which is the most dangerous threat.In DTI survey,72% of all the companies received infected e-mails last year.the internet experienceed three worms in 12 day which causes 1.8bn damages,according to Symantec\'s internet Security Threat Report. 2)Social network attacks which are cyber security threat due to social networks like facebook,twitter.We can access them by even phone. While subscribing to these we provide some information regarding us like phone number to the web thus the hackers can easily get our information through internet.They use one Technic called \"Intelligence Information gathering\" which is more advance than google.They can see our posts also so try to avoid posting sensitive and personal information. 3)By gathering the Celebrity information the hackers sent malware(.pif,.scr,.exe,.com,.vbs,.bat,.cmd or .h.
1)The viruse which is the most dangerous threat.In DTI survey,72 of.pdf
1)The viruse which is the most dangerous threat.In DTI survey,72 of.pdf
anjanaarts2014
HClO3 + NaOH NaClO3 +H2O n(HClO3) = 0.1*0.04 = 4*10-3(mol) When added 10ml NaOH n(NaOH) used = 0.1 * 0.01 = 1*10-3(mol) n(HClO3) left = 3*10-3(mol) [HClO3] = 3*10-3/ 0.05 = 0.06 (M) [ClO3-] = 1*10-3/ 0.05 = 0.02(M) pH=pKa + log[ClO3-]/[HClO3] = 7.42 - 0.477 = 6.943 When added 20ml NaOH n(NaOH) used = 0.1 * 0.02 = 2*10-3(mol) n(HClO3) left = 2*10-3(mol) [HClO3] = 2*10-3/ 0.06 = 0.033 (M) [ClO3-] = 2*10-3/ 0.06 = 0.033(M) pH=pKa + log[ClO3-]/[HClO3] = 7.42 - 0 = 7.42 When added 30ml NaOH n(NaOH) used = 0.1 * 0.03 = 3*10-3(mol) n(HClO3) left = 1*10-3(mol) [HClO3] = 1*10-3/ 0.07 = 0.0143 (M) [ClO3-] = 3*10-3/ 0.07 = 0.0429(M) pH=pKa + log[ClO3-]/[HClO3] = 7.42 + 0.477 = 7.897 When added 40ml NaOH n(NaOH) used = 0.1 * 0.04 = 4*10-3(mol) n(HClO3) left = 0(mol) [ClO3-] = 4*10-3/ 0.08 = 0.05(M) Kb = Kw/Ka = 2.63 *10-7 [OH-] = Kb[ClO3-] =2.63*10-7*0.05 =1.14*10-4pOH = 3.94 pH = 10.06 Solution HClO3 + NaOH NaClO3 +H2O n(HClO3) = 0.1*0.04 = 4*10-3(mol) When added 10ml NaOH n(NaOH) used = 0.1 * 0.01 = 1*10-3(mol) n(HClO3) left = 3*10-3(mol) [HClO3] = 3*10-3/ 0.05 = 0.06 (M) [ClO3-] = 1*10-3/ 0.05 = 0.02(M) pH=pKa + log[ClO3-]/[HClO3] = 7.42 - 0.477 = 6.943 When added 20ml NaOH n(NaOH) used = 0.1 * 0.02 = 2*10-3(mol) n(HClO3) left = 2*10-3(mol) [HClO3] = 2*10-3/ 0.06 = 0.033 (M) [ClO3-] = 2*10-3/ 0.06 = 0.033(M) pH=pKa + log[ClO3-]/[HClO3] = 7.42 - 0 = 7.42 When added 30ml NaOH n(NaOH) used = 0.1 * 0.03 = 3*10-3(mol) n(HClO3) left = 1*10-3(mol) [HClO3] = 1*10-3/ 0.07 = 0.0143 (M) [ClO3-] = 3*10-3/ 0.07 = 0.0429(M) pH=pKa + log[ClO3-]/[HClO3] = 7.42 + 0.477 = 7.897 When added 40ml NaOH n(NaOH) used = 0.1 * 0.04 = 4*10-3(mol) n(HClO3) left = 0(mol) [ClO3-] = 4*10-3/ 0.08 = 0.05(M) Kb = Kw/Ka = 2.63 *10-7 [OH-] = Kb[ClO3-] =2.63*10-7*0.05 =1.14*10-4pOH = 3.94 pH = 10.06.
HClO3 + NaOH NaClO3 +H2O n(HClO3) = 0.10.04 = 4.pdf
HClO3 + NaOH NaClO3 +H2O n(HClO3) = 0.10.04 = 4.pdf
anjanaarts2014
I have already answered it. Check. Solution I have already answered it. Check..
I have already answered it. Check. .pdf
I have already answered it. Check. .pdf
anjanaarts2014
Iodide is a strong nucleophile but a weak base, so SN2 is the preferred reaction. Only the bromine on C1 is eligible to undergo SN2, so that one will be replaced by iodide. The product formed is 3-bromo 1-iodo 3 methylpentane Solution Iodide is a strong nucleophile but a weak base, so SN2 is the preferred reaction. Only the bromine on C1 is eligible to undergo SN2, so that one will be replaced by iodide. The product formed is 3-bromo 1-iodo 3 methylpentane.
Iodide is a strong nucleophile but a weak base, s.pdf
Iodide is a strong nucleophile but a weak base, s.pdf
anjanaarts2014
NaCl : It\'s the coulombic interaction between Na+ and Cl- due to which crystal exists. It\'s not ionic intereaction. It\'s ionic interaction. (very strong). NaCl dissolves in water. It\'s not due to hydrogen bonding. It\'s due to strong ion-dipole interactions. Initially, the positive and negative ion are only attracted to each other. The water molecules are hydrogen bonded to each other. If the crystal is to dissolve, these bonds must be broken. Negative chloride ions on the surface are attracted by neighboring positive sodium ions and by the partially positive hydrogen atom in the polar water molecule (See the graphic on the left). Similarly, the positive sodium ions are attracted by both chloride ions and the partially negative oxygen atom in the polar water molecule. A \"tug-of-war\" occurs for the positive and negative ions between the other ions in the crystal and the water molecules. Several water molecules are attracted to each of the ions. Whether the crystal dissolves is determined by which attractive force is stronger. If the internal ionic forces in the crystal are the strongest, the crystal does not dissolve. This is the situation in reactions where precipitates form. If the attractions for the ions by the polar water molecules are the strongest, the crystal will dissolve. This is the situation in sodium chloride. Once the ions are released from the crystals, the ions are completely surrounded by water molecules. Note that the proper atom in the water molecule must \"point\" toward the correct ion. The charge principle and the partial charges in the polar molecule determine the correct orientation. Partially negative oxygen atoms in the water molecule interact with the positive sodium ion. Partially positive hydrogen atoms in the water molecule interact with the negative chloride ion. Sugar : The interaction between sugar molecule and water is greater than between sugar molecules. So When sugar is put into water, it likes to form hydrogen bonds with water. So it breaks glycosidic linkages and forms hydrogen bonds with water. Initially it separates into polymer sugar chains, but later it dissociates into individual molecules. Solution NaCl : It\'s the coulombic interaction between Na+ and Cl- due to which crystal exists. It\'s not ionic intereaction. It\'s ionic interaction. (very strong). NaCl dissolves in water. It\'s not due to hydrogen bonding. It\'s due to strong ion-dipole interactions. Initially, the positive and negative ion are only attracted to each other. The water molecules are hydrogen bonded to each other. If the crystal is to dissolve, these bonds must be broken. Negative chloride ions on the surface are attracted by neighboring positive sodium ions and by the partially positive hydrogen atom in the polar water molecule (See the graphic on the left). Similarly, the positive sodium ions are attracted by both chloride ions and the partially negative oxygen atom in the polar water molecule. A \"tug-of-war\" occurs for.
NaCl Its the coulombic interaction between Na.pdf
NaCl Its the coulombic interaction between Na.pdf
anjanaarts2014
Here: Jupiter is the fifth planet from the Sun and the largest planet within the Solar System.It is a gas giant with a mass slightly less than one-thousandth of the Sun but is two and a half times the mass of all the other planets in our Solar System combined. Jupiter is classified as a gas giant along with Saturn, Uranus and Neptune. Together, these four planets are sometimes referred to as the Jovian or outer planets. The planet was known by astronomers of ancient times and was associated with the mythology and religious beliefs of many cultures. The Romans named the planet after the Roman god Jupiter.[14] When viewed from Earth, Jupiter can reach an apparent magnitude of -2.94, making it on average the third-brightest object in the night sky after the Moon and Venus. (Mars can briefly match Jupiter\'s brightness at certain points in its orbit.) Jupiter is primarily composed of hydrogen with a quarter of its mass being helium; it may also have a rocky core of heavier elements. Because of its rapid rotation, Jupiter\'s shape is that of an oblate spheroid (it possesses a slight but noticeable bulge around the equator). The outer atmosphere is visibly segregated into several bands at different latitudes, resulting in turbulence and storms along their interacting boundaries. A prominent result is the Great Red Spot, a giant storm that is known to have existed since at least the 17th century when it was first seen by telescope. Surrounding the planet is a faint planetary ring system and a powerful magnetosphere. There are also at least 63 moons, including the four large moons called the Galilean moons that were first discovered by Galileo Galilei in 1610. Ganymede, the largest of these moons, has a diameter greater than that of the planet Mercury. Jupiter has been explored on several occasions by robotic spacecraft, most notably during the early Pioneer and Voyager flyby missions and later by the Galileo orbiter. The most recent probe to visit Jupiter was the Pluto-bound New Horizons spacecraft in late February 2007. The probe used the gravity from Jupiter to increase its speed. Future targets for exploration in the Jovian system include the possible ice-covered liquid ocean on the moon Europa. Jupiter\'s structure: Jupiter is composed primarily of gaseous and liquid matter. It is the largest of four gas giants as well as the largest planet in the solar system with a diameter of 142,984 km at its equator. The density of Jupiter, 1.326 g/cm³, is the second highest of the gas giant planets. However, the density is lower than any of the four terrestrial planets. Solution Here: Jupiter is the fifth planet from the Sun and the largest planet within the Solar System.It is a gas giant with a mass slightly less than one-thousandth of the Sun but is two and a half times the mass of all the other planets in our Solar System combined. Jupiter is classified as a gas giant along with Saturn, Uranus and Neptune. Together, these four planets are sometimes refe.
Here Jupiter is the fifth planet from the Sun a.pdf
Here Jupiter is the fifth planet from the Sun a.pdf
anjanaarts2014
moles of H+ = 42.0 x 0.120 millimoles = 5.04 millimoles final volume = 1.55 L [H+] = (5.04 x 10^-3)/1.55 = 3.25 x 10^-3M pH = -log[H+] = 2.49 Solution moles of H+ = 42.0 x 0.120 millimoles = 5.04 millimoles final volume = 1.55 L [H+] = (5.04 x 10^-3)/1.55 = 3.25 x 10^-3M pH = -log[H+] = 2.49.
moles of H+ = 42.0 x 0.120 millimoles = 5.04 mill.pdf
moles of H+ = 42.0 x 0.120 millimoles = 5.04 mill.pdf
anjanaarts2014
Lithium is group 1 metal, while nitrogen is group 15 nonmetallic element. Thus, the bond that they form is ionic bond. metal + nonmetal = ionic compound Solution Lithium is group 1 metal, while nitrogen is group 15 nonmetallic element. Thus, the bond that they form is ionic bond. metal + nonmetal = ionic compound.
Lithium is group 1 metal, while nitrogen is group.pdf
Lithium is group 1 metal, while nitrogen is group.pdf
anjanaarts2014
there is some special character in ur question like ? pls re post the question properly thanks Solution there is some special character in ur question like ? pls re post the question properly thanks.
there is some special character in ur question li.pdf
there is some special character in ur question li.pdf
anjanaarts2014
a. Times Interest Earned Ratio = EBIT / interest Expense For 2014: $8,799 / $600 = 14.7 times For 2015: $9,454 / $838 = 11.3 times For 2016: $10,338 / $1,058 = 9.8 times The TImes Interest Earned (TIE) Ratio measures the operating income level that can be used for interest payments in the future, i.e. how many times the Company can make the interest payments. It is a solvency ratio since it measures the Company\'s ability to service it\'s debt interest obligations. Hence, the higher the ratio, the more favorable is the company\'s liquidity position. In the given question, we can observe that the ratio has consistently decreased year on year. This implies that the Company\'s debt has increased while the increase in operating income has not been at the same pace. The Company can afford to service interest fewer times each year from 2014 to 2016. This implies that the Company has become comparatively more risky than before from the Creditor\'s standpoint. b. Current Ratio = Current Assets / Current Liabilities For 2014: $25,983 / $19,027 = 1.4 For 2015: $29,158 / $23,169 = 1.3 For 2016: $31,042 / $26,250 = 1.2 The Current Ratio is an indicator of how efficiently the Company manages it\'s operating cycle. A higher Current Ratio implies that the COmpany is in a better position to discharge all it\'s short term obligations without affecting or jeopardising it\'s working capital requirements. On the flipside, a ratio that is too high indicates that the Company\'s current assets are lying idle or that it is not efficiently using short term funds. In the Given case, the Company\'s current ratio shows a steady decline. However, the lowest is 1.2 times, which implies that the Company is still sufficiently liquid. This is becase even at the lowest, the Company\'s Current Assets can pay off the Current Liabilities 1.2 times without affecting working capital cycle for regular business needs. Hence, as per the options given, we can say that CVS’s current ratio has decreased over the past three years, however, it is greater than 1 indicating CVS is liquid. c. Total Liabilities to Equity Ratio = Total Liabilities / Total Equity For 2014: $36,224 / $37,963 = 0.95 For 2015: $55,234 / $37,203 = 1.5 For 2016: $57,628 / $36,834 = 1.6 Total Liabilities include all the claims, long term and short term, against the Company\'s Assets. Total Liabilities to Equity Ratio indicates how much of the Company\'s Assets are funded by creditors versus how much is funded by the equity holders. Broadly, a 1:1 ratio is considered acceptable for most industries. Creditors would prefer a lower ratio as this implies their money is at less risk. However, equity holders would like to benefit from the risk taken by them due to the Company using debt, and would prefer a higher ratio. In the given case, this ratio is increasing year on year. This implies that the COmpany is taking on more debt financing. This implies that the Company has more risk overall, although it benefits the equity .
a.Times Interest Earned Ratio = EBIT interest ExpenseFor 2014.pdf
a.Times Interest Earned Ratio = EBIT interest ExpenseFor 2014.pdf
anjanaarts2014
45 yrs. Solution 45 yrs..
45 yrs.Solution45 yrs..pdf
45 yrs.Solution45 yrs..pdf
anjanaarts2014
the bacteria isolated from the dead mouse is smooth (S) S. pneumoniae bacteria. and the DNA was radio-labeled. the Griffith experiment proves that bacteria are capable of transforming genetic information. in the above experiment Griffith grown the smooth (S) S. pneumoniae bacteria with radio-labelled phosphate (32P), radiolabelled 32P, which is incorporated into the DNA and he had also separately grown rough (R) bacteria in regular (not radio-labelled) phosphate. while injecting the heat killed S bacteria to live R bacteria, the R bacteria had been transformed into lethal S bacteria. the DNA had survived in Heating process of S strain, and this DNA transform into the R strain and synthesizes the polysaccharide capsule that protects form host immune response in the R bacteria and transform the R strain into the S strain which results Mouse dies. in the first step we are labeled the DNA of S Strain with 32P, so the final DNA which is isolating from the died mouse contains both radiolabeled DNA and non radiolabeled DNA in different proportions. Solution the bacteria isolated from the dead mouse is smooth (S) S. pneumoniae bacteria. and the DNA was radio-labeled. the Griffith experiment proves that bacteria are capable of transforming genetic information. in the above experiment Griffith grown the smooth (S) S. pneumoniae bacteria with radio-labelled phosphate (32P), radiolabelled 32P, which is incorporated into the DNA and he had also separately grown rough (R) bacteria in regular (not radio-labelled) phosphate. while injecting the heat killed S bacteria to live R bacteria, the R bacteria had been transformed into lethal S bacteria. the DNA had survived in Heating process of S strain, and this DNA transform into the R strain and synthesizes the polysaccharide capsule that protects form host immune response in the R bacteria and transform the R strain into the S strain which results Mouse dies. in the first step we are labeled the DNA of S Strain with 32P, so the final DNA which is isolating from the died mouse contains both radiolabeled DNA and non radiolabeled DNA in different proportions..
the bacteria isolated from the dead mouse is smooth (S) S. pneumonia.pdf
the bacteria isolated from the dead mouse is smooth (S) S. pneumonia.pdf
anjanaarts2014
ANSWER: A for example: fluoride, oxide.... etc. If you need more help let me know and I can explain further Solution ANSWER: A for example: fluoride, oxide.... etc. If you need more help let me know and I can explain further.
ANSWER A for example fluoride, oxide.... etc. .pdf
ANSWER A for example fluoride, oxide.... etc. .pdf
anjanaarts2014
Java Program: import java.io.*; import java.util.*; public class LongestStreak { public static int[] longestStreak(boolean[] values) { int max=0,index=0,mIndex=0; for(int i=0;i Solution Java Program: import java.io.*; import java.util.*; public class LongestStreak { public static int[] longestStreak(boolean[] values) { int max=0,index=0,mIndex=0; for(int i=0;i.
Java Program import java.io.; import java.util.;public cl.pdf
Java Program import java.io.; import java.util.;public cl.pdf
anjanaarts2014
H2SO4 => 2H+ + SO42- [H+] = 2[H2SO4] = 0.0092 M pH = -log[H+] = 2.04 Solution H2SO4 => 2H+ + SO42- [H+] = 2[H2SO4] = 0.0092 M pH = -log[H+] = 2.04.
H2SO4 = 2H+ + SO42-[H+] = 2[H2SO4] = 0.0092 MpH = -log[H+] = 2..pdf
H2SO4 = 2H+ + SO42-[H+] = 2[H2SO4] = 0.0092 MpH = -log[H+] = 2..pdf
anjanaarts2014
here reaction is occurring when Sn and I2 are refluxed together reduction and oxidation is takes place(called redox ) he would ask the oxidation potential &its reduction potential is asking . Sn + 2I2------>SnI4 here Sn oxidised to Sn2+ ,iodine reduced ,means reduction potential of tin(Sn) is E0= -0.136, I2 reduction potential is =+0.53 ,its indicates reduction potential of iodine is more so it can oxidise the Sn & it can be reduced. Zn(s) + I2(s) -----> Zn2+ (aq) + 2I- (aq) here Zn oxidised to Zn2+ ,iodine reduced ,means reduction potential of tin(Sn) is E0=-0.763 I2 reduction potential is =+0.53 ,its indicates reduction potential of iodine is more so it can oxidise the Zn & it can be reduced Sn(s)+2HCl (aq) -----> Sn2+ (aq) +2Cl- (aq) + H2 (g) here Sn oxidised to Sn2+ ,chlorine reduced ,means reduction potential of tin(Sn) is E0=- 0.136here Sn oxidised to Sn2+ ,chlorine reduced ,means reduction potential of tin(Sn) is E0=- 0.136, cl- reduction potential is =+1.36 ,its indicates reduction potential of chlorine is more so it can oxidise the Sn & it(Cl) can be reduced. Sn2+ (aq) + 2I- (aq) -----> SnI2 (s) I2 reduction potential is =+0.53 ,its indicates reduction potential of iodine is less than the Sn2+ so it can reduce the Sn2+ & it(I-) can be oxidised. Solution here reaction is occurring when Sn and I2 are refluxed together reduction and oxidation is takes place(called redox ) he would ask the oxidation potential &its reduction potential is asking . Sn + 2I2------>SnI4 here Sn oxidised to Sn2+ ,iodine reduced ,means reduction potential of tin(Sn) is E0= -0.136, I2 reduction potential is =+0.53 ,its indicates reduction potential of iodine is more so it can oxidise the Sn & it can be reduced. Zn(s) + I2(s) -----> Zn2+ (aq) + 2I- (aq) here Zn oxidised to Zn2+ ,iodine reduced ,means reduction potential of tin(Sn) is E0=-0.763 I2 reduction potential is =+0.53 ,its indicates reduction potential of iodine is more so it can oxidise the Zn & it can be reduced Sn(s)+2HCl (aq) -----> Sn2+ (aq) +2Cl- (aq) + H2 (g) here Sn oxidised to Sn2+ ,chlorine reduced ,means reduction potential of tin(Sn) is E0=- 0.136here Sn oxidised to Sn2+ ,chlorine reduced ,means reduction potential of tin(Sn) is E0=- 0.136, cl- reduction potential is =+1.36 ,its indicates reduction potential of chlorine is more so it can oxidise the Sn & it(Cl) can be reduced. Sn2+ (aq) + 2I- (aq) -----> SnI2 (s) I2 reduction potential is =+0.53 ,its indicates reduction potential of iodine is less than the Sn2+ so it can reduce the Sn2+ & it(I-) can be oxidised..
here reaction is occurring when Sn and I2 are refluxed together redu.pdf
here reaction is occurring when Sn and I2 are refluxed together redu.pdf
anjanaarts2014
An undirected graph is acyclic (i.e., a forest) iff a DFS yields no back edges. Since back edges are those edges (u, v) connecting a vertex u to an ancestor v in a depth-first tree, so no back edges means there are only tree edges, so there is no cycle. So we can simply fun DFS. If find a back edge, there is a cycle. The complexity is O(V ) instead of O(E + V ). Since if there is a back edge, it must be found before seeing |V | distinct edges. This is because in a acyclic (undirected ) forest, |E| |V | + 1 Solution An undirected graph is acyclic (i.e., a forest) iff a DFS yields no back edges. Since back edges are those edges (u, v) connecting a vertex u to an ancestor v in a depth-first tree, so no back edges means there are only tree edges, so there is no cycle. So we can simply fun DFS. If find a back edge, there is a cycle. The complexity is O(V ) instead of O(E + V ). Since if there is a back edge, it must be found before seeing |V | distinct edges. This is because in a acyclic (undirected ) forest, |E| |V | + 1.
An undirected graph is acyclic (i.e., a forest) iff a DFS yields no .pdf
An undirected graph is acyclic (i.e., a forest) iff a DFS yields no .pdf
anjanaarts2014
A) [Sn2+]^3 / [Fe3+]^2 (don\'t include solids) B) [OH-]^4 [Cu2+]^2 / [O2] (no solids or liquids) C) [Br-]^2 / [I-]^2 Solution A) [Sn2+]^3 / [Fe3+]^2 (don\'t include solids) B) [OH-]^4 [Cu2+]^2 / [O2] (no solids or liquids) C) [Br-]^2 / [I-]^2.
A) [Sn2+]^3 [Fe3+]^2 (dont include solids) B).pdf
A) [Sn2+]^3 [Fe3+]^2 (dont include solids) B).pdf
anjanaarts2014
Given LOSS Sells a variety of equpiment including the executive office chair Particulars Current No.of Units expected to sell the executive office chair 6000 Selling Price per executive office chair in $ 250 Current manufacturing costs in $ per unit 175 Current Fixed cost 3,60,000 Proposed Changes No.of Units expected to sell the executive office chair 6000 Selling Price per executive office chair in $ 250 Alternative manufacturing costs in $ per unit 75 Alternative Fixed cost 9,45,000 Sl Particulars Current Proposed A Sales 15,00,000 15,00,000 B Variable cost 10,50,000 4,50,000 C Contribution ( A - B ) 4,50,000 10,50,000 D Fixed cost 3,60,000 9,45,000 E Profit ( C - D ) 90,000 1,05,000 F price Volume Ratio % ( C/A * 100 ) 30% 70% G Break even Point Volume ( D/F) 12,00,000 13,50,000 H Contribution Per Unit ( C/6000 ) 75 175 I Break even Point Units ( D/H) 4800 5400 J Margin of safety sales ( A -G ) 3,00,000 1,50,000 K Margin of safety in Units ( J / selling Price pu ) 1200 600 L Margin of safety in Percentage ( J / A )*100 20% 10% Indiffernce point between two alternatives Indifference cost = point where there is no difference in cost between two alternatives Indifference point Units Assumptions Q = Quantity / units Cpu 1 = Contribution per unit for current alternative Cpu 2= Contribution per unit for proposed alternative FC 1 = fixed cost for current alternative FC 2 = fixed cost for Proposed alternative alternative Indifference point equation Cpu1 * Q - FC1 = Cpu2 * Q - FC2 = 75 * Q - 360,000 = 175 * Q -945,000 = 945,000 - 360,000 = Q ( 175 - 75 ) = 100 * Q = 585,000 = Q = 585,000/100 = Q = 5850 Units Therefore Indifference point is 5850 Units Given LOSS Sells a variety of equpiment including the executive office chair Particulars Current No.of Units expected to sell the executive office chair 6000 Selling Price per executive office chair in $ 250 Current manufacturing costs in $ per unit 175 Current Fixed cost 3,60,000 Proposed Changes No.of Units expected to sell the executive office chair 6000 Selling Price per executive office chair in $ 250 Alternative manufacturing costs in $ per unit 75 Alternative Fixed cost 9,45,000 Sl Particulars Current Proposed A Sales 15,00,000 15,00,000 B Variable cost 10,50,000 4,50,000 C Contribution ( A - B ) 4,50,000 10,50,000 D Fixed cost 3,60,000 9,45,000 E Profit ( C - D ) 90,000 1,05,000 F price Volume Ratio % ( C/A * 100 ) 30% 70% G Break even Point Volume ( D/F) 12,00,000 13,50,000 H Contribution Per Unit ( C/6000 ) 75 175 I Break even Point Units ( D/H) 4800 5400 J Margin of safety sales ( A -G ) 3,00,000 1,50,000 K Margin of safety in Units ( J / selling Price pu ) 1200 600 L Margin of safety in Percentage ( J / A )*100 20% 10% Indiffernce point between two alternatives Indifference cost = point where there is no difference in cost between two alternatives Indifference point Units Assumptions Q = Quantity / units Cpu 1 = Contribution per unit for current alternative Cpu 2= Contribution per unit for.
GivenLOSS Sells a variety of equpiment including the executive off.pdf
GivenLOSS Sells a variety of equpiment including the executive off.pdf
anjanaarts2014
Answer for question 1 DuPont analysis explains what is affecting company\'s ROE. DuPont analysis can be explained with formula as below ROE= Profit Margin× Total asset turnover× Leverage factor Profit Margin= Net Income/ Total Revenue Total Asset turnover=Revenue/ Total assets Leverage factor= Total assets/ Shareholder equity DuPont analysis points out operating efficiency in form of profit margin, Asset efficiency in form of asset turnover and leverage impact with the help of leverage factor as how much of the equity is used to finance total assets Solution Answer for question 1 DuPont analysis explains what is affecting company\'s ROE. DuPont analysis can be explained with formula as below ROE= Profit Margin× Total asset turnover× Leverage factor Profit Margin= Net Income/ Total Revenue Total Asset turnover=Revenue/ Total assets Leverage factor= Total assets/ Shareholder equity DuPont analysis points out operating efficiency in form of profit margin, Asset efficiency in form of asset turnover and leverage impact with the help of leverage factor as how much of the equity is used to finance total assets.
Answer for question 1DuPont analysis explains what is affecting co.pdf
Answer for question 1DuPont analysis explains what is affecting co.pdf
anjanaarts2014
3360 Solution 3360.
3360Solution3360.pdf
3360Solution3360.pdf
anjanaarts2014
In the world of commerce, precision is paramount. Pro-Forma Invoices serve as the blueprint for these precise financial transactions. These documents encapsulate critical information such as quantity, transportation charges, value, weight, and the range of goods involved in a transaction.
How to Send Pro Forma Invoice to Your Customers in Odoo 17
How to Send Pro Forma Invoice to Your Customers in Odoo 17
Celine George
The Department of Emergency Medicine at Carolinas Medical Center is passionate about education! Dr. Michael Gibbs is a world-renowned clinician and educator and has helped guide numerous young clinicians on the long path of Mastery of Emergency Medical Care. With his oversight, the EMGuideWire team aim to help augment our understanding of emergent imaging. You can follow along with the EMGuideWire.com team as they post these educational, self-guided radiology slides or you can also use this section to learn more in-depth about specific conditions and diseases. This Radiology Reading Room pertains to Sternal Fractures and Dislocations and is brought to you by Carrie Bissell, MD, Aaron Fox, MD, Kendrick Lim, MD, Stephanie Jensen, MD, and Olivia Rice, MD. It is has special guest editor: Sean Dieffenbaugher, MD and Laurence Kempton, MD
Sternal Fractures & Dislocations - EMGuidewire Radiology Reading Room
Sternal Fractures & Dislocations - EMGuidewire Radiology Reading Room
Sean M. Fox
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HClO3 + NaOH NaClO3 +H2O n(HClO3) = 0.1*0.04 = 4*10-3(mol) When added 10ml NaOH n(NaOH) used = 0.1 * 0.01 = 1*10-3(mol) n(HClO3) left = 3*10-3(mol) [HClO3] = 3*10-3/ 0.05 = 0.06 (M) [ClO3-] = 1*10-3/ 0.05 = 0.02(M) pH=pKa + log[ClO3-]/[HClO3] = 7.42 - 0.477 = 6.943 When added 20ml NaOH n(NaOH) used = 0.1 * 0.02 = 2*10-3(mol) n(HClO3) left = 2*10-3(mol) [HClO3] = 2*10-3/ 0.06 = 0.033 (M) [ClO3-] = 2*10-3/ 0.06 = 0.033(M) pH=pKa + log[ClO3-]/[HClO3] = 7.42 - 0 = 7.42 When added 30ml NaOH n(NaOH) used = 0.1 * 0.03 = 3*10-3(mol) n(HClO3) left = 1*10-3(mol) [HClO3] = 1*10-3/ 0.07 = 0.0143 (M) [ClO3-] = 3*10-3/ 0.07 = 0.0429(M) pH=pKa + log[ClO3-]/[HClO3] = 7.42 + 0.477 = 7.897 When added 40ml NaOH n(NaOH) used = 0.1 * 0.04 = 4*10-3(mol) n(HClO3) left = 0(mol) [ClO3-] = 4*10-3/ 0.08 = 0.05(M) Kb = Kw/Ka = 2.63 *10-7 [OH-] = Kb[ClO3-] =2.63*10-7*0.05 =1.14*10-4pOH = 3.94 pH = 10.06 Solution HClO3 + NaOH NaClO3 +H2O n(HClO3) = 0.1*0.04 = 4*10-3(mol) When added 10ml NaOH n(NaOH) used = 0.1 * 0.01 = 1*10-3(mol) n(HClO3) left = 3*10-3(mol) [HClO3] = 3*10-3/ 0.05 = 0.06 (M) [ClO3-] = 1*10-3/ 0.05 = 0.02(M) pH=pKa + log[ClO3-]/[HClO3] = 7.42 - 0.477 = 6.943 When added 20ml NaOH n(NaOH) used = 0.1 * 0.02 = 2*10-3(mol) n(HClO3) left = 2*10-3(mol) [HClO3] = 2*10-3/ 0.06 = 0.033 (M) [ClO3-] = 2*10-3/ 0.06 = 0.033(M) pH=pKa + log[ClO3-]/[HClO3] = 7.42 - 0 = 7.42 When added 30ml NaOH n(NaOH) used = 0.1 * 0.03 = 3*10-3(mol) n(HClO3) left = 1*10-3(mol) [HClO3] = 1*10-3/ 0.07 = 0.0143 (M) [ClO3-] = 3*10-3/ 0.07 = 0.0429(M) pH=pKa + log[ClO3-]/[HClO3] = 7.42 + 0.477 = 7.897 When added 40ml NaOH n(NaOH) used = 0.1 * 0.04 = 4*10-3(mol) n(HClO3) left = 0(mol) [ClO3-] = 4*10-3/ 0.08 = 0.05(M) Kb = Kw/Ka = 2.63 *10-7 [OH-] = Kb[ClO3-] =2.63*10-7*0.05 =1.14*10-4pOH = 3.94 pH = 10.06.
HClO3 + NaOH NaClO3 +H2O n(HClO3) = 0.10.04 = 4.pdf
HClO3 + NaOH NaClO3 +H2O n(HClO3) = 0.10.04 = 4.pdf
anjanaarts2014
I have already answered it. Check. Solution I have already answered it. Check..
I have already answered it. Check. .pdf
I have already answered it. Check. .pdf
anjanaarts2014
Iodide is a strong nucleophile but a weak base, so SN2 is the preferred reaction. Only the bromine on C1 is eligible to undergo SN2, so that one will be replaced by iodide. The product formed is 3-bromo 1-iodo 3 methylpentane Solution Iodide is a strong nucleophile but a weak base, so SN2 is the preferred reaction. Only the bromine on C1 is eligible to undergo SN2, so that one will be replaced by iodide. The product formed is 3-bromo 1-iodo 3 methylpentane.
Iodide is a strong nucleophile but a weak base, s.pdf
Iodide is a strong nucleophile but a weak base, s.pdf
anjanaarts2014
NaCl : It\'s the coulombic interaction between Na+ and Cl- due to which crystal exists. It\'s not ionic intereaction. It\'s ionic interaction. (very strong). NaCl dissolves in water. It\'s not due to hydrogen bonding. It\'s due to strong ion-dipole interactions. Initially, the positive and negative ion are only attracted to each other. The water molecules are hydrogen bonded to each other. If the crystal is to dissolve, these bonds must be broken. Negative chloride ions on the surface are attracted by neighboring positive sodium ions and by the partially positive hydrogen atom in the polar water molecule (See the graphic on the left). Similarly, the positive sodium ions are attracted by both chloride ions and the partially negative oxygen atom in the polar water molecule. A \"tug-of-war\" occurs for the positive and negative ions between the other ions in the crystal and the water molecules. Several water molecules are attracted to each of the ions. Whether the crystal dissolves is determined by which attractive force is stronger. If the internal ionic forces in the crystal are the strongest, the crystal does not dissolve. This is the situation in reactions where precipitates form. If the attractions for the ions by the polar water molecules are the strongest, the crystal will dissolve. This is the situation in sodium chloride. Once the ions are released from the crystals, the ions are completely surrounded by water molecules. Note that the proper atom in the water molecule must \"point\" toward the correct ion. The charge principle and the partial charges in the polar molecule determine the correct orientation. Partially negative oxygen atoms in the water molecule interact with the positive sodium ion. Partially positive hydrogen atoms in the water molecule interact with the negative chloride ion. Sugar : The interaction between sugar molecule and water is greater than between sugar molecules. So When sugar is put into water, it likes to form hydrogen bonds with water. So it breaks glycosidic linkages and forms hydrogen bonds with water. Initially it separates into polymer sugar chains, but later it dissociates into individual molecules. Solution NaCl : It\'s the coulombic interaction between Na+ and Cl- due to which crystal exists. It\'s not ionic intereaction. It\'s ionic interaction. (very strong). NaCl dissolves in water. It\'s not due to hydrogen bonding. It\'s due to strong ion-dipole interactions. Initially, the positive and negative ion are only attracted to each other. The water molecules are hydrogen bonded to each other. If the crystal is to dissolve, these bonds must be broken. Negative chloride ions on the surface are attracted by neighboring positive sodium ions and by the partially positive hydrogen atom in the polar water molecule (See the graphic on the left). Similarly, the positive sodium ions are attracted by both chloride ions and the partially negative oxygen atom in the polar water molecule. A \"tug-of-war\" occurs for.
NaCl Its the coulombic interaction between Na.pdf
NaCl Its the coulombic interaction between Na.pdf
anjanaarts2014
Here: Jupiter is the fifth planet from the Sun and the largest planet within the Solar System.It is a gas giant with a mass slightly less than one-thousandth of the Sun but is two and a half times the mass of all the other planets in our Solar System combined. Jupiter is classified as a gas giant along with Saturn, Uranus and Neptune. Together, these four planets are sometimes referred to as the Jovian or outer planets. The planet was known by astronomers of ancient times and was associated with the mythology and religious beliefs of many cultures. The Romans named the planet after the Roman god Jupiter.[14] When viewed from Earth, Jupiter can reach an apparent magnitude of -2.94, making it on average the third-brightest object in the night sky after the Moon and Venus. (Mars can briefly match Jupiter\'s brightness at certain points in its orbit.) Jupiter is primarily composed of hydrogen with a quarter of its mass being helium; it may also have a rocky core of heavier elements. Because of its rapid rotation, Jupiter\'s shape is that of an oblate spheroid (it possesses a slight but noticeable bulge around the equator). The outer atmosphere is visibly segregated into several bands at different latitudes, resulting in turbulence and storms along their interacting boundaries. A prominent result is the Great Red Spot, a giant storm that is known to have existed since at least the 17th century when it was first seen by telescope. Surrounding the planet is a faint planetary ring system and a powerful magnetosphere. There are also at least 63 moons, including the four large moons called the Galilean moons that were first discovered by Galileo Galilei in 1610. Ganymede, the largest of these moons, has a diameter greater than that of the planet Mercury. Jupiter has been explored on several occasions by robotic spacecraft, most notably during the early Pioneer and Voyager flyby missions and later by the Galileo orbiter. The most recent probe to visit Jupiter was the Pluto-bound New Horizons spacecraft in late February 2007. The probe used the gravity from Jupiter to increase its speed. Future targets for exploration in the Jovian system include the possible ice-covered liquid ocean on the moon Europa. Jupiter\'s structure: Jupiter is composed primarily of gaseous and liquid matter. It is the largest of four gas giants as well as the largest planet in the solar system with a diameter of 142,984 km at its equator. The density of Jupiter, 1.326 g/cm³, is the second highest of the gas giant planets. However, the density is lower than any of the four terrestrial planets. Solution Here: Jupiter is the fifth planet from the Sun and the largest planet within the Solar System.It is a gas giant with a mass slightly less than one-thousandth of the Sun but is two and a half times the mass of all the other planets in our Solar System combined. Jupiter is classified as a gas giant along with Saturn, Uranus and Neptune. Together, these four planets are sometimes refe.
Here Jupiter is the fifth planet from the Sun a.pdf
Here Jupiter is the fifth planet from the Sun a.pdf
anjanaarts2014
moles of H+ = 42.0 x 0.120 millimoles = 5.04 millimoles final volume = 1.55 L [H+] = (5.04 x 10^-3)/1.55 = 3.25 x 10^-3M pH = -log[H+] = 2.49 Solution moles of H+ = 42.0 x 0.120 millimoles = 5.04 millimoles final volume = 1.55 L [H+] = (5.04 x 10^-3)/1.55 = 3.25 x 10^-3M pH = -log[H+] = 2.49.
moles of H+ = 42.0 x 0.120 millimoles = 5.04 mill.pdf
moles of H+ = 42.0 x 0.120 millimoles = 5.04 mill.pdf
anjanaarts2014
Lithium is group 1 metal, while nitrogen is group 15 nonmetallic element. Thus, the bond that they form is ionic bond. metal + nonmetal = ionic compound Solution Lithium is group 1 metal, while nitrogen is group 15 nonmetallic element. Thus, the bond that they form is ionic bond. metal + nonmetal = ionic compound.
Lithium is group 1 metal, while nitrogen is group.pdf
Lithium is group 1 metal, while nitrogen is group.pdf
anjanaarts2014
there is some special character in ur question like ? pls re post the question properly thanks Solution there is some special character in ur question like ? pls re post the question properly thanks.
there is some special character in ur question li.pdf
there is some special character in ur question li.pdf
anjanaarts2014
a. Times Interest Earned Ratio = EBIT / interest Expense For 2014: $8,799 / $600 = 14.7 times For 2015: $9,454 / $838 = 11.3 times For 2016: $10,338 / $1,058 = 9.8 times The TImes Interest Earned (TIE) Ratio measures the operating income level that can be used for interest payments in the future, i.e. how many times the Company can make the interest payments. It is a solvency ratio since it measures the Company\'s ability to service it\'s debt interest obligations. Hence, the higher the ratio, the more favorable is the company\'s liquidity position. In the given question, we can observe that the ratio has consistently decreased year on year. This implies that the Company\'s debt has increased while the increase in operating income has not been at the same pace. The Company can afford to service interest fewer times each year from 2014 to 2016. This implies that the Company has become comparatively more risky than before from the Creditor\'s standpoint. b. Current Ratio = Current Assets / Current Liabilities For 2014: $25,983 / $19,027 = 1.4 For 2015: $29,158 / $23,169 = 1.3 For 2016: $31,042 / $26,250 = 1.2 The Current Ratio is an indicator of how efficiently the Company manages it\'s operating cycle. A higher Current Ratio implies that the COmpany is in a better position to discharge all it\'s short term obligations without affecting or jeopardising it\'s working capital requirements. On the flipside, a ratio that is too high indicates that the Company\'s current assets are lying idle or that it is not efficiently using short term funds. In the Given case, the Company\'s current ratio shows a steady decline. However, the lowest is 1.2 times, which implies that the Company is still sufficiently liquid. This is becase even at the lowest, the Company\'s Current Assets can pay off the Current Liabilities 1.2 times without affecting working capital cycle for regular business needs. Hence, as per the options given, we can say that CVS’s current ratio has decreased over the past three years, however, it is greater than 1 indicating CVS is liquid. c. Total Liabilities to Equity Ratio = Total Liabilities / Total Equity For 2014: $36,224 / $37,963 = 0.95 For 2015: $55,234 / $37,203 = 1.5 For 2016: $57,628 / $36,834 = 1.6 Total Liabilities include all the claims, long term and short term, against the Company\'s Assets. Total Liabilities to Equity Ratio indicates how much of the Company\'s Assets are funded by creditors versus how much is funded by the equity holders. Broadly, a 1:1 ratio is considered acceptable for most industries. Creditors would prefer a lower ratio as this implies their money is at less risk. However, equity holders would like to benefit from the risk taken by them due to the Company using debt, and would prefer a higher ratio. In the given case, this ratio is increasing year on year. This implies that the COmpany is taking on more debt financing. This implies that the Company has more risk overall, although it benefits the equity .
a.Times Interest Earned Ratio = EBIT interest ExpenseFor 2014.pdf
a.Times Interest Earned Ratio = EBIT interest ExpenseFor 2014.pdf
anjanaarts2014
45 yrs. Solution 45 yrs..
45 yrs.Solution45 yrs..pdf
45 yrs.Solution45 yrs..pdf
anjanaarts2014
the bacteria isolated from the dead mouse is smooth (S) S. pneumoniae bacteria. and the DNA was radio-labeled. the Griffith experiment proves that bacteria are capable of transforming genetic information. in the above experiment Griffith grown the smooth (S) S. pneumoniae bacteria with radio-labelled phosphate (32P), radiolabelled 32P, which is incorporated into the DNA and he had also separately grown rough (R) bacteria in regular (not radio-labelled) phosphate. while injecting the heat killed S bacteria to live R bacteria, the R bacteria had been transformed into lethal S bacteria. the DNA had survived in Heating process of S strain, and this DNA transform into the R strain and synthesizes the polysaccharide capsule that protects form host immune response in the R bacteria and transform the R strain into the S strain which results Mouse dies. in the first step we are labeled the DNA of S Strain with 32P, so the final DNA which is isolating from the died mouse contains both radiolabeled DNA and non radiolabeled DNA in different proportions. Solution the bacteria isolated from the dead mouse is smooth (S) S. pneumoniae bacteria. and the DNA was radio-labeled. the Griffith experiment proves that bacteria are capable of transforming genetic information. in the above experiment Griffith grown the smooth (S) S. pneumoniae bacteria with radio-labelled phosphate (32P), radiolabelled 32P, which is incorporated into the DNA and he had also separately grown rough (R) bacteria in regular (not radio-labelled) phosphate. while injecting the heat killed S bacteria to live R bacteria, the R bacteria had been transformed into lethal S bacteria. the DNA had survived in Heating process of S strain, and this DNA transform into the R strain and synthesizes the polysaccharide capsule that protects form host immune response in the R bacteria and transform the R strain into the S strain which results Mouse dies. in the first step we are labeled the DNA of S Strain with 32P, so the final DNA which is isolating from the died mouse contains both radiolabeled DNA and non radiolabeled DNA in different proportions..
the bacteria isolated from the dead mouse is smooth (S) S. pneumonia.pdf
the bacteria isolated from the dead mouse is smooth (S) S. pneumonia.pdf
anjanaarts2014
ANSWER: A for example: fluoride, oxide.... etc. If you need more help let me know and I can explain further Solution ANSWER: A for example: fluoride, oxide.... etc. If you need more help let me know and I can explain further.
ANSWER A for example fluoride, oxide.... etc. .pdf
ANSWER A for example fluoride, oxide.... etc. .pdf
anjanaarts2014
Java Program: import java.io.*; import java.util.*; public class LongestStreak { public static int[] longestStreak(boolean[] values) { int max=0,index=0,mIndex=0; for(int i=0;i Solution Java Program: import java.io.*; import java.util.*; public class LongestStreak { public static int[] longestStreak(boolean[] values) { int max=0,index=0,mIndex=0; for(int i=0;i.
Java Program import java.io.; import java.util.;public cl.pdf
Java Program import java.io.; import java.util.;public cl.pdf
anjanaarts2014
H2SO4 => 2H+ + SO42- [H+] = 2[H2SO4] = 0.0092 M pH = -log[H+] = 2.04 Solution H2SO4 => 2H+ + SO42- [H+] = 2[H2SO4] = 0.0092 M pH = -log[H+] = 2.04.
H2SO4 = 2H+ + SO42-[H+] = 2[H2SO4] = 0.0092 MpH = -log[H+] = 2..pdf
H2SO4 = 2H+ + SO42-[H+] = 2[H2SO4] = 0.0092 MpH = -log[H+] = 2..pdf
anjanaarts2014
here reaction is occurring when Sn and I2 are refluxed together reduction and oxidation is takes place(called redox ) he would ask the oxidation potential &its reduction potential is asking . Sn + 2I2------>SnI4 here Sn oxidised to Sn2+ ,iodine reduced ,means reduction potential of tin(Sn) is E0= -0.136, I2 reduction potential is =+0.53 ,its indicates reduction potential of iodine is more so it can oxidise the Sn & it can be reduced. Zn(s) + I2(s) -----> Zn2+ (aq) + 2I- (aq) here Zn oxidised to Zn2+ ,iodine reduced ,means reduction potential of tin(Sn) is E0=-0.763 I2 reduction potential is =+0.53 ,its indicates reduction potential of iodine is more so it can oxidise the Zn & it can be reduced Sn(s)+2HCl (aq) -----> Sn2+ (aq) +2Cl- (aq) + H2 (g) here Sn oxidised to Sn2+ ,chlorine reduced ,means reduction potential of tin(Sn) is E0=- 0.136here Sn oxidised to Sn2+ ,chlorine reduced ,means reduction potential of tin(Sn) is E0=- 0.136, cl- reduction potential is =+1.36 ,its indicates reduction potential of chlorine is more so it can oxidise the Sn & it(Cl) can be reduced. Sn2+ (aq) + 2I- (aq) -----> SnI2 (s) I2 reduction potential is =+0.53 ,its indicates reduction potential of iodine is less than the Sn2+ so it can reduce the Sn2+ & it(I-) can be oxidised. Solution here reaction is occurring when Sn and I2 are refluxed together reduction and oxidation is takes place(called redox ) he would ask the oxidation potential &its reduction potential is asking . Sn + 2I2------>SnI4 here Sn oxidised to Sn2+ ,iodine reduced ,means reduction potential of tin(Sn) is E0= -0.136, I2 reduction potential is =+0.53 ,its indicates reduction potential of iodine is more so it can oxidise the Sn & it can be reduced. Zn(s) + I2(s) -----> Zn2+ (aq) + 2I- (aq) here Zn oxidised to Zn2+ ,iodine reduced ,means reduction potential of tin(Sn) is E0=-0.763 I2 reduction potential is =+0.53 ,its indicates reduction potential of iodine is more so it can oxidise the Zn & it can be reduced Sn(s)+2HCl (aq) -----> Sn2+ (aq) +2Cl- (aq) + H2 (g) here Sn oxidised to Sn2+ ,chlorine reduced ,means reduction potential of tin(Sn) is E0=- 0.136here Sn oxidised to Sn2+ ,chlorine reduced ,means reduction potential of tin(Sn) is E0=- 0.136, cl- reduction potential is =+1.36 ,its indicates reduction potential of chlorine is more so it can oxidise the Sn & it(Cl) can be reduced. Sn2+ (aq) + 2I- (aq) -----> SnI2 (s) I2 reduction potential is =+0.53 ,its indicates reduction potential of iodine is less than the Sn2+ so it can reduce the Sn2+ & it(I-) can be oxidised..
here reaction is occurring when Sn and I2 are refluxed together redu.pdf
here reaction is occurring when Sn and I2 are refluxed together redu.pdf
anjanaarts2014
An undirected graph is acyclic (i.e., a forest) iff a DFS yields no back edges. Since back edges are those edges (u, v) connecting a vertex u to an ancestor v in a depth-first tree, so no back edges means there are only tree edges, so there is no cycle. So we can simply fun DFS. If find a back edge, there is a cycle. The complexity is O(V ) instead of O(E + V ). Since if there is a back edge, it must be found before seeing |V | distinct edges. This is because in a acyclic (undirected ) forest, |E| |V | + 1 Solution An undirected graph is acyclic (i.e., a forest) iff a DFS yields no back edges. Since back edges are those edges (u, v) connecting a vertex u to an ancestor v in a depth-first tree, so no back edges means there are only tree edges, so there is no cycle. So we can simply fun DFS. If find a back edge, there is a cycle. The complexity is O(V ) instead of O(E + V ). Since if there is a back edge, it must be found before seeing |V | distinct edges. This is because in a acyclic (undirected ) forest, |E| |V | + 1.
An undirected graph is acyclic (i.e., a forest) iff a DFS yields no .pdf
An undirected graph is acyclic (i.e., a forest) iff a DFS yields no .pdf
anjanaarts2014
A) [Sn2+]^3 / [Fe3+]^2 (don\'t include solids) B) [OH-]^4 [Cu2+]^2 / [O2] (no solids or liquids) C) [Br-]^2 / [I-]^2 Solution A) [Sn2+]^3 / [Fe3+]^2 (don\'t include solids) B) [OH-]^4 [Cu2+]^2 / [O2] (no solids or liquids) C) [Br-]^2 / [I-]^2.
A) [Sn2+]^3 [Fe3+]^2 (dont include solids) B).pdf
A) [Sn2+]^3 [Fe3+]^2 (dont include solids) B).pdf
anjanaarts2014
Given LOSS Sells a variety of equpiment including the executive office chair Particulars Current No.of Units expected to sell the executive office chair 6000 Selling Price per executive office chair in $ 250 Current manufacturing costs in $ per unit 175 Current Fixed cost 3,60,000 Proposed Changes No.of Units expected to sell the executive office chair 6000 Selling Price per executive office chair in $ 250 Alternative manufacturing costs in $ per unit 75 Alternative Fixed cost 9,45,000 Sl Particulars Current Proposed A Sales 15,00,000 15,00,000 B Variable cost 10,50,000 4,50,000 C Contribution ( A - B ) 4,50,000 10,50,000 D Fixed cost 3,60,000 9,45,000 E Profit ( C - D ) 90,000 1,05,000 F price Volume Ratio % ( C/A * 100 ) 30% 70% G Break even Point Volume ( D/F) 12,00,000 13,50,000 H Contribution Per Unit ( C/6000 ) 75 175 I Break even Point Units ( D/H) 4800 5400 J Margin of safety sales ( A -G ) 3,00,000 1,50,000 K Margin of safety in Units ( J / selling Price pu ) 1200 600 L Margin of safety in Percentage ( J / A )*100 20% 10% Indiffernce point between two alternatives Indifference cost = point where there is no difference in cost between two alternatives Indifference point Units Assumptions Q = Quantity / units Cpu 1 = Contribution per unit for current alternative Cpu 2= Contribution per unit for proposed alternative FC 1 = fixed cost for current alternative FC 2 = fixed cost for Proposed alternative alternative Indifference point equation Cpu1 * Q - FC1 = Cpu2 * Q - FC2 = 75 * Q - 360,000 = 175 * Q -945,000 = 945,000 - 360,000 = Q ( 175 - 75 ) = 100 * Q = 585,000 = Q = 585,000/100 = Q = 5850 Units Therefore Indifference point is 5850 Units Given LOSS Sells a variety of equpiment including the executive office chair Particulars Current No.of Units expected to sell the executive office chair 6000 Selling Price per executive office chair in $ 250 Current manufacturing costs in $ per unit 175 Current Fixed cost 3,60,000 Proposed Changes No.of Units expected to sell the executive office chair 6000 Selling Price per executive office chair in $ 250 Alternative manufacturing costs in $ per unit 75 Alternative Fixed cost 9,45,000 Sl Particulars Current Proposed A Sales 15,00,000 15,00,000 B Variable cost 10,50,000 4,50,000 C Contribution ( A - B ) 4,50,000 10,50,000 D Fixed cost 3,60,000 9,45,000 E Profit ( C - D ) 90,000 1,05,000 F price Volume Ratio % ( C/A * 100 ) 30% 70% G Break even Point Volume ( D/F) 12,00,000 13,50,000 H Contribution Per Unit ( C/6000 ) 75 175 I Break even Point Units ( D/H) 4800 5400 J Margin of safety sales ( A -G ) 3,00,000 1,50,000 K Margin of safety in Units ( J / selling Price pu ) 1200 600 L Margin of safety in Percentage ( J / A )*100 20% 10% Indiffernce point between two alternatives Indifference cost = point where there is no difference in cost between two alternatives Indifference point Units Assumptions Q = Quantity / units Cpu 1 = Contribution per unit for current alternative Cpu 2= Contribution per unit for.
GivenLOSS Sells a variety of equpiment including the executive off.pdf
GivenLOSS Sells a variety of equpiment including the executive off.pdf
anjanaarts2014
Answer for question 1 DuPont analysis explains what is affecting company\'s ROE. DuPont analysis can be explained with formula as below ROE= Profit Margin× Total asset turnover× Leverage factor Profit Margin= Net Income/ Total Revenue Total Asset turnover=Revenue/ Total assets Leverage factor= Total assets/ Shareholder equity DuPont analysis points out operating efficiency in form of profit margin, Asset efficiency in form of asset turnover and leverage impact with the help of leverage factor as how much of the equity is used to finance total assets Solution Answer for question 1 DuPont analysis explains what is affecting company\'s ROE. DuPont analysis can be explained with formula as below ROE= Profit Margin× Total asset turnover× Leverage factor Profit Margin= Net Income/ Total Revenue Total Asset turnover=Revenue/ Total assets Leverage factor= Total assets/ Shareholder equity DuPont analysis points out operating efficiency in form of profit margin, Asset efficiency in form of asset turnover and leverage impact with the help of leverage factor as how much of the equity is used to finance total assets.
Answer for question 1DuPont analysis explains what is affecting co.pdf
Answer for question 1DuPont analysis explains what is affecting co.pdf
anjanaarts2014
3360 Solution 3360.
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anjanaarts2014
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APM webinar hosted by the Scotland Network on 14 May 2024. Speakers: Chris Drysdale and Peter Huggett An interactive session discussing how Project Managers can identify mental health symptoms, provide tools to help themselves and others, plus also increase the capabilities of the Project Management function. This webinar was held on 14 May 2024. The covid-19 pandemic led to concerns about a worsening of mental health & wellbeing across the world and increased awareness in both society and the workplace. This webinar looks to advise the benefits of having a Mental Health First Aid function in the workplace whilst also providing tools and techniques that can be readily used and applied to yourself and colleagues. Additionally, there are wider benefits to Project Management which will be proposed and discussed.
Including Mental Health Support in Project Delivery, 14 May.pdf
Including Mental Health Support in Project Delivery, 14 May.pdf
Association for Project Management
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