the bacteria isolated from the dead mouse is smooth (S) S. pneumonia.pdf
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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..
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What would the predicted result be if Griffith injected mice with a .pdf
What would the predicted result be if Griffith injected mice with a mixture of heat-killed R
bacteria and living S bacteria?
a. All mice would live, but only if DNAse was added to the mixture prior to injection
b. All mice would die.
c. All mice would live.
d. All mice would die, but only if DNAse was added to the mixture prior to injection.
Solution
Griffith’s Experiment-A Quick Review: Fredrick Griffith, an English physician, first reported
transformation of one strain of Streptococcus pneumoniae into other strain in 1928. The bacteria
cause pneumonia. A strain of this bacterium produces capsule and gives a smooth, shiny colony
on culture media, thus called S-strain (smooth strain). The S-strain exhibits pathogenicity in
laboratory mice. Another strain without capsule produces rough and coarse colony on culture
medium, thus called R-strain (rough strain). The R-strain is avirulent (non-pathogenic).
Mice injected with
Outcome
1
Live S-strain (virulent)
Pneumonia mice dead
2
Live R-strain (avirulent)
No pneumonia mice alive
3
Heat killed S-strain + live R-strain
Pneumonia mice dead
When injected with live S-strain bacteria, mice developed pneumonia and died. When injected
with live R-strain bacteria, mice did not develop pneumonia and survived. When heat killed S-
strain injected to mice, they remain unaffected. However, when heat killed S-strain bacteria
mixed with live R-strain injected to mice, they developed pneumonia and died. He was also able
to isolate and culture S-strain bacteria from affected mice. Since heat killed bacteria could have
no way been revived, he concluded that some molecules from heat killed S-strain had
transformed the R-strain into S-train.
Now,
The Current Perspective: Presence of living S-strain causes pneumonia development in mice
killing them at last. It is noteworthy that transformation of R-strain into S-strain occurs only
when there is alive R-strain and fragments of DNA from heat-killed S-strain.
Option A: Incorrect. DNAse is a large protein incapable of entering into bacterial cytoplasm
through bacterial membrane of living S-strain. So, this treatment further cleaves DNA fragments
of heat killed R-strain BUT leaves the S-strain unaffected. So, injecting the mixture to mice kills
them as live S-strain causes pneumonia.
Option B: Correct. Presence of living S-strain causes pneumonia development in mice killing
them at last.
Option C: Incorrect. All mice die because S-strain cause pneumonia to them.
Option D: Incorrect. All mice would die, BUT NOT because of addition of DNAse to the
mixture. As mentioned in option A explanation, DNAse does not penetrate into live S-strain,
thus leave them unaffected. Therefore, mice die irrespective of DNAse treatment.
Mice injected with
Outcome
1
Live S-strain (virulent)
Pneumonia mice dead
2
Live R-strain (avirulent)
No pneumonia mice alive
3
Heat killed S-strain + live R-strain
Pneumonia mice dead.
Genetic material
Genetic material
Anant Mohan Sharma
All cell have the capability to give rise to the cell and
the encoded information in living cell is passed from
one generation to another. The information encoded
material is the genetic material or hereditary material
of the cell.
The genetic material is long sequence of nucleic
acids that contain the genetic instruction. Nucleic
acid are macromolecules in the form of DNA or
RNA.
Experimental evidences
Griffith’s experiment
Avery, MacLeod &McCarty experiment
Hershey & Chase experiment
RNA as genetic material
DNA structure
Z- DNA
V. Sasisekharan RL model
Types of RNA
A robust whistleblowing regime is now an integral part of governance.pdf
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.
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 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.
Recommended
What would the predicted result be if Griffith injected mice with a .pdf
What would the predicted result be if Griffith injected mice with a mixture of heat-killed R
bacteria and living S bacteria?
a. All mice would live, but only if DNAse was added to the mixture prior to injection
b. All mice would die.
c. All mice would live.
d. All mice would die, but only if DNAse was added to the mixture prior to injection.
Solution
Griffith’s Experiment-A Quick Review: Fredrick Griffith, an English physician, first reported
transformation of one strain of Streptococcus pneumoniae into other strain in 1928. The bacteria
cause pneumonia. A strain of this bacterium produces capsule and gives a smooth, shiny colony
on culture media, thus called S-strain (smooth strain). The S-strain exhibits pathogenicity in
laboratory mice. Another strain without capsule produces rough and coarse colony on culture
medium, thus called R-strain (rough strain). The R-strain is avirulent (non-pathogenic).
Mice injected with
Outcome
1
Live S-strain (virulent)
Pneumonia mice dead
2
Live R-strain (avirulent)
No pneumonia mice alive
3
Heat killed S-strain + live R-strain
Pneumonia mice dead
When injected with live S-strain bacteria, mice developed pneumonia and died. When injected
with live R-strain bacteria, mice did not develop pneumonia and survived. When heat killed S-
strain injected to mice, they remain unaffected. However, when heat killed S-strain bacteria
mixed with live R-strain injected to mice, they developed pneumonia and died. He was also able
to isolate and culture S-strain bacteria from affected mice. Since heat killed bacteria could have
no way been revived, he concluded that some molecules from heat killed S-strain had
transformed the R-strain into S-train.
Now,
The Current Perspective: Presence of living S-strain causes pneumonia development in mice
killing them at last. It is noteworthy that transformation of R-strain into S-strain occurs only
when there is alive R-strain and fragments of DNA from heat-killed S-strain.
Option A: Incorrect. DNAse is a large protein incapable of entering into bacterial cytoplasm
through bacterial membrane of living S-strain. So, this treatment further cleaves DNA fragments
of heat killed R-strain BUT leaves the S-strain unaffected. So, injecting the mixture to mice kills
them as live S-strain causes pneumonia.
Option B: Correct. Presence of living S-strain causes pneumonia development in mice killing
them at last.
Option C: Incorrect. All mice die because S-strain cause pneumonia to them.
Option D: Incorrect. All mice would die, BUT NOT because of addition of DNAse to the
mixture. As mentioned in option A explanation, DNAse does not penetrate into live S-strain,
thus leave them unaffected. Therefore, mice die irrespective of DNAse treatment.
Mice injected with
Outcome
1
Live S-strain (virulent)
Pneumonia mice dead
2
Live R-strain (avirulent)
No pneumonia mice alive
3
Heat killed S-strain + live R-strain
Pneumonia mice dead.
Genetic material
Genetic material
Anant Mohan Sharma
All cell have the capability to give rise to the cell and
the encoded information in living cell is passed from
one generation to another. The information encoded
material is the genetic material or hereditary material
of the cell.
The genetic material is long sequence of nucleic
acids that contain the genetic instruction. Nucleic
acid are macromolecules in the form of DNA or
RNA.
Experimental evidences
Griffith’s experiment
Avery, MacLeod &McCarty experiment
Hershey & Chase experiment
RNA as genetic material
DNA structure
Z- DNA
V. Sasisekharan RL model
Types of RNA
A robust whistleblowing regime is now an integral part of governance.pdf
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.
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 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.
a)dydx + 2yx = xI.F. = x^2solution iy .pdf
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.
23.C. 1 in 424. B. Homozygous25. B. 5026.All females will be .pdf
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..
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 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. Gene expression.B. Transcription factor.C. Gene expression..pdf
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..
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 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).
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 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.
HClO3 + NaOH NaClO3 +H2O n(HClO3) = 0.10.04 = 4.pdf
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.
I have already answered it. Check. .pdf
I have already answered it. Check.
Solution
I have already answered it. Check..
Iodide is a strong nucleophile but a weak base, s.pdf
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.
NaCl Its the coulombic interaction between Na.pdf
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.
Here Jupiter is the fifth planet from the Sun a.pdf
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.
moles of H+ = 42.0 x 0.120 millimoles = 5.04 mill.pdf
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.
Lithium is group 1 metal, while nitrogen is group.pdf
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.
there is some special character in ur question li.pdf
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.
a.Times Interest Earned Ratio = EBIT interest ExpenseFor 2014.pdf
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 .
ANSWER A for example fluoride, oxide.... etc. .pdf
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.
Java Program import java.io.; import java.util.;public cl.pdf
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.
H2SO4 = 2H+ + SO42-[H+] = 2[H2SO4] = 0.0092 MpH = -log[H+] = 2..pdf
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.
"Protectable subject matters, Protection in biotechnology, Protection of othe...
Protectable subject matters, Protection in biotechnology, Protection of other biological materials, Ownership and period of protection
Honest Reviews of Tim Han LMA Course Program.pptx
Personal development courses are widely available today, with each one promising life-changing outcomes. Tim Han’s Life Mastery Achievers (LMA) Course has drawn a lot of interest. In addition to offering my frank assessment of Success Insider’s LMA Course, this piece examines the course’s effects via a variety of Tim Han LMA course reviews and Success Insider comments.
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a)dydx + 2yx = xI.F. = x^2solution iy .pdf
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.
23.C. 1 in 424. B. Homozygous25. B. 5026.All females will be .pdf
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..
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 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. Gene expression.B. Transcription factor.C. Gene expression..pdf
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..
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 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).
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 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.
HClO3 + NaOH NaClO3 +H2O n(HClO3) = 0.10.04 = 4.pdf
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.
I have already answered it. Check. .pdf
I have already answered it. Check.
Solution
I have already answered it. Check..
Iodide is a strong nucleophile but a weak base, s.pdf
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.
NaCl Its the coulombic interaction between Na.pdf
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.
Here Jupiter is the fifth planet from the Sun a.pdf
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.
moles of H+ = 42.0 x 0.120 millimoles = 5.04 mill.pdf
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.
Lithium is group 1 metal, while nitrogen is group.pdf
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.
there is some special character in ur question li.pdf
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.
a.Times Interest Earned Ratio = EBIT interest ExpenseFor 2014.pdf
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 .
ANSWER A for example fluoride, oxide.... etc. .pdf
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.
Java Program import java.io.; import java.util.;public cl.pdf
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.
H2SO4 = 2H+ + SO42-[H+] = 2[H2SO4] = 0.0092 MpH = -log[H+] = 2..pdf
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.
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- 1. 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.