moles of CO2 = moles of NaHCO3 = 284 =0.0238 .pdf
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moles of CO2 = moles of NaHCO3 = 2/84 =0.0238 Solution moles of CO2 = moles of NaHCO3 = 2/84 =0.0238.
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4r = 2 L2r = L2thats none of the above, since you didnt wrote t.pdf
4r = 2 L
2r = L/2
thats none of the above, since you didnt wrote the options properly,
answer is L/2
Solution
4r = 2 L
2r = L/2
thats none of the above, since you didnt wrote the options properly,
answer is L/2.
41.Ans. Ans,Sarcoplasmic reticulumSarcoplasmic reticulum is a type.pdf
41.Ans. Ans,Sarcoplasmic reticulum
Sarcoplasmic reticulum is a type of Endoplasmic Reticulum which present in striated muscle,
does not involves in flagellar movement. the other options are microtubules have important role
in flagella movement. these are the main components of axoneme, which is a unit of falgellar
structure and Dyneins are motor proteins.
42.Ans. Other than the nucleus most organelles are small and thus in a typical cell only about
10% of the cells volume is occupied by membrane enclosed organelles and other 90% of the cell
volume is cytosol.
the other options are wrong. membrane bound organelles have outer layer plasma membrane, so
the endoplasmic reticulum not far exceeds the area of plasma membrane. not only the nucleus
surrounded by double layer, other organells for example Mitochondria also surrounded by
double layers. and not only the nucleus, mitochondria also have the DNA. and chloroplast also
have the DNA.
43.Ans. True
once loaded vesicles are carry the cargo to different location with in the body, and reaches the
site of secretion. once reached some of the vesicles containing material for constitutive -
Produced in constant amount, releases and immediately fused with Plasma membrane. where as
vesicles called secretory vesicles- these are the vesicles in which molecules destined for
secretion are stored, these vesicles are in regulated pathway means once reached to the site of
secretion these vesicles wait at he membrane until it receives a signal to secrete, once cell
receives the signal these are going secrete the content and material fuse with the plasma
membrane
44.Ans.False
Nuclear localization signal needed to the peptide until it crosses the membrane
45Ans.True
Chaperones are needed for the import of the proteins. these are prevent the premature folding of
the proteins. while import into the mitochondria, the Chaperones present in the inner surface of
the membrane of mitochondria able to bind the protruding ends of polypeptide and prevents them
to diffuse back into the cytosol through the pore and carry them further and into the matrix. while
moving the peptides further into the matrix at every stage the se peptides are bounded with the
Chaperones which prevent backward movement of the peptides. this kind of diffusion called as
biased diffusion and Chaperones acted as Brownian ratchet.
46.Ans.E) All of These happen.
the above all happens when haploid cell mates
Solution
41.Ans. Ans,Sarcoplasmic reticulum
Sarcoplasmic reticulum is a type of Endoplasmic Reticulum which present in striated muscle,
does not involves in flagellar movement. the other options are microtubules have important role
in flagella movement. these are the main components of axoneme, which is a unit of falgellar
structure and Dyneins are motor proteins.
42.Ans. Other than the nucleus most organelles are small and thus in a typical cell only about
10% of the cells volume is occupied by membrane enclosed organelles and other 90% of the cell
vol.
1.c. developers2.e. A and B3.c. Land and Built space4. Money a.pdf
1.c. developers
2.e. A and B
3.c. Land and Built space
4. Money and Stock
5.c. construction and land market
6. expect a reduction in property asset value
7.b. Demand
8. converts physical capital into financial capital
9.D
10.A
11.D
12.C
13.A
14.B
15.A
16. rent go up, prices of real estsate asset go up, new development increases
17.C
18.D
19.D
20. location and size
21.B
22.D
23.D
24.B
25.C
26.B
27.D
28.B
29.A
30.D
31.D
32.B
33.C
34.D
35.C
36.D
37.A
38.E
39.A
40.C
41.C
42.A
43.A
44.D
45.C
Solution
1.c. developers
2.e. A and B
3.c. Land and Built space
4. Money and Stock
5.c. construction and land market
6. expect a reduction in property asset value
7.b. Demand
8. converts physical capital into financial capital
9.D
10.A
11.D
12.C
13.A
14.B
15.A
16. rent go up, prices of real estsate asset go up, new development increases
17.C
18.D
19.D
20. location and size
21.B
22.D
23.D
24.B
25.C
26.B
27.D
28.B
29.A
30.D
31.D
32.B
33.C
34.D
35.C
36.D
37.A
38.E
39.A
40.C
41.C
42.A
43.A
44.D
45.C.
package patienttest;import java.util.Comparator; import java..pdf
package patienttest;
import java.util.Comparator;
import java.util.Date;
public class Patient implements Comparator{
String Name;
Date DOB;
Date Admission;
String Complaint;
int Priority;
public Patient(String name,Date d1,Date ad,String complaint,int p)
{
Name=name;
DOB=d1;
Admission=ad;
Complaint=complaint;
Priority=p;
}
public void setName(String name)
{
Name=name;
}
public void setDOB(Date d1)
{
DOB=d1;
}
public void setAdmission(Date d2)
{
Admission=d2;
}
public void setComplaint(String complaint)
{
Complaint=complaint;
}
public void setPriority(int p)
{
Priority=p;
}
public String setName()
{
return Name;
}
public Date setDOB()
{
return DOB;
}
public Date setAdmission()
{
return Admission;
}
public String setComplaint()
{
return Complaint;
}
public int setPriority()
{
return Priority;
}
@Override
public String toString()
{
String msg=\"\";
msg+=\"Name: \"+Name;
msg+=\"DOB: \"+DOB;
msg+=\"Admission: \"+Admission;
msg+=\"Complaint: \"+Complaint;
msg+=\"Priority: \"+Priority;
return msg;
}
@Override
public int compare(Patient p1, Patient p2) {
int value=1;
if(p1.Priority>p2.Priority)
value=1;
else if(p1.Priority0)
value=1;
else
value=-1;
}
return value;
}
}
Solution
package patienttest;
import java.util.Comparator;
import java.util.Date;
public class Patient implements Comparator{
String Name;
Date DOB;
Date Admission;
String Complaint;
int Priority;
public Patient(String name,Date d1,Date ad,String complaint,int p)
{
Name=name;
DOB=d1;
Admission=ad;
Complaint=complaint;
Priority=p;
}
public void setName(String name)
{
Name=name;
}
public void setDOB(Date d1)
{
DOB=d1;
}
public void setAdmission(Date d2)
{
Admission=d2;
}
public void setComplaint(String complaint)
{
Complaint=complaint;
}
public void setPriority(int p)
{
Priority=p;
}
public String setName()
{
return Name;
}
public Date setDOB()
{
return DOB;
}
public Date setAdmission()
{
return Admission;
}
public String setComplaint()
{
return Complaint;
}
public int setPriority()
{
return Priority;
}
@Override
public String toString()
{
String msg=\"\";
msg+=\"Name: \"+Name;
msg+=\"DOB: \"+DOB;
msg+=\"Admission: \"+Admission;
msg+=\"Complaint: \"+Complaint;
msg+=\"Priority: \"+Priority;
return msg;
}
@Override
public int compare(Patient p1, Patient p2) {
int value=1;
if(p1.Priority>p2.Priority)
value=1;
else if(p1.Priority0)
value=1;
else
value=-1;
}
return value;
}
}.
ZnO + H2O = Zn(OH)2 .pdf
This document describes a chemical reaction where zinc oxide reacts with water to form zinc hydroxide. Zinc oxide and water react to form zinc hydroxide according to the chemical equation ZnO + H2O => Zn(OH)2. The reaction shows zinc oxide dissolving in water to form zinc hydroxide as the product.
will result in the formation of a soluble complex.pdf
will result in the formation of a soluble complex ion formed from reaction with
hydroxide ion
Solution
will result in the formation of a soluble complex ion formed from reaction with
hydroxide ion.
What types of elements are involved in metallic b.pdf
What types of elements are involved in metallic bonding? D. metals only Metallic
bonding is the electromagnetic interaction between delocalized electrons, called conduction
electrons and the metallic nuclei within metals.
Solution
What types of elements are involved in metallic bonding? D. metals only Metallic
bonding is the electromagnetic interaction between delocalized electrons, called conduction
electrons and the metallic nuclei within metals..
The correct statements are When two atoms of app.pdf
The correct statements are When two atoms of approach each other, the two nuclei
repel each other. When two atoms of approach each other, each electron is attracted by the other
nucleus
Solution
The correct statements are When two atoms of approach each other, the two nuclei
repel each other. When two atoms of approach each other, each electron is attracted by the other
nucleus.
Recommended
4r = 2 L2r = L2thats none of the above, since you didnt wrote t.pdf
4r = 2 L
2r = L/2
thats none of the above, since you didnt wrote the options properly,
answer is L/2
Solution
4r = 2 L
2r = L/2
thats none of the above, since you didnt wrote the options properly,
answer is L/2.
41.Ans. Ans,Sarcoplasmic reticulumSarcoplasmic reticulum is a type.pdf
41.Ans. Ans,Sarcoplasmic reticulum
Sarcoplasmic reticulum is a type of Endoplasmic Reticulum which present in striated muscle,
does not involves in flagellar movement. the other options are microtubules have important role
in flagella movement. these are the main components of axoneme, which is a unit of falgellar
structure and Dyneins are motor proteins.
42.Ans. Other than the nucleus most organelles are small and thus in a typical cell only about
10% of the cells volume is occupied by membrane enclosed organelles and other 90% of the cell
volume is cytosol.
the other options are wrong. membrane bound organelles have outer layer plasma membrane, so
the endoplasmic reticulum not far exceeds the area of plasma membrane. not only the nucleus
surrounded by double layer, other organells for example Mitochondria also surrounded by
double layers. and not only the nucleus, mitochondria also have the DNA. and chloroplast also
have the DNA.
43.Ans. True
once loaded vesicles are carry the cargo to different location with in the body, and reaches the
site of secretion. once reached some of the vesicles containing material for constitutive -
Produced in constant amount, releases and immediately fused with Plasma membrane. where as
vesicles called secretory vesicles- these are the vesicles in which molecules destined for
secretion are stored, these vesicles are in regulated pathway means once reached to the site of
secretion these vesicles wait at he membrane until it receives a signal to secrete, once cell
receives the signal these are going secrete the content and material fuse with the plasma
membrane
44.Ans.False
Nuclear localization signal needed to the peptide until it crosses the membrane
45Ans.True
Chaperones are needed for the import of the proteins. these are prevent the premature folding of
the proteins. while import into the mitochondria, the Chaperones present in the inner surface of
the membrane of mitochondria able to bind the protruding ends of polypeptide and prevents them
to diffuse back into the cytosol through the pore and carry them further and into the matrix. while
moving the peptides further into the matrix at every stage the se peptides are bounded with the
Chaperones which prevent backward movement of the peptides. this kind of diffusion called as
biased diffusion and Chaperones acted as Brownian ratchet.
46.Ans.E) All of These happen.
the above all happens when haploid cell mates
Solution
41.Ans. Ans,Sarcoplasmic reticulum
Sarcoplasmic reticulum is a type of Endoplasmic Reticulum which present in striated muscle,
does not involves in flagellar movement. the other options are microtubules have important role
in flagella movement. these are the main components of axoneme, which is a unit of falgellar
structure and Dyneins are motor proteins.
42.Ans. Other than the nucleus most organelles are small and thus in a typical cell only about
10% of the cells volume is occupied by membrane enclosed organelles and other 90% of the cell
vol.
1.c. developers2.e. A and B3.c. Land and Built space4. Money a.pdf
1.c. developers
2.e. A and B
3.c. Land and Built space
4. Money and Stock
5.c. construction and land market
6. expect a reduction in property asset value
7.b. Demand
8. converts physical capital into financial capital
9.D
10.A
11.D
12.C
13.A
14.B
15.A
16. rent go up, prices of real estsate asset go up, new development increases
17.C
18.D
19.D
20. location and size
21.B
22.D
23.D
24.B
25.C
26.B
27.D
28.B
29.A
30.D
31.D
32.B
33.C
34.D
35.C
36.D
37.A
38.E
39.A
40.C
41.C
42.A
43.A
44.D
45.C
Solution
1.c. developers
2.e. A and B
3.c. Land and Built space
4. Money and Stock
5.c. construction and land market
6. expect a reduction in property asset value
7.b. Demand
8. converts physical capital into financial capital
9.D
10.A
11.D
12.C
13.A
14.B
15.A
16. rent go up, prices of real estsate asset go up, new development increases
17.C
18.D
19.D
20. location and size
21.B
22.D
23.D
24.B
25.C
26.B
27.D
28.B
29.A
30.D
31.D
32.B
33.C
34.D
35.C
36.D
37.A
38.E
39.A
40.C
41.C
42.A
43.A
44.D
45.C.
package patienttest;import java.util.Comparator; import java..pdf
package patienttest;
import java.util.Comparator;
import java.util.Date;
public class Patient implements Comparator{
String Name;
Date DOB;
Date Admission;
String Complaint;
int Priority;
public Patient(String name,Date d1,Date ad,String complaint,int p)
{
Name=name;
DOB=d1;
Admission=ad;
Complaint=complaint;
Priority=p;
}
public void setName(String name)
{
Name=name;
}
public void setDOB(Date d1)
{
DOB=d1;
}
public void setAdmission(Date d2)
{
Admission=d2;
}
public void setComplaint(String complaint)
{
Complaint=complaint;
}
public void setPriority(int p)
{
Priority=p;
}
public String setName()
{
return Name;
}
public Date setDOB()
{
return DOB;
}
public Date setAdmission()
{
return Admission;
}
public String setComplaint()
{
return Complaint;
}
public int setPriority()
{
return Priority;
}
@Override
public String toString()
{
String msg=\"\";
msg+=\"Name: \"+Name;
msg+=\"DOB: \"+DOB;
msg+=\"Admission: \"+Admission;
msg+=\"Complaint: \"+Complaint;
msg+=\"Priority: \"+Priority;
return msg;
}
@Override
public int compare(Patient p1, Patient p2) {
int value=1;
if(p1.Priority>p2.Priority)
value=1;
else if(p1.Priority0)
value=1;
else
value=-1;
}
return value;
}
}
Solution
package patienttest;
import java.util.Comparator;
import java.util.Date;
public class Patient implements Comparator{
String Name;
Date DOB;
Date Admission;
String Complaint;
int Priority;
public Patient(String name,Date d1,Date ad,String complaint,int p)
{
Name=name;
DOB=d1;
Admission=ad;
Complaint=complaint;
Priority=p;
}
public void setName(String name)
{
Name=name;
}
public void setDOB(Date d1)
{
DOB=d1;
}
public void setAdmission(Date d2)
{
Admission=d2;
}
public void setComplaint(String complaint)
{
Complaint=complaint;
}
public void setPriority(int p)
{
Priority=p;
}
public String setName()
{
return Name;
}
public Date setDOB()
{
return DOB;
}
public Date setAdmission()
{
return Admission;
}
public String setComplaint()
{
return Complaint;
}
public int setPriority()
{
return Priority;
}
@Override
public String toString()
{
String msg=\"\";
msg+=\"Name: \"+Name;
msg+=\"DOB: \"+DOB;
msg+=\"Admission: \"+Admission;
msg+=\"Complaint: \"+Complaint;
msg+=\"Priority: \"+Priority;
return msg;
}
@Override
public int compare(Patient p1, Patient p2) {
int value=1;
if(p1.Priority>p2.Priority)
value=1;
else if(p1.Priority0)
value=1;
else
value=-1;
}
return value;
}
}.
ZnO + H2O = Zn(OH)2 .pdf
This document describes a chemical reaction where zinc oxide reacts with water to form zinc hydroxide. Zinc oxide and water react to form zinc hydroxide according to the chemical equation ZnO + H2O => Zn(OH)2. The reaction shows zinc oxide dissolving in water to form zinc hydroxide as the product.
will result in the formation of a soluble complex.pdf
will result in the formation of a soluble complex ion formed from reaction with
hydroxide ion
Solution
will result in the formation of a soluble complex ion formed from reaction with
hydroxide ion.
What types of elements are involved in metallic b.pdf
What types of elements are involved in metallic bonding? D. metals only Metallic
bonding is the electromagnetic interaction between delocalized electrons, called conduction
electrons and the metallic nuclei within metals.
Solution
What types of elements are involved in metallic bonding? D. metals only Metallic
bonding is the electromagnetic interaction between delocalized electrons, called conduction
electrons and the metallic nuclei within metals..
The correct statements are When two atoms of app.pdf
The correct statements are When two atoms of approach each other, the two nuclei
repel each other. When two atoms of approach each other, each electron is attracted by the other
nucleus
Solution
The correct statements are When two atoms of approach each other, the two nuclei
repel each other. When two atoms of approach each other, each electron is attracted by the other
nucleus.
Sure can There you go. .pdf
This very short document contains only two sentences that state "Sure can: There you go." without providing any other context or information. It does not have enough content to create a meaningful multi-sentence summary.
One P2O5 contains five oxygen 3.45 x 5 = 17.3 mol.pdf
One P2O5 contains five oxygen 3.45 x 5 = 17.3 mol oxygen atoms
Solution
One P2O5 contains five oxygen 3.45 x 5 = 17.3 mol oxygen atoms.
Na2O is basic NO2 is acidic SO2 is acidic CO2 is .pdf
Na2O is basic NO2 is acidic SO2 is acidic CO2 is acidic CaO is basic
Solution
Na2O is basic NO2 is acidic SO2 is acidic CO2 is acidic CaO is basic.
Eukaryotic Chromosome Structure The length of DN.pdf
Eukaryotic Chromosome Structure The length of DNA in the nucleus is far greater
than the size of the compartment in which it is contained. To fit into this compartment the DNA
has to be condensed in some manner. The degree to which DNA is condensed is expressed as its
packing ratio. Packing ratio - the length of DNA divided by the length into which it is packaged
For example, the shortest human chromosome contains 4.6 x 107 bp of DNA (about 10 times the
genome size of E. coli). This is equivalent to 14,000 µm of extended DNA. In its most condensed
state during mitosis, the chromosome is about 2 µm long. This gives a packing ratio of 7000
(14,000/2). To achieve the overall packing ratio, DNA is not packaged directly into final
structure of chromatin. Instead, it contains several hierarchies of organization. The first level of
packing is achieved by the winding of DNA around a protein core to produce a \"bead-like\"
structure called a nucleosome. This gives a packing ratio of about 6. This structure is invariant in
both the euchromatin and heterochromatin of all chromosomes. The second level of packing is
the coiling of beads in a helical structure called the 30 nm fiber that is found in both interphase
chromatin and mitotic chromosomes. This structure increases the packing ratio to about 40. The
final packaging occurs when the fiber is organized in loops, scaffolds and domains that give a
final packing ratio of about 1000 in interphase chromosomes and about 10,000 in mitotic
chromosomes. Eukaryotic chromosomes consist of a DNA-protein complex that is organized in
a compact manner which permits the large amount of DNA to be stored in the nucleus of the cell.
The subunit designation of the chromosome is chromatin. The fundamental unit of chromatin is
the nucleosome. Chromatin - the unit of analysis of the chromosome; chromatin reflects the
general structure of the chromosome but is not unique to any particular chromosome
Nucleosome - simplest packaging structure of DNA that is found in all eukaryotic chromosomes;
DNA is wrapped around an octamer of small basic proteins called histones; 146 bp is wrapped
around the core and the remaining bases link to the next nucleosome; this structure causes
negative supercoiling The nucleosome consists of about 200 bp wrapped around a histone
octamer that contains two copies of histone proteins H2A, H2B, H3 and H4. These are known as
the core histones. Histones are basic proteins that have an affinity for DNA and are the most
abundant proteins associated with DNA. The amino acid sequence of these four histones is
conserved suggesting a similar function for all. The length of DNA that is associated with the
nucleosome unit varies between species. But regardless of the size, two DNA components are
involved. Core DNA is the DNA that is actually associated with the histone octamer. This value
is invariant and is 146 base pairs. The core DNA forms two loops around the octamer, and this
permits two regions that are .
Fe + S ---- FeS moles of Fe = 9.4256 = 0.1682 m.pdf
Fe + S ----> FeS moles of Fe = 9.42/56 = 0.1682 moles of S = 8.5/32 = 0.265625
here Fe is limiting so, all the Fe will react to form same amount of FeS as Fe. S0, amount of FeS
= 0.1682moles mass of FeS = 0.1682*88 = 14.8g
Solution
Fe + S ----> FeS moles of Fe = 9.42/56 = 0.1682 moles of S = 8.5/32 = 0.265625
here Fe is limiting so, all the Fe will react to form same amount of FeS as Fe. S0, amount of FeS
= 0.1682moles mass of FeS = 0.1682*88 = 14.8g.
D. is correct. Solution .pdf
The document states that option D is the correct answer. It provides no other context or details about what question or problem is being addressed.
The various ways to finance a project are 1)Debt financing-One ca.pdf
The various ways to finance a project are:
1)Debt financing-One can take out a loan and finance the project costs
Pros:
Ownership is maintained
Lower interest rate
Lower cost of capital
Cons:
Repayment needed even if business goes bankrupt
High rates for old debt even if macroeconomic conditions change
Collateral needed
2) Equity financing is when project financing provides share in a firm in lieu of financing
Pros- Easily availaible than debt
Not to be repaid in case of bankrupcy
Periodic payment as in case of debt is not required. Dividends can be paid as when the owner
feels so
Cons-
have to share profit
High return required by investor
May lose control of company
Private equity funding is when a large investor invests in company and his share cannot be
publicly tarded:
Pros- Capital is easily availaible and all capital can be raised at once
The private equity investor provides his insights
No return requirement and no cash flow has to be paid out as in case of equity
Cons-
More interference in day to day matters by investor
investor may want to recoup his investment when he feels so but company may still have better
profits
Solution
The various ways to finance a project are:
1)Debt financing-One can take out a loan and finance the project costs
Pros:
Ownership is maintained
Lower interest rate
Lower cost of capital
Cons:
Repayment needed even if business goes bankrupt
High rates for old debt even if macroeconomic conditions change
Collateral needed
2) Equity financing is when project financing provides share in a firm in lieu of financing
Pros- Easily availaible than debt
Not to be repaid in case of bankrupcy
Periodic payment as in case of debt is not required. Dividends can be paid as when the owner
feels so
Cons-
have to share profit
High return required by investor
May lose control of company
Private equity funding is when a large investor invests in company and his share cannot be
publicly tarded:
Pros- Capital is easily availaible and all capital can be raised at once
The private equity investor provides his insights
No return requirement and no cash flow has to be paid out as in case of equity
Cons-
More interference in day to day matters by investor
investor may want to recoup his investment when he feels so but company may still have better
profits.
boiling point 1-propanol propanone note there.pdf
boiling point: 1-propanol > propanone note: there exists intermolecular hydrogen
bonding interaction between 1-propanol (i.e., H-O...H bond), whereas no such interaction
between propanone molecules. Thus, bp of 1-propanol is higher than propanone.
Solution
boiling point: 1-propanol > propanone note: there exists intermolecular hydrogen
bonding interaction between 1-propanol (i.e., H-O...H bond), whereas no such interaction
between propanone molecules. Thus, bp of 1-propanol is higher than propanone..
The organs of urinary system includesPaired kidneys, paired ureter.pdf
The organs of urinary system includes
Paired kidneys, paired ureters, a urinary bladder and a urethra.
So the answer is option 4.
Solution
The organs of urinary system includes
Paired kidneys, paired ureters, a urinary bladder and a urethra.
So the answer is option 4..
The incorrect statement ise. Glucose is a stereoisomer of ribose..pdf
The incorrect statement is:
e. Glucose is a stereoisomer of ribose.
Glucose is C6H12O6 while ribose is C5H10O5. They are not ismoers.
Solution
The incorrect statement is:
e. Glucose is a stereoisomer of ribose.
Glucose is C6H12O6 while ribose is C5H10O5. They are not ismoers..
The answer is a. The intermolecular forces are much greater in wate.pdf
The answer is: a. The intermolecular forces are much greater in water than in hydrogen sulfide
and hydrogen selenide.
Boiling point increases with strength of intermolecular forces as these intermolecular bonds need
to be broken for the molecules to escape into the gas phase. Water has strong intermolecular
hydrogen bonds not present in hydrogen sulfide and hydrogen selenide. Thus water has a much
higher boiling point.
Solution
The answer is: a. The intermolecular forces are much greater in water than in hydrogen sulfide
and hydrogen selenide.
Boiling point increases with strength of intermolecular forces as these intermolecular bonds need
to be broken for the molecules to escape into the gas phase. Water has strong intermolecular
hydrogen bonds not present in hydrogen sulfide and hydrogen selenide. Thus water has a much
higher boiling point..
potassium and sulfur - K2S Ionic Bondcalcium and chlorine - CaCl2 .pdf
potassium and sulfur - K2S Ionic Bond
calcium and chlorine - CaCl2 Ionic Bond
hydrogen and bromine - HBr Covalent Bond
nitrogen and nitrogen - N2 Covalent Bond
zinc and zinc - No bond
Solution
potassium and sulfur - K2S Ionic Bond
calcium and chlorine - CaCl2 Ionic Bond
hydrogen and bromine - HBr Covalent Bond
nitrogen and nitrogen - N2 Covalent Bond
zinc and zinc - No bond.
About elements, atoms and compounds definition an.pdf
About elements, atoms and compounds definition and example.
Solution
About elements, atoms and compounds definition and example..
Part AJean Baptiste de Lamarck is best known for his Theory of In.pdf
Part A:
Jean Baptiste de Lamarck is best known for his Theory of Inheritance of Acquired
Characteristics, first presented in 1801. Here are some of the basic theories of Lamarckism:
The theory of evolution by natural selection, first formulated in Darwin\'s book \"On the Origin
of Species\" in 1859, is the process by which organisms change over time as a result of changes
in heritable physical or behavioral traits. The main points regarding Darwinism are as follows:
Darwin\'s theory has been supported by a lot of evidence. Hence, it was accepted by the
scientific community during his time. Lamarck\'s Theory of Inheritance of Acquired
Characteristics has been disproved. This was done in two major ways. The first is by experiment.
We have seen through many real examples and observations that changes that occur in an animal
during life are not passed on to the animal\'s offspring. If a dog\'s ears are cropped short, its
puppies are still born with long ears. The other way that Lamarck\'s theory has been proven
wrong is the study of genetics. Darwin knew that traits are passed on, but he never understood
how they are passed on. During the time when Darwin\'s first book first came out, Gregor
Mendel, who discovered genetics, was just starting his experiments. These and other examples
show that Lamarck\'s theory does not explain how life formed and became the way it is.
Part B:
(i) Alfred Russel Wallace: He discovered the concept of evolution by natural selection. Although
he is now rarely mentioned as the discoverer (Darwin, who discovered the theory independently,
is usually cited) Wallace enjoyed a very high reputation in his lifetime and was awarded many of
science’s most prestigious awards.Alfred Russel Wallace did not exactly influence Charles
Darwin, but rather was his contemporary and collaborated with Darwin on solidifying his Theory
of Evolution through Natural Selection.The two pooled their data to present the idea jointly to
the Linnaean Society of London.
(ii)Stanley Miller: He was an American chemist who designed the first experiment to produce
organic molecules from some of the inorganic components of the Earth’s prebiotic atmosphere.
Miller’s procedure (which was co-designed by Harold Urey and is known as the Miller-Urey
experiment) contained three key elements: a heated pool of water meant to simulate the primitive
Earth’s ocean; an atmosphere of water vapour, methane, ammonia, and molecular hydrogen; and
storms of “lightning” in the form of continuous electric discharges. After one week, 10–15% of
the system’s carbon was found in organic molecules such as amino acids, the building blocks of
proteins. Miller’s experiment was not only a groundbreaking moment for research into the origin
of life on Earth but also a breakthrough that captured the popular imagination and gave rise to
the term prebiotic soup. Miller\'s experiment was important because it showed that organic
compounds such as amino acids, which are essential to cellular .
a) The solubility of the ionic compound is reduce.pdf
a) The solubility of the ionic compound is reduced. due to the common ion effect
and the reverse reaction to the ionization takes place.
Solution
a) The solubility of the ionic compound is reduced. due to the common ion effect
and the reverse reaction to the ionization takes place..
Hyperventilation is “clinical reduction in PCO2 caused by increasing.pdf
Hyperventilation is “clinical reduction in PCO2 caused by increasing the alveolar minute volume
of ventilation”. Caused by rapid/deep breathing.
Tidal volume (TV) – air that moves into and out of the lungs with each breath (approximately
500 ml)
Alveolar ventilation rate (AVR) – measures the flow of fresh gases into and out of the alveoli
during a particular time
When alveolar ventilation is excessive, more carbon dioxide will be removed from the blood
stream than the body can produce. This causes the concentration of carbon dioxide in the blood
stream to fall and produces a state known as hypocapnia (reduced Carbon dioxide). The body
normally attempts to compensate for this metabolically.
Thus there is decrease in respiration rate and tidal volume as metabolic measure to avoid side
effects from reduction of carbon dioxide by naturally trying to bring the breathing pattern back to
normal.
Solution
Hyperventilation is “clinical reduction in PCO2 caused by increasing the alveolar minute volume
of ventilation”. Caused by rapid/deep breathing.
Tidal volume (TV) – air that moves into and out of the lungs with each breath (approximately
500 ml)
Alveolar ventilation rate (AVR) – measures the flow of fresh gases into and out of the alveoli
during a particular time
When alveolar ventilation is excessive, more carbon dioxide will be removed from the blood
stream than the body can produce. This causes the concentration of carbon dioxide in the blood
stream to fall and produces a state known as hypocapnia (reduced Carbon dioxide). The body
normally attempts to compensate for this metabolically.
Thus there is decrease in respiration rate and tidal volume as metabolic measure to avoid side
effects from reduction of carbon dioxide by naturally trying to bring the breathing pattern back to
normal..
import java.util.Scanner; import java.math.BigInteger; public cl.pdf
import java.util.Scanner;
import java.math.BigInteger;
public class Money{
private int dollars;
private int cents;
public Money(){
dollars=0;
cents=0;
}
public Money(int d,int c){
dollars=d+c/100;
cents=c%100;
}
public Money(int c){
dollars=c/100;
cents=c%100;
}
public void add(int c){
cents+=c;
dollars+=cents/100;
cents=cents%100;
}
public void add(int d,int c){
cents+=c;
dollars=dollars+d+cents/100;
cents=cents%100;
}
public void remove(int c){
cents=dollars*100+cents;
if(c>cents){
cents=0;
}
else{
cents-=c;
}
dollars=cents/100;
cents=cents%100;
}
public String getMoney(){
return dollars+\" dollars and \"+cents+\" cents\";
}
public static void main(String []args){
Money h=new Money(1,75);
h.add(190);
System.out.println(h.getMoney());
}
}
Solution
import java.util.Scanner;
import java.math.BigInteger;
public class Money{
private int dollars;
private int cents;
public Money(){
dollars=0;
cents=0;
}
public Money(int d,int c){
dollars=d+c/100;
cents=c%100;
}
public Money(int c){
dollars=c/100;
cents=c%100;
}
public void add(int c){
cents+=c;
dollars+=cents/100;
cents=cents%100;
}
public void add(int d,int c){
cents+=c;
dollars=dollars+d+cents/100;
cents=cents%100;
}
public void remove(int c){
cents=dollars*100+cents;
if(c>cents){
cents=0;
}
else{
cents-=c;
}
dollars=cents/100;
cents=cents%100;
}
public String getMoney(){
return dollars+\" dollars and \"+cents+\" cents\";
}
public static void main(String []args){
Money h=new Money(1,75);
h.add(190);
System.out.println(h.getMoney());
}
}.
#include iostream #includeData.h #includePerson.h#in.pdf
#include
#include\"Data.h\"
#include\"Person.h\"
#include\"Report.h\"
#include
#include
#include \"Menu.h\"
using namespace std;
/*
Programmers: Javier A. Molina, Eduardo Bonnet
Students #:79249
*/
int main()
{
int cnt;
int idx;
string fileName;
Report * telefonia;
Menu menu(\"Reports\");
bool fileNameExists(string fileName);
Report * addReport(Report * reports, int & totalReports);
Report * delReport(Report * reports, int & totalReports, int idx);
int displayReports(Report * reports, int totalReports);
menu.addItem(\"New report\");
menu.addItem(\"Delete a report\");
menu.addItem(\"Display a report\");
cnt = 0;
telefonia = NULL;
do {
cout << menu;
cin >> menu;
//cls();
switch (menu.getOption()) {
case 1:
cout << \"Enter report name (TE?????): \";
cin >> fileName;
if (fileNameExists(fileName + \".tel\")) {
telefonia = addReport(telefonia, cnt);
idx = cnt - 1;
telefonia[idx].setFileName(fileName);
// telefonia[idx].loadReport();
}
else {
cls();
cerr << \"Error: Report name \'\" << fileName << \"\' does not exists!\ \ \ \";
//pause();
}
break;
case 2:
if (telefonia != NULL) {
do {
cls();
cout << \"Delete a Report:\ \ \";
idx = displayReports(telefonia, cnt);
if (between(idx, 0, cnt - 1)) {
cls();
telefonia = delReport(telefonia, cnt, idx);
idx = 0;
// pause();
}
} while (idx != cnt);
}
else {
cerr << \"No reports available for deleting!\ \ \ \";
//pause();
}
break;
case 3:
char ans;
if (telefonia != NULL) {
do {
// cls();
cout << \"Display/Filed a Report:\ \ \";
idx = displayReports(telefonia, cnt);
if (between(idx, 0, cnt - 1)) {
// cls();
cout << \"Would you like the report to be display otherwise save into a file? \";
cin >> ans;
cls();
cin.ignore();
if (upper(ans) == \'Y\') {
cout << telefonia[idx] << endl;
}
else {
//telefonia[idx].saveReport();
cout << \"Report \'\" << telefonia[idx].getFileName() << \"\' saved as \'\" <<
telefonia[idx].getFileName() << \".rep\'.\ \ \";
}
//pause();
}
} while (idx != cnt);
}
else {
cerr << \"No reports available for displaying!\ \ \ \";
// pause();
}
break;
case 4:
cout << \"Thanks for using Telefonia Reports System!!!\ \ \ \";
break;
}
} while (menu.getLast() != menu.getOption());
delete[] telefonia;
return 0;
}
//Funcion que verifica si existe un archivo en disco
bool fileNameExists(string fileName) {
ifstream fin(fileName.c_str());
bool exists = !(fin.fail());
if (exists) fin.close();
return(exists);
}
//Anade dinamicamente objetos tipo Report al arreglo reports
Report * addReport(Report * reports, int & totalReports) {
if (totalReports == 0) {
reports = new Report[1];
}
else {
Report * tmp = new Report[totalReports];
for (int i = 0; i < totalReports; ++i) {
tmp[i] = reports[i];
}
delete[] reports;
reports = new Report[totalReports + 1];
for (int i = 0; i < totalReports; ++i) {
reports[i] = tmp[i];
}
delete[] tmp;
}
totalReports++;
return(reports);
}
//Borra objetos del arreglo reports segun el idx enviado
Report * delReport(Report * reports, int & totalReports, int idx) {
Report * tmp = new Report[tota.
GDP is defined as an aggregate measure of production equal to the .pdf
GDP is defined as \"an aggregate measure of production equal to the sum of the gross values
added of all resident, institutional units engaged in production (plus any taxes, and minus any
subsidies, on products not included in the value of their outputs)\" by the OECD. It can be
measured through any of the following three ways:
In general the various components of GDP includes the following:
Y = C + I + G + (X M)
where Y = GDP
C = Consumption
I = INvestment
G = Govt. Spending
X-M = Net Export
When talking about US\'s GDP, the major contribution is made by Consumption. Nearly 70% of
what the United States produces is for personal consumption. In 2014, $11.929 trillion of the the
total $17.418 trillion produced in the U.S, where goods contributed almost 1/4 of total GDP and
service nearly half (45.7%) of GDP. Because of post recession period, investment is not that
much contributing. Although fixed investment (specially real estate development) has a better
role to play than private inventory change. Government spending added $3.176 trillion to the
economy in 2014, 18.2% of total GDP. In the end, the United States imports more than it
exports, creating a trade deficit hence having negative effect on GDP. In 2014, imports were
$2.875 trillion, while exports were $2.335 trillion. As a result, international tradesubtracted $540
billion from GDP.
All data are sourced from U.S. Bureau of Economic Analysis, National Income and Product
Accounts Tables.
Differences between real GDP and nominal GDP are given in the following table:
Real GDP = pbqt
where b denotes the base year.
Nominal GDP = ptqt
where p refers to price, q is quantity, and t indicates the year in question (usually the current
year).
Business Cycle refers to economic contractions, troughs, expansions and peaks that are
unpredictable phases of economic activity. As the economy goes through business cycle changes,
these positively or negatively affect the GDP.
During a contraction, economic output slows, usually due to decreased demand for products and
services, an increase in the cost of raw materials or both. An economic trough occurs after a
contraction. A historically high national unemployment rate and low economic output usually
mark this trough, which often signals that the economy is already in or heading toward a
recession. Unlike a contractionary phase in which the GDP decreases but is still positive, during
a trough the GDP is negative.After the \"rock bottom\" of an economic trough, expansion is its
recovery. If the economy grows for two or three consecutive calendar quarters, it indicates that it
is beginning its recovery and GDP begins to increase. An economic peak is like a mountain
summit. Once the economy reaches this peak, it must come down. This economic peak is the
highest point of economic growth and output, resulting in an increase in the GDP.BasisReal
GDPNominal GDPMeaning/MeasureIt measures the value of output economy, adjusted for price
changes.It measures the mark.
code ----- public void add(int key, T value) { Link previou.pdf
code:
-----
public void add(int key, T value)
{
Link previous = null; // start at first
Link current = first;
// until end of list,
while( current != null && key > current.getKey() )
{ // or current > key,
previous = current;
current = current.next; // go to next item
}
if(previous==null) // if beginning of list,
first = theLink; // first --> new link
else // not at beginning,
previous.next = theLink; // prev --> new link
theLink.next = current; // new link --> current
}
Pseudo code:
------------
1. start from first node
Link current=first;
2. Traverse the hash table till the end of the list and
the key value becomes greater than the current key value,
and locate a position,called previous.
3. If the obtained position is null, make the new node as
first node. Place the value in that node.
4. if the position is not null,insert the new node as next
to previous node.
Solution
code:
-----
public void add(int key, T value)
{
Link previous = null; // start at first
Link current = first;
// until end of list,
while( current != null && key > current.getKey() )
{ // or current > key,
previous = current;
current = current.next; // go to next item
}
if(previous==null) // if beginning of list,
first = theLink; // first --> new link
else // not at beginning,
previous.next = theLink; // prev --> new link
theLink.next = current; // new link --> current
}
Pseudo code:
------------
1. start from first node
Link current=first;
2. Traverse the hash table till the end of the list and
the key value becomes greater than the current key value,
and locate a position,called previous.
3. If the obtained position is null, make the new node as
first node. Place the value in that node.
4. if the position is not null,insert the new node as next
to previous node..
How to Fix the Import Error in the Odoo 17
An import error occurs when a program fails to import a module or library, disrupting its execution. In languages like Python, this issue arises when the specified module cannot be found or accessed, hindering the program's functionality. Resolving import errors is crucial for maintaining smooth software operation and uninterrupted development processes.
Main Java[All of the Base Concepts}.docx
This is part 1 of my Java Learning Journey. This Contains Custom methods, classes, constructors, packages, multithreading , try- catch block, finally block and more.
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Sure can There you go. .pdf
This very short document contains only two sentences that state "Sure can: There you go." without providing any other context or information. It does not have enough content to create a meaningful multi-sentence summary.
One P2O5 contains five oxygen 3.45 x 5 = 17.3 mol.pdf
One P2O5 contains five oxygen 3.45 x 5 = 17.3 mol oxygen atoms
Solution
One P2O5 contains five oxygen 3.45 x 5 = 17.3 mol oxygen atoms.
Na2O is basic NO2 is acidic SO2 is acidic CO2 is .pdf
Na2O is basic NO2 is acidic SO2 is acidic CO2 is acidic CaO is basic
Solution
Na2O is basic NO2 is acidic SO2 is acidic CO2 is acidic CaO is basic.
Eukaryotic Chromosome Structure The length of DN.pdf
Eukaryotic Chromosome Structure The length of DNA in the nucleus is far greater
than the size of the compartment in which it is contained. To fit into this compartment the DNA
has to be condensed in some manner. The degree to which DNA is condensed is expressed as its
packing ratio. Packing ratio - the length of DNA divided by the length into which it is packaged
For example, the shortest human chromosome contains 4.6 x 107 bp of DNA (about 10 times the
genome size of E. coli). This is equivalent to 14,000 µm of extended DNA. In its most condensed
state during mitosis, the chromosome is about 2 µm long. This gives a packing ratio of 7000
(14,000/2). To achieve the overall packing ratio, DNA is not packaged directly into final
structure of chromatin. Instead, it contains several hierarchies of organization. The first level of
packing is achieved by the winding of DNA around a protein core to produce a \"bead-like\"
structure called a nucleosome. This gives a packing ratio of about 6. This structure is invariant in
both the euchromatin and heterochromatin of all chromosomes. The second level of packing is
the coiling of beads in a helical structure called the 30 nm fiber that is found in both interphase
chromatin and mitotic chromosomes. This structure increases the packing ratio to about 40. The
final packaging occurs when the fiber is organized in loops, scaffolds and domains that give a
final packing ratio of about 1000 in interphase chromosomes and about 10,000 in mitotic
chromosomes. Eukaryotic chromosomes consist of a DNA-protein complex that is organized in
a compact manner which permits the large amount of DNA to be stored in the nucleus of the cell.
The subunit designation of the chromosome is chromatin. The fundamental unit of chromatin is
the nucleosome. Chromatin - the unit of analysis of the chromosome; chromatin reflects the
general structure of the chromosome but is not unique to any particular chromosome
Nucleosome - simplest packaging structure of DNA that is found in all eukaryotic chromosomes;
DNA is wrapped around an octamer of small basic proteins called histones; 146 bp is wrapped
around the core and the remaining bases link to the next nucleosome; this structure causes
negative supercoiling The nucleosome consists of about 200 bp wrapped around a histone
octamer that contains two copies of histone proteins H2A, H2B, H3 and H4. These are known as
the core histones. Histones are basic proteins that have an affinity for DNA and are the most
abundant proteins associated with DNA. The amino acid sequence of these four histones is
conserved suggesting a similar function for all. The length of DNA that is associated with the
nucleosome unit varies between species. But regardless of the size, two DNA components are
involved. Core DNA is the DNA that is actually associated with the histone octamer. This value
is invariant and is 146 base pairs. The core DNA forms two loops around the octamer, and this
permits two regions that are .
Fe + S ---- FeS moles of Fe = 9.4256 = 0.1682 m.pdf
Fe + S ----> FeS moles of Fe = 9.42/56 = 0.1682 moles of S = 8.5/32 = 0.265625
here Fe is limiting so, all the Fe will react to form same amount of FeS as Fe. S0, amount of FeS
= 0.1682moles mass of FeS = 0.1682*88 = 14.8g
Solution
Fe + S ----> FeS moles of Fe = 9.42/56 = 0.1682 moles of S = 8.5/32 = 0.265625
here Fe is limiting so, all the Fe will react to form same amount of FeS as Fe. S0, amount of FeS
= 0.1682moles mass of FeS = 0.1682*88 = 14.8g.
D. is correct. Solution .pdf
The document states that option D is the correct answer. It provides no other context or details about what question or problem is being addressed.
The various ways to finance a project are 1)Debt financing-One ca.pdf
The various ways to finance a project are:
1)Debt financing-One can take out a loan and finance the project costs
Pros:
Ownership is maintained
Lower interest rate
Lower cost of capital
Cons:
Repayment needed even if business goes bankrupt
High rates for old debt even if macroeconomic conditions change
Collateral needed
2) Equity financing is when project financing provides share in a firm in lieu of financing
Pros- Easily availaible than debt
Not to be repaid in case of bankrupcy
Periodic payment as in case of debt is not required. Dividends can be paid as when the owner
feels so
Cons-
have to share profit
High return required by investor
May lose control of company
Private equity funding is when a large investor invests in company and his share cannot be
publicly tarded:
Pros- Capital is easily availaible and all capital can be raised at once
The private equity investor provides his insights
No return requirement and no cash flow has to be paid out as in case of equity
Cons-
More interference in day to day matters by investor
investor may want to recoup his investment when he feels so but company may still have better
profits
Solution
The various ways to finance a project are:
1)Debt financing-One can take out a loan and finance the project costs
Pros:
Ownership is maintained
Lower interest rate
Lower cost of capital
Cons:
Repayment needed even if business goes bankrupt
High rates for old debt even if macroeconomic conditions change
Collateral needed
2) Equity financing is when project financing provides share in a firm in lieu of financing
Pros- Easily availaible than debt
Not to be repaid in case of bankrupcy
Periodic payment as in case of debt is not required. Dividends can be paid as when the owner
feels so
Cons-
have to share profit
High return required by investor
May lose control of company
Private equity funding is when a large investor invests in company and his share cannot be
publicly tarded:
Pros- Capital is easily availaible and all capital can be raised at once
The private equity investor provides his insights
No return requirement and no cash flow has to be paid out as in case of equity
Cons-
More interference in day to day matters by investor
investor may want to recoup his investment when he feels so but company may still have better
profits.
boiling point 1-propanol propanone note there.pdf
boiling point: 1-propanol > propanone note: there exists intermolecular hydrogen
bonding interaction between 1-propanol (i.e., H-O...H bond), whereas no such interaction
between propanone molecules. Thus, bp of 1-propanol is higher than propanone.
Solution
boiling point: 1-propanol > propanone note: there exists intermolecular hydrogen
bonding interaction between 1-propanol (i.e., H-O...H bond), whereas no such interaction
between propanone molecules. Thus, bp of 1-propanol is higher than propanone..
The organs of urinary system includesPaired kidneys, paired ureter.pdf
The organs of urinary system includes
Paired kidneys, paired ureters, a urinary bladder and a urethra.
So the answer is option 4.
Solution
The organs of urinary system includes
Paired kidneys, paired ureters, a urinary bladder and a urethra.
So the answer is option 4..
The incorrect statement ise. Glucose is a stereoisomer of ribose..pdf
The incorrect statement is:
e. Glucose is a stereoisomer of ribose.
Glucose is C6H12O6 while ribose is C5H10O5. They are not ismoers.
Solution
The incorrect statement is:
e. Glucose is a stereoisomer of ribose.
Glucose is C6H12O6 while ribose is C5H10O5. They are not ismoers..
The answer is a. The intermolecular forces are much greater in wate.pdf
The answer is: a. The intermolecular forces are much greater in water than in hydrogen sulfide
and hydrogen selenide.
Boiling point increases with strength of intermolecular forces as these intermolecular bonds need
to be broken for the molecules to escape into the gas phase. Water has strong intermolecular
hydrogen bonds not present in hydrogen sulfide and hydrogen selenide. Thus water has a much
higher boiling point.
Solution
The answer is: a. The intermolecular forces are much greater in water than in hydrogen sulfide
and hydrogen selenide.
Boiling point increases with strength of intermolecular forces as these intermolecular bonds need
to be broken for the molecules to escape into the gas phase. Water has strong intermolecular
hydrogen bonds not present in hydrogen sulfide and hydrogen selenide. Thus water has a much
higher boiling point..
potassium and sulfur - K2S Ionic Bondcalcium and chlorine - CaCl2 .pdf
potassium and sulfur - K2S Ionic Bond
calcium and chlorine - CaCl2 Ionic Bond
hydrogen and bromine - HBr Covalent Bond
nitrogen and nitrogen - N2 Covalent Bond
zinc and zinc - No bond
Solution
potassium and sulfur - K2S Ionic Bond
calcium and chlorine - CaCl2 Ionic Bond
hydrogen and bromine - HBr Covalent Bond
nitrogen and nitrogen - N2 Covalent Bond
zinc and zinc - No bond.
About elements, atoms and compounds definition an.pdf
About elements, atoms and compounds definition and example.
Solution
About elements, atoms and compounds definition and example..
Part AJean Baptiste de Lamarck is best known for his Theory of In.pdf
Part A:
Jean Baptiste de Lamarck is best known for his Theory of Inheritance of Acquired
Characteristics, first presented in 1801. Here are some of the basic theories of Lamarckism:
The theory of evolution by natural selection, first formulated in Darwin\'s book \"On the Origin
of Species\" in 1859, is the process by which organisms change over time as a result of changes
in heritable physical or behavioral traits. The main points regarding Darwinism are as follows:
Darwin\'s theory has been supported by a lot of evidence. Hence, it was accepted by the
scientific community during his time. Lamarck\'s Theory of Inheritance of Acquired
Characteristics has been disproved. This was done in two major ways. The first is by experiment.
We have seen through many real examples and observations that changes that occur in an animal
during life are not passed on to the animal\'s offspring. If a dog\'s ears are cropped short, its
puppies are still born with long ears. The other way that Lamarck\'s theory has been proven
wrong is the study of genetics. Darwin knew that traits are passed on, but he never understood
how they are passed on. During the time when Darwin\'s first book first came out, Gregor
Mendel, who discovered genetics, was just starting his experiments. These and other examples
show that Lamarck\'s theory does not explain how life formed and became the way it is.
Part B:
(i) Alfred Russel Wallace: He discovered the concept of evolution by natural selection. Although
he is now rarely mentioned as the discoverer (Darwin, who discovered the theory independently,
is usually cited) Wallace enjoyed a very high reputation in his lifetime and was awarded many of
science’s most prestigious awards.Alfred Russel Wallace did not exactly influence Charles
Darwin, but rather was his contemporary and collaborated with Darwin on solidifying his Theory
of Evolution through Natural Selection.The two pooled their data to present the idea jointly to
the Linnaean Society of London.
(ii)Stanley Miller: He was an American chemist who designed the first experiment to produce
organic molecules from some of the inorganic components of the Earth’s prebiotic atmosphere.
Miller’s procedure (which was co-designed by Harold Urey and is known as the Miller-Urey
experiment) contained three key elements: a heated pool of water meant to simulate the primitive
Earth’s ocean; an atmosphere of water vapour, methane, ammonia, and molecular hydrogen; and
storms of “lightning” in the form of continuous electric discharges. After one week, 10–15% of
the system’s carbon was found in organic molecules such as amino acids, the building blocks of
proteins. Miller’s experiment was not only a groundbreaking moment for research into the origin
of life on Earth but also a breakthrough that captured the popular imagination and gave rise to
the term prebiotic soup. Miller\'s experiment was important because it showed that organic
compounds such as amino acids, which are essential to cellular .
a) The solubility of the ionic compound is reduce.pdf
a) The solubility of the ionic compound is reduced. due to the common ion effect
and the reverse reaction to the ionization takes place.
Solution
a) The solubility of the ionic compound is reduced. due to the common ion effect
and the reverse reaction to the ionization takes place..
Hyperventilation is “clinical reduction in PCO2 caused by increasing.pdf
Hyperventilation is “clinical reduction in PCO2 caused by increasing the alveolar minute volume
of ventilation”. Caused by rapid/deep breathing.
Tidal volume (TV) – air that moves into and out of the lungs with each breath (approximately
500 ml)
Alveolar ventilation rate (AVR) – measures the flow of fresh gases into and out of the alveoli
during a particular time
When alveolar ventilation is excessive, more carbon dioxide will be removed from the blood
stream than the body can produce. This causes the concentration of carbon dioxide in the blood
stream to fall and produces a state known as hypocapnia (reduced Carbon dioxide). The body
normally attempts to compensate for this metabolically.
Thus there is decrease in respiration rate and tidal volume as metabolic measure to avoid side
effects from reduction of carbon dioxide by naturally trying to bring the breathing pattern back to
normal.
Solution
Hyperventilation is “clinical reduction in PCO2 caused by increasing the alveolar minute volume
of ventilation”. Caused by rapid/deep breathing.
Tidal volume (TV) – air that moves into and out of the lungs with each breath (approximately
500 ml)
Alveolar ventilation rate (AVR) – measures the flow of fresh gases into and out of the alveoli
during a particular time
When alveolar ventilation is excessive, more carbon dioxide will be removed from the blood
stream than the body can produce. This causes the concentration of carbon dioxide in the blood
stream to fall and produces a state known as hypocapnia (reduced Carbon dioxide). The body
normally attempts to compensate for this metabolically.
Thus there is decrease in respiration rate and tidal volume as metabolic measure to avoid side
effects from reduction of carbon dioxide by naturally trying to bring the breathing pattern back to
normal..
import java.util.Scanner; import java.math.BigInteger; public cl.pdf
import java.util.Scanner;
import java.math.BigInteger;
public class Money{
private int dollars;
private int cents;
public Money(){
dollars=0;
cents=0;
}
public Money(int d,int c){
dollars=d+c/100;
cents=c%100;
}
public Money(int c){
dollars=c/100;
cents=c%100;
}
public void add(int c){
cents+=c;
dollars+=cents/100;
cents=cents%100;
}
public void add(int d,int c){
cents+=c;
dollars=dollars+d+cents/100;
cents=cents%100;
}
public void remove(int c){
cents=dollars*100+cents;
if(c>cents){
cents=0;
}
else{
cents-=c;
}
dollars=cents/100;
cents=cents%100;
}
public String getMoney(){
return dollars+\" dollars and \"+cents+\" cents\";
}
public static void main(String []args){
Money h=new Money(1,75);
h.add(190);
System.out.println(h.getMoney());
}
}
Solution
import java.util.Scanner;
import java.math.BigInteger;
public class Money{
private int dollars;
private int cents;
public Money(){
dollars=0;
cents=0;
}
public Money(int d,int c){
dollars=d+c/100;
cents=c%100;
}
public Money(int c){
dollars=c/100;
cents=c%100;
}
public void add(int c){
cents+=c;
dollars+=cents/100;
cents=cents%100;
}
public void add(int d,int c){
cents+=c;
dollars=dollars+d+cents/100;
cents=cents%100;
}
public void remove(int c){
cents=dollars*100+cents;
if(c>cents){
cents=0;
}
else{
cents-=c;
}
dollars=cents/100;
cents=cents%100;
}
public String getMoney(){
return dollars+\" dollars and \"+cents+\" cents\";
}
public static void main(String []args){
Money h=new Money(1,75);
h.add(190);
System.out.println(h.getMoney());
}
}.
#include iostream #includeData.h #includePerson.h#in.pdf
#include
#include\"Data.h\"
#include\"Person.h\"
#include\"Report.h\"
#include
#include
#include \"Menu.h\"
using namespace std;
/*
Programmers: Javier A. Molina, Eduardo Bonnet
Students #:79249
*/
int main()
{
int cnt;
int idx;
string fileName;
Report * telefonia;
Menu menu(\"Reports\");
bool fileNameExists(string fileName);
Report * addReport(Report * reports, int & totalReports);
Report * delReport(Report * reports, int & totalReports, int idx);
int displayReports(Report * reports, int totalReports);
menu.addItem(\"New report\");
menu.addItem(\"Delete a report\");
menu.addItem(\"Display a report\");
cnt = 0;
telefonia = NULL;
do {
cout << menu;
cin >> menu;
//cls();
switch (menu.getOption()) {
case 1:
cout << \"Enter report name (TE?????): \";
cin >> fileName;
if (fileNameExists(fileName + \".tel\")) {
telefonia = addReport(telefonia, cnt);
idx = cnt - 1;
telefonia[idx].setFileName(fileName);
// telefonia[idx].loadReport();
}
else {
cls();
cerr << \"Error: Report name \'\" << fileName << \"\' does not exists!\ \ \ \";
//pause();
}
break;
case 2:
if (telefonia != NULL) {
do {
cls();
cout << \"Delete a Report:\ \ \";
idx = displayReports(telefonia, cnt);
if (between(idx, 0, cnt - 1)) {
cls();
telefonia = delReport(telefonia, cnt, idx);
idx = 0;
// pause();
}
} while (idx != cnt);
}
else {
cerr << \"No reports available for deleting!\ \ \ \";
//pause();
}
break;
case 3:
char ans;
if (telefonia != NULL) {
do {
// cls();
cout << \"Display/Filed a Report:\ \ \";
idx = displayReports(telefonia, cnt);
if (between(idx, 0, cnt - 1)) {
// cls();
cout << \"Would you like the report to be display otherwise save into a file? \";
cin >> ans;
cls();
cin.ignore();
if (upper(ans) == \'Y\') {
cout << telefonia[idx] << endl;
}
else {
//telefonia[idx].saveReport();
cout << \"Report \'\" << telefonia[idx].getFileName() << \"\' saved as \'\" <<
telefonia[idx].getFileName() << \".rep\'.\ \ \";
}
//pause();
}
} while (idx != cnt);
}
else {
cerr << \"No reports available for displaying!\ \ \ \";
// pause();
}
break;
case 4:
cout << \"Thanks for using Telefonia Reports System!!!\ \ \ \";
break;
}
} while (menu.getLast() != menu.getOption());
delete[] telefonia;
return 0;
}
//Funcion que verifica si existe un archivo en disco
bool fileNameExists(string fileName) {
ifstream fin(fileName.c_str());
bool exists = !(fin.fail());
if (exists) fin.close();
return(exists);
}
//Anade dinamicamente objetos tipo Report al arreglo reports
Report * addReport(Report * reports, int & totalReports) {
if (totalReports == 0) {
reports = new Report[1];
}
else {
Report * tmp = new Report[totalReports];
for (int i = 0; i < totalReports; ++i) {
tmp[i] = reports[i];
}
delete[] reports;
reports = new Report[totalReports + 1];
for (int i = 0; i < totalReports; ++i) {
reports[i] = tmp[i];
}
delete[] tmp;
}
totalReports++;
return(reports);
}
//Borra objetos del arreglo reports segun el idx enviado
Report * delReport(Report * reports, int & totalReports, int idx) {
Report * tmp = new Report[tota.
GDP is defined as an aggregate measure of production equal to the .pdf
GDP is defined as \"an aggregate measure of production equal to the sum of the gross values
added of all resident, institutional units engaged in production (plus any taxes, and minus any
subsidies, on products not included in the value of their outputs)\" by the OECD. It can be
measured through any of the following three ways:
In general the various components of GDP includes the following:
Y = C + I + G + (X M)
where Y = GDP
C = Consumption
I = INvestment
G = Govt. Spending
X-M = Net Export
When talking about US\'s GDP, the major contribution is made by Consumption. Nearly 70% of
what the United States produces is for personal consumption. In 2014, $11.929 trillion of the the
total $17.418 trillion produced in the U.S, where goods contributed almost 1/4 of total GDP and
service nearly half (45.7%) of GDP. Because of post recession period, investment is not that
much contributing. Although fixed investment (specially real estate development) has a better
role to play than private inventory change. Government spending added $3.176 trillion to the
economy in 2014, 18.2% of total GDP. In the end, the United States imports more than it
exports, creating a trade deficit hence having negative effect on GDP. In 2014, imports were
$2.875 trillion, while exports were $2.335 trillion. As a result, international tradesubtracted $540
billion from GDP.
All data are sourced from U.S. Bureau of Economic Analysis, National Income and Product
Accounts Tables.
Differences between real GDP and nominal GDP are given in the following table:
Real GDP = pbqt
where b denotes the base year.
Nominal GDP = ptqt
where p refers to price, q is quantity, and t indicates the year in question (usually the current
year).
Business Cycle refers to economic contractions, troughs, expansions and peaks that are
unpredictable phases of economic activity. As the economy goes through business cycle changes,
these positively or negatively affect the GDP.
During a contraction, economic output slows, usually due to decreased demand for products and
services, an increase in the cost of raw materials or both. An economic trough occurs after a
contraction. A historically high national unemployment rate and low economic output usually
mark this trough, which often signals that the economy is already in or heading toward a
recession. Unlike a contractionary phase in which the GDP decreases but is still positive, during
a trough the GDP is negative.After the \"rock bottom\" of an economic trough, expansion is its
recovery. If the economy grows for two or three consecutive calendar quarters, it indicates that it
is beginning its recovery and GDP begins to increase. An economic peak is like a mountain
summit. Once the economy reaches this peak, it must come down. This economic peak is the
highest point of economic growth and output, resulting in an increase in the GDP.BasisReal
GDPNominal GDPMeaning/MeasureIt measures the value of output economy, adjusted for price
changes.It measures the mark.
code ----- public void add(int key, T value) { Link previou.pdf
code:
-----
public void add(int key, T value)
{
Link previous = null; // start at first
Link current = first;
// until end of list,
while( current != null && key > current.getKey() )
{ // or current > key,
previous = current;
current = current.next; // go to next item
}
if(previous==null) // if beginning of list,
first = theLink; // first --> new link
else // not at beginning,
previous.next = theLink; // prev --> new link
theLink.next = current; // new link --> current
}
Pseudo code:
------------
1. start from first node
Link current=first;
2. Traverse the hash table till the end of the list and
the key value becomes greater than the current key value,
and locate a position,called previous.
3. If the obtained position is null, make the new node as
first node. Place the value in that node.
4. if the position is not null,insert the new node as next
to previous node.
Solution
code:
-----
public void add(int key, T value)
{
Link previous = null; // start at first
Link current = first;
// until end of list,
while( current != null && key > current.getKey() )
{ // or current > key,
previous = current;
current = current.next; // go to next item
}
if(previous==null) // if beginning of list,
first = theLink; // first --> new link
else // not at beginning,
previous.next = theLink; // prev --> new link
theLink.next = current; // new link --> current
}
Pseudo code:
------------
1. start from first node
Link current=first;
2. Traverse the hash table till the end of the list and
the key value becomes greater than the current key value,
and locate a position,called previous.
3. If the obtained position is null, make the new node as
first node. Place the value in that node.
4. if the position is not null,insert the new node as next
to previous node..
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