Addition of NH3(aq) produced a white gelatinous precipitate in a col.pdf
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Addition of NH3(aq) produced a white gelatinous precipitate in a colorless solution. note: The bismuth III ions are separated by precipitation as white gelatinous bismuth (III) hydroxide upon the addition of an aqueous solution of ammonia. Bi definitely present. The addition of the ammonia water to precipitate the bismuth hydroxide also converts any copper(II) ion present to the deep blue complex ion, Cu(NH3)4^2+. since the solution is colorless, Cu definitely absent. When solution (Z) was treated with K3Fe(CN)6, a dark blue suspension was produced. KFe[Fe(CN)6] this indicates Fe3+ definitely present Co2+,Ni2+ cannot decide if it exists or not. Solution Addition of NH3(aq) produced a white gelatinous precipitate in a colorless solution. note: The bismuth III ions are separated by precipitation as white gelatinous bismuth (III) hydroxide upon the addition of an aqueous solution of ammonia. Bi definitely present. The addition of the ammonia water to precipitate the bismuth hydroxide also converts any copper(II) ion present to the deep blue complex ion, Cu(NH3)4^2+. since the solution is colorless, Cu definitely absent. When solution (Z) was treated with K3Fe(CN)6, a dark blue suspension was produced. KFe[Fe(CN)6] this indicates Fe3+ definitely present Co2+,Ni2+ cannot decide if it exists or not..
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![Addition of NH3(aq) produced a white gelatinous precipitate in a colorless solution.
note:
The bismuth III ions are separated by precipitation as white gelatinous
bismuth (III) hydroxide upon the addition of an aqueous solution of ammonia.
Bi definitely present.
The addition of the ammonia water to precipitate the bismuth hydroxide also converts any
copper(II) ion present to the deep blue complex ion, Cu(NH3)4^2+.
since the solution is colorless, Cu definitely absent.
When solution (Z) was treated with K3Fe(CN)6, a dark blue suspension was
produced. KFe[Fe(CN)6]
this indicates Fe3+ definitely present
Co2+,Ni2+ cannot decide if it exists or not.
Solution
Addition of NH3(aq) produced a white gelatinous precipitate in a colorless solution.
note:
The bismuth III ions are separated by precipitation as white gelatinous
bismuth (III) hydroxide upon the addition of an aqueous solution of ammonia.
Bi definitely present.
The addition of the ammonia water to precipitate the bismuth hydroxide also converts any
copper(II) ion present to the deep blue complex ion, Cu(NH3)4^2+.
since the solution is colorless, Cu definitely absent.
When solution (Z) was treated with K3Fe(CN)6, a dark blue suspension was
produced. KFe[Fe(CN)6]
this indicates Fe3+ definitely present
Co2+,Ni2+ cannot decide if it exists or not.](https://image.slidesharecdn.com/additionofnh3aqproducedawhitegelatinousprecipitateinacol-230413054402-808f7778/85/Addition-of-NH3-aq-produced-a-white-gelatinous-precipitate-in-a-col-pdf-1-320.jpg)
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Recommended
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Alternative 1 Continuing present jobPV = Pmt x (1 + i) x (1 - (1 .pdf
Alternative 1: Continuing present job
PV = Pmt x (1 + i) x (1 - (1 + g) n x (1 + i)-n) / (i - g)
Pmt = 60,000
Discount rate (i) = 12%
Growth rate (g) = 14%
No of years (n) = 10
= 60000 × (1+0.12) × (1- (1+0.14)10 × (1+0.12)-10/ (0.12-0.14)
= 650583.51
(Or)
Year
Cashflow
Present value@12%
Discounted cashflow
1
60000.00
1
60000.00
2
68400.00
0.8929
61071.43
3
77976.00
0.7972
62161.99
4
88892.64
0.7118
63272.03
5
101337.61
0.6355
64401.88
6
115524.87
0.5674
65551.92
7
131698.36
0.5066
66722.49
8
150136.13
0.4523
67913.96
9
171155.19
0.4039
69126.71
10
195116.91
0.3606
70361.11
Net Present value
650583.51
Alternative2: Going for MBA
Year
Cash flow
Present value@12%
Discounted cash flow
1
-45000.00
1
-45000.00
2
-45000.00
0.8929
-40178.57
3
90000.00
0.7972
71747.45
4
99000.00
0.7118
70466.24
5
108900.00
0.6355
69207.92
6
119790.00
0.5674
67972.06
7
131769.00
0.5066
66758.28
8
144945.90
0.4523
65566.16
9
159440.49
0.4039
64395.34
10
175384.54
0.3606
63245.42
Net present value
454180.31
The NPV of most attractive choice of continuing current job is 650583.51
Year
Cashflow
Present value@12%
Discounted cashflow
1
60000.00
1
60000.00
2
68400.00
0.8929
61071.43
3
77976.00
0.7972
62161.99
4
88892.64
0.7118
63272.03
5
101337.61
0.6355
64401.88
6
115524.87
0.5674
65551.92
7
131698.36
0.5066
66722.49
8
150136.13
0.4523
67913.96
9
171155.19
0.4039
69126.71
10
195116.91
0.3606
70361.11
Net Present value
650583.51
Solution
Alternative 1: Continuing present job
PV = Pmt x (1 + i) x (1 - (1 + g) n x (1 + i)-n) / (i - g)
Pmt = 60,000
Discount rate (i) = 12%
Growth rate (g) = 14%
No of years (n) = 10
= 60000 × (1+0.12) × (1- (1+0.14)10 × (1+0.12)-10/ (0.12-0.14)
= 650583.51
(Or)
Year
Cashflow
Present value@12%
Discounted cashflow
1
60000.00
1
60000.00
2
68400.00
0.8929
61071.43
3
77976.00
0.7972
62161.99
4
88892.64
0.7118
63272.03
5
101337.61
0.6355
64401.88
6
115524.87
0.5674
65551.92
7
131698.36
0.5066
66722.49
8
150136.13
0.4523
67913.96
9
171155.19
0.4039
69126.71
10
195116.91
0.3606
70361.11
Net Present value
650583.51
Alternative2: Going for MBA
Year
Cash flow
Present value@12%
Discounted cash flow
1
-45000.00
1
-45000.00
2
-45000.00
0.8929
-40178.57
3
90000.00
0.7972
71747.45
4
99000.00
0.7118
70466.24
5
108900.00
0.6355
69207.92
6
119790.00
0.5674
67972.06
7
131769.00
0.5066
66758.28
8
144945.90
0.4523
65566.16
9
159440.49
0.4039
64395.34
10
175384.54
0.3606
63245.42
Net present value
454180.31
The NPV of most attractive choice of continuing current job is 650583.51
Year
Cashflow
Present value@12%
Discounted cashflow
1
60000.00
1
60000.00
2
68400.00
0.8929
61071.43
3
77976.00
0.7972
62161.99
4
88892.64
0.7118
63272.03
5
101337.61
0.6355
64401.88
6
115524.87
0.5674
65551.92
7
131698.36
0.5066
66722.49
8
150136.13
0.4523
67913.96
9
171155.19
0.4039
69126.71
10
195116.91
0.3606
70361.11
Net Present value
650583.51.
#include iostream #include fstreamusing namespace std;void.pdf
#include
#include
using namespace std;
void getData(ifstream &in, int highLowTemperatures[][2]){
for(int i = 0; i < 12; ++i){
in >> highLowTemperatures[i][0];
}
for(int i = 0; i < 12; ++i){
in >> highLowTemperatures[i][1];
}
}
int indexHighTemp(int temperatures[][2]){
int val = 0;
for(int i = 1; i < 12; ++i){
if(temperatures[val][0] < temperatures[i][0]){
val = i;
}
}
return val;
}
int indexLowTemp(int temperatures[][2]){
int val = 0;
for(int i = 1; i < 12; ++i){
if(temperatures[val][1] > temperatures[i][1]){
val = i;
}
}
return val;
}
int main(){
ifstream in;
string fName;
cout << \"Enter file name: \";
cin >> fName;
in.open(fName.c_str());
int highLowTemperatures[12][2];
if(in.is_open()){
getData(in, highLowTemperatures);
cout << \"Highest temperature of the year is \" <<
highLowTemperatures[indexHighTemp(highLowTemperatures)][0] << endl;
cout << \"Lowest temperature of the year is \" <<
highLowTemperatures[indexLowTemp(highLowTemperatures)][1] << endl;
}
else{
cout << \"Can not open the file\" << endl;
}
return 0;
}
Solution
#include
#include
using namespace std;
void getData(ifstream &in, int highLowTemperatures[][2]){
for(int i = 0; i < 12; ++i){
in >> highLowTemperatures[i][0];
}
for(int i = 0; i < 12; ++i){
in >> highLowTemperatures[i][1];
}
}
int indexHighTemp(int temperatures[][2]){
int val = 0;
for(int i = 1; i < 12; ++i){
if(temperatures[val][0] < temperatures[i][0]){
val = i;
}
}
return val;
}
int indexLowTemp(int temperatures[][2]){
int val = 0;
for(int i = 1; i < 12; ++i){
if(temperatures[val][1] > temperatures[i][1]){
val = i;
}
}
return val;
}
int main(){
ifstream in;
string fName;
cout << \"Enter file name: \";
cin >> fName;
in.open(fName.c_str());
int highLowTemperatures[12][2];
if(in.is_open()){
getData(in, highLowTemperatures);
cout << \"Highest temperature of the year is \" <<
highLowTemperatures[indexHighTemp(highLowTemperatures)][0] << endl;
cout << \"Lowest temperature of the year is \" <<
highLowTemperatures[indexLowTemp(highLowTemperatures)][1] << endl;
}
else{
cout << \"Can not open the file\" << endl;
}
return 0;
}.
Wilhelm Johannsen first proposed the distinction between genotype a.pdf
Wilhelm Johannsen first proposed distinguishing between an organism's genotype and phenotype in 1909 while working in Denmark. The genotype refers to an organism's hereditary dispositions, while the phenotype describes how those dispositions are manifested physically. Johannsen developed this distinction between an organism's genetic makeup and observable traits as an outgrowth of his work studying heredity.
True, they are identical Thats just the bottom vi.pdf
True, they are identical Thats just the bottom view and top view of the molecule
Man please rate.
Solution
True, they are identical Thats just the bottom view and top view of the molecule
Man please rate..
The NH2OH is in acidic solution as a reactant and.pdf
The NH2OH is in acidic solution as a reactant and copper is reduced to Cu+. The
rest is confusing so here you go. 2Cu2+ + 2NH3OH+ -> 2Cu+ + 2H20 + N2 + 4H+......your
welcome.
Solution
The NH2OH is in acidic solution as a reactant and copper is reduced to Cu+. The
rest is confusing so here you go. 2Cu2+ + 2NH3OH+ -> 2Cu+ + 2H20 + N2 + 4H+......your
welcome..
The atom is a basic unit of matter consisting of .pdf
The atom is a basic unit of matter consisting of central nucleus surrounded by a
cloud of negatively charged electrons. The atomic nucleus contains a mix of positively charged
protons and electrically neutral neutrons (except in the case of hydrogen-1, which is the only
stable nuclide with no neutrons). The electrons of an atom are bound to the nucleus by the
electromagnetic force. Likewise, a group of atoms can remain bound to each other, forming a
molecule. An atom containing an equal number of protons and electrons is electrically neutral,
otherwise it has a positive charge if there are fewer electrons (electron deficiency) or negative
charge if there are more electrons (electron excess). A positively or negatively charged atom is
known as an ion. An atom is classified according to the number of protons and neutrons in its
nucleus: the number of protons determines the chemical element, and the number of neutrons
determines the isotope of the element.[1] The name atom comes from the Greek ?t?µ?? (atomos,
\"indivisible\") from ?- (a-, \"not\") and t?µ?? (temno, \"I cut\")[2], which means uncuttable, or
indivisible, something that cannot be divided further.[3] The concept of an atom as an indivisible
component of matter was first proposed by early Indian and Greek philosophers. In the 17th and
18th centuries, chemists provided a physical basis for this idea by showing that certain
substances could not be further broken down by chemical methods. During the late 19th and
early 20th centuries, physicists discovered subatomic components and structure inside the atom,
thereby demonstrating that the \'atom\' was divisible. The principles of quantum mechanics were
used to successfully model the atom.[4][5] Atoms are minuscule objects with proportionately
tiny masses. Atoms can only be observed individually using special instruments such as the
scanning tunneling microscope. Over 99.94% of an atom\'s mass is concentrated in the
nucleus,[note 1] with protons and neutrons having roughly equal mass. Each element has at least
one isotope with unstable nuclei that can undergo radioactive decay. This can result in a
transmutation that changes the number of protons or neutrons in a nucleus.[6] Electrons that are
bound to atoms possess a set of stable energy levels, or orbitals, and can undergo transitions
between them by absorbing or emitting photons that match the energy differences between the
levels. The electrons determine the chemical properties of an element, and strongly influence an
atom\'s magnetic properties.
Solution
The atom is a basic unit of matter consisting of central nucleus surrounded by a
cloud of negatively charged electrons. The atomic nucleus contains a mix of positively charged
protons and electrically neutral neutrons (except in the case of hydrogen-1, which is the only
stable nuclide with no neutrons). The electrons of an atom are bound to the nucleus by the
electromagnetic force. Likewise, a group of atoms can remain bound to each .
see the inner one is union of events from 1-inf. .pdf
see the inner one is union of events from 1-inf. 2-inf. 3-inf and so on
Solution
see the inner one is union of events from 1-inf. 2-inf. 3-inf and so on.
No you wouldnt include the intermediates in the.pdf
No you wouldn\'t include the intermediates in the rate law but you need to identify
them
Solution
No you wouldn\'t include the intermediates in the rate law but you need to identify
them.
mole of NaOH= 1240 =0.3 so molarity = 0.31.8 = .pdf
mole of NaOH= 12/40 =0.3 so molarity = 0.3/1.8 = 0.167 so ans B) 0.167
Solution
mole of NaOH= 12/40 =0.3 so molarity = 0.3/1.8 = 0.167 so ans B) 0.167.
millimoles of HBr=500.04=2 millimoles of KOH=25.pdf
millimoles of HBr=50*0.04=2 millimoles of KOH=25*.107=2.675 millimoles of
acid remaining=0 millimoles of base remaining=0.675 Kb=[OH-][K+]/[KOH] 1=[OH-]*2/0.675
[oh-]=0.675/2 =0.3375 pH=14+log(0.3375) =13.53
Solution
millimoles of HBr=50*0.04=2 millimoles of KOH=25*.107=2.675 millimoles of
acid remaining=0 millimoles of base remaining=0.675 Kb=[OH-][K+]/[KOH] 1=[OH-]*2/0.675
[oh-]=0.675/2 =0.3375 pH=14+log(0.3375) =13.53.
initially 2 moles of HCl per L of water suppose x.pdf
initially 2 moles of HCl per L of water suppose x L of water is added more final
Molarity =2/(x+1)=.1 =>x=20-1=19L
Solution
initially 2 moles of HCl per L of water suppose x L of water is added more final
Molarity =2/(x+1)=.1 =>x=20-1=19L.
i cant see the question.. please post the complet.pdf
i cant see the question.. please post the complete question so dat i can provide uthe
best possible answer.....
Solution
i cant see the question.. please post the complete question so dat i can provide uthe
best possible answer......
hey pls give the reactions.... .pdf
This short document appears to be a request for chemical reactions but provides no other context or details. It does not contain enough information to generate a meaningful 3 sentence summary.
FT-Raman spectra of some calcium phosphates, calc.pdf
FT-Raman spectra of some calcium phosphates, calcium carbonate and calcium
hydroxide are reported. Comparing with dispersive Raman spectra, it is shown that some strong
bands in the FT-Raman spectra of these compounds which have been erroneously assigned to
Ramanbands, are in fact fluorescence emission bands. The origin of these fluorescencebands
have been discussed in terms of rare earth impurities in a particular structure. It is further
concluded that in using FT-Raman spectroscopy of minerals, care should be taken in not
confusing the fluorescencebands with Raman scattering bands. Among instrumental techniques,
fluorescence spectroscopy is recognized as one of the more sensitive. In fluorescence, the
intensity of the emission of the sample is measured. The reason for the high sensitivity of
fluorescence techniques is that the emission signal is measured above a low background level.
This is inherently more sensitive than comparing two relatively large signals as in absorption
spectroscopy. The sensitivity of fluorescence techniques is as much as 1000 times more sensitive
than absorption spectroscopy. A spectrofluorometer with high sensitivity is an asset for the
researcher. It is obvious that an instrument with high sensitivity will be able to perform
experiments with the fluorescing species at a low concentration. For analytical measurements,
sensitivity determines the detection limit of a material. For research in life science, it may be
important to work at concentration levels that are very low. This is almost always the case in
biomedical research. A more subtle requirement is the need to measure changes in fluorescence
during the course of an experiment. Even though the gross signal levels may be high, the changes
in the signal may be quite small. The ability to reliably measure these small changes is directly
dependent upon the sensitivity of the fluorometer. The measurement of sensitivity has been
discussed in the literature [1]. Unfortunately, there is little consensus on a standard procedure for
determining this important capability. This paper discusses the measurement of sensitivity and
details two methods that are applicable to any instrument. The Spectrofluorometer The
sensitivity of fluorescence is dependent on both the fluorophore and the instrument. The response
of a fluorophore will depend on the molar absorptivity and the quantum yield. These factors are,
in general, beyond the control of the analyst. The sensitivity of a spectrofluorometer depends on
a number of factors. Instrumental contributions to sensitivity are described below. 1. Source
intensity. In general, a brighter excitation source will result in brighter emission. The source for
most fluorometers is a xenon arc, which has a high intensity between 200-900 nm, the spectral
region where most fluorescence experiments are performed. While a high power arc lamp is
good for highest intensity, a more important criteria is the brightness of the lamp. The brightness.
H is oxidised here no O2 is not oxidised as it g.pdf
H is oxidised here no O2 is not oxidised as it gains electrons yes Cu is reduced
here
Solution
H is oxidised here no O2 is not oxidised as it gains electrons yes Cu is reduced
here.
Current Stock price = Expected dividend (Required rate of return -.pdf
Current Stock price = Expected dividend / (Required rate of return - Growth rate)
= $2.54 / (11% - 3.50%)
= $2.54 / 7.50%
= $33.87
Current Stock price is $33.87.
Solution
Current Stock price = Expected dividend / (Required rate of return - Growth rate)
= $2.54 / (11% - 3.50%)
= $2.54 / 7.50%
= $33.87
Current Stock price is $33.87..
YES the news stories in the Illustrated American sympathetic or non-.pdf
YES the news stories in the Illustrated American sympathetic or non-sympathetic to the cause of
the strikers.
The Homestead Strike - this was at the Homestead Steel Works, owned by Andrew Carnegie but
run by Henry Clay Frick. Labor went on strike because of pay rates that were continually
adjusted down as well as poor working conditions. Carnegie was in Scotland at this time and
Frick was the man on the spot and in charge. A miscommunication between these two resulted in
Frick using too much force to suppress the strikers. Labor was responsible for the strike, but
Frick (and maybe Carnegie) were resposible for subsequent events and deaths.
The Homestead strike was organized and purposeful, a harbinger of the type of strike which
would mark the modern age of labor relations in the United States. The AA strike at the
Homestead steel mill in 1892 was different from previous large-scale strikes in American history
such as the Great railroad strike of 1877 or the Great Southwest Railroad Strike of 1886. Earlier
strikes had been largely leaderless and disorganized mass uprisings of workers.
The AFL refused to call for a boycott of Carnegie products in September 1892. Wholesale
crossing of the picket line occurred, first among Eastern European immigrants and then among
all workers. The strike had collapsed so much that the state militia pulled out on October 13,
ending the 95-day occupation. The AA was nearly bankrupted by the job action. Weekly Union
relief for a member averaged $6.25 but totalled a staggering $10,000 per week when including
1,600 strikers. With only 192 out of more than 3,800 strikers in attendance, the Homestead
chapter of the AA voted, 101 to 91, to return to work on November 20, 1892.
Solution
YES the news stories in the Illustrated American sympathetic or non-sympathetic to the cause of
the strikers.
The Homestead Strike - this was at the Homestead Steel Works, owned by Andrew Carnegie but
run by Henry Clay Frick. Labor went on strike because of pay rates that were continually
adjusted down as well as poor working conditions. Carnegie was in Scotland at this time and
Frick was the man on the spot and in charge. A miscommunication between these two resulted in
Frick using too much force to suppress the strikers. Labor was responsible for the strike, but
Frick (and maybe Carnegie) were resposible for subsequent events and deaths.
The Homestead strike was organized and purposeful, a harbinger of the type of strike which
would mark the modern age of labor relations in the United States. The AA strike at the
Homestead steel mill in 1892 was different from previous large-scale strikes in American history
such as the Great railroad strike of 1877 or the Great Southwest Railroad Strike of 1886. Earlier
strikes had been largely leaderless and disorganized mass uprisings of workers.
The AFL refused to call for a boycott of Carnegie products in September 1892. Wholesale
crossing of the picket line occurred, first among Eastern European i.
Well its quite an easy one. The given curve is a cardioid. The limit.pdf
Well its quite an easy one. The given curve is a cardioid. The limits of theta, unless otherwise
mentioned is always from 0 to . The secret to finding it is that, cos=(r-4)/3 . So for any value of r
in polar coordinate, (r-4)/3 will give some value. But we know cos=cos(2-), so the same value
wil be given by and 2- . So if we put =2-, we get = which is maximum range for non repeatation
of r values. A simpler way to find the range is:
You get the minimum value of r at cos=-1, so minimum r=1. So range of r must be from 1 to 4 +
3cos.
r is minimm at = and maximum of r is when cos=1 that is =0. So your can range from 0 to .
However, integrating from 0 to will only give you half the area ( upper part of curve). So
multiply the answer with 2 to get the full area.
Solution
Well its quite an easy one. The given curve is a cardioid. The limits of theta, unless otherwise
mentioned is always from 0 to . The secret to finding it is that, cos=(r-4)/3 . So for any value of r
in polar coordinate, (r-4)/3 will give some value. But we know cos=cos(2-), so the same value
wil be given by and 2- . So if we put =2-, we get = which is maximum range for non repeatation
of r values. A simpler way to find the range is:
You get the minimum value of r at cos=-1, so minimum r=1. So range of r must be from 1 to 4 +
3cos.
r is minimm at = and maximum of r is when cos=1 that is =0. So your can range from 0 to .
However, integrating from 0 to will only give you half the area ( upper part of curve). So
multiply the answer with 2 to get the full area..
Total ionic Equation is 2Cu2+ +4 Cl - + 4H+ +SO42- --- Cu2SiO4+ 4.pdf
Total ionic Equation is : 2Cu2+ +4 Cl - + 4H+ +SO42- ---> Cu2SiO4+ 4H+ + 4Cl-
Spectator ions are : H+ & Cl -
Spectator ions are : H+ & Cl -
Net ionic Equation is : 2Cu2+ + SO42- --->Cu2SiO4
Solution
Total ionic Equation is : 2Cu2+ +4 Cl - + 4H+ +SO42- ---> Cu2SiO4+ 4H+ + 4Cl-
Spectator ions are : H+ & Cl -
Spectator ions are : H+ & Cl -
Net ionic Equation is : 2Cu2+ + SO42- --->Cu2SiO4.
AnswerSOCS are the supressor of cytokine signaling proteins and a.pdf
Answer:
SOCS are the supressor of cytokine signaling proteins and are negative feedback regulators of
JAK-STAT signaling pathway. It means the SOCS are inhibitors of activation of JAK-STAT
pathway proteins.
As JAK-STAT pathway plays essential roles in initiation and development of cancers, mutation
in SOCS family proteins could be one of the mechanism of abnormal activation of STAT .
The expression of SOCS are inversly related to the STAT-3 activation. Hence overexpression of
SOCS may not be lead to cancer risk. Overexpression of SOCS may supress the cytokine
signaling or inhibit cytokine signaling by
(i) Blocking STAT recruitment to the cytokine receptor
(ii) may target protein for degradation
(iii) bind to JAK and inhibit their kinase
(iv) target JAK for degradation via proteosome pathway
Solution
Answer:
SOCS are the supressor of cytokine signaling proteins and are negative feedback regulators of
JAK-STAT signaling pathway. It means the SOCS are inhibitors of activation of JAK-STAT
pathway proteins.
As JAK-STAT pathway plays essential roles in initiation and development of cancers, mutation
in SOCS family proteins could be one of the mechanism of abnormal activation of STAT .
The expression of SOCS are inversly related to the STAT-3 activation. Hence overexpression of
SOCS may not be lead to cancer risk. Overexpression of SOCS may supress the cytokine
signaling or inhibit cytokine signaling by
(i) Blocking STAT recruitment to the cytokine receptor
(ii) may target protein for degradation
(iii) bind to JAK and inhibit their kinase
(iv) target JAK for degradation via proteosome pathway.
Advanced Java[Extra Concepts, Not Difficult].docx
This is part 2 of my Java Learning Journey. This contains Hashing, ArrayList, LinkedList, Date and Time Classes, Calendar Class and more.
More Related Content
More from anupamele
Alternative 1 Continuing present jobPV = Pmt x (1 + i) x (1 - (1 .pdf
Alternative 1: Continuing present job
PV = Pmt x (1 + i) x (1 - (1 + g) n x (1 + i)-n) / (i - g)
Pmt = 60,000
Discount rate (i) = 12%
Growth rate (g) = 14%
No of years (n) = 10
= 60000 × (1+0.12) × (1- (1+0.14)10 × (1+0.12)-10/ (0.12-0.14)
= 650583.51
(Or)
Year
Cashflow
Present value@12%
Discounted cashflow
1
60000.00
1
60000.00
2
68400.00
0.8929
61071.43
3
77976.00
0.7972
62161.99
4
88892.64
0.7118
63272.03
5
101337.61
0.6355
64401.88
6
115524.87
0.5674
65551.92
7
131698.36
0.5066
66722.49
8
150136.13
0.4523
67913.96
9
171155.19
0.4039
69126.71
10
195116.91
0.3606
70361.11
Net Present value
650583.51
Alternative2: Going for MBA
Year
Cash flow
Present value@12%
Discounted cash flow
1
-45000.00
1
-45000.00
2
-45000.00
0.8929
-40178.57
3
90000.00
0.7972
71747.45
4
99000.00
0.7118
70466.24
5
108900.00
0.6355
69207.92
6
119790.00
0.5674
67972.06
7
131769.00
0.5066
66758.28
8
144945.90
0.4523
65566.16
9
159440.49
0.4039
64395.34
10
175384.54
0.3606
63245.42
Net present value
454180.31
The NPV of most attractive choice of continuing current job is 650583.51
Year
Cashflow
Present value@12%
Discounted cashflow
1
60000.00
1
60000.00
2
68400.00
0.8929
61071.43
3
77976.00
0.7972
62161.99
4
88892.64
0.7118
63272.03
5
101337.61
0.6355
64401.88
6
115524.87
0.5674
65551.92
7
131698.36
0.5066
66722.49
8
150136.13
0.4523
67913.96
9
171155.19
0.4039
69126.71
10
195116.91
0.3606
70361.11
Net Present value
650583.51
Solution
Alternative 1: Continuing present job
PV = Pmt x (1 + i) x (1 - (1 + g) n x (1 + i)-n) / (i - g)
Pmt = 60,000
Discount rate (i) = 12%
Growth rate (g) = 14%
No of years (n) = 10
= 60000 × (1+0.12) × (1- (1+0.14)10 × (1+0.12)-10/ (0.12-0.14)
= 650583.51
(Or)
Year
Cashflow
Present value@12%
Discounted cashflow
1
60000.00
1
60000.00
2
68400.00
0.8929
61071.43
3
77976.00
0.7972
62161.99
4
88892.64
0.7118
63272.03
5
101337.61
0.6355
64401.88
6
115524.87
0.5674
65551.92
7
131698.36
0.5066
66722.49
8
150136.13
0.4523
67913.96
9
171155.19
0.4039
69126.71
10
195116.91
0.3606
70361.11
Net Present value
650583.51
Alternative2: Going for MBA
Year
Cash flow
Present value@12%
Discounted cash flow
1
-45000.00
1
-45000.00
2
-45000.00
0.8929
-40178.57
3
90000.00
0.7972
71747.45
4
99000.00
0.7118
70466.24
5
108900.00
0.6355
69207.92
6
119790.00
0.5674
67972.06
7
131769.00
0.5066
66758.28
8
144945.90
0.4523
65566.16
9
159440.49
0.4039
64395.34
10
175384.54
0.3606
63245.42
Net present value
454180.31
The NPV of most attractive choice of continuing current job is 650583.51
Year
Cashflow
Present value@12%
Discounted cashflow
1
60000.00
1
60000.00
2
68400.00
0.8929
61071.43
3
77976.00
0.7972
62161.99
4
88892.64
0.7118
63272.03
5
101337.61
0.6355
64401.88
6
115524.87
0.5674
65551.92
7
131698.36
0.5066
66722.49
8
150136.13
0.4523
67913.96
9
171155.19
0.4039
69126.71
10
195116.91
0.3606
70361.11
Net Present value
650583.51.
#include iostream #include fstreamusing namespace std;void.pdf
#include
#include
using namespace std;
void getData(ifstream &in, int highLowTemperatures[][2]){
for(int i = 0; i < 12; ++i){
in >> highLowTemperatures[i][0];
}
for(int i = 0; i < 12; ++i){
in >> highLowTemperatures[i][1];
}
}
int indexHighTemp(int temperatures[][2]){
int val = 0;
for(int i = 1; i < 12; ++i){
if(temperatures[val][0] < temperatures[i][0]){
val = i;
}
}
return val;
}
int indexLowTemp(int temperatures[][2]){
int val = 0;
for(int i = 1; i < 12; ++i){
if(temperatures[val][1] > temperatures[i][1]){
val = i;
}
}
return val;
}
int main(){
ifstream in;
string fName;
cout << \"Enter file name: \";
cin >> fName;
in.open(fName.c_str());
int highLowTemperatures[12][2];
if(in.is_open()){
getData(in, highLowTemperatures);
cout << \"Highest temperature of the year is \" <<
highLowTemperatures[indexHighTemp(highLowTemperatures)][0] << endl;
cout << \"Lowest temperature of the year is \" <<
highLowTemperatures[indexLowTemp(highLowTemperatures)][1] << endl;
}
else{
cout << \"Can not open the file\" << endl;
}
return 0;
}
Solution
#include
#include
using namespace std;
void getData(ifstream &in, int highLowTemperatures[][2]){
for(int i = 0; i < 12; ++i){
in >> highLowTemperatures[i][0];
}
for(int i = 0; i < 12; ++i){
in >> highLowTemperatures[i][1];
}
}
int indexHighTemp(int temperatures[][2]){
int val = 0;
for(int i = 1; i < 12; ++i){
if(temperatures[val][0] < temperatures[i][0]){
val = i;
}
}
return val;
}
int indexLowTemp(int temperatures[][2]){
int val = 0;
for(int i = 1; i < 12; ++i){
if(temperatures[val][1] > temperatures[i][1]){
val = i;
}
}
return val;
}
int main(){
ifstream in;
string fName;
cout << \"Enter file name: \";
cin >> fName;
in.open(fName.c_str());
int highLowTemperatures[12][2];
if(in.is_open()){
getData(in, highLowTemperatures);
cout << \"Highest temperature of the year is \" <<
highLowTemperatures[indexHighTemp(highLowTemperatures)][0] << endl;
cout << \"Lowest temperature of the year is \" <<
highLowTemperatures[indexLowTemp(highLowTemperatures)][1] << endl;
}
else{
cout << \"Can not open the file\" << endl;
}
return 0;
}.
Wilhelm Johannsen first proposed the distinction between genotype a.pdf
Wilhelm Johannsen first proposed distinguishing between an organism's genotype and phenotype in 1909 while working in Denmark. The genotype refers to an organism's hereditary dispositions, while the phenotype describes how those dispositions are manifested physically. Johannsen developed this distinction between an organism's genetic makeup and observable traits as an outgrowth of his work studying heredity.
True, they are identical Thats just the bottom vi.pdf
True, they are identical Thats just the bottom view and top view of the molecule
Man please rate.
Solution
True, they are identical Thats just the bottom view and top view of the molecule
Man please rate..
The NH2OH is in acidic solution as a reactant and.pdf
The NH2OH is in acidic solution as a reactant and copper is reduced to Cu+. The
rest is confusing so here you go. 2Cu2+ + 2NH3OH+ -> 2Cu+ + 2H20 + N2 + 4H+......your
welcome.
Solution
The NH2OH is in acidic solution as a reactant and copper is reduced to Cu+. The
rest is confusing so here you go. 2Cu2+ + 2NH3OH+ -> 2Cu+ + 2H20 + N2 + 4H+......your
welcome..
The atom is a basic unit of matter consisting of .pdf
The atom is a basic unit of matter consisting of central nucleus surrounded by a
cloud of negatively charged electrons. The atomic nucleus contains a mix of positively charged
protons and electrically neutral neutrons (except in the case of hydrogen-1, which is the only
stable nuclide with no neutrons). The electrons of an atom are bound to the nucleus by the
electromagnetic force. Likewise, a group of atoms can remain bound to each other, forming a
molecule. An atom containing an equal number of protons and electrons is electrically neutral,
otherwise it has a positive charge if there are fewer electrons (electron deficiency) or negative
charge if there are more electrons (electron excess). A positively or negatively charged atom is
known as an ion. An atom is classified according to the number of protons and neutrons in its
nucleus: the number of protons determines the chemical element, and the number of neutrons
determines the isotope of the element.[1] The name atom comes from the Greek ?t?µ?? (atomos,
\"indivisible\") from ?- (a-, \"not\") and t?µ?? (temno, \"I cut\")[2], which means uncuttable, or
indivisible, something that cannot be divided further.[3] The concept of an atom as an indivisible
component of matter was first proposed by early Indian and Greek philosophers. In the 17th and
18th centuries, chemists provided a physical basis for this idea by showing that certain
substances could not be further broken down by chemical methods. During the late 19th and
early 20th centuries, physicists discovered subatomic components and structure inside the atom,
thereby demonstrating that the \'atom\' was divisible. The principles of quantum mechanics were
used to successfully model the atom.[4][5] Atoms are minuscule objects with proportionately
tiny masses. Atoms can only be observed individually using special instruments such as the
scanning tunneling microscope. Over 99.94% of an atom\'s mass is concentrated in the
nucleus,[note 1] with protons and neutrons having roughly equal mass. Each element has at least
one isotope with unstable nuclei that can undergo radioactive decay. This can result in a
transmutation that changes the number of protons or neutrons in a nucleus.[6] Electrons that are
bound to atoms possess a set of stable energy levels, or orbitals, and can undergo transitions
between them by absorbing or emitting photons that match the energy differences between the
levels. The electrons determine the chemical properties of an element, and strongly influence an
atom\'s magnetic properties.
Solution
The atom is a basic unit of matter consisting of central nucleus surrounded by a
cloud of negatively charged electrons. The atomic nucleus contains a mix of positively charged
protons and electrically neutral neutrons (except in the case of hydrogen-1, which is the only
stable nuclide with no neutrons). The electrons of an atom are bound to the nucleus by the
electromagnetic force. Likewise, a group of atoms can remain bound to each .
see the inner one is union of events from 1-inf. .pdf
see the inner one is union of events from 1-inf. 2-inf. 3-inf and so on
Solution
see the inner one is union of events from 1-inf. 2-inf. 3-inf and so on.
No you wouldnt include the intermediates in the.pdf
No you wouldn\'t include the intermediates in the rate law but you need to identify
them
Solution
No you wouldn\'t include the intermediates in the rate law but you need to identify
them.
mole of NaOH= 1240 =0.3 so molarity = 0.31.8 = .pdf
mole of NaOH= 12/40 =0.3 so molarity = 0.3/1.8 = 0.167 so ans B) 0.167
Solution
mole of NaOH= 12/40 =0.3 so molarity = 0.3/1.8 = 0.167 so ans B) 0.167.
millimoles of HBr=500.04=2 millimoles of KOH=25.pdf
millimoles of HBr=50*0.04=2 millimoles of KOH=25*.107=2.675 millimoles of
acid remaining=0 millimoles of base remaining=0.675 Kb=[OH-][K+]/[KOH] 1=[OH-]*2/0.675
[oh-]=0.675/2 =0.3375 pH=14+log(0.3375) =13.53
Solution
millimoles of HBr=50*0.04=2 millimoles of KOH=25*.107=2.675 millimoles of
acid remaining=0 millimoles of base remaining=0.675 Kb=[OH-][K+]/[KOH] 1=[OH-]*2/0.675
[oh-]=0.675/2 =0.3375 pH=14+log(0.3375) =13.53.
initially 2 moles of HCl per L of water suppose x.pdf
initially 2 moles of HCl per L of water suppose x L of water is added more final
Molarity =2/(x+1)=.1 =>x=20-1=19L
Solution
initially 2 moles of HCl per L of water suppose x L of water is added more final
Molarity =2/(x+1)=.1 =>x=20-1=19L.
i cant see the question.. please post the complet.pdf
i cant see the question.. please post the complete question so dat i can provide uthe
best possible answer.....
Solution
i cant see the question.. please post the complete question so dat i can provide uthe
best possible answer......
hey pls give the reactions.... .pdf
This short document appears to be a request for chemical reactions but provides no other context or details. It does not contain enough information to generate a meaningful 3 sentence summary.
FT-Raman spectra of some calcium phosphates, calc.pdf
FT-Raman spectra of some calcium phosphates, calcium carbonate and calcium
hydroxide are reported. Comparing with dispersive Raman spectra, it is shown that some strong
bands in the FT-Raman spectra of these compounds which have been erroneously assigned to
Ramanbands, are in fact fluorescence emission bands. The origin of these fluorescencebands
have been discussed in terms of rare earth impurities in a particular structure. It is further
concluded that in using FT-Raman spectroscopy of minerals, care should be taken in not
confusing the fluorescencebands with Raman scattering bands. Among instrumental techniques,
fluorescence spectroscopy is recognized as one of the more sensitive. In fluorescence, the
intensity of the emission of the sample is measured. The reason for the high sensitivity of
fluorescence techniques is that the emission signal is measured above a low background level.
This is inherently more sensitive than comparing two relatively large signals as in absorption
spectroscopy. The sensitivity of fluorescence techniques is as much as 1000 times more sensitive
than absorption spectroscopy. A spectrofluorometer with high sensitivity is an asset for the
researcher. It is obvious that an instrument with high sensitivity will be able to perform
experiments with the fluorescing species at a low concentration. For analytical measurements,
sensitivity determines the detection limit of a material. For research in life science, it may be
important to work at concentration levels that are very low. This is almost always the case in
biomedical research. A more subtle requirement is the need to measure changes in fluorescence
during the course of an experiment. Even though the gross signal levels may be high, the changes
in the signal may be quite small. The ability to reliably measure these small changes is directly
dependent upon the sensitivity of the fluorometer. The measurement of sensitivity has been
discussed in the literature [1]. Unfortunately, there is little consensus on a standard procedure for
determining this important capability. This paper discusses the measurement of sensitivity and
details two methods that are applicable to any instrument. The Spectrofluorometer The
sensitivity of fluorescence is dependent on both the fluorophore and the instrument. The response
of a fluorophore will depend on the molar absorptivity and the quantum yield. These factors are,
in general, beyond the control of the analyst. The sensitivity of a spectrofluorometer depends on
a number of factors. Instrumental contributions to sensitivity are described below. 1. Source
intensity. In general, a brighter excitation source will result in brighter emission. The source for
most fluorometers is a xenon arc, which has a high intensity between 200-900 nm, the spectral
region where most fluorescence experiments are performed. While a high power arc lamp is
good for highest intensity, a more important criteria is the brightness of the lamp. The brightness.
H is oxidised here no O2 is not oxidised as it g.pdf
H is oxidised here no O2 is not oxidised as it gains electrons yes Cu is reduced
here
Solution
H is oxidised here no O2 is not oxidised as it gains electrons yes Cu is reduced
here.
Current Stock price = Expected dividend (Required rate of return -.pdf
Current Stock price = Expected dividend / (Required rate of return - Growth rate)
= $2.54 / (11% - 3.50%)
= $2.54 / 7.50%
= $33.87
Current Stock price is $33.87.
Solution
Current Stock price = Expected dividend / (Required rate of return - Growth rate)
= $2.54 / (11% - 3.50%)
= $2.54 / 7.50%
= $33.87
Current Stock price is $33.87..
YES the news stories in the Illustrated American sympathetic or non-.pdf
YES the news stories in the Illustrated American sympathetic or non-sympathetic to the cause of
the strikers.
The Homestead Strike - this was at the Homestead Steel Works, owned by Andrew Carnegie but
run by Henry Clay Frick. Labor went on strike because of pay rates that were continually
adjusted down as well as poor working conditions. Carnegie was in Scotland at this time and
Frick was the man on the spot and in charge. A miscommunication between these two resulted in
Frick using too much force to suppress the strikers. Labor was responsible for the strike, but
Frick (and maybe Carnegie) were resposible for subsequent events and deaths.
The Homestead strike was organized and purposeful, a harbinger of the type of strike which
would mark the modern age of labor relations in the United States. The AA strike at the
Homestead steel mill in 1892 was different from previous large-scale strikes in American history
such as the Great railroad strike of 1877 or the Great Southwest Railroad Strike of 1886. Earlier
strikes had been largely leaderless and disorganized mass uprisings of workers.
The AFL refused to call for a boycott of Carnegie products in September 1892. Wholesale
crossing of the picket line occurred, first among Eastern European immigrants and then among
all workers. The strike had collapsed so much that the state militia pulled out on October 13,
ending the 95-day occupation. The AA was nearly bankrupted by the job action. Weekly Union
relief for a member averaged $6.25 but totalled a staggering $10,000 per week when including
1,600 strikers. With only 192 out of more than 3,800 strikers in attendance, the Homestead
chapter of the AA voted, 101 to 91, to return to work on November 20, 1892.
Solution
YES the news stories in the Illustrated American sympathetic or non-sympathetic to the cause of
the strikers.
The Homestead Strike - this was at the Homestead Steel Works, owned by Andrew Carnegie but
run by Henry Clay Frick. Labor went on strike because of pay rates that were continually
adjusted down as well as poor working conditions. Carnegie was in Scotland at this time and
Frick was the man on the spot and in charge. A miscommunication between these two resulted in
Frick using too much force to suppress the strikers. Labor was responsible for the strike, but
Frick (and maybe Carnegie) were resposible for subsequent events and deaths.
The Homestead strike was organized and purposeful, a harbinger of the type of strike which
would mark the modern age of labor relations in the United States. The AA strike at the
Homestead steel mill in 1892 was different from previous large-scale strikes in American history
such as the Great railroad strike of 1877 or the Great Southwest Railroad Strike of 1886. Earlier
strikes had been largely leaderless and disorganized mass uprisings of workers.
The AFL refused to call for a boycott of Carnegie products in September 1892. Wholesale
crossing of the picket line occurred, first among Eastern European i.
Well its quite an easy one. The given curve is a cardioid. The limit.pdf
Well its quite an easy one. The given curve is a cardioid. The limits of theta, unless otherwise
mentioned is always from 0 to . The secret to finding it is that, cos=(r-4)/3 . So for any value of r
in polar coordinate, (r-4)/3 will give some value. But we know cos=cos(2-), so the same value
wil be given by and 2- . So if we put =2-, we get = which is maximum range for non repeatation
of r values. A simpler way to find the range is:
You get the minimum value of r at cos=-1, so minimum r=1. So range of r must be from 1 to 4 +
3cos.
r is minimm at = and maximum of r is when cos=1 that is =0. So your can range from 0 to .
However, integrating from 0 to will only give you half the area ( upper part of curve). So
multiply the answer with 2 to get the full area.
Solution
Well its quite an easy one. The given curve is a cardioid. The limits of theta, unless otherwise
mentioned is always from 0 to . The secret to finding it is that, cos=(r-4)/3 . So for any value of r
in polar coordinate, (r-4)/3 will give some value. But we know cos=cos(2-), so the same value
wil be given by and 2- . So if we put =2-, we get = which is maximum range for non repeatation
of r values. A simpler way to find the range is:
You get the minimum value of r at cos=-1, so minimum r=1. So range of r must be from 1 to 4 +
3cos.
r is minimm at = and maximum of r is when cos=1 that is =0. So your can range from 0 to .
However, integrating from 0 to will only give you half the area ( upper part of curve). So
multiply the answer with 2 to get the full area..
Total ionic Equation is 2Cu2+ +4 Cl - + 4H+ +SO42- --- Cu2SiO4+ 4.pdf
Total ionic Equation is : 2Cu2+ +4 Cl - + 4H+ +SO42- ---> Cu2SiO4+ 4H+ + 4Cl-
Spectator ions are : H+ & Cl -
Spectator ions are : H+ & Cl -
Net ionic Equation is : 2Cu2+ + SO42- --->Cu2SiO4
Solution
Total ionic Equation is : 2Cu2+ +4 Cl - + 4H+ +SO42- ---> Cu2SiO4+ 4H+ + 4Cl-
Spectator ions are : H+ & Cl -
Spectator ions are : H+ & Cl -
Net ionic Equation is : 2Cu2+ + SO42- --->Cu2SiO4.
AnswerSOCS are the supressor of cytokine signaling proteins and a.pdf
Answer:
SOCS are the supressor of cytokine signaling proteins and are negative feedback regulators of
JAK-STAT signaling pathway. It means the SOCS are inhibitors of activation of JAK-STAT
pathway proteins.
As JAK-STAT pathway plays essential roles in initiation and development of cancers, mutation
in SOCS family proteins could be one of the mechanism of abnormal activation of STAT .
The expression of SOCS are inversly related to the STAT-3 activation. Hence overexpression of
SOCS may not be lead to cancer risk. Overexpression of SOCS may supress the cytokine
signaling or inhibit cytokine signaling by
(i) Blocking STAT recruitment to the cytokine receptor
(ii) may target protein for degradation
(iii) bind to JAK and inhibit their kinase
(iv) target JAK for degradation via proteosome pathway
Solution
Answer:
SOCS are the supressor of cytokine signaling proteins and are negative feedback regulators of
JAK-STAT signaling pathway. It means the SOCS are inhibitors of activation of JAK-STAT
pathway proteins.
As JAK-STAT pathway plays essential roles in initiation and development of cancers, mutation
in SOCS family proteins could be one of the mechanism of abnormal activation of STAT .
The expression of SOCS are inversly related to the STAT-3 activation. Hence overexpression of
SOCS may not be lead to cancer risk. Overexpression of SOCS may supress the cytokine
signaling or inhibit cytokine signaling by
(i) Blocking STAT recruitment to the cytokine receptor
(ii) may target protein for degradation
(iii) bind to JAK and inhibit their kinase
(iv) target JAK for degradation via proteosome pathway.
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