TlI -- Tl+ + I- Ksp = 8.9E-8 = [Tl+][I-] (eqn. 1) PbI2 .pdf
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TlI <--> Tl+ + I- Ksp = 8.9E-8 = [Tl+][I-] (eqn. 1) PbI2 <--> Pb2+ + 2I- Ksp = 8.7E-9 = [Pb2+][I-]2 (eqn. 2) In a solution, [Tl+] = 0.50 M, [Pb2+] = 0.50 M When I- is steadily added to the solution, at a particular point, TlI begins to precipitate. And at another point, PbI2 begins to precipitate. Find a point of TlI precipitation: [Tl+][I-] = Ksp = 8.9E-8 (0.50)[I-] = 8.9E-8 [I-] = 1.78E-7 M Find a point of PbI2 precipitation: [Pb2+][I-]2 = 8.7E-9 (0.50)[I-]2 = 8.7E-9 [I-] = 1.32E-4 M Therefore, TlI begins to precipitate before PbI2 (only when [I-] is just 1.78E-7 M) When KI is added, TlI precipitates more and more until when [I-] reaches 1.32E-4, PbI2 just begins to precipitates. At this point, we stop the ions separation. Find remaining [Tl+]: [Tl+][I-] = 8.9E-8 [Tl+](1.32E-4) = 8.9E-8 [Tl+] = 6.74E-4 M The remaining Tl+ is 6.74E-4/0.50 x 100% = 0.13% Tl+ is separated by 99.87% of original amount Solution TlI <--> Tl+ + I- Ksp = 8.9E-8 = [Tl+][I-] (eqn. 1) PbI2 <--> Pb2+ + 2I- Ksp = 8.7E-9 = [Pb2+][I-]2 (eqn. 2) In a solution, [Tl+] = 0.50 M, [Pb2+] = 0.50 M When I- is steadily added to the solution, at a particular point, TlI begins to precipitate. And at another point, PbI2 begins to precipitate. Find a point of TlI precipitation: [Tl+][I-] = Ksp = 8.9E-8 (0.50)[I-] = 8.9E-8 [I-] = 1.78E-7 M Find a point of PbI2 precipitation: [Pb2+][I-]2 = 8.7E-9 (0.50)[I-]2 = 8.7E-9 [I-] = 1.32E-4 M Therefore, TlI begins to precipitate before PbI2 (only when [I-] is just 1.78E-7 M) When KI is added, TlI precipitates more and more until when [I-] reaches 1.32E-4, PbI2 just begins to precipitates. At this point, we stop the ions separation. Find remaining [Tl+]: [Tl+][I-] = 8.9E-8 [Tl+](1.32E-4) = 8.9E-8 [Tl+] = 6.74E-4 M The remaining Tl+ is 6.74E-4/0.50 x 100% = 0.13% Tl+ is separated by 99.87% of original amount.
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![TlI <--> Tl+ + I- Ksp = 8.9E-8 = [Tl+][I-] (eqn. 1)
PbI2 <--> Pb2+ + 2I- Ksp = 8.7E-9 = [Pb2+][I-]2 (eqn. 2)
In a solution, [Tl+] = 0.50 M, [Pb2+] = 0.50 M
When I- is steadily added to the solution, at a particular point, TlI begins to precipitate. And at
another point, PbI2 begins to precipitate.
Find a point of TlI precipitation:
[Tl+][I-] = Ksp = 8.9E-8
(0.50)[I-] = 8.9E-8
[I-] = 1.78E-7 M
Find a point of PbI2 precipitation:
[Pb2+][I-]2 = 8.7E-9
(0.50)[I-]2 = 8.7E-9
[I-] = 1.32E-4 M
Therefore, TlI begins to precipitate before PbI2 (only when [I-] is just 1.78E-7 M)
When KI is added, TlI precipitates more and more until when [I-] reaches 1.32E-4, PbI2 just
begins to precipitates. At this point, we stop the ions separation.
Find remaining [Tl+]:
[Tl+][I-] = 8.9E-8
[Tl+](1.32E-4) = 8.9E-8
[Tl+] = 6.74E-4 M
The remaining Tl+ is 6.74E-4/0.50 x 100% = 0.13%
Tl+ is separated by 99.87% of original amount
Solution
TlI <--> Tl+ + I- Ksp = 8.9E-8 = [Tl+][I-] (eqn. 1)
PbI2 <--> Pb2+ + 2I- Ksp = 8.7E-9 = [Pb2+][I-]2 (eqn. 2)
In a solution, [Tl+] = 0.50 M, [Pb2+] = 0.50 M
When I- is steadily added to the solution, at a particular point, TlI begins to precipitate. And at
another point, PbI2 begins to precipitate.
Find a point of TlI precipitation:
[Tl+][I-] = Ksp = 8.9E-8
(0.50)[I-] = 8.9E-8
[I-] = 1.78E-7 M
Find a point of PbI2 precipitation:](https://image.slidesharecdn.com/tli-tli-ksp8-230628111047-6708f86c/85/TlI-Tl-I-Ksp-8-9E-8-Tl-I-eqn-1-PbI2-pdf-1-320.jpg)
![[Pb2+][I-]2 = 8.7E-9
(0.50)[I-]2 = 8.7E-9
[I-] = 1.32E-4 M
Therefore, TlI begins to precipitate before PbI2 (only when [I-] is just 1.78E-7 M)
When KI is added, TlI precipitates more and more until when [I-] reaches 1.32E-4, PbI2 just
begins to precipitates. At this point, we stop the ions separation.
Find remaining [Tl+]:
[Tl+][I-] = 8.9E-8
[Tl+](1.32E-4) = 8.9E-8
[Tl+] = 6.74E-4 M
The remaining Tl+ is 6.74E-4/0.50 x 100% = 0.13%
Tl+ is separated by 99.87% of original amount](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)
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e Sort the array using Selection Sort
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10
5
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65
3
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45
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bool is_sorted(const std::vector& array) //Conforming for display
{
for (typename std::vector::size_type i = 1; i < array.size(); ++i)
{
if (array[i] < array[i-1])
{
return false;
}
}
return true;
}
std::vector a (10); //Creates a vector of size 10
int arr[10];
template
void accept(std::vector& array) //Accepts data to a vector
{
int c;
for(c = 0; c < 10; c++)
cin>>arr[c];
a.assign (arr,arr+10); //Assign array data to the vector
}
template
void display(std::vector& array) //Display the vector contents
{
for (std::vector::iterator it = array.begin(); it != array.end(); ++it)
std::cout << \' \' << *it;
}
template
void selection(std::vector& arr) //Selection soring operation
{
int mini, loc, temp, x, y;
mini = arr[0];
for(x = 0; x <= arr.size()-1; x++)
{
mini = arr[x]; //x position value of array is considered as minimum value
loc = x; //Stores the location of the minimum value
for(y = x + 1; y <= arr.size()-1; y++)
{
if(arr[y] < mini) //Checks the already minimum value with the array contents
{
mini = arr[y]; //Stores the new minimum value
loc = y; //Stores the location of the minimum value
}
}
if(loc != x) //If it is not the same location
{
temp = arr[x];
arr[x] = arr[loc];
arr[loc] = temp;
}
}
}
template
void bubble(std::vector &ar) //Bubble sort operation
{
bool swapp = true;
while(swapp)
{
swapp = false;
for (int i = 0; i < ar.size()-1; i++)
{
if (ar[i]>ar[i+1] )
{
ar[i] += ar[i+1];
ar[i+1] = ar[i] - ar[i+1];
ar[i] -=ar[i+1];
swapp = true;
}
}
}
}
int main()
{
char ch;
do
{
cout<<\"\ MENU\";
cout<<\"\ a Fill the array with user input\";
cout<<\"\ b Display the array\";
cout<<\"\ c Suffel the array element randomly\";
cout<<\"\ d Sort the array using bubble sort\";
cout<<\"\ e Sort the array using Selection Sort\";
cout<<\"\ f Exit Program\";
cout<<\"\ Enter your choice: \";
cin>>ch;
switch(ch)
{
case \'a\':
case \'A\':
accept(a);
cout<<\"\ Accept data operation completed \ \";
break;
case \'b\':
case \'B\':
display(a);
cout<<\"\ Display data operation completed \ \";
break;
case \'c\':
case \'C\':
bogosort(a);
bool b = is_sorted(a);
if(b == true)
display(a);
cout<<\"\ Bugg Sorting operation completed \ \";
break;
case \'d\':
case \'D\':
bubble(a);
display(a);
cout<<\"\ Bubble Sort operation completed \ \";
break;
case \'e\':
case \'E\':
selection(a);
display(a);
cout<<\"\ Selection Sort operation completed \ \";
break;
case \'f\':
case \'F\':
exit(0);
default:
cout<<\"\ Wrong Choice\";
}
}while(1); //Loop will continue infinitely till F or f is pressed
}
Output
MENU
a Fill the array with user input
b Display the array
c Suffel the array element randomly
d Sort the array using bubble sort
e Sort the array using Selection Sort
f Exit Program
Enter your choice: a
10
5
4
78
9
65
3
1
45
66
Accept data operation completed
MENU
a Fill th.
A) I will look for LP (limited partner) such as Yale Investment fund.pdf
A) I will look for LP (limited partner) such as Yale Investment fund or pension funds or
insurance companies.
B) LP are technically are partners in the fund with limited rights and obligations. LP will fund
the business for some fees and profit sharing but will not participate in day to day affairs leaving
aside the managment part on promoters. This enables promotores to do what they are best at
without worrying about what investors tells them to do.
Solution
A) I will look for LP (limited partner) such as Yale Investment fund or pension funds or
insurance companies.
B) LP are technically are partners in the fund with limited rights and obligations. LP will fund
the business for some fees and profit sharing but will not participate in day to day affairs leaving
aside the managment part on promoters. This enables promotores to do what they are best at
without worrying about what investors tells them to do..
1. The Kepler space telescope detects planets and planet candidates .pdf
1. The Kepler space telescope detects planets and planet candidates by measuring dips in the
brightness of stars to search for planets. The two new planets, named Kepler-34b and Kepler-35b
are both gaseous Saturn-like planets. Kepler-34b orbits its two Sun-like stars every 289 days, and
the stars orbit one another every 28 days. Located in the constellation Cygnus, Kepler-34b and
Kepler-35b are discovered at 4,900 and 5,400 light-years from Earth.Kepler-35b orbits its
smaller and cooler hot stars every 131 days.
2. The milky way galaxy may be home to at least two billion earthlike planets. Up to 2.7% of a
sunlike stars in the milky way are Earth analogs.However, the number is actually far lower than
many scientists anticipated. The Keplar telescope has been scanning a patch of sky near Cygnus
constellation and analyzed the light from the 156,000 stars in its field of view. The published
data showed that 1.4% of sunlike stars should have Earth analogs in the Goldllocks zone. The
Goldilocks zone is dependent on the star\'s luminosity - brighter, hotter stars have habitable
zones with a larger radius. The possibility of liquid water is the main criterion for whether a
planet is suitable for life. The detection of water around a planet is detected via spectroscopy.
Solution
1. The Kepler space telescope detects planets and planet candidates by measuring dips in the
brightness of stars to search for planets. The two new planets, named Kepler-34b and Kepler-35b
are both gaseous Saturn-like planets. Kepler-34b orbits its two Sun-like stars every 289 days, and
the stars orbit one another every 28 days. Located in the constellation Cygnus, Kepler-34b and
Kepler-35b are discovered at 4,900 and 5,400 light-years from Earth.Kepler-35b orbits its
smaller and cooler hot stars every 131 days.
2. The milky way galaxy may be home to at least two billion earthlike planets. Up to 2.7% of a
sunlike stars in the milky way are Earth analogs.However, the number is actually far lower than
many scientists anticipated. The Keplar telescope has been scanning a patch of sky near Cygnus
constellation and analyzed the light from the 156,000 stars in its field of view. The published
data showed that 1.4% of sunlike stars should have Earth analogs in the Goldllocks zone. The
Goldilocks zone is dependent on the star\'s luminosity - brighter, hotter stars have habitable
zones with a larger radius. The possibility of liquid water is the main criterion for whether a
planet is suitable for life. The detection of water around a planet is detected via spectroscopy..
1. When cell in S phase is fused with cell in G1 phase, the cell of .pdf
1. When cell in S phase is fused with cell in G1 phase, the cell of G1 phase is stimulated to
proceed back to S pahse. Stage 1 matches with 2nd cell\'s nucleus is activated to go prematurely,
into DNA synthesis. This is because G1 is premature stage and enter back to S pahse to continue
the synthesis of DNA.
2. When cell in S phase is fused with cell in G2 phase, the cell of G2 phase has no effect of S
pahse on it. Stage 2 matches with 2nd cell\'s nucleus does not respond. This is because G2 cell
contains all things provided by S phase.
3. When cell in M phase is fused with cell in S phase, the cell of S phase undergo chromatin
condensation and chromosomes are visible with single chromatid. Stage 3 matches with 2n cell\'s
nucleus undergoes chromatin condensation, giving visible chromosomes with single chromatid.
This is because, S pahse is synthesis phase and the synthesized DNA is condensed and only one
chromatid of each chromosome is visible.
4. When cell in M phase is fused with cell in G1 phase, The cell of G1 phase undergo
chromosome condensation and chromatin damage. Stage 4 matches with, 2nd cell nucleus
undergo chromosome condensation causing massive chromatin damage. This is because, the G1
pahse is preparatory phase of cell division and M phase is mitotic phase containing disassembly
of chromatids and deacetylation of histone proteins.
Solution
1. When cell in S phase is fused with cell in G1 phase, the cell of G1 phase is stimulated to
proceed back to S pahse. Stage 1 matches with 2nd cell\'s nucleus is activated to go prematurely,
into DNA synthesis. This is because G1 is premature stage and enter back to S pahse to continue
the synthesis of DNA.
2. When cell in S phase is fused with cell in G2 phase, the cell of G2 phase has no effect of S
pahse on it. Stage 2 matches with 2nd cell\'s nucleus does not respond. This is because G2 cell
contains all things provided by S phase.
3. When cell in M phase is fused with cell in S phase, the cell of S phase undergo chromatin
condensation and chromosomes are visible with single chromatid. Stage 3 matches with 2n cell\'s
nucleus undergoes chromatin condensation, giving visible chromosomes with single chromatid.
This is because, S pahse is synthesis phase and the synthesized DNA is condensed and only one
chromatid of each chromosome is visible.
4. When cell in M phase is fused with cell in G1 phase, The cell of G1 phase undergo
chromosome condensation and chromatin damage. Stage 4 matches with, 2nd cell nucleus
undergo chromosome condensation causing massive chromatin damage. This is because, the G1
pahse is preparatory phase of cell division and M phase is mitotic phase containing disassembly
of chromatids and deacetylation of histone proteins..
1. A Holds individuals accountable for their actions.2. E. A,B and.pdf
1. A Holds individuals accountable for their actions.
2. E. A,B and C
HIV Retroviruses have an enzyme, called reverse transcriptase is a retrovirus meaning that it is
an RNA virus . Retroviruses have an enzyme, called reverse transcriptase which gives them the
property to transcribe their RNA into DNA when they enter the host cell.THe binding of the
virus is to the B cells of the immune system and prevent their proliferation and it integrates at
specific site of the host genome.
3.A. Scientists have identified a single gene on chr 22 that, when deleted produces schizophernia
phenotype.
4.E.Williams syndrome is a neurodevelopmental disorder caused due to the deletion of 26 genes
from the long arm of chromosome number 7.The children affected have elfin charcetrs, strong
social and verbal skills but ghave severe heart defects and low muscle tone.
Solution
1. A Holds individuals accountable for their actions.
2. E. A,B and C
HIV Retroviruses have an enzyme, called reverse transcriptase is a retrovirus meaning that it is
an RNA virus . Retroviruses have an enzyme, called reverse transcriptase which gives them the
property to transcribe their RNA into DNA when they enter the host cell.THe binding of the
virus is to the B cells of the immune system and prevent their proliferation and it integrates at
specific site of the host genome.
3.A. Scientists have identified a single gene on chr 22 that, when deleted produces schizophernia
phenotype.
4.E.Williams syndrome is a neurodevelopmental disorder caused due to the deletion of 26 genes
from the long arm of chromosome number 7.The children affected have elfin charcetrs, strong
social and verbal skills but ghave severe heart defects and low muscle tone..
EtSH is the strongest acid, so EtS- is the weakest conjugate base. .pdf
EtSH is the strongest acid, so EtS- is the weakest conjugate base. Two trends operate -
electronegativity increases as you move to the right in the periodic table (thus O and S are more
electronegative than N and P). Thus these atoms are better able to stabilize (-) charge. The other
trend is that atomic size increases moving down in the table, so S is biggest than O. The larger an
atom is, the more it is able to \"spread out\" the negative charge, stabilizing it (in other words, a
charge of (-) is much smaller in relation to the total number of charges present, and the electron
cloud can better accomodate the imbalance).
Solution
EtSH is the strongest acid, so EtS- is the weakest conjugate base. Two trends operate -
electronegativity increases as you move to the right in the periodic table (thus O and S are more
electronegative than N and P). Thus these atoms are better able to stabilize (-) charge. The other
trend is that atomic size increases moving down in the table, so S is biggest than O. The larger an
atom is, the more it is able to \"spread out\" the negative charge, stabilizing it (in other words, a
charge of (-) is much smaller in relation to the total number of charges present, and the electron
cloud can better accomodate the imbalance)..
Particulars Amount $ Share Holders Funds Com.pdf
Particulars Amount $ Share Holders Funds Common Stock : Equity share
capital as on 01.01.2013 500000 Issued During the Year 50000 550000 Retained
Earnings : Previous Years 100000 Current Period Profit/ Loss: Revenue 150000
Expense (90000) 60000 160000 Less: Dividends declared -10000 Total
700000
Solution
Particulars Amount $ Share Holders Funds Common Stock : Equity share
capital as on 01.01.2013 500000 Issued During the Year 50000 550000 Retained
Earnings : Previous Years 100000 Current Period Profit/ Loss: Revenue 150000
Expense (90000) 60000 160000 Less: Dividends declared -10000 Total
700000.
The Nitro group is electron withdrawing, and ther.pdf
The Nitro group is electron withdrawing, and therefore enhances the reactivity of
the carbonyl group in the ester function towards nucleophiles, such as amines and hydrazine. The
Nitro group ensures that any derived compounds will be crystalline solids. The Nitro group itself
can be converted to a variety of other functional groups, such as hydroxylamine (RNHOH) and
amine (RNH2) which can be used as building blocks for more complicated structures: anilines
(derived from reducing NO2 to NH2) can be converted easily to quinolones with antibiotic
properties generally.
Solution
The Nitro group is electron withdrawing, and therefore enhances the reactivity of
the carbonyl group in the ester function towards nucleophiles, such as amines and hydrazine. The
Nitro group ensures that any derived compounds will be crystalline solids. The Nitro group itself
can be converted to a variety of other functional groups, such as hydroxylamine (RNHOH) and
amine (RNH2) which can be used as building blocks for more complicated structures: anilines
(derived from reducing NO2 to NH2) can be converted easily to quinolones with antibiotic
properties generally..
The Systems Development Life Cycle Moderate and large firms with uni.pdf
The Systems Development Life Cycle Moderate and large firms with unique information needs
often develop information systems in-house. That is to say that information technology (IT)
professional within the firm design and program the systems. A greater number of smaller
companies and large firms with relatively standardized information needs opt to purchase
information systems from software vendors. Both approaches represent significant financial and
operational risks. a model for reducing this risk through careful planning, execution, control, and
documentation of key activities.
The five phases of this model are:
1) Business Needs and Strategy
Systems Strategy –Assess Strategic Information Needs –Develop a Strategic Systems Plan
–Create an Action Plan
2) Project Initiation –
Systems Analysis –Conceptualization of Alternative Designs –Systems Evaluation and Selection
3.) In-House Systems Development –Construct the System –Deliver the System
4). Commercial Packages –Trends in Commercial Packages –Choosing a Package
5) Maintenance and Support
The participants in systems development can be classified into three broad groups: systems
professionals, end users, and stakeholders. Systems professionals are systems analysts, systems
designers, and programmers. These individuals actually build the system. They gather facts
about problems with the current system, analyze these facts, and formulate a solution to solve the
problems. The product of their efforts is a new system. End users are those for whom the system
is built. Many users exist at all levels in an organization. These include managers, operations
personnel, accountants, and internal auditors. In some organizations, it is difficult to find
someone who is not a user. During systems development, systems professionals work with the
primary users to obtain an understanding of the users’ problems and a clear statement of their
needs. As defined in Chapter 1, stakeholders are individuals either within or outside the
organization who have an interest in the system but are not end users. These include accountants,
internal auditors, external auditors, and the internal steering committee that oversees systems
development.
Cost/Time Analysis:
As stated before, the cost/time analysis is an attempt to calculate to what degree the project and
system will meet the objectives. The SDLC must address two topics in its support of this area:
the scope of the analysis and the algorithm for doing it. Cost/Benefit Scope The benefit side of
the analysis should be expressed in quantitative terms wherever possible. Qualitative or
intangible benefits usually are reflections of poorly analyzed tangible benefits. The SDLC should
support the process of quantifying all benefits. On the costs side, the SDLC must address
development costs, installation costs and ongoing operational costs. In doing these calculations it
should differentiate between capital costs and expense costs. Cost/Benefit Algorithm The method
of calcu.
You are asked to determine DNA content of a liver tissue. Describe h.pdf
You are asked to determine DNA content of a liver tissue. Describe how you will porceed in
finding amount of DNA in the sample
Take a piece of live tissue and subject it to detergent solution i.e. SDS (sodiumdodecyl sulfate)
so that the cell membrane of the cells will be removed out through micelle formation & then
alcohol is added as DNA soluble in water. This is followed by the formation of aqueous phase
with DNA, precipitate layer with protein and bottom organic layer. Now the purity of DNA
sample is calculated to know the “total amount of DNA in the sample as ng DNA / µg dry
weight”. There are various processes applicable to determine total DNA content using either
cloning, qPCR, sequencing. However, initial purity of DNA can be calculated by measuring
absorbance or OD260/OD280 ratio at 280nm using “spectrophotomemeter” (UV-Visible). The
range of OD with absorption of UV light between 1.8 to 2.0 denotes the presence of DNA.
Lower than this OD ratio, denotes the presence of proteins. The another method of examining the
sample is through & this method is useful to deterimine total DNA concentration by use of the
fluorochrome Hoechst 33258 & filters, & scanning fluorescence spectrophotometer at emission
at 458 nm.
Solution
You are asked to determine DNA content of a liver tissue. Describe how you will porceed in
finding amount of DNA in the sample
Take a piece of live tissue and subject it to detergent solution i.e. SDS (sodiumdodecyl sulfate)
so that the cell membrane of the cells will be removed out through micelle formation & then
alcohol is added as DNA soluble in water. This is followed by the formation of aqueous phase
with DNA, precipitate layer with protein and bottom organic layer. Now the purity of DNA
sample is calculated to know the “total amount of DNA in the sample as ng DNA / µg dry
weight”. There are various processes applicable to determine total DNA content using either
cloning, qPCR, sequencing. However, initial purity of DNA can be calculated by measuring
absorbance or OD260/OD280 ratio at 280nm using “spectrophotomemeter” (UV-Visible). The
range of OD with absorption of UV light between 1.8 to 2.0 denotes the presence of DNA.
Lower than this OD ratio, denotes the presence of proteins. The another method of examining the
sample is through & this method is useful to deterimine total DNA concentration by use of the
fluorochrome Hoechst 33258 & filters, & scanning fluorescence spectrophotometer at emission
at 458 nm..
when we draw a vertical line it will cross the graph twice(that is m.pdf
when we draw a vertical line it will cross the graph twice(that is more than once) if vertical line
cross the graph more than once then relation is not a function .so FALSE
It is not a function
Solution
when we draw a vertical line it will cross the graph twice(that is more than once) if vertical line
cross the graph more than once then relation is not a function .so FALSE
It is not a function.
We Know that Baking powder and Baking soda isused as a leave.pdf
We Know that:
Baking powder and Baking soda isused as a leavening agent helps for the neutralization. and
alsoused in the food recipes to make the substance puffy.
Solution
We Know that:
Baking powder and Baking soda isused as a leavening agent helps for the neutralization. and
alsoused in the food recipes to make the substance puffy..
The fallowing program shows the simple transformation #define GLEW.pdf
The fallowing program shows the simple transformation
#define GLEW_STATIC
#include
#include
#include
#include
#include
#include
#include
const GLchar* vertexSource =
\"#version 150 core\ \"
\"in vec2 position;\"
\"in vec3 color;\"
\"in vec2 texcoord;\"
\"out vec3 Color;\"
\"out vec2 Texcoord;\"
\"uniform mat4 trans;\"
\"void main()\"
\"{\"
\" Color = color;\"
\" Texcoord = texcoord;\"
\" gl_Position = trans * vec4(position, 0.0, 1.0);\"
\"}\";
const GLchar* fragmentSource =
\"#version 150 core\ \"
\"in vec3 Color;\"
\"in vec2 Texcoord;\"
\"out vec4 outColor;\"
\"uniform sampler2D texKitten;\"
\"uniform sampler2D texPuppy;\"
\"void main()\"
\"{\"
\" outColor = mix(texture(texKitten, Texcoord), texture(texPuppy, Texcoord), 0.5);\"
\"}\";
int main()
{
auto t_start = std::chrono::high_resolution_clock::now();
sf::ContextSettings settings;
settings.depthBits = 24;
settings.stencilBits = 8;
sf::Window window(sf::VideoMode(800, 600, 32), \"OpenGL\", sf::Style::Titlebar |
sf::Style::Close, settings);
glewExperimental = GL_TRUE;
glewInit();
GLuint vao;
glGenVertexArrays(1, &vao);
glBindVertexArray(vao);
GLuint vbo;
glGenBuffers(1, &vbo);
GLfloat vertices[] = {
-0.5f, 0.5f, 1.0f, 0.0f, 0.0f, 0.0f, 0.0f,
0.5f, 0.5f, 0.0f, 1.0f, 0.0f, 1.0f, 0.0f,
0.5f, -0.5f, 0.0f, 0.0f, 1.0f, 1.0f, 1.0f,
-0.5f, -0.5f, 1.0f, 1.0f, 1.0f, 0.0f, 1.0f
};
glBindBuffer(GL_ARRAY_BUFFER, vbo);
glBufferData(GL_ARRAY_BUFFER, sizeof(vertices), vertices, GL_STATIC_DRAW);
GLuint ebo;
glGenBuffers(1, &ebo);
GLuint elements[] = {
0, 1, 2,
2, 3, 0
};
glBindBuffer(GL_ELEMENT_ARRAY_BUFFER, ebo);
glBufferData(GL_ELEMENT_ARRAY_BUFFER, sizeof(elements), elements,
GL_STATIC_DRAW);
GLuint vertexShader = glCreateShader(GL_VERTEX_SHADER);
glShaderSource(vertexShader, 1, &vertexSource, NULL);
glCompileShader(vertexShader);
GLuint fragmentShader = glCreateShader(GL_FRAGMENT_SHADER);
glShaderSource(fragmentShader, 1, &fragmentSource, NULL);
glCompileShader(fragmentShader);
GLuint shaderProgram = glCreateProgram();
glAttachShader(shaderProgram, vertexShader);
glAttachShader(shaderProgram, fragmentShader);
glBindFragDataLocation(shaderProgram, 0, \"outColor\");
glLinkProgram(shaderProgram);
glUseProgram(shaderProgram);
GLint posAttrib = glGetAttribLocation(shaderProgram, \"position\");
glEnableVertexAttribArray(posAttrib);
glVertexAttribPointer(posAttrib, 2, GL_FLOAT, GL_FALSE, 7 * sizeof(GLfloat), 0);
GLint colAttrib = glGetAttribLocation(shaderProgram, \"color\");
glEnableVertexAttribArray(colAttrib);
glVertexAttribPointer(colAttrib, 3, GL_FLOAT, GL_FALSE, 7 * sizeof(GLfloat), (void*)(2
* sizeof(GLfloat)));
GLint texAttrib = glGetAttribLocation(shaderProgram, \"texcoord\");
glEnableVertexAttribArray(texAttrib);
glVertexAttribPointer(texAttrib, 2, GL_FLOAT, GL_FALSE, 7 * sizeof(GLfloat), (void*)(5
* sizeof(GLfloat)));
GLuint textures[2];
glGenTextures(2, textures);
int width, height;
unsigned char* image;
glActiveTexture(GL_TEXTURE0);
glBindTexture(GL_TEXTURE_2D, textures[0]);
image = SOIL_load_imag.
Two types of malware -1. Virus A virus is a contagious program o.pdf
Two types of malware :-
1. Virus: A virus is a contagious program or code that attaches itself to another piece of software,
and then reproduces itself when that software is run. Most often this is spread by sharing
software or files between computers.
Virus cannot be spread without a human action. So when you share infected files or send email
with viruses as attachment in email, virus spread in another system also.
If you have turned off your anti-virus protection your system is vulnerable to virus and other
threats.
2. Trojan: This is the most dangerous Malware. Trojans are written with the purpose of
discovering your financial information, taking over your computer’s system resources, and in
larger systems creating a “denial-of-service attack ”.
Denial-of-service attack: an attempt to make a machine or network resource unavailable to those
attempting to reach it. Example: AOL, Yahoo or your business network becoming unavailable.
70% malware are trojan horse. When you open a file that have trojan horse or other kind of
unscrupulous virus program then it replicate in your system and your system is also get infected.
Solution
Two types of malware :-
1. Virus: A virus is a contagious program or code that attaches itself to another piece of software,
and then reproduces itself when that software is run. Most often this is spread by sharing
software or files between computers.
Virus cannot be spread without a human action. So when you share infected files or send email
with viruses as attachment in email, virus spread in another system also.
If you have turned off your anti-virus protection your system is vulnerable to virus and other
threats.
2. Trojan: This is the most dangerous Malware. Trojans are written with the purpose of
discovering your financial information, taking over your computer’s system resources, and in
larger systems creating a “denial-of-service attack ”.
Denial-of-service attack: an attempt to make a machine or network resource unavailable to those
attempting to reach it. Example: AOL, Yahoo or your business network becoming unavailable.
70% malware are trojan horse. When you open a file that have trojan horse or other kind of
unscrupulous virus program then it replicate in your system and your system is also get infected..
The cells of the reproductive organs (Eg ovaries ad testes) undergo.pdf
The cells of the reproductive organs (Eg: ovaries ad testes) undergo meiosis. Meiosis is a type of
nuclear division, which produces four haploid cells from a single diploid cell. Meiosis is divided
into two phases, meiosis I (one diploid nucleus to two haploid nuclei) and meiosis II (two
haploid nuclei to four haploid nuclei). Meiosis I and meiosis II are subdivided into four phases
again namely, prophase, metaphase, anaphase and telophase. Thus, meiosis is divided into a total
of eight phases. The phases of meiosis I are prohase-i, metaphase-I, anaphase-I and telophase-I.
Similarly, the phases of meiosis-II are prohase-II, metaphase-II, anaphase-II and telophase-II.
Prophase is the first phase of cell division in both mitosis and meiosis. During this phase,
cohesion that is present throughout the length of the chromatids is removed except at the
centromere. The chromatids become more visible during this phase. In meiosis, the substages of
prophase-I include, leptonema, zygonema, pachynema, diplonema, diakinesis.
During the metaphase, two chromatids are formed from each chromosome by longitudinal
splitting. These chromatids form the chromosomes of daughter cells.
During anaphase, centromeres are pulled and the chromatids become separated and move
towards the opposite poles. These chromatids now are the chromosomes of the daughter cells,
and are pulled towards the centrioles by means of spindle fibres.
During “metaphase I and II” the chromosomes align at the midway between the two opposite
spindle poles.
In the stage of telophase, cytoplasmic division of the cell takes place (cytokinesis) during which
each of the daughter cell receives one of the daughter nuclei. The interval between each division
is called interphase.
During meiosis, the homologous chromosomes align during the prophase. These chromosomes
crossover at some point and rejoin to form recombinant chromatids, and move towards the
opposite poles during anaphase. The chromosomal crossover ocurs during metaphase-I of the
meiosis,
In meiosis, Alignment of homologous chromosomes at Mid-Prophase –I is characteristic of
meiosis. The two chromosomes sharing the same quality are called “homologous chromosomes.”
These chromosomes crossover at some point and rejoin to form recombinant chromatids. Now,
these chromosomes are separated at Anaphase-I and move to their opposite poles.
During anaphase of mitosis, the sister chromatids get separated and move to the opposite poles.
In contrast to mitosis, the “homologues chromosomes” (but not sister chromatids) move to the
opposite poles in anaphase –I of meiosis. In meiosis, the separation of sister chromatids occur in
anaphase –II.
Solution
The cells of the reproductive organs (Eg: ovaries ad testes) undergo meiosis. Meiosis is a type of
nuclear division, which produces four haploid cells from a single diploid cell. Meiosis is divided
into two phases, meiosis I (one diploid nucleus to two haploid nuclei) and meiosis II (two
haploid nuclei to four h.
Scientific evidences I would give in support of evolution1. Paleo.pdf
Scientific evidences I would give in support of evolution:
1. Paleontological evidence: Fossils are those remains of plants and animals that could not be
degraded. Some of them bear resemblances to present day animals and serve as connecting links,
eg. Duck billed platypus is connecting link between reptiles and mammals. Some of them
represent extinct organisms, eg. dinosaurs. Cross sections of the earth\'s crust from where the
fossils have been isolated determines the geological time scale during which the organism
existed.
2. Comparative anatomy/physiology: Homologous organs of animals such as forelimb bones of
humans, tigers, whales and bats share similar type of arrangement and indicates common
ancestry. Such type of evolution is called \"divergent\". Analogous organs are opposite to
homologous organs, such as wings of butterfly and birds. Although they perform the same
function, they are not anatomically similar. Thus, these are results of \"convergent \"evolution.
3. Similarities in genes and proteins among genetically different organisms also support for the
existence of common ancestry.
4. Natural selection/Industrial melanism: The dominance of dark coloured moths in industrial
areas compared to the white forms is an example of industrial melanism. This type of selection
resulted in the dark coloured moths from becoming a prey of birds compared to the paler ones as
the darker forms camouflage in the dark coloured soots. Another example of natural selection is
antibiotic resistance in bacteria. The resistant bacteria can be isolated by plating the organisms in
a LB agar plate containing the corresponding antibiotic.
5. Adaptive radiation: Darwin\'s finches in Galapagos island is an example of adaptive radiation.
It was found that the different species of finches (with different beak structure) arose from an
initial seed eating finch, enabling them to become vegetarian and insectivorous finches.
Solution
Scientific evidences I would give in support of evolution:
1. Paleontological evidence: Fossils are those remains of plants and animals that could not be
degraded. Some of them bear resemblances to present day animals and serve as connecting links,
eg. Duck billed platypus is connecting link between reptiles and mammals. Some of them
represent extinct organisms, eg. dinosaurs. Cross sections of the earth\'s crust from where the
fossils have been isolated determines the geological time scale during which the organism
existed.
2. Comparative anatomy/physiology: Homologous organs of animals such as forelimb bones of
humans, tigers, whales and bats share similar type of arrangement and indicates common
ancestry. Such type of evolution is called \"divergent\". Analogous organs are opposite to
homologous organs, such as wings of butterfly and birds. Although they perform the same
function, they are not anatomically similar. Thus, these are results of \"convergent \"evolution.
3. Similarities in genes and proteins among genetically different o.
Package scopeis the range of visibility for aparticular element or c.pdf
Package scopeis the range of visibility for aparticular element or construct contained in
apackage.–“range of visibility” describe whether a packageitem (such as function) can be
accessed fromoutside the package or only from other programunits within the same
package.–Any elements declared in a package specificationare consideredpublic.–The elements
are considered private if they can be called from other program units within the samepackage.
Solution
Package scopeis the range of visibility for aparticular element or construct contained in
apackage.–“range of visibility” describe whether a packageitem (such as function) can be
accessed fromoutside the package or only from other programunits within the same
package.–Any elements declared in a package specificationare consideredpublic.–The elements
are considered private if they can be called from other program units within the samepackage..
Point of zero charge related to the adsorption phenomenon I surface .pdf
Point of zero charge related to the adsorption phenomenon I surface chemistry. Here it is the
point if achievement of zero charge density on a surface. If pHzpc is 8.5 that means at this pH
solid submerged in electrolyte will show zero electrical charge on the surface.
P sorption refers to binding of phosphates in other forms to soil particles.
At high pH, P availability is low because of reaction of phosphates to calcium which result in
formation of apatite.
At low pH, phosphate reacts with aluminum, iron etc and results in formation of various other
minerals.
Solution
Point of zero charge related to the adsorption phenomenon I surface chemistry. Here it is the
point if achievement of zero charge density on a surface. If pHzpc is 8.5 that means at this pH
solid submerged in electrolyte will show zero electrical charge on the surface.
P sorption refers to binding of phosphates in other forms to soil particles.
At high pH, P availability is low because of reaction of phosphates to calcium which result in
formation of apatite.
At low pH, phosphate reacts with aluminum, iron etc and results in formation of various other
minerals..
Maternal and parental chromosomes are separated in Anaphase I of mei.pdf
Maternal and parental chromosomes are separated in Anaphase I of meiosis cell division. So
sister chromatids are hold together in the Anaphase I. The Anaphase II is responsible for the
separation of sister chromatids to form chromosomes. Sister chromatids are chromosomes of
either maternal or paternal.
Solution
Maternal and parental chromosomes are separated in Anaphase I of meiosis cell division. So
sister chromatids are hold together in the Anaphase I. The Anaphase II is responsible for the
separation of sister chromatids to form chromosomes. Sister chromatids are chromosomes of
either maternal or paternal..
main.cpp #include iostream #include iomanip #include fs.pdf
main.cpp
#include
#include
#include
#include
#include \"Bank.hpp\"
using namespace std;
int read_accts(Bank& bank, int max_accts)
{
int i = 0;
//ifstream infile(\"C:\\\\Users\\\\Smart PC\\\\Desktop\\\\Assignment 3 (3110)\\\\myinput.txt\");
ifstream infile(\"myinput\");
string whiteSpace;
getline(infile, whiteSpace);
// check is file can be opened
if (infile)
{
// read only first max_accts accounts,
// in order to avoid overflow
for (i = 0; i> firstName;
// check is end of file reached
if (!infile.eof())
{
infile >> lastName;
infile >> ssn;
infile >> accountNumber;
infile >> accountType;
infile >> balance;
infile >> status;
infile >> transactions;
bank.addAccount(firstName, lastName, ssn, accountNumber, accountType, balance,
status);
int index = bank.findAccount(accountNumber);
Account* acc = bank.getAccount(index);
for(int i=0; i> transactionType;
infile >> amount;
if (acc)
acc->addTransaction(Transaction(transactionType, amount));
}
}
else {
break;
}
}
infile.close();
}
else {
cout << \"cannot open inpout file\" << endl;
}
return i;
}
*/
void menu()
{
cout << \"Select one of the following:\" << endl << endl;
cout << \" W - Withdrawal\" << endl;
cout << \" D - Deposit\" << endl;
cout << \" N - New account\" << endl;
cout << \" B - Balance\" << endl;
cout << \" I - Account Info\" << endl;
cout << \" H - Account Info plus Account Transaction History\" << endl;
cout << \" C - Close Account (close but do not delete the account)\" << endl;
cout << \" R - Reopen a closed account\" << endl;
cout << \" X - Delete Account (close and delete the account from the database))\" << endl;
cout << \" Q - Quit\" << endl;
}
void withdrawal(Bank& bank)
{
int index, requested_account;
//prompt for account number
cout << endl << \"Enter the account number: \";
cin >> requested_account;
// account #0 is not valid
if (requested_account != 0)
{
index = bank.findAccount(requested_account);
if (index >= 0)
{
double amount;
cout << \"Enter amount to withdraw: \";
cin >> amount;
// check is there possible to withdraw
if (amount>0)
{
// check is sufficient balance at account
if (bank.getAccount(index)->makeWithdrawal(amount))
{
cout << \"Withdraw is completed.\" << endl;
}
else {
cout << \"Error. Insuffucient funds.\" << endl;
}
}
else {
cout << \"Error. Invalid amount, needed to entyer positiove number.\" << endl;
}
}
else {
cout << \"Error. Account number \\\"\" << requested_account << \"\\\" is not found.\" <<
endl;
}
}
else {
cout << \"Error. Invalid account number.\" << endl;
}
}
void deposit(Bank& bank)
{
int index, requested_account;
//prompt for account number
cout << endl << \"Enter the account number: \";
cin >> requested_account;
// account #0 is not valid
if (requested_account != 0)
{
index = bank.findAccount(requested_account);
if (index >= 0)
{
double amount;
cout << \"Enter amount to deposit: \";
cin >> amount;
// check is there possible to withdraw
if (amount>0)
{
bank.getAccount(index)->makeDeposit(amount); //deposit operation
cout << \"Deposit is complet.
loan amount = 17500present value of annuity = payment per period .pdf
loan amount = 17500
present value of annuity = payment per period * [ 1 - (1+i)^-n ]/i
=>
monthly payment *[1-(1+10.4%/12)^-39]/10.4%/12 = 17500
=>
monthly payment = 530.74
remaining balance = PV of remaining paymens
= 530.74 * [1-(1+10.4%/12)^-28]/10.4%/12
= 13144.78
Solution
loan amount = 17500
present value of annuity = payment per period * [ 1 - (1+i)^-n ]/i
=>
monthly payment *[1-(1+10.4%/12)^-39]/10.4%/12 = 17500
=>
monthly payment = 530.74
remaining balance = PV of remaining paymens
= 530.74 * [1-(1+10.4%/12)^-28]/10.4%/12
= 13144.78.
import javax.swing.; import java.awt.; import java.awt.event..pdf
import javax.swing.*;
import java.awt.*;
import java.awt.event.*;
import java.text.*;
public class convertTemperature extends JApplet implements ActionListener {
JTextField inputField;
JLabel label;
JRadioButton celcius, fahrenheit;
public void init(){
Container contain = getContentPane();
contain.setLayout (new FlowLayout());
inputField = new JTextField(\"\",10);
contain.add(inputField);
celcius= new JRadioButton (\"to Celcius\", true);
contain.add(celcius);
fahrenheit = new JRadioButton(\"to fahrenheit\", false);
contain.add(fahrenheit);
ButtonGroup selection = new ButtonGroup();
selection.add(celcius);
selection.add(fahrenheit);
JButton button1 = new JButton (\"Show Result\");
button1.addActionListener(this);
contain.add(button1);
label= new JLabel (\"Enter temperature and choose an option\");
contain.add(label);
}
public void actionPerformed(ActionEvent e) {
DecimalFormat df = new DecimalFormat (\"#.##\");
double input = Double.parseDouble(inputField.getText());
double convertedTemp = 0.0;
if (celcius.isSelected())
convertedTemp = ((5.0/9.0)*(input - 32.0));
if (fahrenheit.isSelected())
convertedTemp = ((9.0/5.0)*(input + 32.0));
label.setText(String.valueOf(df.format(convertedTemp)));
}
}
Solution
import javax.swing.*;
import java.awt.*;
import java.awt.event.*;
import java.text.*;
public class convertTemperature extends JApplet implements ActionListener {
JTextField inputField;
JLabel label;
JRadioButton celcius, fahrenheit;
public void init(){
Container contain = getContentPane();
contain.setLayout (new FlowLayout());
inputField = new JTextField(\"\",10);
contain.add(inputField);
celcius= new JRadioButton (\"to Celcius\", true);
contain.add(celcius);
fahrenheit = new JRadioButton(\"to fahrenheit\", false);
contain.add(fahrenheit);
ButtonGroup selection = new ButtonGroup();
selection.add(celcius);
selection.add(fahrenheit);
JButton button1 = new JButton (\"Show Result\");
button1.addActionListener(this);
contain.add(button1);
label= new JLabel (\"Enter temperature and choose an option\");
contain.add(label);
}
public void actionPerformed(ActionEvent e) {
DecimalFormat df = new DecimalFormat (\"#.##\");
double input = Double.parseDouble(inputField.getText());
double convertedTemp = 0.0;
if (celcius.isSelected())
convertedTemp = ((5.0/9.0)*(input - 32.0));
if (fahrenheit.isSelected())
convertedTemp = ((9.0/5.0)*(input + 32.0));
label.setText(String.valueOf(df.format(convertedTemp)));
}
}.
Hg is an atom , because an atom means that a smallest individual spe.pdf
Hg is an atom , because an atom means that a smallest individual spece
and it cannot be undivisible further.
H+ is an ion, it is a spece was formed by the H atom after lost one electron from it.
H2O is a molecule ,and it has obtained by the reaction of H2 atoms and O atoms.
O2 and N2 , Ar and CO2 both are called as mixures becaues those were formed
by the mixing from two individual species or molecules.
NaCl is a compound , it has the fixed proportion of both Na and Cl in its compound.
Solution
Hg is an atom , because an atom means that a smallest individual spece
and it cannot be undivisible further.
H+ is an ion, it is a spece was formed by the H atom after lost one electron from it.
H2O is a molecule ,and it has obtained by the reaction of H2 atoms and O atoms.
O2 and N2 , Ar and CO2 both are called as mixures becaues those were formed
by the mixing from two individual species or molecules.
NaCl is a compound , it has the fixed proportion of both Na and Cl in its compound..
Hello! Please see the correction below! False Hydrogen atoms ar.pdf
Hello! Please see the correction below!
False: Hydrogen atoms are often the central atom of a lewis structure. (This doesn\'t make sense
since hydrogen can only form 1 bond!)
True: For a neutral molecule, the number of electrons in the Lewis structure is the sum of the
valence electrons for the atoms.
FALSE: Each atom of a Lewis structure must have eight electrons. (This is the \"octet rule.)
Typically you would have each atom with eight electrons, but as your hint indicated, Hydrogen
only gets 2 electrons. This exception would make this statement false.
True: Electrons of covalent compounds may be shared between atoms.
Sorry about that - Hope this helps! Please rate if it does : )
Solution
Hello! Please see the correction below!
False: Hydrogen atoms are often the central atom of a lewis structure. (This doesn\'t make sense
since hydrogen can only form 1 bond!)
True: For a neutral molecule, the number of electrons in the Lewis structure is the sum of the
valence electrons for the atoms.
FALSE: Each atom of a Lewis structure must have eight electrons. (This is the \"octet rule.)
Typically you would have each atom with eight electrons, but as your hint indicated, Hydrogen
only gets 2 electrons. This exception would make this statement false.
True: Electrons of covalent compounds may be shared between atoms.
Sorry about that - Hope this helps! Please rate if it does : ).
From LLMm has possible four gamates those are LM ,Lm,LM and Lm in me.pdf
From LLMm has possible four gamates those are LM ,Lm,LM and Lm in meiosis 1.then it will
be doubled in meiosis2. But when the secondary non disjunction occurs at M locus that means
than gene is not segregated during meiosis 2. Then you get four gametes for L gene after meiosis
2 and in case of M gene you will have M and m allele which will not segregated in meiosis 2. So
you will have Four gametes. (LM, Lm, LmM and L).
Solution
From LLMm has possible four gamates those are LM ,Lm,LM and Lm in meiosis 1.then it will
be doubled in meiosis2. But when the secondary non disjunction occurs at M locus that means
than gene is not segregated during meiosis 2. Then you get four gametes for L gene after meiosis
2 and in case of M gene you will have M and m allele which will not segregated in meiosis 2. So
you will have Four gametes. (LM, Lm, LmM and L)..
D is correct. Formatting and code conversion are Presentation layer .pdf
D is correct. Formatting and code conversion are Presentation layer functions in the OSI model.
The Application layer in TCP/IP maps to the top three layers of the OSI model.
A, B, and C are incorrect.
Solution
D is correct. Formatting and code conversion are Presentation layer functions in the OSI model.
The Application layer in TCP/IP maps to the top three layers of the OSI model.
A, B, and C are incorrect..
Biological Screening of Herbal Drugs in detailed.
Biological screening of herbal drugs: Introduction and Need for
Phyto-Pharmacological Screening, New Strategies for evaluating
Natural Products, In vitro evaluation techniques for Antioxidants, Antimicrobial and Anticancer drugs. In vivo evaluation techniques
for Anti-inflammatory, Antiulcer, Anticancer, Wound healing, Antidiabetic, Hepatoprotective, Cardio protective, Diuretics and
Antifertility, Toxicity studies as per OECD guidelines
special B.ed 2nd year old paper_20240531.pdf
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Particulars Amount $ Share Holders Funds Common Stock : Equity share
capital as on 01.01.2013 500000 Issued During the Year 50000 550000 Retained
Earnings : Previous Years 100000 Current Period Profit/ Loss: Revenue 150000
Expense (90000) 60000 160000 Less: Dividends declared -10000 Total
700000
Solution
Particulars Amount $ Share Holders Funds Common Stock : Equity share
capital as on 01.01.2013 500000 Issued During the Year 50000 550000 Retained
Earnings : Previous Years 100000 Current Period Profit/ Loss: Revenue 150000
Expense (90000) 60000 160000 Less: Dividends declared -10000 Total
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The Nitro group is electron withdrawing, and ther.pdf
The Nitro group is electron withdrawing, and therefore enhances the reactivity of
the carbonyl group in the ester function towards nucleophiles, such as amines and hydrazine. The
Nitro group ensures that any derived compounds will be crystalline solids. The Nitro group itself
can be converted to a variety of other functional groups, such as hydroxylamine (RNHOH) and
amine (RNH2) which can be used as building blocks for more complicated structures: anilines
(derived from reducing NO2 to NH2) can be converted easily to quinolones with antibiotic
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Solution
The Nitro group is electron withdrawing, and therefore enhances the reactivity of
the carbonyl group in the ester function towards nucleophiles, such as amines and hydrazine. The
Nitro group ensures that any derived compounds will be crystalline solids. The Nitro group itself
can be converted to a variety of other functional groups, such as hydroxylamine (RNHOH) and
amine (RNH2) which can be used as building blocks for more complicated structures: anilines
(derived from reducing NO2 to NH2) can be converted easily to quinolones with antibiotic
properties generally..
The Systems Development Life Cycle Moderate and large firms with uni.pdf
The Systems Development Life Cycle Moderate and large firms with unique information needs
often develop information systems in-house. That is to say that information technology (IT)
professional within the firm design and program the systems. A greater number of smaller
companies and large firms with relatively standardized information needs opt to purchase
information systems from software vendors. Both approaches represent significant financial and
operational risks. a model for reducing this risk through careful planning, execution, control, and
documentation of key activities.
The five phases of this model are:
1) Business Needs and Strategy
Systems Strategy –Assess Strategic Information Needs –Develop a Strategic Systems Plan
–Create an Action Plan
2) Project Initiation –
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3.) In-House Systems Development –Construct the System –Deliver the System
4). Commercial Packages –Trends in Commercial Packages –Choosing a Package
5) Maintenance and Support
The participants in systems development can be classified into three broad groups: systems
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designers, and programmers. These individuals actually build the system. They gather facts
about problems with the current system, analyze these facts, and formulate a solution to solve the
problems. The product of their efforts is a new system. End users are those for whom the system
is built. Many users exist at all levels in an organization. These include managers, operations
personnel, accountants, and internal auditors. In some organizations, it is difficult to find
someone who is not a user. During systems development, systems professionals work with the
primary users to obtain an understanding of the users’ problems and a clear statement of their
needs. As defined in Chapter 1, stakeholders are individuals either within or outside the
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internal auditors, external auditors, and the internal steering committee that oversees systems
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Cost/Time Analysis:
As stated before, the cost/time analysis is an attempt to calculate to what degree the project and
system will meet the objectives. The SDLC must address two topics in its support of this area:
the scope of the analysis and the algorithm for doing it. Cost/Benefit Scope The benefit side of
the analysis should be expressed in quantitative terms wherever possible. Qualitative or
intangible benefits usually are reflections of poorly analyzed tangible benefits. The SDLC should
support the process of quantifying all benefits. On the costs side, the SDLC must address
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You are asked to determine DNA content of a liver tissue. Describe h.pdf
You are asked to determine DNA content of a liver tissue. Describe how you will porceed in
finding amount of DNA in the sample
Take a piece of live tissue and subject it to detergent solution i.e. SDS (sodiumdodecyl sulfate)
so that the cell membrane of the cells will be removed out through micelle formation & then
alcohol is added as DNA soluble in water. This is followed by the formation of aqueous phase
with DNA, precipitate layer with protein and bottom organic layer. Now the purity of DNA
sample is calculated to know the “total amount of DNA in the sample as ng DNA / µg dry
weight”. There are various processes applicable to determine total DNA content using either
cloning, qPCR, sequencing. However, initial purity of DNA can be calculated by measuring
absorbance or OD260/OD280 ratio at 280nm using “spectrophotomemeter” (UV-Visible). The
range of OD with absorption of UV light between 1.8 to 2.0 denotes the presence of DNA.
Lower than this OD ratio, denotes the presence of proteins. The another method of examining the
sample is through & this method is useful to deterimine total DNA concentration by use of the
fluorochrome Hoechst 33258 & filters, & scanning fluorescence spectrophotometer at emission
at 458 nm.
Solution
You are asked to determine DNA content of a liver tissue. Describe how you will porceed in
finding amount of DNA in the sample
Take a piece of live tissue and subject it to detergent solution i.e. SDS (sodiumdodecyl sulfate)
so that the cell membrane of the cells will be removed out through micelle formation & then
alcohol is added as DNA soluble in water. This is followed by the formation of aqueous phase
with DNA, precipitate layer with protein and bottom organic layer. Now the purity of DNA
sample is calculated to know the “total amount of DNA in the sample as ng DNA / µg dry
weight”. There are various processes applicable to determine total DNA content using either
cloning, qPCR, sequencing. However, initial purity of DNA can be calculated by measuring
absorbance or OD260/OD280 ratio at 280nm using “spectrophotomemeter” (UV-Visible). The
range of OD with absorption of UV light between 1.8 to 2.0 denotes the presence of DNA.
Lower than this OD ratio, denotes the presence of proteins. The another method of examining the
sample is through & this method is useful to deterimine total DNA concentration by use of the
fluorochrome Hoechst 33258 & filters, & scanning fluorescence spectrophotometer at emission
at 458 nm..
when we draw a vertical line it will cross the graph twice(that is m.pdf
when we draw a vertical line it will cross the graph twice(that is more than once) if vertical line
cross the graph more than once then relation is not a function .so FALSE
It is not a function
Solution
when we draw a vertical line it will cross the graph twice(that is more than once) if vertical line
cross the graph more than once then relation is not a function .so FALSE
It is not a function.
We Know that Baking powder and Baking soda isused as a leave.pdf
We Know that:
Baking powder and Baking soda isused as a leavening agent helps for the neutralization. and
alsoused in the food recipes to make the substance puffy.
Solution
We Know that:
Baking powder and Baking soda isused as a leavening agent helps for the neutralization. and
alsoused in the food recipes to make the substance puffy..
The fallowing program shows the simple transformation #define GLEW.pdf
The fallowing program shows the simple transformation
#define GLEW_STATIC
#include
#include
#include
#include
#include
#include
#include
const GLchar* vertexSource =
\"#version 150 core\ \"
\"in vec2 position;\"
\"in vec3 color;\"
\"in vec2 texcoord;\"
\"out vec3 Color;\"
\"out vec2 Texcoord;\"
\"uniform mat4 trans;\"
\"void main()\"
\"{\"
\" Color = color;\"
\" Texcoord = texcoord;\"
\" gl_Position = trans * vec4(position, 0.0, 1.0);\"
\"}\";
const GLchar* fragmentSource =
\"#version 150 core\ \"
\"in vec3 Color;\"
\"in vec2 Texcoord;\"
\"out vec4 outColor;\"
\"uniform sampler2D texKitten;\"
\"uniform sampler2D texPuppy;\"
\"void main()\"
\"{\"
\" outColor = mix(texture(texKitten, Texcoord), texture(texPuppy, Texcoord), 0.5);\"
\"}\";
int main()
{
auto t_start = std::chrono::high_resolution_clock::now();
sf::ContextSettings settings;
settings.depthBits = 24;
settings.stencilBits = 8;
sf::Window window(sf::VideoMode(800, 600, 32), \"OpenGL\", sf::Style::Titlebar |
sf::Style::Close, settings);
glewExperimental = GL_TRUE;
glewInit();
GLuint vao;
glGenVertexArrays(1, &vao);
glBindVertexArray(vao);
GLuint vbo;
glGenBuffers(1, &vbo);
GLfloat vertices[] = {
-0.5f, 0.5f, 1.0f, 0.0f, 0.0f, 0.0f, 0.0f,
0.5f, 0.5f, 0.0f, 1.0f, 0.0f, 1.0f, 0.0f,
0.5f, -0.5f, 0.0f, 0.0f, 1.0f, 1.0f, 1.0f,
-0.5f, -0.5f, 1.0f, 1.0f, 1.0f, 0.0f, 1.0f
};
glBindBuffer(GL_ARRAY_BUFFER, vbo);
glBufferData(GL_ARRAY_BUFFER, sizeof(vertices), vertices, GL_STATIC_DRAW);
GLuint ebo;
glGenBuffers(1, &ebo);
GLuint elements[] = {
0, 1, 2,
2, 3, 0
};
glBindBuffer(GL_ELEMENT_ARRAY_BUFFER, ebo);
glBufferData(GL_ELEMENT_ARRAY_BUFFER, sizeof(elements), elements,
GL_STATIC_DRAW);
GLuint vertexShader = glCreateShader(GL_VERTEX_SHADER);
glShaderSource(vertexShader, 1, &vertexSource, NULL);
glCompileShader(vertexShader);
GLuint fragmentShader = glCreateShader(GL_FRAGMENT_SHADER);
glShaderSource(fragmentShader, 1, &fragmentSource, NULL);
glCompileShader(fragmentShader);
GLuint shaderProgram = glCreateProgram();
glAttachShader(shaderProgram, vertexShader);
glAttachShader(shaderProgram, fragmentShader);
glBindFragDataLocation(shaderProgram, 0, \"outColor\");
glLinkProgram(shaderProgram);
glUseProgram(shaderProgram);
GLint posAttrib = glGetAttribLocation(shaderProgram, \"position\");
glEnableVertexAttribArray(posAttrib);
glVertexAttribPointer(posAttrib, 2, GL_FLOAT, GL_FALSE, 7 * sizeof(GLfloat), 0);
GLint colAttrib = glGetAttribLocation(shaderProgram, \"color\");
glEnableVertexAttribArray(colAttrib);
glVertexAttribPointer(colAttrib, 3, GL_FLOAT, GL_FALSE, 7 * sizeof(GLfloat), (void*)(2
* sizeof(GLfloat)));
GLint texAttrib = glGetAttribLocation(shaderProgram, \"texcoord\");
glEnableVertexAttribArray(texAttrib);
glVertexAttribPointer(texAttrib, 2, GL_FLOAT, GL_FALSE, 7 * sizeof(GLfloat), (void*)(5
* sizeof(GLfloat)));
GLuint textures[2];
glGenTextures(2, textures);
int width, height;
unsigned char* image;
glActiveTexture(GL_TEXTURE0);
glBindTexture(GL_TEXTURE_2D, textures[0]);
image = SOIL_load_imag.
Two types of malware -1. Virus A virus is a contagious program o.pdf
Two types of malware :-
1. Virus: A virus is a contagious program or code that attaches itself to another piece of software,
and then reproduces itself when that software is run. Most often this is spread by sharing
software or files between computers.
Virus cannot be spread without a human action. So when you share infected files or send email
with viruses as attachment in email, virus spread in another system also.
If you have turned off your anti-virus protection your system is vulnerable to virus and other
threats.
2. Trojan: This is the most dangerous Malware. Trojans are written with the purpose of
discovering your financial information, taking over your computer’s system resources, and in
larger systems creating a “denial-of-service attack ”.
Denial-of-service attack: an attempt to make a machine or network resource unavailable to those
attempting to reach it. Example: AOL, Yahoo or your business network becoming unavailable.
70% malware are trojan horse. When you open a file that have trojan horse or other kind of
unscrupulous virus program then it replicate in your system and your system is also get infected.
Solution
Two types of malware :-
1. Virus: A virus is a contagious program or code that attaches itself to another piece of software,
and then reproduces itself when that software is run. Most often this is spread by sharing
software or files between computers.
Virus cannot be spread without a human action. So when you share infected files or send email
with viruses as attachment in email, virus spread in another system also.
If you have turned off your anti-virus protection your system is vulnerable to virus and other
threats.
2. Trojan: This is the most dangerous Malware. Trojans are written with the purpose of
discovering your financial information, taking over your computer’s system resources, and in
larger systems creating a “denial-of-service attack ”.
Denial-of-service attack: an attempt to make a machine or network resource unavailable to those
attempting to reach it. Example: AOL, Yahoo or your business network becoming unavailable.
70% malware are trojan horse. When you open a file that have trojan horse or other kind of
unscrupulous virus program then it replicate in your system and your system is also get infected..
The cells of the reproductive organs (Eg ovaries ad testes) undergo.pdf
The cells of the reproductive organs (Eg: ovaries ad testes) undergo meiosis. Meiosis is a type of
nuclear division, which produces four haploid cells from a single diploid cell. Meiosis is divided
into two phases, meiosis I (one diploid nucleus to two haploid nuclei) and meiosis II (two
haploid nuclei to four haploid nuclei). Meiosis I and meiosis II are subdivided into four phases
again namely, prophase, metaphase, anaphase and telophase. Thus, meiosis is divided into a total
of eight phases. The phases of meiosis I are prohase-i, metaphase-I, anaphase-I and telophase-I.
Similarly, the phases of meiosis-II are prohase-II, metaphase-II, anaphase-II and telophase-II.
Prophase is the first phase of cell division in both mitosis and meiosis. During this phase,
cohesion that is present throughout the length of the chromatids is removed except at the
centromere. The chromatids become more visible during this phase. In meiosis, the substages of
prophase-I include, leptonema, zygonema, pachynema, diplonema, diakinesis.
During the metaphase, two chromatids are formed from each chromosome by longitudinal
splitting. These chromatids form the chromosomes of daughter cells.
During anaphase, centromeres are pulled and the chromatids become separated and move
towards the opposite poles. These chromatids now are the chromosomes of the daughter cells,
and are pulled towards the centrioles by means of spindle fibres.
During “metaphase I and II” the chromosomes align at the midway between the two opposite
spindle poles.
In the stage of telophase, cytoplasmic division of the cell takes place (cytokinesis) during which
each of the daughter cell receives one of the daughter nuclei. The interval between each division
is called interphase.
During meiosis, the homologous chromosomes align during the prophase. These chromosomes
crossover at some point and rejoin to form recombinant chromatids, and move towards the
opposite poles during anaphase. The chromosomal crossover ocurs during metaphase-I of the
meiosis,
In meiosis, Alignment of homologous chromosomes at Mid-Prophase –I is characteristic of
meiosis. The two chromosomes sharing the same quality are called “homologous chromosomes.”
These chromosomes crossover at some point and rejoin to form recombinant chromatids. Now,
these chromosomes are separated at Anaphase-I and move to their opposite poles.
During anaphase of mitosis, the sister chromatids get separated and move to the opposite poles.
In contrast to mitosis, the “homologues chromosomes” (but not sister chromatids) move to the
opposite poles in anaphase –I of meiosis. In meiosis, the separation of sister chromatids occur in
anaphase –II.
Solution
The cells of the reproductive organs (Eg: ovaries ad testes) undergo meiosis. Meiosis is a type of
nuclear division, which produces four haploid cells from a single diploid cell. Meiosis is divided
into two phases, meiosis I (one diploid nucleus to two haploid nuclei) and meiosis II (two
haploid nuclei to four h.
Scientific evidences I would give in support of evolution1. Paleo.pdf
Scientific evidences I would give in support of evolution:
1. Paleontological evidence: Fossils are those remains of plants and animals that could not be
degraded. Some of them bear resemblances to present day animals and serve as connecting links,
eg. Duck billed platypus is connecting link between reptiles and mammals. Some of them
represent extinct organisms, eg. dinosaurs. Cross sections of the earth\'s crust from where the
fossils have been isolated determines the geological time scale during which the organism
existed.
2. Comparative anatomy/physiology: Homologous organs of animals such as forelimb bones of
humans, tigers, whales and bats share similar type of arrangement and indicates common
ancestry. Such type of evolution is called \"divergent\". Analogous organs are opposite to
homologous organs, such as wings of butterfly and birds. Although they perform the same
function, they are not anatomically similar. Thus, these are results of \"convergent \"evolution.
3. Similarities in genes and proteins among genetically different organisms also support for the
existence of common ancestry.
4. Natural selection/Industrial melanism: The dominance of dark coloured moths in industrial
areas compared to the white forms is an example of industrial melanism. This type of selection
resulted in the dark coloured moths from becoming a prey of birds compared to the paler ones as
the darker forms camouflage in the dark coloured soots. Another example of natural selection is
antibiotic resistance in bacteria. The resistant bacteria can be isolated by plating the organisms in
a LB agar plate containing the corresponding antibiotic.
5. Adaptive radiation: Darwin\'s finches in Galapagos island is an example of adaptive radiation.
It was found that the different species of finches (with different beak structure) arose from an
initial seed eating finch, enabling them to become vegetarian and insectivorous finches.
Solution
Scientific evidences I would give in support of evolution:
1. Paleontological evidence: Fossils are those remains of plants and animals that could not be
degraded. Some of them bear resemblances to present day animals and serve as connecting links,
eg. Duck billed platypus is connecting link between reptiles and mammals. Some of them
represent extinct organisms, eg. dinosaurs. Cross sections of the earth\'s crust from where the
fossils have been isolated determines the geological time scale during which the organism
existed.
2. Comparative anatomy/physiology: Homologous organs of animals such as forelimb bones of
humans, tigers, whales and bats share similar type of arrangement and indicates common
ancestry. Such type of evolution is called \"divergent\". Analogous organs are opposite to
homologous organs, such as wings of butterfly and birds. Although they perform the same
function, they are not anatomically similar. Thus, these are results of \"convergent \"evolution.
3. Similarities in genes and proteins among genetically different o.
Package scopeis the range of visibility for aparticular element or c.pdf
Package scopeis the range of visibility for aparticular element or construct contained in
apackage.–“range of visibility” describe whether a packageitem (such as function) can be
accessed fromoutside the package or only from other programunits within the same
package.–Any elements declared in a package specificationare consideredpublic.–The elements
are considered private if they can be called from other program units within the samepackage.
Solution
Package scopeis the range of visibility for aparticular element or construct contained in
apackage.–“range of visibility” describe whether a packageitem (such as function) can be
accessed fromoutside the package or only from other programunits within the same
package.–Any elements declared in a package specificationare consideredpublic.–The elements
are considered private if they can be called from other program units within the samepackage..
Point of zero charge related to the adsorption phenomenon I surface .pdf
Point of zero charge related to the adsorption phenomenon I surface chemistry. Here it is the
point if achievement of zero charge density on a surface. If pHzpc is 8.5 that means at this pH
solid submerged in electrolyte will show zero electrical charge on the surface.
P sorption refers to binding of phosphates in other forms to soil particles.
At high pH, P availability is low because of reaction of phosphates to calcium which result in
formation of apatite.
At low pH, phosphate reacts with aluminum, iron etc and results in formation of various other
minerals.
Solution
Point of zero charge related to the adsorption phenomenon I surface chemistry. Here it is the
point if achievement of zero charge density on a surface. If pHzpc is 8.5 that means at this pH
solid submerged in electrolyte will show zero electrical charge on the surface.
P sorption refers to binding of phosphates in other forms to soil particles.
At high pH, P availability is low because of reaction of phosphates to calcium which result in
formation of apatite.
At low pH, phosphate reacts with aluminum, iron etc and results in formation of various other
minerals..
Maternal and parental chromosomes are separated in Anaphase I of mei.pdf
Maternal and parental chromosomes are separated in Anaphase I of meiosis cell division. So
sister chromatids are hold together in the Anaphase I. The Anaphase II is responsible for the
separation of sister chromatids to form chromosomes. Sister chromatids are chromosomes of
either maternal or paternal.
Solution
Maternal and parental chromosomes are separated in Anaphase I of meiosis cell division. So
sister chromatids are hold together in the Anaphase I. The Anaphase II is responsible for the
separation of sister chromatids to form chromosomes. Sister chromatids are chromosomes of
either maternal or paternal..
main.cpp #include iostream #include iomanip #include fs.pdf
main.cpp
#include
#include
#include
#include
#include \"Bank.hpp\"
using namespace std;
int read_accts(Bank& bank, int max_accts)
{
int i = 0;
//ifstream infile(\"C:\\\\Users\\\\Smart PC\\\\Desktop\\\\Assignment 3 (3110)\\\\myinput.txt\");
ifstream infile(\"myinput\");
string whiteSpace;
getline(infile, whiteSpace);
// check is file can be opened
if (infile)
{
// read only first max_accts accounts,
// in order to avoid overflow
for (i = 0; i> firstName;
// check is end of file reached
if (!infile.eof())
{
infile >> lastName;
infile >> ssn;
infile >> accountNumber;
infile >> accountType;
infile >> balance;
infile >> status;
infile >> transactions;
bank.addAccount(firstName, lastName, ssn, accountNumber, accountType, balance,
status);
int index = bank.findAccount(accountNumber);
Account* acc = bank.getAccount(index);
for(int i=0; i> transactionType;
infile >> amount;
if (acc)
acc->addTransaction(Transaction(transactionType, amount));
}
}
else {
break;
}
}
infile.close();
}
else {
cout << \"cannot open inpout file\" << endl;
}
return i;
}
*/
void menu()
{
cout << \"Select one of the following:\" << endl << endl;
cout << \" W - Withdrawal\" << endl;
cout << \" D - Deposit\" << endl;
cout << \" N - New account\" << endl;
cout << \" B - Balance\" << endl;
cout << \" I - Account Info\" << endl;
cout << \" H - Account Info plus Account Transaction History\" << endl;
cout << \" C - Close Account (close but do not delete the account)\" << endl;
cout << \" R - Reopen a closed account\" << endl;
cout << \" X - Delete Account (close and delete the account from the database))\" << endl;
cout << \" Q - Quit\" << endl;
}
void withdrawal(Bank& bank)
{
int index, requested_account;
//prompt for account number
cout << endl << \"Enter the account number: \";
cin >> requested_account;
// account #0 is not valid
if (requested_account != 0)
{
index = bank.findAccount(requested_account);
if (index >= 0)
{
double amount;
cout << \"Enter amount to withdraw: \";
cin >> amount;
// check is there possible to withdraw
if (amount>0)
{
// check is sufficient balance at account
if (bank.getAccount(index)->makeWithdrawal(amount))
{
cout << \"Withdraw is completed.\" << endl;
}
else {
cout << \"Error. Insuffucient funds.\" << endl;
}
}
else {
cout << \"Error. Invalid amount, needed to entyer positiove number.\" << endl;
}
}
else {
cout << \"Error. Account number \\\"\" << requested_account << \"\\\" is not found.\" <<
endl;
}
}
else {
cout << \"Error. Invalid account number.\" << endl;
}
}
void deposit(Bank& bank)
{
int index, requested_account;
//prompt for account number
cout << endl << \"Enter the account number: \";
cin >> requested_account;
// account #0 is not valid
if (requested_account != 0)
{
index = bank.findAccount(requested_account);
if (index >= 0)
{
double amount;
cout << \"Enter amount to deposit: \";
cin >> amount;
// check is there possible to withdraw
if (amount>0)
{
bank.getAccount(index)->makeDeposit(amount); //deposit operation
cout << \"Deposit is complet.
loan amount = 17500present value of annuity = payment per period .pdf
loan amount = 17500
present value of annuity = payment per period * [ 1 - (1+i)^-n ]/i
=>
monthly payment *[1-(1+10.4%/12)^-39]/10.4%/12 = 17500
=>
monthly payment = 530.74
remaining balance = PV of remaining paymens
= 530.74 * [1-(1+10.4%/12)^-28]/10.4%/12
= 13144.78
Solution
loan amount = 17500
present value of annuity = payment per period * [ 1 - (1+i)^-n ]/i
=>
monthly payment *[1-(1+10.4%/12)^-39]/10.4%/12 = 17500
=>
monthly payment = 530.74
remaining balance = PV of remaining paymens
= 530.74 * [1-(1+10.4%/12)^-28]/10.4%/12
= 13144.78.
import javax.swing.; import java.awt.; import java.awt.event..pdf
import javax.swing.*;
import java.awt.*;
import java.awt.event.*;
import java.text.*;
public class convertTemperature extends JApplet implements ActionListener {
JTextField inputField;
JLabel label;
JRadioButton celcius, fahrenheit;
public void init(){
Container contain = getContentPane();
contain.setLayout (new FlowLayout());
inputField = new JTextField(\"\",10);
contain.add(inputField);
celcius= new JRadioButton (\"to Celcius\", true);
contain.add(celcius);
fahrenheit = new JRadioButton(\"to fahrenheit\", false);
contain.add(fahrenheit);
ButtonGroup selection = new ButtonGroup();
selection.add(celcius);
selection.add(fahrenheit);
JButton button1 = new JButton (\"Show Result\");
button1.addActionListener(this);
contain.add(button1);
label= new JLabel (\"Enter temperature and choose an option\");
contain.add(label);
}
public void actionPerformed(ActionEvent e) {
DecimalFormat df = new DecimalFormat (\"#.##\");
double input = Double.parseDouble(inputField.getText());
double convertedTemp = 0.0;
if (celcius.isSelected())
convertedTemp = ((5.0/9.0)*(input - 32.0));
if (fahrenheit.isSelected())
convertedTemp = ((9.0/5.0)*(input + 32.0));
label.setText(String.valueOf(df.format(convertedTemp)));
}
}
Solution
import javax.swing.*;
import java.awt.*;
import java.awt.event.*;
import java.text.*;
public class convertTemperature extends JApplet implements ActionListener {
JTextField inputField;
JLabel label;
JRadioButton celcius, fahrenheit;
public void init(){
Container contain = getContentPane();
contain.setLayout (new FlowLayout());
inputField = new JTextField(\"\",10);
contain.add(inputField);
celcius= new JRadioButton (\"to Celcius\", true);
contain.add(celcius);
fahrenheit = new JRadioButton(\"to fahrenheit\", false);
contain.add(fahrenheit);
ButtonGroup selection = new ButtonGroup();
selection.add(celcius);
selection.add(fahrenheit);
JButton button1 = new JButton (\"Show Result\");
button1.addActionListener(this);
contain.add(button1);
label= new JLabel (\"Enter temperature and choose an option\");
contain.add(label);
}
public void actionPerformed(ActionEvent e) {
DecimalFormat df = new DecimalFormat (\"#.##\");
double input = Double.parseDouble(inputField.getText());
double convertedTemp = 0.0;
if (celcius.isSelected())
convertedTemp = ((5.0/9.0)*(input - 32.0));
if (fahrenheit.isSelected())
convertedTemp = ((9.0/5.0)*(input + 32.0));
label.setText(String.valueOf(df.format(convertedTemp)));
}
}.
Hg is an atom , because an atom means that a smallest individual spe.pdf
Hg is an atom , because an atom means that a smallest individual spece
and it cannot be undivisible further.
H+ is an ion, it is a spece was formed by the H atom after lost one electron from it.
H2O is a molecule ,and it has obtained by the reaction of H2 atoms and O atoms.
O2 and N2 , Ar and CO2 both are called as mixures becaues those were formed
by the mixing from two individual species or molecules.
NaCl is a compound , it has the fixed proportion of both Na and Cl in its compound.
Solution
Hg is an atom , because an atom means that a smallest individual spece
and it cannot be undivisible further.
H+ is an ion, it is a spece was formed by the H atom after lost one electron from it.
H2O is a molecule ,and it has obtained by the reaction of H2 atoms and O atoms.
O2 and N2 , Ar and CO2 both are called as mixures becaues those were formed
by the mixing from two individual species or molecules.
NaCl is a compound , it has the fixed proportion of both Na and Cl in its compound..
Hello! Please see the correction below! False Hydrogen atoms ar.pdf
Hello! Please see the correction below!
False: Hydrogen atoms are often the central atom of a lewis structure. (This doesn\'t make sense
since hydrogen can only form 1 bond!)
True: For a neutral molecule, the number of electrons in the Lewis structure is the sum of the
valence electrons for the atoms.
FALSE: Each atom of a Lewis structure must have eight electrons. (This is the \"octet rule.)
Typically you would have each atom with eight electrons, but as your hint indicated, Hydrogen
only gets 2 electrons. This exception would make this statement false.
True: Electrons of covalent compounds may be shared between atoms.
Sorry about that - Hope this helps! Please rate if it does : )
Solution
Hello! Please see the correction below!
False: Hydrogen atoms are often the central atom of a lewis structure. (This doesn\'t make sense
since hydrogen can only form 1 bond!)
True: For a neutral molecule, the number of electrons in the Lewis structure is the sum of the
valence electrons for the atoms.
FALSE: Each atom of a Lewis structure must have eight electrons. (This is the \"octet rule.)
Typically you would have each atom with eight electrons, but as your hint indicated, Hydrogen
only gets 2 electrons. This exception would make this statement false.
True: Electrons of covalent compounds may be shared between atoms.
Sorry about that - Hope this helps! Please rate if it does : ).
From LLMm has possible four gamates those are LM ,Lm,LM and Lm in me.pdf
From LLMm has possible four gamates those are LM ,Lm,LM and Lm in meiosis 1.then it will
be doubled in meiosis2. But when the secondary non disjunction occurs at M locus that means
than gene is not segregated during meiosis 2. Then you get four gametes for L gene after meiosis
2 and in case of M gene you will have M and m allele which will not segregated in meiosis 2. So
you will have Four gametes. (LM, Lm, LmM and L).
Solution
From LLMm has possible four gamates those are LM ,Lm,LM and Lm in meiosis 1.then it will
be doubled in meiosis2. But when the secondary non disjunction occurs at M locus that means
than gene is not segregated during meiosis 2. Then you get four gametes for L gene after meiosis
2 and in case of M gene you will have M and m allele which will not segregated in meiosis 2. So
you will have Four gametes. (LM, Lm, LmM and L)..
D is correct. Formatting and code conversion are Presentation layer .pdf
D is correct. Formatting and code conversion are Presentation layer functions in the OSI model.
The Application layer in TCP/IP maps to the top three layers of the OSI model.
A, B, and C are incorrect.
Solution
D is correct. Formatting and code conversion are Presentation layer functions in the OSI model.
The Application layer in TCP/IP maps to the top three layers of the OSI model.
A, B, and C are incorrect..
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A Strategic Approach: GenAI in Education
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
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- 1. TlI <--> Tl+ + I- Ksp = 8.9E-8 = [Tl+][I-] (eqn. 1) PbI2 <--> Pb2+ + 2I- Ksp = 8.7E-9 = [Pb2+][I-]2 (eqn. 2) In a solution, [Tl+] = 0.50 M, [Pb2+] = 0.50 M When I- is steadily added to the solution, at a particular point, TlI begins to precipitate. And at another point, PbI2 begins to precipitate. Find a point of TlI precipitation: [Tl+][I-] = Ksp = 8.9E-8 (0.50)[I-] = 8.9E-8 [I-] = 1.78E-7 M Find a point of PbI2 precipitation: [Pb2+][I-]2 = 8.7E-9 (0.50)[I-]2 = 8.7E-9 [I-] = 1.32E-4 M Therefore, TlI begins to precipitate before PbI2 (only when [I-] is just 1.78E-7 M) When KI is added, TlI precipitates more and more until when [I-] reaches 1.32E-4, PbI2 just begins to precipitates. At this point, we stop the ions separation. Find remaining [Tl+]: [Tl+][I-] = 8.9E-8 [Tl+](1.32E-4) = 8.9E-8 [Tl+] = 6.74E-4 M The remaining Tl+ is 6.74E-4/0.50 x 100% = 0.13% Tl+ is separated by 99.87% of original amount Solution TlI <--> Tl+ + I- Ksp = 8.9E-8 = [Tl+][I-] (eqn. 1) PbI2 <--> Pb2+ + 2I- Ksp = 8.7E-9 = [Pb2+][I-]2 (eqn. 2) In a solution, [Tl+] = 0.50 M, [Pb2+] = 0.50 M When I- is steadily added to the solution, at a particular point, TlI begins to precipitate. And at another point, PbI2 begins to precipitate. Find a point of TlI precipitation: [Tl+][I-] = Ksp = 8.9E-8 (0.50)[I-] = 8.9E-8 [I-] = 1.78E-7 M Find a point of PbI2 precipitation:
- 2. [Pb2+][I-]2 = 8.7E-9 (0.50)[I-]2 = 8.7E-9 [I-] = 1.32E-4 M Therefore, TlI begins to precipitate before PbI2 (only when [I-] is just 1.78E-7 M) When KI is added, TlI precipitates more and more until when [I-] reaches 1.32E-4, PbI2 just begins to precipitates. At this point, we stop the ions separation. Find remaining [Tl+]: [Tl+][I-] = 8.9E-8 [Tl+](1.32E-4) = 8.9E-8 [Tl+] = 6.74E-4 M The remaining Tl+ is 6.74E-4/0.50 x 100% = 0.13% Tl+ is separated by 99.87% of original amount