With the HNMR the two CH3\'s are chemically equivalent so they only give off one
peak for both. The peak will be a singlet indicating they have no other neighboring hydrogens
near them. I would expect it to look like
this...http://www.muhlenberg.edu/depts/chemistry/chem201woh/1Hacetone.html With the
CNMR I would expect to see a signal in the range of 205-220 because of a chemical shift for this
specific type of carbon ( R-C=O-R). The two R\'s in this instant are chemically equivalent when
in comes to environment so they will give off one signal for two carbons and since they are Sp3
they should be within the range of 0-50ppm and this signal will be much taller than the C=O
carbon.
Solution
With the HNMR the two CH3\'s are chemically equivalent so they only give off one
peak for both. The peak will be a singlet indicating they have no other neighboring hydrogens
near them. I would expect it to look like
this...http://www.muhlenberg.edu/depts/chemistry/chem201woh/1Hacetone.html With the
CNMR I would expect to see a signal in the range of 205-220 because of a chemical shift for this
specific type of carbon ( R-C=O-R). The two R\'s in this instant are chemically equivalent when
in comes to environment so they will give off one signal for two carbons and since they are Sp3
they should be within the range of 0-50ppm and this signal will be much taller than the C=O
carbon..
Ok, so the number of double bond equivalents in the compound is [C4H.pdfannaelctronics
Ok, so the number of double bond equivalents in the compound is [C4H10)-(C4H8)]/2 = 1 DBE
1) IR
so the large trough from 3400cm-1 to 2400cm-1 as well as the peak at ~1700cm-1 indicates an
acid. That accounts for 1 carbon, both oxygens, and one hydrogen. That also accounts for the one
double bond equivalent.
That leaves 3 carbons and 7 hydrogens.
2) NMR
Methyl groups usually give proton peaks around 0.9ppm and shift further downfield when
adding electron withdrawing groups nearby. It\'s a pretty good guess that the triplet at ~1.1ppm is
a methyl group. The fact that it\'s a triplet indicates the carbon next to it has two protons attached
to it.
That gives a CH3-CH2- (as well as the -COOH group)
all that\'s left now is one carbon and two hydrogens (CH2)
Looking at the other two proton peaks in the nmr ... there\'s one triplet and one sextet.
The triplet would be a CH2 group that is next to another CH2 group (and not anything else), so it
must be a CH2 next to the COOH
Now you would have two pieces:
CH3-CH2- and -CH2-COOH
put them together and check the coupling of the protons...
you should end up with two triplets (proton split by two others), one sextet (proton split by 5
others), and a small peak somewhere around 9-13ppm. (that\'s the peak indicated by the offset of
2.0ppm on the spectrum)
So the structure would be:
CH3-CH2-CH2-COOH (C4H8O2)
Solution
Ok, so the number of double bond equivalents in the compound is [C4H10)-(C4H8)]/2 = 1 DBE
1) IR
so the large trough from 3400cm-1 to 2400cm-1 as well as the peak at ~1700cm-1 indicates an
acid. That accounts for 1 carbon, both oxygens, and one hydrogen. That also accounts for the one
double bond equivalent.
That leaves 3 carbons and 7 hydrogens.
2) NMR
Methyl groups usually give proton peaks around 0.9ppm and shift further downfield when
adding electron withdrawing groups nearby. It\'s a pretty good guess that the triplet at ~1.1ppm is
a methyl group. The fact that it\'s a triplet indicates the carbon next to it has two protons attached
to it.
That gives a CH3-CH2- (as well as the -COOH group)
all that\'s left now is one carbon and two hydrogens (CH2)
Looking at the other two proton peaks in the nmr ... there\'s one triplet and one sextet.
The triplet would be a CH2 group that is next to another CH2 group (and not anything else), so it
must be a CH2 next to the COOH
Now you would have two pieces:
CH3-CH2- and -CH2-COOH
put them together and check the coupling of the protons...
you should end up with two triplets (proton split by two others), one sextet (proton split by 5
others), and a small peak somewhere around 9-13ppm. (that\'s the peak indicated by the offset of
2.0ppm on the spectrum)
So the structure would be:
CH3-CH2-CH2-COOH (C4H8O2).
1) Current assets = Net working capital + Current liabilities = $1,0.pdfanonakitchen
1) Current assets = Net working capital + Current liabilities = $1,015 + $6,725 = $7,740
Current ratio = Current assets / Current liabilities = $7,740 / $6,725 = 1.15
2) Quick ratio = (Current assets - Inventory) / Current liabilities = ($7,740 - $1,135) / $6,725 =
0.98
Solution
1) Current assets = Net working capital + Current liabilities = $1,015 + $6,725 = $7,740
Current ratio = Current assets / Current liabilities = $7,740 / $6,725 = 1.15
2) Quick ratio = (Current assets - Inventory) / Current liabilities = ($7,740 - $1,135) / $6,725 =
0.98.
Ok, so the number of double bond equivalents in the compound is [C4H.pdfannaelctronics
Ok, so the number of double bond equivalents in the compound is [C4H10)-(C4H8)]/2 = 1 DBE
1) IR
so the large trough from 3400cm-1 to 2400cm-1 as well as the peak at ~1700cm-1 indicates an
acid. That accounts for 1 carbon, both oxygens, and one hydrogen. That also accounts for the one
double bond equivalent.
That leaves 3 carbons and 7 hydrogens.
2) NMR
Methyl groups usually give proton peaks around 0.9ppm and shift further downfield when
adding electron withdrawing groups nearby. It\'s a pretty good guess that the triplet at ~1.1ppm is
a methyl group. The fact that it\'s a triplet indicates the carbon next to it has two protons attached
to it.
That gives a CH3-CH2- (as well as the -COOH group)
all that\'s left now is one carbon and two hydrogens (CH2)
Looking at the other two proton peaks in the nmr ... there\'s one triplet and one sextet.
The triplet would be a CH2 group that is next to another CH2 group (and not anything else), so it
must be a CH2 next to the COOH
Now you would have two pieces:
CH3-CH2- and -CH2-COOH
put them together and check the coupling of the protons...
you should end up with two triplets (proton split by two others), one sextet (proton split by 5
others), and a small peak somewhere around 9-13ppm. (that\'s the peak indicated by the offset of
2.0ppm on the spectrum)
So the structure would be:
CH3-CH2-CH2-COOH (C4H8O2)
Solution
Ok, so the number of double bond equivalents in the compound is [C4H10)-(C4H8)]/2 = 1 DBE
1) IR
so the large trough from 3400cm-1 to 2400cm-1 as well as the peak at ~1700cm-1 indicates an
acid. That accounts for 1 carbon, both oxygens, and one hydrogen. That also accounts for the one
double bond equivalent.
That leaves 3 carbons and 7 hydrogens.
2) NMR
Methyl groups usually give proton peaks around 0.9ppm and shift further downfield when
adding electron withdrawing groups nearby. It\'s a pretty good guess that the triplet at ~1.1ppm is
a methyl group. The fact that it\'s a triplet indicates the carbon next to it has two protons attached
to it.
That gives a CH3-CH2- (as well as the -COOH group)
all that\'s left now is one carbon and two hydrogens (CH2)
Looking at the other two proton peaks in the nmr ... there\'s one triplet and one sextet.
The triplet would be a CH2 group that is next to another CH2 group (and not anything else), so it
must be a CH2 next to the COOH
Now you would have two pieces:
CH3-CH2- and -CH2-COOH
put them together and check the coupling of the protons...
you should end up with two triplets (proton split by two others), one sextet (proton split by 5
others), and a small peak somewhere around 9-13ppm. (that\'s the peak indicated by the offset of
2.0ppm on the spectrum)
So the structure would be:
CH3-CH2-CH2-COOH (C4H8O2).
1) Current assets = Net working capital + Current liabilities = $1,0.pdfanonakitchen
1) Current assets = Net working capital + Current liabilities = $1,015 + $6,725 = $7,740
Current ratio = Current assets / Current liabilities = $7,740 / $6,725 = 1.15
2) Quick ratio = (Current assets - Inventory) / Current liabilities = ($7,740 - $1,135) / $6,725 =
0.98
Solution
1) Current assets = Net working capital + Current liabilities = $1,015 + $6,725 = $7,740
Current ratio = Current assets / Current liabilities = $7,740 / $6,725 = 1.15
2) Quick ratio = (Current assets - Inventory) / Current liabilities = ($7,740 - $1,135) / $6,725 =
0.98.
(a). The posterior pole of the embryo of drosophila contains essenti.pdfanonakitchen
(a). The posterior pole of the embryo of drosophila contains essential substances for the fertility
of the fly. The cytoplasm contains pole plasm, which consists of particular materials called polar
granules and the pole cells are the ancestors to primordial germ cells.Pole plasm contains the
mRNA and proteins of the posterior group genes are oskar, nanos gene, tudor, vasa, and vlois.
These genes play an important role in localize germ cell determinants and germ line development
to localize nanos mRNA to the posterior.
1.Oskar is required to recruit the other genes to produce functional germ plasm.
2.Nanos is required to prevent somatic differentiation and mitosis and for the pole cells to
migrate to function as PGCs.
3.Gcls gene is required for pole cell formation.
(b). The proneural genes are important regulators of neurogenesis, which are involved in the
coordinating the acquisition of a generic neuronal fate and of specific subtype identities that are
appropriate for the location and time of neuronal generation.
(c). The selector gene, engrailed (en), is an important key determinant in boundary formation
among the anterior and posterior compartments. The wing imaginal disc expands, posterior, but
not anterior cells express engrailed and maintain expression state as they expand and form the
disc.
Solution
(a). The posterior pole of the embryo of drosophila contains essential substances for the fertility
of the fly. The cytoplasm contains pole plasm, which consists of particular materials called polar
granules and the pole cells are the ancestors to primordial germ cells.Pole plasm contains the
mRNA and proteins of the posterior group genes are oskar, nanos gene, tudor, vasa, and vlois.
These genes play an important role in localize germ cell determinants and germ line development
to localize nanos mRNA to the posterior.
1.Oskar is required to recruit the other genes to produce functional germ plasm.
2.Nanos is required to prevent somatic differentiation and mitosis and for the pole cells to
migrate to function as PGCs.
3.Gcls gene is required for pole cell formation.
(b). The proneural genes are important regulators of neurogenesis, which are involved in the
coordinating the acquisition of a generic neuronal fate and of specific subtype identities that are
appropriate for the location and time of neuronal generation.
(c). The selector gene, engrailed (en), is an important key determinant in boundary formation
among the anterior and posterior compartments. The wing imaginal disc expands, posterior, but
not anterior cells express engrailed and maintain expression state as they expand and form the
disc..
As per interest rate parity, Expected.pdfanonakitchen
As per interest rate parity, Expected Spot rate = Current Spot rate *
((1+p)/(1+b)) Where, = 1.1*((1+0.02)/(1+0.01)) p = Interest rate of quote currency = 2%
= 1.11 b = Interest rate of base currency = 1% $/€=1.1 means one euro is
exchanged for 1.1 dollar. Thus, expected spot rate in 30 days is E$/€=1.11
Solution
As per interest rate parity, Expected Spot rate = Current Spot rate *
((1+p)/(1+b)) Where, = 1.1*((1+0.02)/(1+0.01)) p = Interest rate of quote currency = 2%
= 1.11 b = Interest rate of base currency = 1% $/€=1.1 means one euro is
exchanged for 1.1 dollar. Thus, expected spot rate in 30 days is E$/€=1.11.
resonance structures 1 and 3 from the top reson.pdfanonakitchen
resonance structures: 1 and 3 from the top resonance hybride: the last one (at the
bottom)
Solution
resonance structures: 1 and 3 from the top resonance hybride: the last one (at the
bottom).
It is methane since there is only 1 Carbon and it.pdfanonakitchen
It is methane since there is only 1 Carbon and it has only single bonds
Solution
It is methane since there is only 1 Carbon and it has only single bonds.
Thermochemistry is the study of the energy and heat associated with .pdfanonakitchen
Thermochemistry is the study of the energy and heat associated with chemical reactions. It
depends on two main laws: Laplace and Hess laws. It also studies the exchange or transfer of
heat energy between a system and its surroundings.
Please be more specific in your question for a more specific answer.
Solution
Thermochemistry is the study of the energy and heat associated with chemical reactions. It
depends on two main laws: Laplace and Hess laws. It also studies the exchange or transfer of
heat energy between a system and its surroundings.
Please be more specific in your question for a more specific answer..
Vibration and WavesT = 2pisqrt(mk)v = lambdafp = A sin(kx.pdfanonakitchen
Vibration and Waves
T = 2pi*sqrt(m/k)
v = lambda*f
p = A sin(kx-wt) w=2*pi*f (pressure in travelling sound wave), k = 2pi/lambda, A = amplitude,
x = displacement.
1. Periodic motion: used to describe a vibration or an oscillation that repeats itself over the same
path.. Displacement: The position of an object. Amplitude: The maximum displacement. Period:
The time in seconds required for one complete cycle. Frequency: The number of cycles, per
second. Simple Harmonic Motion: Sinusoidal motion of a single frequency.
2. Total E = 1/2kA^2. = 1/2kx^2+1/2mv^2. EPE = 1/2kx^2.
3. The equation for the period of a spring is T = 2pi*sqrt(m/k). For example, how long does it
take for a spring with a hanging mass of 5 kg and spring constant 10N/m to exhibit one complete
cycle? T = 2pi*sqrt(5/10). T = 2pi*sqrt(1/2).
4. Since a period is the time it takes for one complete cycle in seconds, frequency is the inverse
of the period. f = 1/T. T = 1/f.
5. For describing the motion generated SHM, you can use x = A sin 2pi*f*t. This is if the motion
starts at equilibrium point. If it starts at the crest/trough of its motion, then use x = A cos 2pi*f*t.
6. For describing the motion generated SHM, you can use x = A sin 2pi*f*t. This is if the motion
starts at equilibrium point. If it starts at the crest/trough of its motion, then use x = A cos 2pi*f*t.
7. The maximum velocity of a particle on SHM can be found by v = 2pi*A*f. Now, for any other
velocity at a point in time, it can be found by v = v0(sqrt(1-x^2/A^2)).
8. The equation for the period of a pendulum is T = 2*pi*sqrt(L/g). L = length of the pendulum.
This only works if its amplitude is small and friction can be ignored.
9. Dampened harmonic motion is the gradual decrease of maximum displacement. This is caused
by the presence of frictional forces. The energy is eventually all transformed to heat.
10. Forced vibration is due to an external source, which applies its own frequency to the spring.
The natural frequency of the spring is actually 1/2pi * sqrt(k/m), which is what it will actually
vibrate at without external force.
11. Amplitude denotes the distance between the equilibrium and the crest or trough of the wave.
Wavelength is the distance from one point of the wave to the corresponding point in the next
cycle. (crest-crest). This can be found be lambda = v/f. Wave velocity is the velocity at which
wave crests move. V = lambda*f. Node is referring to the point of destructive interference.
Antinode is referring to the point of constructive interference.
12. The relationship between velocity, wavelength, and frequency is v = lambda*f, where
lambda is the wavelength and f is the frequency, v is the wave velocity.
13. A transverse wave refers to a wave that causes the particles of the medium it travels in to
move a direction perpendicular to the motion of the wave itself. On the other hand, in a
longitudinal wave, the vibration of the particles of the medium is along the same direction as the
motion of the wa.
The following steps illustrate how to write a simple Level-2 MATLAB .pdfanonakitchen
The following steps illustrate how to write a simple Level-2 MATLAB S-function. When
applicable, the steps include examples from the S-function demo msfcn_unit_delay.m used in the
model msfcndemo_sfundsc2.mdl. All lines of code use the variable name block for the S-
function run-time object.
Step 1
Copy the Level-2 MATLAB S-function template msfuntmpl.m to your working folder. If you
change the file name when you copy the file, change the function name in the function line to the
same name.
Step 2
Modify the setup method to initialize the S-function’s attributes. For this example:
Set the run-time object’s NumInputPorts and NumOutputPorts properties to 1 in order to
initialize one input port and one output port.
Invoke the run-time object’s SetPreCompInpPortInfoToDynamic and
SetPreCompOutPortInfoToDynamic methods to indicate that the input and output ports inherit
their compiled properties (dimensions, data type, complexity, and sampling mode) from the
model.
Set the DirectFeedthrough property of the run-time object’s InputPort to false in order to indicate
the input port does not have direct feedthrough.
Retain the default values for all other input and output port properties that are set in your copy of
the template file. The values set for the Dimensions, DatatypeID, and Complexity properties
override the values inherited using the SetPreCompInpPortInfoToDynamic and
SetPreCompOutPortInfoToDynamic methods.
Set the run-time object’s NumDialogPrms property to 1 in order to initialize one S-function
dialog parameter.
Specify that the S-function has an inherited sample time by setting the value of the runtime
object’s SampleTimes property to [-1 0].
Call the run-time object’s RegBlockMethod method to register the following four callback
methods used in this S-function.
Remove any other registered callback methods from your copy of the template file. In the calls to
RegBlockMethod, the first input argument is the name of the S-function API method and the
second input argument is the function handle to the associated local function in the MATLAB S-
function.
The following setup method from msfcn_unit_delay.m performs the previous list of steps:
function setup(block)
%% Register a single dialog parameter
block.NumDialogPrms = 1;
%% Register number of input and output ports
block.NumInputPorts = 1;
block.NumOutputPorts = 1
%% Setup functional port properties to dynamically
%% inherited.
block.SetPreCompInpPortInfoToDynamic;
block.SetPreCompOutPortInfoToDynamic;
%% Hard-code certain port properties
block.InputPort(1).Dimensions = 1;
block.InputPort(1).DirectFeedthrough = false;
block.OutputPort(1).Dimensions = 1;
%% Set block sample time to inherited
block.SampleTimes = [-1 0];
%% Register methods
block.RegBlockMethod(\'PostPropagationSetup\',@DoPostPropSetup);
block.RegBlockMethod(\'InitializeConditions\',@InitConditions);
block.RegBlockMethod(\'Outputs\', @Output);
block.RegBlockMethod(\'Update\', @Update);
If your S-function needs continuous stat.
CO + 2H2 =CH3OH moles of CO = 2528 moles of H2 =.pdfanonakitchen
CO + 2H2 =CH3OH moles of CO = 25/28 moles of H2 = 6/2 =3 CO is limiting
reactant
Solution
CO + 2H2 =CH3OH moles of CO = 25/28 moles of H2 = 6/2 =3 CO is limiting
reactant.
Solution for Q-23a) Pay interest on bank reserve ( The main tool a.pdfanonakitchen
Solution
for Q-23
a) Pay interest on bank reserve ( The main tool availble with FED is to maintain the
reserve/margin requirement and accordingly fixed interest rate).
Please find the required program below with output#include stdio.pdfanonakitchen
Please find the required program below with output:
#include
void transposeMatrix(int arr[][5], int trans[][4]){
for (int j = 0; j < 5; ++j)
{
for (int i = 0; i < 4; ++i)
{
trans[j][i] = arr[i][j]; // just reversing the address i,j to j,i so that the transpose will be made
}
}
}
void main()
{
int array[4][5], trans[5][4];
int i, j;
printf(\"Enter the elements of matrix\ \");
for (i = 0; i < 4; ++i)
{
for (j = 0; j < 5; ++j)
{
scanf(\"%d\", &array[i][j]); // reading the 4x5 matrix array
}
}
printf(\"The matrix is \ \"); //just printing the user input array
for (i = 0; i < 4; ++i)
{
for (j = 0; j < 5; ++j)
{
printf(\" %d\", array[i][j]);
}
printf(\"\ \");
}
transposeMatrix(array,trans); //calling the transposeMatrix method, so that the transpose of the
matrix array will be stored in trans
printf(\"Transpose of matrix is \ \"); //just printing the returned transpose matrix trans
for (i = 0; i < 5; ++i)
{
for (j = 0; j < 4; ++j)
{
printf(\" %d\", trans[i][j]);
}
printf(\"\ \");
}
}
---------------------------------------------------------------------------------
OUTPUT:
Enter the elements of
matrix
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
The matrix
is
1 2 3 4
5
6 7 8 9
10
11 12 13 14 15
16 17 18 19
20
Transpose of matrix
is
1 6 11
16
2 7 12
17
3 8 13
18
4 9 14
19
5 10 15 20
Solution
Please find the required program below with output:
#include
void transposeMatrix(int arr[][5], int trans[][4]){
for (int j = 0; j < 5; ++j)
{
for (int i = 0; i < 4; ++i)
{
trans[j][i] = arr[i][j]; // just reversing the address i,j to j,i so that the transpose will be made
}
}
}
void main()
{
int array[4][5], trans[5][4];
int i, j;
printf(\"Enter the elements of matrix\ \");
for (i = 0; i < 4; ++i)
{
for (j = 0; j < 5; ++j)
{
scanf(\"%d\", &array[i][j]); // reading the 4x5 matrix array
}
}
printf(\"The matrix is \ \"); //just printing the user input array
for (i = 0; i < 4; ++i)
{
for (j = 0; j < 5; ++j)
{
printf(\" %d\", array[i][j]);
}
printf(\"\ \");
}
transposeMatrix(array,trans); //calling the transposeMatrix method, so that the transpose of the
matrix array will be stored in trans
printf(\"Transpose of matrix is \ \"); //just printing the returned transpose matrix trans
for (i = 0; i < 5; ++i)
{
for (j = 0; j < 4; ++j)
{
printf(\" %d\", trans[i][j]);
}
printf(\"\ \");
}
}
---------------------------------------------------------------------------------
OUTPUT:
Enter the elements of
matrix
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
The matrix
is
1 2 3 4
5
6 7 8 9
10
11 12 13 14 15
16 17 18 19
20
Transpose of matrix
is
1 6 11
16
2 7 12
17
3 8 13
18
4 9 14
19
5 10 15 20.
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With the HNMR the two CH3s are chemically equiv.pdf
1. With the HNMR the two CH3's are chemically equivalent so they only give off one
peak for both. The peak will be a singlet indicating they have no other neighboring hydrogens
near them. I would expect it to look like
this...http://www.muhlenberg.edu/depts/chemistry/chem201woh/1Hacetone.html With the
CNMR I would expect to see a signal in the range of 205-220 because of a chemical shift for this
specific type of carbon ( R-C=O-R). The two R's in this instant are chemically equivalent when
in comes to environment so they will give off one signal for two carbons and since they are Sp3
they should be within the range of 0-50ppm and this signal will be much taller than the C=O
carbon.
Solution
With the HNMR the two CH3's are chemically equivalent so they only give off one
peak for both. The peak will be a singlet indicating they have no other neighboring hydrogens
near them. I would expect it to look like
this...http://www.muhlenberg.edu/depts/chemistry/chem201woh/1Hacetone.html With the
CNMR I would expect to see a signal in the range of 205-220 because of a chemical shift for this
specific type of carbon ( R-C=O-R). The two R's in this instant are chemically equivalent when
in comes to environment so they will give off one signal for two carbons and since they are Sp3
they should be within the range of 0-50ppm and this signal will be much taller than the C=O
carbon.
