A. the chemical formula for the base calcium (II) hydroxide is Ca(OH.pdf
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The document lists the chemical formulas for calcium hydroxide as Ca(OH)2, selenic acid as H2SeO4, and provides the names of several acids based on their chemical formulas - nitric acid is HNO3, nitrous acid is HNO2, hydrochloric acid is HCl, and chloric acid is HClO3.
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First, lets find out what we need.The question asks for the empi.pdf
First, let\'s find out what we need.
The question asks for the empirical formula of some compound. To do this, we are going to need
to find the amount of moles of each substance.
To find the moles of Sb, all we will need to do is divide the mass of Sb by the molar mass of Sb.
To find the moles of S will be a bit more complicated. Since we know that Sulfur is left over
after the reaction, we know that sulfur is in excess (meaning Sb is the only thing that will limit
our reaction). Assuming this is an ideal gas, we can take the original amount of moles of Sulfur
and subtract it from the moles we get later (after it is transferred to the other container). This will
give us the amount of moles of Sulfur that reacted with the moles of Sb.
So this is what we need:
moles Sb = ?
moles S (reacted with Sb) = ?
Calculate moles Sb.
moles Sb = (mass Sb) / (molar mass Sb) = (0.350g) / (121.76g/mol) = 2.87 x 10^-3 moles
Calculate moles S.
For this, we are going to have to use the PV = nRT equation two times. One for the S that is
placed with the Sb, and one for the S that is placed in the container after the reaction.
Moles of S BEFORE reaction:
PV = nRT
(10.72 atm)(0.100L) = n(0.082)(373.15 K)
n = (10.72 atm)(0.100L) / (0.082)(373.15 K) = 0.035 moles S
Moles of S AFTER reaction:
PV = nRT
(14.8 atm)(0.050L) = n(0.082)(323.15 K)
n = (14.8 atm)(0.050L) / (0.082)(323.15 K) = 0.028 moles S
Calculate moles of S that reacted with the Sb:
Since the moles of S went down after the reaction, we know that the moles of S lost must have
reacted with the Sb. So, we can simply find the change in moles of S to find the amount of S that
reacted with the Sb.
Moles of S before reaction - Moles of S after reaction = Moles of S reacted
0.035 moles S - 0.028 moles S = 7.07 x 10^-3 moles S
Now that we have our moles of each element that reacted, we can find the empirical formula.
To do this, we need to divide through by the element with the least amount of moles. So, in this
case, we must divide through by the moles of Sb.
(2.87 x 10^-3 moles Sb) / (2.87 x 10^-3 moles Sb) = 1
(7.07 x 10^-3 moles S) / (2.87 x 10^-3 moles Sb) = 2.46 = 2.5
We need subscripts that are integers in empirical formulas. To make 2.5 an integer, we can
simply multiply it by 2 to make it 5. When we do this, we also need to multiply 1 by 2.
2.5 x 2 = 5
1 x 2 = 2
These numbers represent the subscripts for each of our elements in our compound.
So the empirical formula would be:
Sb2S5
Solution
First, let\'s find out what we need.
The question asks for the empirical formula of some compound. To do this, we are going to need
to find the amount of moles of each substance.
To find the moles of Sb, all we will need to do is divide the mass of Sb by the molar mass of Sb.
To find the moles of S will be a bit more complicated. Since we know that Sulfur is left over
after the reaction, we know that sulfur is in excess (meaning Sb is the only thing that will limit
our reaction). Assuming this is an ideal gas, we can take.
DiversityDiversity scheme refers to a method for improving the re.pdf
Diversity:
Diversity scheme refers to a method for improving the reliability of a message signal by using
two or more communication channels with different characteristics. Diversity is mainly used in
radio communication and is a common technique for combatting fading and co-channel
interference and avoiding error bursts. It is based on the fact that individual channels experience
different levels of fading and interference. Multiple versions of the same signal may be
transmitted and/or received and combined in the receiver
Types:
In Micro diversity The spacing between transmitters are shorter than wavelength of transmitted
signal
In Macro diversity The spacing between transmitters are longer than wavelength of transmitted
signal
Types of microdiversity:
Time diversity: Multiple versions of the same signal are transmitted at different time instants.
Alternatively, a redundant forward error correction codeis added and the message is spread in
time by means of bit-interleaving before it is transmitted. Thus, error bursts are avoided, which
simplifies the error correction.
Frequency diversity: The signal is transmitted using several frequency channels or spread over a
wide spectrum that is affected by frequency-selective fading. Middle-late 20th century
microwave radio relay lines often used several regular wideband radio channels, and one
protection channel for automatic use by any faded channel. Later examples include:
OFDM modulation in combination with subcarrier interleaving and forward error correction
Spread spectrum, for example frequency hopping or DS-CDMA.
Space diversity: The signal is transmitted over several different propagation paths. In the case of
wired transmission, this can be achieved by transmitting via multiple wires. In the case of
wireless transmission, it can be achieved by antenna diversity using multiple transmitter antennas
(transmit diversity) and/or multiple receiving antennas (reception diversity). In the latter case, a
diversity combining technique is applied before further signal processing takes place. If the
antennas are far apart, for example at different cellular base station sites or WLAN access points,
this is calledmacrodiversity or site diversity. If the antennas are at a distance in the order of one
wavelength, this is called microdiversity. A special case is phasedantenna arrays, which also can
be used for beamforming, MIMO channels and space–time coding (STC).
Polarization diversity: Multiple versions of a signal are transmitted and received via antennas
with different polarization. A diversity combiningtechnique is applied on the receiver side.
Solution
Diversity:
Diversity scheme refers to a method for improving the reliability of a message signal by using
two or more communication channels with different characteristics. Diversity is mainly used in
radio communication and is a common technique for combatting fading and co-channel
interference and avoiding error bursts. It is based on the fact that i.
due to repulsions between lone pair and bond pair of electrons on Te.pdf
due to repulsions between lone pair and bond pair of electrons on Te ,TeCl2 adopt bent shape
.but Be does not have lone pair ,so BeCl2 adopt linear structure.
Solution
due to repulsions between lone pair and bond pair of electrons on Te ,TeCl2 adopt bent shape
.but Be does not have lone pair ,so BeCl2 adopt linear structure..
AnswerKidneys are the functional units of renal system of rom uri.pdf
Answer:
Kidneys are the functional units of renal system of rom urine and it is essential to maintain
homeostasis. Blood is going to enter into the kidney through renal arteries followed by entry of
afferent arterioles into the \"glomerulus of nephrons\" & leave the filtrate (according to
concentration gradient) finally exists as efferent arteriole
Kidneys are the major organs of urinary system. The filtration of blood takes place in the kidneys
through the nephrons. Two ureters carry the urine from the kidneys to the urinary bladder. From
the bladder, urine is excreted out through the urethra. Relatable analogy is presence of \"similar
ureters\" and two bean shaped kidenys with nephrons
The different stages in glomerular filtration or urine formation are.
Filtration: The movement of water and solutes from the plasma into the renal tubules is known as
filtration. The major driving force of the filtration is hydrostatic pressure in the glomerulus
capillaries.
Blood passes through the capillaries in the glomerulus, due to the hydrostatic pressure in the
Bowman’s capsule, plasma is filtered into the Bowman\'s capsule except blood proteins (this is
called ultra-filtration). This filtrate is called glomerular filtrate. Hydrostatic pressure is the
pressure exerted by the filtrate in Bowman\'s capsule (15 mm Hg).
Absorption (or reabsorption):
It is the movement of water and solutes from the renal tubule into the plasma. The process of
absorption mostly occurs in the proximal tubules (approximately 70% of the filtrate is
reabsorbed from the proximal tubules). The process of reabsorption is driven by hydrostatic and
oncotic pressure, which is strictly under hormonal regulation.
Oncotic pressure is the pressure exerted by the blood proteins, which draws water in to the
circulation.
Secretion:
Tubular reabsorption of water and other substances back to the blood occur while the glomerular
filtrate passes through the tubular portion of nephron. About 99% of water and electrolytes are
reabsorbed during this process by both active and passive reabsorption process. During tubular
secretion, substances pass from the blood into renal tubules.
Excretion:
Urine from the renal tubules enter into the collecting duct à Papillary duct à Calyx à Renal pelvis
à Ureters à Urinary bladder à Urethra (excreted out of the body.).
Normal glomerular filtration rate in the Bowman’s capsule is from 90 to 120 ml / min/1.73 m2.
Usually older people do have a bit lower normal GFR levels as it is naturally reduce with age.
Given GFR is 35mL/min/1.73 m2 (moderate decrease), it tells about the patient’s kidney
function that “a sign of chronic kidney disease” because of diabetes, high blood pressure
followed by glomerulonephritis. It is better control blood pressure by hormonal release.
The three stages of urine formation are 1. Filtration; 2. Tubular reabsorption & secretion; 3.
Water conservation
Glomerular filtration (GFR):
The formation of urine includes
1. Glomerular filtration
.
a. In puple bacteria P870 is present. External electron donors are h.pdf
a. In puple bacteria P870 is present. External electron donors are hydrogen, hydrogen sulphide,
organic molecules like succinate, malate etc. are there. First electron is withdrawn from
Pheophytin.
b. In Green sulfur bacteria source of electron is H2S, when No NADP+ is reduced.
Solution
a. In puple bacteria P870 is present. External electron donors are hydrogen, hydrogen sulphide,
organic molecules like succinate, malate etc. are there. First electron is withdrawn from
Pheophytin.
b. In Green sulfur bacteria source of electron is H2S, when No NADP+ is reduced..
A)ANS)In this TCPIP - System Access Layer, you will find out ab.pdf
A)
ANS)
In this TCP/IP - System Access Layer, you will find out about the base most layer of TCP/IP
convention suit and what are the diverse elements of TCP/IP System Access Layer.
The System Access Layer of the TCP/IP model is connected with the Physical (Layer 1) and the
Information Join (Layer 2) of the OSI model. The System Get to Layer\'s capacity is to move bits
(0s and 1s) over the system medium.
The OSI Physical layer is in charge of changing over the casing into a flood of bits reasonable
for the transmission medium. The OSI Physical layer oversees and synchronizes signals for the
real transmission. On the goal gadget, the Physical layer reassembles these signs into an
information outline.
The OSI Information Join layer is again subdivided into the accompanying two sub layers as
indicated by their capacity:
Media Access Control (Macintosh) Sublayer :— Macintosh sublayer furnishes an interface with
the system connector.
Sensible Connection Control (LLC) Sublayer :— LLC sublayer is in charge of mistake checking
capacities for casings conveyed additionally in charge of overseeing connections between
imparting gadgets.
Structure of an Ethernet Outline
The information parcels from Web Layer is moved to Network Access Layer as it moves down
the TCP/IP convention stack. There is a size restriction for Ethernet Outline. The aggregate size
of the ethernet outline must be between 64 bytes and 1,518 bytes (excluding the preface). System
Access Layer Breaks Web Layer information (IP Datagram) into littler pieces, if essential, which
will end up being the payload of ethernet casings. A Casing incorporates information to be
transmitted furthermore a header and a trailer which contain data that the system connectors on
the ethernet need to handle the edge.
The aggregate size of the ethernet outline must be between 64 bytes and 1,518 bytes (excluding
the introduction). An edge shorter than the base 64 bytes however with a legitimate CRC is
called as a runt. Much of the time, such casings emerge from an impact. Any casing which is
gotten and which is more prominent than the most extreme casing size, is known as a
\"monster\". A \"monster\" is longer than 1518 bytes yet have a substantial CRC. Both runts and
mammoths are considered as invalid.
B.SFD
ANS)
Solution
A)
ANS)
In this TCP/IP - System Access Layer, you will find out about the base most layer of TCP/IP
convention suit and what are the diverse elements of TCP/IP System Access Layer.
The System Access Layer of the TCP/IP model is connected with the Physical (Layer 1) and the
Information Join (Layer 2) of the OSI model. The System Get to Layer\'s capacity is to move bits
(0s and 1s) over the system medium.
The OSI Physical layer is in charge of changing over the casing into a flood of bits reasonable
for the transmission medium. The OSI Physical layer oversees and synchronizes signals for the
real transmission. On the goal gadget, the Physical layer reassembles these signs into an
information ou.
The ice allows the refluxed materials to crystall.pdf
The ice allows the refluxed materials to crystallize, allowing for formation of
crystals and solid product. Also, the ice allows purification via washing, since it gets rid of
contaminants.
Solution
The ice allows the refluxed materials to crystallize, allowing for formation of
crystals and solid product. Also, the ice allows purification via washing, since it gets rid of
contaminants..
this is my code to count the frequency of words in a text file#.pdf
this is my code to count the frequency of words in a text file
#define _GNU_SOURCE
#include
#include
#include
#include
/******************************
here is the Structure for storing words
received from each thread
******************************/
struct return_vall{
char wordlist[100][100];
int count[100];
} *arr;
pthread_t *threads;
int lines;
/*****************************
here are Aggregates Frequency Sum of
different threads
*****************************/
int aggregate(char *a)
{
int sum = 0;
int i=0;
while(1)
{
if(i==lines) break;
int count = 0;
while(1)
{
if(arr[i].count[count] == -1)
{
break;
}
if(strcmp(a,arr[i].wordlist[count]) == 0)
{
sum += arr[i].count[count];
arr[i].count[count] = 0;
}
count++;
}
i++;
}
return sum;
}
/**************************
Function thta are executed by each
thread on seperate line.
**************************/
void *word_count(void* num)
{
int *ln = num;
unsigned int line_number = *ln;
line_number++;
//printf(\"Thread %d\ \",line_number);
char cmd_p1[9] = \"sed -n \'\\0\";
char cmd_p2[2];
sprintf(cmd_p2,\"%d\",line_number);
char cmd_p3[14] = \"p\' \'file.txt\'\\0\";
char command[100];
command[0] = \'\\0\';
strcat(command,cmd_p1);
strcat(command,cmd_p2);
strcat(command,cmd_p3);
//usleep(line_number);
char cmd[100] = \" | tr [:space:] \'\\\ \' | grep -v \'^\\\\s*$\' | sort | uniq -c | sort\\0\";
strcat(command,cmd);
//printf(\"%s\ \",command);
//fflush(stdout);
FILE *in;
in= popen(command, \"r\");
rewind(in);
char buff[50];
int counter = 0;
while(fgets(buff,sizeof(buff),in))
{
char c=\' \';
int i = 0;
int cnt = atoi(buff);
arr[line_number-1].count[counter] = cnt;
while(c!=\'\\0\')
{
c=buff[i];
buff[i]=buff[i+6];
i++;
}
int cnnt = 0;
while(c!=\' \')
{
c = buff[cnnt];
cnnt++;
}
i=0;
while(c!=\'\\0\')
{
c=buff[i];
buff[i]=buff[i+cnnt];
i++;
}
sprintf(arr[line_number-1].wordlist[counter],\"%s\",buff);
//printf(\"%d %s\",arr[line_number-1].count[counter],arr[line_number-
1].wordlist[counter]);
counter++;
}
arr[line_number-1].count[counter] = -1;
fclose(in);
return NULL;
}
int main(void)
{
FILE *fp;
fp = fopen(\"file.txt\",\"r\");
char a[200];
lines = 0;
void *status = NULL;
while(fgets(a,sizeof(a),fp) !=NULL)
{
lines++;
}
fclose(fp);
arr = (struct return_vall*) calloc(lines, sizeof(struct return_vall) * lines);
threads = malloc(sizeof(pthread_t)*lines);
int i;
for(i=0;i= 0)
{
if(arr[i].count[ctr] > 0)
{
strcpy(final[final_ctr],arr[i].wordlist[ctr]);
final_cnt[final_ctr] = arr[i].count[ctr];
arr[i].count[ctr] = 0;
int sum = aggregate(arr[i].wordlist[ctr]);
final_cnt[final_ctr] += sum;
printf(\"%d %s\",final_cnt[final_ctr],final[final_ctr]);
}
ctr++; final_ctr++;
}
i++;
if(i==lines)
{
break;
}
}
free(arr);
return 0;
}
Solution
this is my code to count the frequency of words in a text file
#define _GNU_SOURCE
#include
#include
#include
#include
/******************************
here is the Structure for storing words
received from each thread
******************************/
struct return_vall{
char wordlist[100][100];
.
Recommended
First, lets find out what we need.The question asks for the empi.pdf
First, let\'s find out what we need.
The question asks for the empirical formula of some compound. To do this, we are going to need
to find the amount of moles of each substance.
To find the moles of Sb, all we will need to do is divide the mass of Sb by the molar mass of Sb.
To find the moles of S will be a bit more complicated. Since we know that Sulfur is left over
after the reaction, we know that sulfur is in excess (meaning Sb is the only thing that will limit
our reaction). Assuming this is an ideal gas, we can take the original amount of moles of Sulfur
and subtract it from the moles we get later (after it is transferred to the other container). This will
give us the amount of moles of Sulfur that reacted with the moles of Sb.
So this is what we need:
moles Sb = ?
moles S (reacted with Sb) = ?
Calculate moles Sb.
moles Sb = (mass Sb) / (molar mass Sb) = (0.350g) / (121.76g/mol) = 2.87 x 10^-3 moles
Calculate moles S.
For this, we are going to have to use the PV = nRT equation two times. One for the S that is
placed with the Sb, and one for the S that is placed in the container after the reaction.
Moles of S BEFORE reaction:
PV = nRT
(10.72 atm)(0.100L) = n(0.082)(373.15 K)
n = (10.72 atm)(0.100L) / (0.082)(373.15 K) = 0.035 moles S
Moles of S AFTER reaction:
PV = nRT
(14.8 atm)(0.050L) = n(0.082)(323.15 K)
n = (14.8 atm)(0.050L) / (0.082)(323.15 K) = 0.028 moles S
Calculate moles of S that reacted with the Sb:
Since the moles of S went down after the reaction, we know that the moles of S lost must have
reacted with the Sb. So, we can simply find the change in moles of S to find the amount of S that
reacted with the Sb.
Moles of S before reaction - Moles of S after reaction = Moles of S reacted
0.035 moles S - 0.028 moles S = 7.07 x 10^-3 moles S
Now that we have our moles of each element that reacted, we can find the empirical formula.
To do this, we need to divide through by the element with the least amount of moles. So, in this
case, we must divide through by the moles of Sb.
(2.87 x 10^-3 moles Sb) / (2.87 x 10^-3 moles Sb) = 1
(7.07 x 10^-3 moles S) / (2.87 x 10^-3 moles Sb) = 2.46 = 2.5
We need subscripts that are integers in empirical formulas. To make 2.5 an integer, we can
simply multiply it by 2 to make it 5. When we do this, we also need to multiply 1 by 2.
2.5 x 2 = 5
1 x 2 = 2
These numbers represent the subscripts for each of our elements in our compound.
So the empirical formula would be:
Sb2S5
Solution
First, let\'s find out what we need.
The question asks for the empirical formula of some compound. To do this, we are going to need
to find the amount of moles of each substance.
To find the moles of Sb, all we will need to do is divide the mass of Sb by the molar mass of Sb.
To find the moles of S will be a bit more complicated. Since we know that Sulfur is left over
after the reaction, we know that sulfur is in excess (meaning Sb is the only thing that will limit
our reaction). Assuming this is an ideal gas, we can take.
DiversityDiversity scheme refers to a method for improving the re.pdf
Diversity:
Diversity scheme refers to a method for improving the reliability of a message signal by using
two or more communication channels with different characteristics. Diversity is mainly used in
radio communication and is a common technique for combatting fading and co-channel
interference and avoiding error bursts. It is based on the fact that individual channels experience
different levels of fading and interference. Multiple versions of the same signal may be
transmitted and/or received and combined in the receiver
Types:
In Micro diversity The spacing between transmitters are shorter than wavelength of transmitted
signal
In Macro diversity The spacing between transmitters are longer than wavelength of transmitted
signal
Types of microdiversity:
Time diversity: Multiple versions of the same signal are transmitted at different time instants.
Alternatively, a redundant forward error correction codeis added and the message is spread in
time by means of bit-interleaving before it is transmitted. Thus, error bursts are avoided, which
simplifies the error correction.
Frequency diversity: The signal is transmitted using several frequency channels or spread over a
wide spectrum that is affected by frequency-selective fading. Middle-late 20th century
microwave radio relay lines often used several regular wideband radio channels, and one
protection channel for automatic use by any faded channel. Later examples include:
OFDM modulation in combination with subcarrier interleaving and forward error correction
Spread spectrum, for example frequency hopping or DS-CDMA.
Space diversity: The signal is transmitted over several different propagation paths. In the case of
wired transmission, this can be achieved by transmitting via multiple wires. In the case of
wireless transmission, it can be achieved by antenna diversity using multiple transmitter antennas
(transmit diversity) and/or multiple receiving antennas (reception diversity). In the latter case, a
diversity combining technique is applied before further signal processing takes place. If the
antennas are far apart, for example at different cellular base station sites or WLAN access points,
this is calledmacrodiversity or site diversity. If the antennas are at a distance in the order of one
wavelength, this is called microdiversity. A special case is phasedantenna arrays, which also can
be used for beamforming, MIMO channels and space–time coding (STC).
Polarization diversity: Multiple versions of a signal are transmitted and received via antennas
with different polarization. A diversity combiningtechnique is applied on the receiver side.
Solution
Diversity:
Diversity scheme refers to a method for improving the reliability of a message signal by using
two or more communication channels with different characteristics. Diversity is mainly used in
radio communication and is a common technique for combatting fading and co-channel
interference and avoiding error bursts. It is based on the fact that i.
due to repulsions between lone pair and bond pair of electrons on Te.pdf
due to repulsions between lone pair and bond pair of electrons on Te ,TeCl2 adopt bent shape
.but Be does not have lone pair ,so BeCl2 adopt linear structure.
Solution
due to repulsions between lone pair and bond pair of electrons on Te ,TeCl2 adopt bent shape
.but Be does not have lone pair ,so BeCl2 adopt linear structure..
AnswerKidneys are the functional units of renal system of rom uri.pdf
Answer:
Kidneys are the functional units of renal system of rom urine and it is essential to maintain
homeostasis. Blood is going to enter into the kidney through renal arteries followed by entry of
afferent arterioles into the \"glomerulus of nephrons\" & leave the filtrate (according to
concentration gradient) finally exists as efferent arteriole
Kidneys are the major organs of urinary system. The filtration of blood takes place in the kidneys
through the nephrons. Two ureters carry the urine from the kidneys to the urinary bladder. From
the bladder, urine is excreted out through the urethra. Relatable analogy is presence of \"similar
ureters\" and two bean shaped kidenys with nephrons
The different stages in glomerular filtration or urine formation are.
Filtration: The movement of water and solutes from the plasma into the renal tubules is known as
filtration. The major driving force of the filtration is hydrostatic pressure in the glomerulus
capillaries.
Blood passes through the capillaries in the glomerulus, due to the hydrostatic pressure in the
Bowman’s capsule, plasma is filtered into the Bowman\'s capsule except blood proteins (this is
called ultra-filtration). This filtrate is called glomerular filtrate. Hydrostatic pressure is the
pressure exerted by the filtrate in Bowman\'s capsule (15 mm Hg).
Absorption (or reabsorption):
It is the movement of water and solutes from the renal tubule into the plasma. The process of
absorption mostly occurs in the proximal tubules (approximately 70% of the filtrate is
reabsorbed from the proximal tubules). The process of reabsorption is driven by hydrostatic and
oncotic pressure, which is strictly under hormonal regulation.
Oncotic pressure is the pressure exerted by the blood proteins, which draws water in to the
circulation.
Secretion:
Tubular reabsorption of water and other substances back to the blood occur while the glomerular
filtrate passes through the tubular portion of nephron. About 99% of water and electrolytes are
reabsorbed during this process by both active and passive reabsorption process. During tubular
secretion, substances pass from the blood into renal tubules.
Excretion:
Urine from the renal tubules enter into the collecting duct à Papillary duct à Calyx à Renal pelvis
à Ureters à Urinary bladder à Urethra (excreted out of the body.).
Normal glomerular filtration rate in the Bowman’s capsule is from 90 to 120 ml / min/1.73 m2.
Usually older people do have a bit lower normal GFR levels as it is naturally reduce with age.
Given GFR is 35mL/min/1.73 m2 (moderate decrease), it tells about the patient’s kidney
function that “a sign of chronic kidney disease” because of diabetes, high blood pressure
followed by glomerulonephritis. It is better control blood pressure by hormonal release.
The three stages of urine formation are 1. Filtration; 2. Tubular reabsorption & secretion; 3.
Water conservation
Glomerular filtration (GFR):
The formation of urine includes
1. Glomerular filtration
.
a. In puple bacteria P870 is present. External electron donors are h.pdf
a. In puple bacteria P870 is present. External electron donors are hydrogen, hydrogen sulphide,
organic molecules like succinate, malate etc. are there. First electron is withdrawn from
Pheophytin.
b. In Green sulfur bacteria source of electron is H2S, when No NADP+ is reduced.
Solution
a. In puple bacteria P870 is present. External electron donors are hydrogen, hydrogen sulphide,
organic molecules like succinate, malate etc. are there. First electron is withdrawn from
Pheophytin.
b. In Green sulfur bacteria source of electron is H2S, when No NADP+ is reduced..
A)ANS)In this TCPIP - System Access Layer, you will find out ab.pdf
A)
ANS)
In this TCP/IP - System Access Layer, you will find out about the base most layer of TCP/IP
convention suit and what are the diverse elements of TCP/IP System Access Layer.
The System Access Layer of the TCP/IP model is connected with the Physical (Layer 1) and the
Information Join (Layer 2) of the OSI model. The System Get to Layer\'s capacity is to move bits
(0s and 1s) over the system medium.
The OSI Physical layer is in charge of changing over the casing into a flood of bits reasonable
for the transmission medium. The OSI Physical layer oversees and synchronizes signals for the
real transmission. On the goal gadget, the Physical layer reassembles these signs into an
information outline.
The OSI Information Join layer is again subdivided into the accompanying two sub layers as
indicated by their capacity:
Media Access Control (Macintosh) Sublayer :— Macintosh sublayer furnishes an interface with
the system connector.
Sensible Connection Control (LLC) Sublayer :— LLC sublayer is in charge of mistake checking
capacities for casings conveyed additionally in charge of overseeing connections between
imparting gadgets.
Structure of an Ethernet Outline
The information parcels from Web Layer is moved to Network Access Layer as it moves down
the TCP/IP convention stack. There is a size restriction for Ethernet Outline. The aggregate size
of the ethernet outline must be between 64 bytes and 1,518 bytes (excluding the preface). System
Access Layer Breaks Web Layer information (IP Datagram) into littler pieces, if essential, which
will end up being the payload of ethernet casings. A Casing incorporates information to be
transmitted furthermore a header and a trailer which contain data that the system connectors on
the ethernet need to handle the edge.
The aggregate size of the ethernet outline must be between 64 bytes and 1,518 bytes (excluding
the introduction). An edge shorter than the base 64 bytes however with a legitimate CRC is
called as a runt. Much of the time, such casings emerge from an impact. Any casing which is
gotten and which is more prominent than the most extreme casing size, is known as a
\"monster\". A \"monster\" is longer than 1518 bytes yet have a substantial CRC. Both runts and
mammoths are considered as invalid.
B.SFD
ANS)
Solution
A)
ANS)
In this TCP/IP - System Access Layer, you will find out about the base most layer of TCP/IP
convention suit and what are the diverse elements of TCP/IP System Access Layer.
The System Access Layer of the TCP/IP model is connected with the Physical (Layer 1) and the
Information Join (Layer 2) of the OSI model. The System Get to Layer\'s capacity is to move bits
(0s and 1s) over the system medium.
The OSI Physical layer is in charge of changing over the casing into a flood of bits reasonable
for the transmission medium. The OSI Physical layer oversees and synchronizes signals for the
real transmission. On the goal gadget, the Physical layer reassembles these signs into an
information ou.
The ice allows the refluxed materials to crystall.pdf
The ice allows the refluxed materials to crystallize, allowing for formation of
crystals and solid product. Also, the ice allows purification via washing, since it gets rid of
contaminants.
Solution
The ice allows the refluxed materials to crystallize, allowing for formation of
crystals and solid product. Also, the ice allows purification via washing, since it gets rid of
contaminants..
this is my code to count the frequency of words in a text file#.pdf
this is my code to count the frequency of words in a text file
#define _GNU_SOURCE
#include
#include
#include
#include
/******************************
here is the Structure for storing words
received from each thread
******************************/
struct return_vall{
char wordlist[100][100];
int count[100];
} *arr;
pthread_t *threads;
int lines;
/*****************************
here are Aggregates Frequency Sum of
different threads
*****************************/
int aggregate(char *a)
{
int sum = 0;
int i=0;
while(1)
{
if(i==lines) break;
int count = 0;
while(1)
{
if(arr[i].count[count] == -1)
{
break;
}
if(strcmp(a,arr[i].wordlist[count]) == 0)
{
sum += arr[i].count[count];
arr[i].count[count] = 0;
}
count++;
}
i++;
}
return sum;
}
/**************************
Function thta are executed by each
thread on seperate line.
**************************/
void *word_count(void* num)
{
int *ln = num;
unsigned int line_number = *ln;
line_number++;
//printf(\"Thread %d\ \",line_number);
char cmd_p1[9] = \"sed -n \'\\0\";
char cmd_p2[2];
sprintf(cmd_p2,\"%d\",line_number);
char cmd_p3[14] = \"p\' \'file.txt\'\\0\";
char command[100];
command[0] = \'\\0\';
strcat(command,cmd_p1);
strcat(command,cmd_p2);
strcat(command,cmd_p3);
//usleep(line_number);
char cmd[100] = \" | tr [:space:] \'\\\ \' | grep -v \'^\\\\s*$\' | sort | uniq -c | sort\\0\";
strcat(command,cmd);
//printf(\"%s\ \",command);
//fflush(stdout);
FILE *in;
in= popen(command, \"r\");
rewind(in);
char buff[50];
int counter = 0;
while(fgets(buff,sizeof(buff),in))
{
char c=\' \';
int i = 0;
int cnt = atoi(buff);
arr[line_number-1].count[counter] = cnt;
while(c!=\'\\0\')
{
c=buff[i];
buff[i]=buff[i+6];
i++;
}
int cnnt = 0;
while(c!=\' \')
{
c = buff[cnnt];
cnnt++;
}
i=0;
while(c!=\'\\0\')
{
c=buff[i];
buff[i]=buff[i+cnnt];
i++;
}
sprintf(arr[line_number-1].wordlist[counter],\"%s\",buff);
//printf(\"%d %s\",arr[line_number-1].count[counter],arr[line_number-
1].wordlist[counter]);
counter++;
}
arr[line_number-1].count[counter] = -1;
fclose(in);
return NULL;
}
int main(void)
{
FILE *fp;
fp = fopen(\"file.txt\",\"r\");
char a[200];
lines = 0;
void *status = NULL;
while(fgets(a,sizeof(a),fp) !=NULL)
{
lines++;
}
fclose(fp);
arr = (struct return_vall*) calloc(lines, sizeof(struct return_vall) * lines);
threads = malloc(sizeof(pthread_t)*lines);
int i;
for(i=0;i= 0)
{
if(arr[i].count[ctr] > 0)
{
strcpy(final[final_ctr],arr[i].wordlist[ctr]);
final_cnt[final_ctr] = arr[i].count[ctr];
arr[i].count[ctr] = 0;
int sum = aggregate(arr[i].wordlist[ctr]);
final_cnt[final_ctr] += sum;
printf(\"%d %s\",final_cnt[final_ctr],final[final_ctr]);
}
ctr++; final_ctr++;
}
i++;
if(i==lines)
{
break;
}
}
free(arr);
return 0;
}
Solution
this is my code to count the frequency of words in a text file
#define _GNU_SOURCE
#include
#include
#include
#include
/******************************
here is the Structure for storing words
received from each thread
******************************/
struct return_vall{
char wordlist[100][100];
.
#In this project you will write a program play TicTacToe #using tw.pdf
#In this project you will write a program play TicTacToe
#using two players (labels 0,1) or one play (label 0) playing with the machine (label 1).
#The TicTacToe board has 9 integers board = [1,2,3,4,5,6,7,8,9]. The following
# are the modules for the program
#
#def reset() resets the board to the original values
# board = [1,2,3,4,5,6,7,8,9]
#
#def printBoard() print the current state of the board using the format
#
#The current TicTacToe Board
# | 1 | 2 | 3 |
# | 4 | 5 | 6 |
# | 7 | O | 9 |
#
#The current TicTacToe Board
# | X | 2 | 3 |
# | 4 | 5 | 6 |
# | 7 | O | 9 |
#Note from the above that player 0 and 1 have played numbers 8 and 1
#respectively and the board display O for player 0 and X from player 1
#
#def changeBoard(num1, player) using the chosen box number to change
#the value of the box to 0 or -1 depending on whether the player is 0 or 1,
#respectively.
#
#def play(player) prints the player number (0 or 1) and prompts the player
# to enter a box value that have not changed to \'O\' or \'X\'
#
#def checkRows(value) checks to see which of the rows of the board
# has the same value and returns True, otherwise, returns False
#
#def checkCols(value) checks to see which of the cols of the board
# has the same value and returns True, otherwise, returns False
#
#def checkDiagonal(value) checks to see which of the diagonals of the board
# has the same value and returns True, otherwise, returns False
#
#def win(player) checks if a player wins the game, returns True of the player wins
# and False otherwise
#
#def machinePlay(player) plays the role of the player using random number.
# this function generates numbers in the interval [1,9] and uses the first
# generated random number that has not been used to play the game. The
#
#def ticTacToe(numPlayers) accepts the number of players and simulates the
#ticTapToe, asking players to enter unused box numbers.
#
#def main() is the driver module that accepts the number of players from the user
# and calls the ticTacToe module
#Assignment: Complete the follwoing modules:
#checkCols
#checkRows
#checkDiagonal
#win
#Sample of the output is
#
#Project
#This game can be played by one or two players
#Enter the number of players, 1/2 for one/two players: 1
#
#The current TicTapToe Board
#| 1 | 2 | 3 |
#| 4 | 5 | 6 |
#| 7 | 8 | 9 |
#Player 0 Enter a box value: 9
#
#The current TicTapToe Board
#| 1 | 2 | 3 |
#| 4 | 5 | 6 |
#| 7 | 8 | O |
#Player 1 ***Computer*** playing
#The current TicTapToe Board
#| 1 | 2 | 3 |
#| 4 | 5 | 6 |
#| 7 | X | O |
#Player 0 Enter a box value: 1
#
#The current TicTapToe Board
#| O | 2 | 3 |
#| 4 | 5 | 6 |
#| 7 | X | O |
#Player 1 ***Computer*** playing
#
#The current TicTapToe Board
#| O | X | 3 |
#| 4 | 5 | 6 |
#| 7 | X | O |
#Player 0 Enter a box value: 5
#
#The current TicTapToe Board
#| O | X | 3 |
#| 4 | O | 6 |
#| 7 | X | O |
#Player 0 Wins
#Do you want to quit?
#Begin Program
#import random number generator
#from random library
#from random import randint, seed
#Global va.
1.1Yearcash flowpresent value of cash inflow = cash flow(1+r).pdf
1.1
Year
cash flow
present value of cash inflow = cash flow/(1+r)^n r= 5%
1
75
71.42857
2
75
68.02721
3
75
64.78782
4
1075
884.4052
value of bond
sum of present value of cash flow
1088.649
1.2
Year
cash flow
present value of cash inflow = cash flow/(1+r)^n r= 5.2%
1
75
71.29278
2
75
67.7688
3
75
64.41901
4
1075
877.6988
value of bond
sum of present value of cash flow
1081.179
1.3
% change in bond price if YTM increases by .2%
(1081.18-1088.65)/1088.65
-0.69%
1.4
Duration is a measure of a bond\'s sensitivity to interest rate changes. The higher the bond\'s
duration, the greater its sensitivity to the change (also know as volatility) and vice versa.
1.4
Year
cash flow
Year*cash flow
present value of Year*cash inflow = cash flow/(1+r)^n r= 5%
1
75
75
71.42857
2
75
150
136.0544
3
75
225
194.3635
4
1075
4300
3537.621
total
3939.467
Macculay duration in years
total/face value of bond
3939.467/1000
3.939467
1.4
modified duration
macculay duration/(1+YTM)
3.94/1.05
3.752381
1.5
change in price if YTM increases by .2%
modified duration*change in YTM
3.75*.2%
0.75%
1.1
Year
cash flow
present value of cash inflow = cash flow/(1+r)^n r= 5%
1
75
71.42857
2
75
68.02721
3
75
64.78782
4
1075
884.4052
value of bond
sum of present value of cash flow
1088.649
1.2
Year
cash flow
present value of cash inflow = cash flow/(1+r)^n r= 5.2%
1
75
71.29278
2
75
67.7688
3
75
64.41901
4
1075
877.6988
value of bond
sum of present value of cash flow
1081.179
1.3
% change in bond price if YTM increases by .2%
(1081.18-1088.65)/1088.65
-0.69%
1.4
Duration is a measure of a bond\'s sensitivity to interest rate changes. The higher the bond\'s
duration, the greater its sensitivity to the change (also know as volatility) and vice versa.
1.4
Year
cash flow
Year*cash flow
present value of Year*cash inflow = cash flow/(1+r)^n r= 5%
1
75
75
71.42857
2
75
150
136.0544
3
75
225
194.3635
4
1075
4300
3537.621
total
3939.467
Macculay duration in years
total/face value of bond
3939.467/1000
3.939467
1.4
modified duration
macculay duration/(1+YTM)
3.94/1.05
3.752381
1.5
change in price if YTM increases by .2%
modified duration*change in YTM
3.75*.2%
0.75%
Solution
1.1
Year
cash flow
present value of cash inflow = cash flow/(1+r)^n r= 5%
1
75
71.42857
2
75
68.02721
3
75
64.78782
4
1075
884.4052
value of bond
sum of present value of cash flow
1088.649
1.2
Year
cash flow
present value of cash inflow = cash flow/(1+r)^n r= 5.2%
1
75
71.29278
2
75
67.7688
3
75
64.41901
4
1075
877.6988
value of bond
sum of present value of cash flow
1081.179
1.3
% change in bond price if YTM increases by .2%
(1081.18-1088.65)/1088.65
-0.69%
1.4
Duration is a measure of a bond\'s sensitivity to interest rate changes. The higher the bond\'s
duration, the greater its sensitivity to the change (also know as volatility) and vice versa.
1.4
Year
cash flow
Year*cash flow
present value of Year*cash inflow = cash flow/(1+r)^n r= 5%
1
75
75
71.42857
2
75
150
136.0544
3
75
225
194.3635
4
1075
4300
3537.621
total
3939.467
M.
HNO2 is a much weaker acid then HNO3 (strong acid.pdf
HNO2 is a much weaker acid then HNO3 (strong acid dissociates completely)
H2SeO3 is weaker than H3AsO4 but only slightly as both are triprotic acids H2SO4 is weaker
than HClO4 but bother are very strong acids.
Solution
HNO2 is a much weaker acid then HNO3 (strong acid dissociates completely)
H2SeO3 is weaker than H3AsO4 but only slightly as both are triprotic acids H2SO4 is weaker
than HClO4 but bother are very strong acids..
1. Identify five differences between DNA replication and gene transc.pdf
1. Identify five differences between DNA replication and gene transcription.
DNA replication
gene transcription.
occurs in preparation for cell division,
Produces two two copies of DNA, double stranded
Uses DNA as template to form DNA
Catalyzed by DNA polymerase
the end result is two daughter cells.
happens in preparation for protein translation.
Produces mRNA, single stranded
Uses template DNA as a template and forms RNA
Catalyzed by RNA polymerase
End result is mRNA which is then exported out of the nucleus for translation
2. Identify four differences between DNA and RNA.
DNA
RNA
Deoxyribonucleic acid is the hereditary material in most of the organisms
Made of 2-deoxyribose sugar
Contains thymine as one of the 4 nitrogen base
It is double stranded
Ribonucleic acid transfers genetic code needed for the creation of proteins from the nucleus to
the ribosome.
Made of ribose sugar
Has Uracil as the nitrogen base instead of thymine
Single stranded
DNA replication
gene transcription.
occurs in preparation for cell division,
Produces two two copies of DNA, double stranded
Uses DNA as template to form DNA
Catalyzed by DNA polymerase
the end result is two daughter cells.
happens in preparation for protein translation.
Produces mRNA, single stranded
Uses template DNA as a template and forms RNA
Catalyzed by RNA polymerase
End result is mRNA which is then exported out of the nucleus for translation
Solution
1. Identify five differences between DNA replication and gene transcription.
DNA replication
gene transcription.
occurs in preparation for cell division,
Produces two two copies of DNA, double stranded
Uses DNA as template to form DNA
Catalyzed by DNA polymerase
the end result is two daughter cells.
happens in preparation for protein translation.
Produces mRNA, single stranded
Uses template DNA as a template and forms RNA
Catalyzed by RNA polymerase
End result is mRNA which is then exported out of the nucleus for translation
2. Identify four differences between DNA and RNA.
DNA
RNA
Deoxyribonucleic acid is the hereditary material in most of the organisms
Made of 2-deoxyribose sugar
Contains thymine as one of the 4 nitrogen base
It is double stranded
Ribonucleic acid transfers genetic code needed for the creation of proteins from the nucleus to
the ribosome.
Made of ribose sugar
Has Uracil as the nitrogen base instead of thymine
Single stranded
DNA replication
gene transcription.
occurs in preparation for cell division,
Produces two two copies of DNA, double stranded
Uses DNA as template to form DNA
Catalyzed by DNA polymerase
the end result is two daughter cells.
happens in preparation for protein translation.
Produces mRNA, single stranded
Uses template DNA as a template and forms RNA
Catalyzed by RNA polymerase
End result is mRNA which is then exported out of the nucleus for translation.
1. b2. c (bed bug)3. d (caterpillars)4. c (lyme disease)5. e.pdf
1. b
2. c (bed bug)
3. d (caterpillars)
4. c (lyme disease)
5. e (chagas disease)
6. a (hives)
7. e (short lived)
8. d (plants)
9. a (1892)
10. e (parvoviridae)
11. mosquito
12. a (flies)
13. c (lyme disease)
14.
15. --- holometabolous is the type of metamorphosis in beetles
16. oligocene
17. e (6 or more)
18. e
Solution
1. b
2. c (bed bug)
3. d (caterpillars)
4. c (lyme disease)
5. e (chagas disease)
6. a (hives)
7. e (short lived)
8. d (plants)
9. a (1892)
10. e (parvoviridae)
11. mosquito
12. a (flies)
13. c (lyme disease)
14.
15. --- holometabolous is the type of metamorphosis in beetles
16. oligocene
17. e (6 or more)
18. e.
1) MODERN TIMES is a very well prepared movie that describes the mis.pdf
1) MODERN TIMES is a very well prepared movie that describes the miserable conditions of
humar resource during the period of industrialization era or machine age worlwide. It is a mix of
social drama and comedy. The content is presented in a satire. How did people survived the great
depression of the world is being documented here. The conditions werre miserable like acute
shortage of work, factory closings, labor strikes and political unrest etc. It is a symbolic
representation of what happens to a human being caught up in the mechanical bureaucracy of the
intellectual state.
2) Economically it could be understood that what miser situations could depression bring up.
Food and jobs comes hard in hand. To which a labor is convict to do under skill jobs as well for
survival. At the time of the great depression Americans had to fight for what little they had, and
as the movie showed many people went hungry. Modern Times looks toward the future, but not
with enthusiasm. Often described as a satire of the machine age, It has in fact a broader theme:
the dehumanizing effects of many aspects of modernity, including industrialization, bureaucracy,
urbanization, and law enforcement.
3) From the movie a prospect employee should not forget about the social ills and failings of
own day even if present age is seems far better than that was shown in the film. A dilbert-like
early scene shows the Tramp trying to take a break during work hours in the washroom — only
to have a big-screen image of the boss’s head suddenly appear on the washroom wall and order
him to quit stalling and get back to work. Modern audiences watching this scene may reflect
ruefully on the everyday reality of electronic surveillance in the workplace. Now, obviously
there’s nothing new about widespread poverty and want, but unemployment as we know it today
is a comparatively recent development — a byproduct of the modern workforce economy that
requires everyone to \"get a job.\"
4) The major ethical dilemma consists whether out of povertty stealing is right or wrong.
However this movie glorifies the struggle of the poor rather than vilifying them for stealing food
to survive. By glorifying the struggle the movie takes on yet another social issue: when the poor
steal to eat it is not criminal, just survival. Minor ones include
Solution
1) MODERN TIMES is a very well prepared movie that describes the miserable conditions of
humar resource during the period of industrialization era or machine age worlwide. It is a mix of
social drama and comedy. The content is presented in a satire. How did people survived the great
depression of the world is being documented here. The conditions werre miserable like acute
shortage of work, factory closings, labor strikes and political unrest etc. It is a symbolic
representation of what happens to a human being caught up in the mechanical bureaucracy of the
intellectual state.
2) Economically it could be understood that what miser situations could depress.
An ion is an atom that has a charge. These charges can be postive or.pdf
An ion is an atom that has a charge. These charges can be postive or negative and depend upon
the amount of electrions lost or gained. Ions are formed when they gain or lose an election ( a
change in valence electrons) An example is Li+. That is a Lithium atom with a positive charge, it
has lost one electron.
A cation is a positively charged ion. Li+
An anion is a negatively charged ion. F-. A Fluorine atom that has gained an electron.
A polyatomic ion is a multi-atom molecule with a charge. Such as ammonia NH4+.
Solution
An ion is an atom that has a charge. These charges can be postive or negative and depend upon
the amount of electrions lost or gained. Ions are formed when they gain or lose an election ( a
change in valence electrons) An example is Li+. That is a Lithium atom with a positive charge, it
has lost one electron.
A cation is a positively charged ion. Li+
An anion is a negatively charged ion. F-. A Fluorine atom that has gained an electron.
A polyatomic ion is a multi-atom molecule with a charge. Such as ammonia NH4+..
Acids are substances that form hydrogen ions, H+(.pdf
Acids are substances that form hydrogen ions, H+(aq), when dissolved in water In
acid solutions there are more H+ ions than OH- ions. Bases which are soluble in water are called
alkalis e.g. NaOH sodium hydroxide, KOH potassium hydroxide or Ca(OH)2 calcium hydroxide.
Bases which are water insoluble include CuO copper(II) oxide, MgO magnesium oxide. After a
neutralisation, the salt solutions consist of a mixture of positive and negative ions (and their
names are in the salt name!) e.g. sodium chloride (NaCl) is a mixture of Na+ and Cl- ions,
calcium chloride (CaCl2) is a mix of Ca2+ and Cl- ions; magnesium nitrate (Mg(NO3)2) is a
mix of Mg2+ and NO3- ions, aluminium sulphate (Al2(SO4)3) consists of Al3+ and SO42- ions
etc. You can test for chloride ions by adding a few drops of dilute nitric acid followed by a few
drops of silver nitrate solution. If chloride ions are present you will see a white precipitate of
silver chloride. However, chlorine water itself will contain chloride ions as chlorine in aqueous
solution disproportionates to form chloride and chlorate(I) ions. Cl2 + H2O --> Cl- + ClO- +
2H+ Testing for the presence of nitrate via wet chemistry is generally difficult compared with
testing for other anions, as almost all nitrates are soluble in water. In contrast, many common
ions give insoluble salts, e.g. halides precipitate with silver, and sulfates precipitate with barium.
The nitrate anion is an oxidizer, and many tests for the nitrate anion are based on this property.
Unfortunately, other oxidants present in the analyte may interfere and give erroneous results.
Solution
Acids are substances that form hydrogen ions, H+(aq), when dissolved in water In
acid solutions there are more H+ ions than OH- ions. Bases which are soluble in water are called
alkalis e.g. NaOH sodium hydroxide, KOH potassium hydroxide or Ca(OH)2 calcium hydroxide.
Bases which are water insoluble include CuO copper(II) oxide, MgO magnesium oxide. After a
neutralisation, the salt solutions consist of a mixture of positive and negative ions (and their
names are in the salt name!) e.g. sodium chloride (NaCl) is a mixture of Na+ and Cl- ions,
calcium chloride (CaCl2) is a mix of Ca2+ and Cl- ions; magnesium nitrate (Mg(NO3)2) is a
mix of Mg2+ and NO3- ions, aluminium sulphate (Al2(SO4)3) consists of Al3+ and SO42- ions
etc. You can test for chloride ions by adding a few drops of dilute nitric acid followed by a few
drops of silver nitrate solution. If chloride ions are present you will see a white precipitate of
silver chloride. However, chlorine water itself will contain chloride ions as chlorine in aqueous
solution disproportionates to form chloride and chlorate(I) ions. Cl2 + H2O --> Cl- + ClO- +
2H+ Testing for the presence of nitrate via wet chemistry is generally difficult compared with
testing for other anions, as almost all nitrates are soluble in water. In contrast, many common
ions give insoluble salts, e.g. halides precipitate with silver, and sulfates .
ANSWER1.Apocalypse Now film overall formIt is the height o.pdf
ANSWER:
1.\"Apocalypse Now\" film overall form:
It is the height of the war in Vietnam, and U.S. Army Captain Willard is sent by Colonel Lucas
and a General to carry out a mission that, officially, \'does not exist - nor will it ever exist\'. The
mission: To seek out a mysterious Green Beret Colonel, Walter Kurtz, whose army has crossed
the border into Cambodia and is conducting hit-and-run missions against the Viet Cong and
NVA. The army believes Kurtz has gone completely insane and Willard\'s job is to eliminate
him! Willard, sent up the Nung River on a U.S. Navy patrol boat, discovers that his target is one
of the most decorated officers in the U.S. Army. His crew meets up with surfer-type Lt-Colonel
Kilgore, head of a U.S Army helicopter cavalry group which eliminates a Viet Cong outpost to
provide an entry point into the Nung River. After some hair-raising encounters, in which some of
his crew are killed, Willard, Lance and Chef reach Colonel Kurtz\'s outpost, beyond the Do Lung
Bridge.
“This is the end, beautiful friend…
This is the end, my only friend, the end
Of our elaborate plans, the end
Of everything that stands, the end
No safety or surprise, the end
I\'ll never look into your eyes...again
The End...”
2.Primary Techniques in Apocalypse Film:
Runtime
3.Techniques:
Although many scenes are unusual since the movie begin an unusual journey in the midst of an
alien world, some of the scenes still stand out.One of the most effective techniques used in the
movie is the usage of very long dissolves. Many scenes are dissolved on each other until they
become a montage of their own in one shot itself and say something completely distinct from
their individual meaning. This was used in the beginning of the movie, in the middle when the
journey enters the final realm and in the end again giving the seemingly untying themes a
thematic unity.
Many times, the war is superimposed on Willard’s face as if he is living it & re-living it long
after it has gone or perhaps long before it has happened...! (Slow Dissolves)
When the boat is finally entering the waters of Cambodia and we know that it is a point of no
return for this boat and especially Willard, a few images are superimposed on each other to
create a heightened sense of an altered reality. The boat is passing through the gates of the
underworld and right in the center of all of this is Willard trying to look, listen and comprehend.
(Slow Dissolve To Gates Of Hell1, Slow Dissolve To Gates Of Hell2, Slow Dissolve To Gates
Of Hell3, Slow Dissolve To Gates Of Hell4, Slow Dissolve To Gates Of Hell5)
Another highly unusual shot worth mentioning is Visual Constancy Or Horror. When I saw this
shot, I thought that if my eyes did not have visual constancy and did not fill in Willard’s body
automatically, the body-less face in this scene would be part of the true horror genre...!
4.Techniques and patterns:
a)Lighting:
Lighting is the key element in the film. Along with fogs and shadows. The general lo.
15.0mL = 0.015L phenylacetic acid 0.50molL 0.01.pdf
15.0mL = 0.015L phenylacetic acid 0.50mol/L *0.015= 0.0075 mol phenylacetic
acid 20.0mL= 0.020L sodium phenylacetate 0.40mol/L * 0.020L=.008 mol sodium
phenylacetate due to the fact that this acts as a buffer Using henderson hasselblach pH= pKa +
log(base/acid) and finding that the pKa of phenylaceticacid is 4.28 pH= 4.28 + log (.008/.0075)
pH= 4.28 + 0.093 pH= 4.373
Solution
15.0mL = 0.015L phenylacetic acid 0.50mol/L *0.015= 0.0075 mol phenylacetic
acid 20.0mL= 0.020L sodium phenylacetate 0.40mol/L * 0.020L=.008 mol sodium
phenylacetate due to the fact that this acts as a buffer Using henderson hasselblach pH= pKa +
log(base/acid) and finding that the pKa of phenylaceticacid is 4.28 pH= 4.28 + log (.008/.0075)
pH= 4.28 + 0.093 pH= 4.373.
4 water molecules are like a magnet because beca.pdf
4: water molecules are like a magnet because because it acts like water 5: As
mentioned above, humans have sweat glands over their entire bodies; the only place dogs (and
cats, and many other animals with fur over their whole bodies) can get rid of excess heat is
through the tongue. By the way, sweating/panting is NOT a means of getting rid of excess water;
it is a way of getting rid of heat. We do this because of the process of evaporation. In changing
from liquid to vapor, the water absorbs heat from the surface on which it is laying. Actually,
most organisms do not want to get rid of excess water; they need to conserve it, but on very hot
days animals \"spend\" the water to achieve the cooling. 6. It can dissolve the most types of
substances. it can dissolve most substances This is because solutes for the most part are able to
dissolve in water with the exception of some solutes like oil. Most of the solids dissolve in water.
Solution
4: water molecules are like a magnet because because it acts like water 5: As
mentioned above, humans have sweat glands over their entire bodies; the only place dogs (and
cats, and many other animals with fur over their whole bodies) can get rid of excess heat is
through the tongue. By the way, sweating/panting is NOT a means of getting rid of excess water;
it is a way of getting rid of heat. We do this because of the process of evaporation. In changing
from liquid to vapor, the water absorbs heat from the surface on which it is laying. Actually,
most organisms do not want to get rid of excess water; they need to conserve it, but on very hot
days animals \"spend\" the water to achieve the cooling. 6. It can dissolve the most types of
substances. it can dissolve most substances This is because solutes for the most part are able to
dissolve in water with the exception of some solutes like oil. Most of the solids dissolve in water..
MARUTI SUZUKI- A Successful Joint Venture in India.pptx
Let us know about Maruti Suzuki, a successful Joint venture in India.
Students, digital devices and success - Andreas Schleicher - 27 May 2024..pptx
Andreas Schleicher presents at the OECD webinar ‘Digital devices in schools: detrimental distraction or secret to success?’ on 27 May 2024. The presentation was based on findings from PISA 2022 results and the webinar helped launch the PISA in Focus ‘Managing screen time: How to protect and equip students against distraction’ https://www.oecd-ilibrary.org/education/managing-screen-time_7c225af4-en and the OECD Education Policy Perspective ‘Students, digital devices and success’ can be found here - https://oe.cd/il/5yV
How to Break the cycle of negative Thoughts
We all have good and bad thoughts from time to time and situation to situation. We are bombarded daily with spiraling thoughts(both negative and positive) creating all-consuming feel , making us difficult to manage with associated suffering. Good thoughts are like our Mob Signal (Positive thought) amidst noise(negative thought) in the atmosphere. Negative thoughts like noise outweigh positive thoughts. These thoughts often create unwanted confusion, trouble, stress and frustration in our mind as well as chaos in our physical world. Negative thoughts are also known as “distorted thinking”.
Polish students' mobility in the Czech Republic
Polish students mobility to the Czech Republic within eTwinning project "Medieval adventures with Marco Polo"
1.4 modern child centered education - mahatma gandhi-2.pptx
Child centred education is an educational approach that priorities the interest, needs and abilities of the child in the learning process.
Unit 2- Research Aptitude (UGC NET Paper I).pdf
This slide describes the research aptitude of unit 2 in the UGC NET paper I.
The geography of Taylor Swift - some ideas
Geographical themes connected with Taylor Swift's ERAS tour - coming to the UK in June 2024
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Event Link:-
https://meetups.mulesoft.com/events/details/mulesoft-mysore-presents-exploring-gemini-ai-and-integration-with-mulesoft/
Agenda
● Java 17 Upgrade Overview
● Why and by when do customers need to upgrade to Java 17?
● Is there any immediate impact to upgrading to Mule Runtime 4.6 and beyond?
● Which MuleSoft products are in scope?
For Upcoming Meetups Join Mysore Meetup Group - https://meetups.mulesoft.com/mysore/
YouTube:- youtube.com/@mulesoftmysore
Mysore WhatsApp group:- https://chat.whatsapp.com/EhqtHtCC75vCAX7gaO842N
Speaker:-
Shubham Chaurasia - https://www.linkedin.com/in/shubhamchaurasia1/
Priya Shaw - https://www.linkedin.com/in/priya-shaw
Organizers:-
Shubham Chaurasia - https://www.linkedin.com/in/shubhamchaurasia1/
Giridhar Meka - https://www.linkedin.com/in/giridharmeka
Priya Shaw - https://www.linkedin.com/in/priya-shaw
Shyam Raj Prasad-
https://www.linkedin.com/in/shyam-raj-prasad/
Unit 8 - Information and Communication Technology (Paper I).pdf
This slides describes the basic concepts of ICT, basics of Email, Emerging Technology and Digital Initiatives in Education. This presentations aligns with the UGC Paper I syllabus.
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#In this project you will write a program play TicTacToe #using tw.pdf
#In this project you will write a program play TicTacToe
#using two players (labels 0,1) or one play (label 0) playing with the machine (label 1).
#The TicTacToe board has 9 integers board = [1,2,3,4,5,6,7,8,9]. The following
# are the modules for the program
#
#def reset() resets the board to the original values
# board = [1,2,3,4,5,6,7,8,9]
#
#def printBoard() print the current state of the board using the format
#
#The current TicTacToe Board
# | 1 | 2 | 3 |
# | 4 | 5 | 6 |
# | 7 | O | 9 |
#
#The current TicTacToe Board
# | X | 2 | 3 |
# | 4 | 5 | 6 |
# | 7 | O | 9 |
#Note from the above that player 0 and 1 have played numbers 8 and 1
#respectively and the board display O for player 0 and X from player 1
#
#def changeBoard(num1, player) using the chosen box number to change
#the value of the box to 0 or -1 depending on whether the player is 0 or 1,
#respectively.
#
#def play(player) prints the player number (0 or 1) and prompts the player
# to enter a box value that have not changed to \'O\' or \'X\'
#
#def checkRows(value) checks to see which of the rows of the board
# has the same value and returns True, otherwise, returns False
#
#def checkCols(value) checks to see which of the cols of the board
# has the same value and returns True, otherwise, returns False
#
#def checkDiagonal(value) checks to see which of the diagonals of the board
# has the same value and returns True, otherwise, returns False
#
#def win(player) checks if a player wins the game, returns True of the player wins
# and False otherwise
#
#def machinePlay(player) plays the role of the player using random number.
# this function generates numbers in the interval [1,9] and uses the first
# generated random number that has not been used to play the game. The
#
#def ticTacToe(numPlayers) accepts the number of players and simulates the
#ticTapToe, asking players to enter unused box numbers.
#
#def main() is the driver module that accepts the number of players from the user
# and calls the ticTacToe module
#Assignment: Complete the follwoing modules:
#checkCols
#checkRows
#checkDiagonal
#win
#Sample of the output is
#
#Project
#This game can be played by one or two players
#Enter the number of players, 1/2 for one/two players: 1
#
#The current TicTapToe Board
#| 1 | 2 | 3 |
#| 4 | 5 | 6 |
#| 7 | 8 | 9 |
#Player 0 Enter a box value: 9
#
#The current TicTapToe Board
#| 1 | 2 | 3 |
#| 4 | 5 | 6 |
#| 7 | 8 | O |
#Player 1 ***Computer*** playing
#The current TicTapToe Board
#| 1 | 2 | 3 |
#| 4 | 5 | 6 |
#| 7 | X | O |
#Player 0 Enter a box value: 1
#
#The current TicTapToe Board
#| O | 2 | 3 |
#| 4 | 5 | 6 |
#| 7 | X | O |
#Player 1 ***Computer*** playing
#
#The current TicTapToe Board
#| O | X | 3 |
#| 4 | 5 | 6 |
#| 7 | X | O |
#Player 0 Enter a box value: 5
#
#The current TicTapToe Board
#| O | X | 3 |
#| 4 | O | 6 |
#| 7 | X | O |
#Player 0 Wins
#Do you want to quit?
#Begin Program
#import random number generator
#from random library
#from random import randint, seed
#Global va.
1.1Yearcash flowpresent value of cash inflow = cash flow(1+r).pdf
1.1
Year
cash flow
present value of cash inflow = cash flow/(1+r)^n r= 5%
1
75
71.42857
2
75
68.02721
3
75
64.78782
4
1075
884.4052
value of bond
sum of present value of cash flow
1088.649
1.2
Year
cash flow
present value of cash inflow = cash flow/(1+r)^n r= 5.2%
1
75
71.29278
2
75
67.7688
3
75
64.41901
4
1075
877.6988
value of bond
sum of present value of cash flow
1081.179
1.3
% change in bond price if YTM increases by .2%
(1081.18-1088.65)/1088.65
-0.69%
1.4
Duration is a measure of a bond\'s sensitivity to interest rate changes. The higher the bond\'s
duration, the greater its sensitivity to the change (also know as volatility) and vice versa.
1.4
Year
cash flow
Year*cash flow
present value of Year*cash inflow = cash flow/(1+r)^n r= 5%
1
75
75
71.42857
2
75
150
136.0544
3
75
225
194.3635
4
1075
4300
3537.621
total
3939.467
Macculay duration in years
total/face value of bond
3939.467/1000
3.939467
1.4
modified duration
macculay duration/(1+YTM)
3.94/1.05
3.752381
1.5
change in price if YTM increases by .2%
modified duration*change in YTM
3.75*.2%
0.75%
1.1
Year
cash flow
present value of cash inflow = cash flow/(1+r)^n r= 5%
1
75
71.42857
2
75
68.02721
3
75
64.78782
4
1075
884.4052
value of bond
sum of present value of cash flow
1088.649
1.2
Year
cash flow
present value of cash inflow = cash flow/(1+r)^n r= 5.2%
1
75
71.29278
2
75
67.7688
3
75
64.41901
4
1075
877.6988
value of bond
sum of present value of cash flow
1081.179
1.3
% change in bond price if YTM increases by .2%
(1081.18-1088.65)/1088.65
-0.69%
1.4
Duration is a measure of a bond\'s sensitivity to interest rate changes. The higher the bond\'s
duration, the greater its sensitivity to the change (also know as volatility) and vice versa.
1.4
Year
cash flow
Year*cash flow
present value of Year*cash inflow = cash flow/(1+r)^n r= 5%
1
75
75
71.42857
2
75
150
136.0544
3
75
225
194.3635
4
1075
4300
3537.621
total
3939.467
Macculay duration in years
total/face value of bond
3939.467/1000
3.939467
1.4
modified duration
macculay duration/(1+YTM)
3.94/1.05
3.752381
1.5
change in price if YTM increases by .2%
modified duration*change in YTM
3.75*.2%
0.75%
Solution
1.1
Year
cash flow
present value of cash inflow = cash flow/(1+r)^n r= 5%
1
75
71.42857
2
75
68.02721
3
75
64.78782
4
1075
884.4052
value of bond
sum of present value of cash flow
1088.649
1.2
Year
cash flow
present value of cash inflow = cash flow/(1+r)^n r= 5.2%
1
75
71.29278
2
75
67.7688
3
75
64.41901
4
1075
877.6988
value of bond
sum of present value of cash flow
1081.179
1.3
% change in bond price if YTM increases by .2%
(1081.18-1088.65)/1088.65
-0.69%
1.4
Duration is a measure of a bond\'s sensitivity to interest rate changes. The higher the bond\'s
duration, the greater its sensitivity to the change (also know as volatility) and vice versa.
1.4
Year
cash flow
Year*cash flow
present value of Year*cash inflow = cash flow/(1+r)^n r= 5%
1
75
75
71.42857
2
75
150
136.0544
3
75
225
194.3635
4
1075
4300
3537.621
total
3939.467
M.
HNO2 is a much weaker acid then HNO3 (strong acid.pdf
HNO2 is a much weaker acid then HNO3 (strong acid dissociates completely)
H2SeO3 is weaker than H3AsO4 but only slightly as both are triprotic acids H2SO4 is weaker
than HClO4 but bother are very strong acids.
Solution
HNO2 is a much weaker acid then HNO3 (strong acid dissociates completely)
H2SeO3 is weaker than H3AsO4 but only slightly as both are triprotic acids H2SO4 is weaker
than HClO4 but bother are very strong acids..
1. Identify five differences between DNA replication and gene transc.pdf
1. Identify five differences between DNA replication and gene transcription.
DNA replication
gene transcription.
occurs in preparation for cell division,
Produces two two copies of DNA, double stranded
Uses DNA as template to form DNA
Catalyzed by DNA polymerase
the end result is two daughter cells.
happens in preparation for protein translation.
Produces mRNA, single stranded
Uses template DNA as a template and forms RNA
Catalyzed by RNA polymerase
End result is mRNA which is then exported out of the nucleus for translation
2. Identify four differences between DNA and RNA.
DNA
RNA
Deoxyribonucleic acid is the hereditary material in most of the organisms
Made of 2-deoxyribose sugar
Contains thymine as one of the 4 nitrogen base
It is double stranded
Ribonucleic acid transfers genetic code needed for the creation of proteins from the nucleus to
the ribosome.
Made of ribose sugar
Has Uracil as the nitrogen base instead of thymine
Single stranded
DNA replication
gene transcription.
occurs in preparation for cell division,
Produces two two copies of DNA, double stranded
Uses DNA as template to form DNA
Catalyzed by DNA polymerase
the end result is two daughter cells.
happens in preparation for protein translation.
Produces mRNA, single stranded
Uses template DNA as a template and forms RNA
Catalyzed by RNA polymerase
End result is mRNA which is then exported out of the nucleus for translation
Solution
1. Identify five differences between DNA replication and gene transcription.
DNA replication
gene transcription.
occurs in preparation for cell division,
Produces two two copies of DNA, double stranded
Uses DNA as template to form DNA
Catalyzed by DNA polymerase
the end result is two daughter cells.
happens in preparation for protein translation.
Produces mRNA, single stranded
Uses template DNA as a template and forms RNA
Catalyzed by RNA polymerase
End result is mRNA which is then exported out of the nucleus for translation
2. Identify four differences between DNA and RNA.
DNA
RNA
Deoxyribonucleic acid is the hereditary material in most of the organisms
Made of 2-deoxyribose sugar
Contains thymine as one of the 4 nitrogen base
It is double stranded
Ribonucleic acid transfers genetic code needed for the creation of proteins from the nucleus to
the ribosome.
Made of ribose sugar
Has Uracil as the nitrogen base instead of thymine
Single stranded
DNA replication
gene transcription.
occurs in preparation for cell division,
Produces two two copies of DNA, double stranded
Uses DNA as template to form DNA
Catalyzed by DNA polymerase
the end result is two daughter cells.
happens in preparation for protein translation.
Produces mRNA, single stranded
Uses template DNA as a template and forms RNA
Catalyzed by RNA polymerase
End result is mRNA which is then exported out of the nucleus for translation.
1. b2. c (bed bug)3. d (caterpillars)4. c (lyme disease)5. e.pdf
1. b
2. c (bed bug)
3. d (caterpillars)
4. c (lyme disease)
5. e (chagas disease)
6. a (hives)
7. e (short lived)
8. d (plants)
9. a (1892)
10. e (parvoviridae)
11. mosquito
12. a (flies)
13. c (lyme disease)
14.
15. --- holometabolous is the type of metamorphosis in beetles
16. oligocene
17. e (6 or more)
18. e
Solution
1. b
2. c (bed bug)
3. d (caterpillars)
4. c (lyme disease)
5. e (chagas disease)
6. a (hives)
7. e (short lived)
8. d (plants)
9. a (1892)
10. e (parvoviridae)
11. mosquito
12. a (flies)
13. c (lyme disease)
14.
15. --- holometabolous is the type of metamorphosis in beetles
16. oligocene
17. e (6 or more)
18. e.
1) MODERN TIMES is a very well prepared movie that describes the mis.pdf
1) MODERN TIMES is a very well prepared movie that describes the miserable conditions of
humar resource during the period of industrialization era or machine age worlwide. It is a mix of
social drama and comedy. The content is presented in a satire. How did people survived the great
depression of the world is being documented here. The conditions werre miserable like acute
shortage of work, factory closings, labor strikes and political unrest etc. It is a symbolic
representation of what happens to a human being caught up in the mechanical bureaucracy of the
intellectual state.
2) Economically it could be understood that what miser situations could depression bring up.
Food and jobs comes hard in hand. To which a labor is convict to do under skill jobs as well for
survival. At the time of the great depression Americans had to fight for what little they had, and
as the movie showed many people went hungry. Modern Times looks toward the future, but not
with enthusiasm. Often described as a satire of the machine age, It has in fact a broader theme:
the dehumanizing effects of many aspects of modernity, including industrialization, bureaucracy,
urbanization, and law enforcement.
3) From the movie a prospect employee should not forget about the social ills and failings of
own day even if present age is seems far better than that was shown in the film. A dilbert-like
early scene shows the Tramp trying to take a break during work hours in the washroom — only
to have a big-screen image of the boss’s head suddenly appear on the washroom wall and order
him to quit stalling and get back to work. Modern audiences watching this scene may reflect
ruefully on the everyday reality of electronic surveillance in the workplace. Now, obviously
there’s nothing new about widespread poverty and want, but unemployment as we know it today
is a comparatively recent development — a byproduct of the modern workforce economy that
requires everyone to \"get a job.\"
4) The major ethical dilemma consists whether out of povertty stealing is right or wrong.
However this movie glorifies the struggle of the poor rather than vilifying them for stealing food
to survive. By glorifying the struggle the movie takes on yet another social issue: when the poor
steal to eat it is not criminal, just survival. Minor ones include
Solution
1) MODERN TIMES is a very well prepared movie that describes the miserable conditions of
humar resource during the period of industrialization era or machine age worlwide. It is a mix of
social drama and comedy. The content is presented in a satire. How did people survived the great
depression of the world is being documented here. The conditions werre miserable like acute
shortage of work, factory closings, labor strikes and political unrest etc. It is a symbolic
representation of what happens to a human being caught up in the mechanical bureaucracy of the
intellectual state.
2) Economically it could be understood that what miser situations could depress.
An ion is an atom that has a charge. These charges can be postive or.pdf
An ion is an atom that has a charge. These charges can be postive or negative and depend upon
the amount of electrions lost or gained. Ions are formed when they gain or lose an election ( a
change in valence electrons) An example is Li+. That is a Lithium atom with a positive charge, it
has lost one electron.
A cation is a positively charged ion. Li+
An anion is a negatively charged ion. F-. A Fluorine atom that has gained an electron.
A polyatomic ion is a multi-atom molecule with a charge. Such as ammonia NH4+.
Solution
An ion is an atom that has a charge. These charges can be postive or negative and depend upon
the amount of electrions lost or gained. Ions are formed when they gain or lose an election ( a
change in valence electrons) An example is Li+. That is a Lithium atom with a positive charge, it
has lost one electron.
A cation is a positively charged ion. Li+
An anion is a negatively charged ion. F-. A Fluorine atom that has gained an electron.
A polyatomic ion is a multi-atom molecule with a charge. Such as ammonia NH4+..
Acids are substances that form hydrogen ions, H+(.pdf
Acids are substances that form hydrogen ions, H+(aq), when dissolved in water In
acid solutions there are more H+ ions than OH- ions. Bases which are soluble in water are called
alkalis e.g. NaOH sodium hydroxide, KOH potassium hydroxide or Ca(OH)2 calcium hydroxide.
Bases which are water insoluble include CuO copper(II) oxide, MgO magnesium oxide. After a
neutralisation, the salt solutions consist of a mixture of positive and negative ions (and their
names are in the salt name!) e.g. sodium chloride (NaCl) is a mixture of Na+ and Cl- ions,
calcium chloride (CaCl2) is a mix of Ca2+ and Cl- ions; magnesium nitrate (Mg(NO3)2) is a
mix of Mg2+ and NO3- ions, aluminium sulphate (Al2(SO4)3) consists of Al3+ and SO42- ions
etc. You can test for chloride ions by adding a few drops of dilute nitric acid followed by a few
drops of silver nitrate solution. If chloride ions are present you will see a white precipitate of
silver chloride. However, chlorine water itself will contain chloride ions as chlorine in aqueous
solution disproportionates to form chloride and chlorate(I) ions. Cl2 + H2O --> Cl- + ClO- +
2H+ Testing for the presence of nitrate via wet chemistry is generally difficult compared with
testing for other anions, as almost all nitrates are soluble in water. In contrast, many common
ions give insoluble salts, e.g. halides precipitate with silver, and sulfates precipitate with barium.
The nitrate anion is an oxidizer, and many tests for the nitrate anion are based on this property.
Unfortunately, other oxidants present in the analyte may interfere and give erroneous results.
Solution
Acids are substances that form hydrogen ions, H+(aq), when dissolved in water In
acid solutions there are more H+ ions than OH- ions. Bases which are soluble in water are called
alkalis e.g. NaOH sodium hydroxide, KOH potassium hydroxide or Ca(OH)2 calcium hydroxide.
Bases which are water insoluble include CuO copper(II) oxide, MgO magnesium oxide. After a
neutralisation, the salt solutions consist of a mixture of positive and negative ions (and their
names are in the salt name!) e.g. sodium chloride (NaCl) is a mixture of Na+ and Cl- ions,
calcium chloride (CaCl2) is a mix of Ca2+ and Cl- ions; magnesium nitrate (Mg(NO3)2) is a
mix of Mg2+ and NO3- ions, aluminium sulphate (Al2(SO4)3) consists of Al3+ and SO42- ions
etc. You can test for chloride ions by adding a few drops of dilute nitric acid followed by a few
drops of silver nitrate solution. If chloride ions are present you will see a white precipitate of
silver chloride. However, chlorine water itself will contain chloride ions as chlorine in aqueous
solution disproportionates to form chloride and chlorate(I) ions. Cl2 + H2O --> Cl- + ClO- +
2H+ Testing for the presence of nitrate via wet chemistry is generally difficult compared with
testing for other anions, as almost all nitrates are soluble in water. In contrast, many common
ions give insoluble salts, e.g. halides precipitate with silver, and sulfates .
ANSWER1.Apocalypse Now film overall formIt is the height o.pdf
ANSWER:
1.\"Apocalypse Now\" film overall form:
It is the height of the war in Vietnam, and U.S. Army Captain Willard is sent by Colonel Lucas
and a General to carry out a mission that, officially, \'does not exist - nor will it ever exist\'. The
mission: To seek out a mysterious Green Beret Colonel, Walter Kurtz, whose army has crossed
the border into Cambodia and is conducting hit-and-run missions against the Viet Cong and
NVA. The army believes Kurtz has gone completely insane and Willard\'s job is to eliminate
him! Willard, sent up the Nung River on a U.S. Navy patrol boat, discovers that his target is one
of the most decorated officers in the U.S. Army. His crew meets up with surfer-type Lt-Colonel
Kilgore, head of a U.S Army helicopter cavalry group which eliminates a Viet Cong outpost to
provide an entry point into the Nung River. After some hair-raising encounters, in which some of
his crew are killed, Willard, Lance and Chef reach Colonel Kurtz\'s outpost, beyond the Do Lung
Bridge.
“This is the end, beautiful friend…
This is the end, my only friend, the end
Of our elaborate plans, the end
Of everything that stands, the end
No safety or surprise, the end
I\'ll never look into your eyes...again
The End...”
2.Primary Techniques in Apocalypse Film:
Runtime
3.Techniques:
Although many scenes are unusual since the movie begin an unusual journey in the midst of an
alien world, some of the scenes still stand out.One of the most effective techniques used in the
movie is the usage of very long dissolves. Many scenes are dissolved on each other until they
become a montage of their own in one shot itself and say something completely distinct from
their individual meaning. This was used in the beginning of the movie, in the middle when the
journey enters the final realm and in the end again giving the seemingly untying themes a
thematic unity.
Many times, the war is superimposed on Willard’s face as if he is living it & re-living it long
after it has gone or perhaps long before it has happened...! (Slow Dissolves)
When the boat is finally entering the waters of Cambodia and we know that it is a point of no
return for this boat and especially Willard, a few images are superimposed on each other to
create a heightened sense of an altered reality. The boat is passing through the gates of the
underworld and right in the center of all of this is Willard trying to look, listen and comprehend.
(Slow Dissolve To Gates Of Hell1, Slow Dissolve To Gates Of Hell2, Slow Dissolve To Gates
Of Hell3, Slow Dissolve To Gates Of Hell4, Slow Dissolve To Gates Of Hell5)
Another highly unusual shot worth mentioning is Visual Constancy Or Horror. When I saw this
shot, I thought that if my eyes did not have visual constancy and did not fill in Willard’s body
automatically, the body-less face in this scene would be part of the true horror genre...!
4.Techniques and patterns:
a)Lighting:
Lighting is the key element in the film. Along with fogs and shadows. The general lo.
15.0mL = 0.015L phenylacetic acid 0.50molL 0.01.pdf
15.0mL = 0.015L phenylacetic acid 0.50mol/L *0.015= 0.0075 mol phenylacetic
acid 20.0mL= 0.020L sodium phenylacetate 0.40mol/L * 0.020L=.008 mol sodium
phenylacetate due to the fact that this acts as a buffer Using henderson hasselblach pH= pKa +
log(base/acid) and finding that the pKa of phenylaceticacid is 4.28 pH= 4.28 + log (.008/.0075)
pH= 4.28 + 0.093 pH= 4.373
Solution
15.0mL = 0.015L phenylacetic acid 0.50mol/L *0.015= 0.0075 mol phenylacetic
acid 20.0mL= 0.020L sodium phenylacetate 0.40mol/L * 0.020L=.008 mol sodium
phenylacetate due to the fact that this acts as a buffer Using henderson hasselblach pH= pKa +
log(base/acid) and finding that the pKa of phenylaceticacid is 4.28 pH= 4.28 + log (.008/.0075)
pH= 4.28 + 0.093 pH= 4.373.
4 water molecules are like a magnet because beca.pdf
4: water molecules are like a magnet because because it acts like water 5: As
mentioned above, humans have sweat glands over their entire bodies; the only place dogs (and
cats, and many other animals with fur over their whole bodies) can get rid of excess heat is
through the tongue. By the way, sweating/panting is NOT a means of getting rid of excess water;
it is a way of getting rid of heat. We do this because of the process of evaporation. In changing
from liquid to vapor, the water absorbs heat from the surface on which it is laying. Actually,
most organisms do not want to get rid of excess water; they need to conserve it, but on very hot
days animals \"spend\" the water to achieve the cooling. 6. It can dissolve the most types of
substances. it can dissolve most substances This is because solutes for the most part are able to
dissolve in water with the exception of some solutes like oil. Most of the solids dissolve in water.
Solution
4: water molecules are like a magnet because because it acts like water 5: As
mentioned above, humans have sweat glands over their entire bodies; the only place dogs (and
cats, and many other animals with fur over their whole bodies) can get rid of excess heat is
through the tongue. By the way, sweating/panting is NOT a means of getting rid of excess water;
it is a way of getting rid of heat. We do this because of the process of evaporation. In changing
from liquid to vapor, the water absorbs heat from the surface on which it is laying. Actually,
most organisms do not want to get rid of excess water; they need to conserve it, but on very hot
days animals \"spend\" the water to achieve the cooling. 6. It can dissolve the most types of
substances. it can dissolve most substances This is because solutes for the most part are able to
dissolve in water with the exception of some solutes like oil. Most of the solids dissolve in water..
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