SolutionDomain name system is used to determine the IP address of.pdf
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Solution : Domain name system is used to determine the IP address of the host/domain name. DNS is very useful to every user where it helps retrieve the website IP address. Without DNS we cannot retrieve the website IP address. This process is known as forward DNS resolution. DNS should be registered by some companies. If any attacker has details then the dns will be controlled by the attacker. when we are sending the e-mail the dns data will be used then it can be catches anywhere in the network. So it is very important that a DNS require reliable communications..
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Using gordon growth model Value of stock=Dividend paid(1+growt.pdf
Using gordon growth model Value of stock=Dividend paid*(1+growth)/required return-
growth rate Value of stock=5*1.05/(.15-.05) Value of stock= $52.5
Solution
Using gordon growth model Value of stock=Dividend paid*(1+growth)/required return-
growth rate Value of stock=5*1.05/(.15-.05) Value of stock= $52.5.
not able to sign in .pdf
The document discusses an issue with signing in. It provides a brief description of the problem but no solution. The document is only two sentences and does not contain enough information to generate a multi-sentence summary.
moles of NH3 = 3417 = 2 moles of O2 = 4032 = 1..pdf
moles of NH3 = 34/17 = 2 moles of O2 = 40/32 = 1.25 Thus, L.R. = O2 So, 2.5
moles or 45 g H2O d. 45.0
Solution
moles of NH3 = 34/17 = 2 moles of O2 = 40/32 = 1.25 Thus, L.R. = O2 So, 2.5
moles or 45 g H2O d. 45.0.
It would be . like for hydrated compound of.pdf
It would be \"*\" . like for hydrated compound of copper(II) hydroxide would be
Cu(OH)2*2(H2O)
Solution
It would be \"*\" . like for hydrated compound of copper(II) hydroxide would be
Cu(OH)2*2(H2O).
NO has got a strong bond because it has got an co.pdf
NO has got a strong bond because it has got an coordinate covalent bond....
Solution
NO has got a strong bond because it has got an coordinate covalent bond.....
total number of moles of liquid increase, .pdf
This short document discusses how the total number of moles of liquid can increase. It states that the total number of moles of liquid increase but provides no further details or context about what is causing the increase or any other related information.
Yearscost of machin7 Year MACRS rateannual depreciation = cost.pdf
Years
cost of machin
7 Year MACRS rate
annual depreciation = cost of machine* MACRS rate
1
190000
0.143
27170
2
190000
0.245
46550
3
190000
0.175
33250
4
190000
0.125
23750
5
190000
0.089
16910
6
190000
0.089
16910
7
190000
0.089
16910
8
190000
0.045
8550
total
1
190000
Year
EBIT
less depreciation
EBT
less tax-30%
EAT
Add depreciation
EATBD
present value of EATBD = EATBD/1+r)^n r= 12%
1
58000
27170
30830
9249
21581
27170
48751
43527.68
2
58000
46550
11450
3435
8015
46550
54565
43498.88
3
58000
33250
24750
7425
17325
33250
50575
35998.29
4
29000
23750
5250
1575
3675
23750
27425
17429.08
5
29000
16910
12090
3627
8463
16910
25373
14397.32
6
29000
16910
12090
3627
8463
16910
25373
12854.75
7
29000
16910
12090
3627
8463
16910
25373
11477.46
8
29000
8550
20450
6135
14315
8550
22865
9234.79
9
29000
0
29000
8700
20300
0
20300
7320.384
10
29000
0
29000
8700
20300
0
20300
6536.057
sum of present value of EATBD
202274.7
less present value of cash outflow
190000
net present value
12274.69
Machine should be purchased as it results In positive NPV
Years
cost of machin
7 Year MACRS rate
annual depreciation = cost of machine* MACRS rate
1
190000
0.143
27170
2
190000
0.245
46550
3
190000
0.175
33250
4
190000
0.125
23750
5
190000
0.089
16910
6
190000
0.089
16910
7
190000
0.089
16910
8
190000
0.045
8550
total
1
190000
Year
EBIT
less depreciation
EBT
less tax-30%
EAT
Add depreciation
EATBD
present value of EATBD = EATBD/1+r)^n r= 12%
1
58000
27170
30830
9249
21581
27170
48751
43527.68
2
58000
46550
11450
3435
8015
46550
54565
43498.88
3
58000
33250
24750
7425
17325
33250
50575
35998.29
4
29000
23750
5250
1575
3675
23750
27425
17429.08
5
29000
16910
12090
3627
8463
16910
25373
14397.32
6
29000
16910
12090
3627
8463
16910
25373
12854.75
7
29000
16910
12090
3627
8463
16910
25373
11477.46
8
29000
8550
20450
6135
14315
8550
22865
9234.79
9
29000
0
29000
8700
20300
0
20300
7320.384
10
29000
0
29000
8700
20300
0
20300
6536.057
sum of present value of EATBD
202274.7
less present value of cash outflow
190000
net present value
12274.69
Machine should be purchased as it results In positive NPV
Solution
Years
cost of machin
7 Year MACRS rate
annual depreciation = cost of machine* MACRS rate
1
190000
0.143
27170
2
190000
0.245
46550
3
190000
0.175
33250
4
190000
0.125
23750
5
190000
0.089
16910
6
190000
0.089
16910
7
190000
0.089
16910
8
190000
0.045
8550
total
1
190000
Year
EBIT
less depreciation
EBT
less tax-30%
EAT
Add depreciation
EATBD
present value of EATBD = EATBD/1+r)^n r= 12%
1
58000
27170
30830
9249
21581
27170
48751
43527.68
2
58000
46550
11450
3435
8015
46550
54565
43498.88
3
58000
33250
24750
7425
17325
33250
50575
35998.29
4
29000
23750
5250
1575
3675
23750
27425
17429.08
5
29000
16910
12090
3627
8463
16910
25373
14397.32
6
29000
16910
12090
3627
8463
16910
25373
12854.75
7
29000
16910
12090
3627
8463
16910
25373
11477.46
8
29000
8550
20450
6135
14315
8550
22865
9234.79
9
29000
0
29000
8700
20300
0
20300
7320.384
10
29000
0
29000
8700
20300
0
20300
6536.057
sum.
what this meansSolutionwhat this means.pdf
This very short document does not provide enough contextual information to generate a meaningful 3 sentence summary. It contains only a heading and subheading with no further details or body content to summarize.
Recommended
Using gordon growth model Value of stock=Dividend paid(1+growt.pdf
Using gordon growth model Value of stock=Dividend paid*(1+growth)/required return-
growth rate Value of stock=5*1.05/(.15-.05) Value of stock= $52.5
Solution
Using gordon growth model Value of stock=Dividend paid*(1+growth)/required return-
growth rate Value of stock=5*1.05/(.15-.05) Value of stock= $52.5.
not able to sign in .pdf
The document discusses an issue with signing in. It provides a brief description of the problem but no solution. The document is only two sentences and does not contain enough information to generate a multi-sentence summary.
moles of NH3 = 3417 = 2 moles of O2 = 4032 = 1..pdf
moles of NH3 = 34/17 = 2 moles of O2 = 40/32 = 1.25 Thus, L.R. = O2 So, 2.5
moles or 45 g H2O d. 45.0
Solution
moles of NH3 = 34/17 = 2 moles of O2 = 40/32 = 1.25 Thus, L.R. = O2 So, 2.5
moles or 45 g H2O d. 45.0.
It would be . like for hydrated compound of.pdf
It would be \"*\" . like for hydrated compound of copper(II) hydroxide would be
Cu(OH)2*2(H2O)
Solution
It would be \"*\" . like for hydrated compound of copper(II) hydroxide would be
Cu(OH)2*2(H2O).
NO has got a strong bond because it has got an co.pdf
NO has got a strong bond because it has got an coordinate covalent bond....
Solution
NO has got a strong bond because it has got an coordinate covalent bond.....
total number of moles of liquid increase, .pdf
This short document discusses how the total number of moles of liquid can increase. It states that the total number of moles of liquid increase but provides no further details or context about what is causing the increase or any other related information.
Yearscost of machin7 Year MACRS rateannual depreciation = cost.pdf
Years
cost of machin
7 Year MACRS rate
annual depreciation = cost of machine* MACRS rate
1
190000
0.143
27170
2
190000
0.245
46550
3
190000
0.175
33250
4
190000
0.125
23750
5
190000
0.089
16910
6
190000
0.089
16910
7
190000
0.089
16910
8
190000
0.045
8550
total
1
190000
Year
EBIT
less depreciation
EBT
less tax-30%
EAT
Add depreciation
EATBD
present value of EATBD = EATBD/1+r)^n r= 12%
1
58000
27170
30830
9249
21581
27170
48751
43527.68
2
58000
46550
11450
3435
8015
46550
54565
43498.88
3
58000
33250
24750
7425
17325
33250
50575
35998.29
4
29000
23750
5250
1575
3675
23750
27425
17429.08
5
29000
16910
12090
3627
8463
16910
25373
14397.32
6
29000
16910
12090
3627
8463
16910
25373
12854.75
7
29000
16910
12090
3627
8463
16910
25373
11477.46
8
29000
8550
20450
6135
14315
8550
22865
9234.79
9
29000
0
29000
8700
20300
0
20300
7320.384
10
29000
0
29000
8700
20300
0
20300
6536.057
sum of present value of EATBD
202274.7
less present value of cash outflow
190000
net present value
12274.69
Machine should be purchased as it results In positive NPV
Years
cost of machin
7 Year MACRS rate
annual depreciation = cost of machine* MACRS rate
1
190000
0.143
27170
2
190000
0.245
46550
3
190000
0.175
33250
4
190000
0.125
23750
5
190000
0.089
16910
6
190000
0.089
16910
7
190000
0.089
16910
8
190000
0.045
8550
total
1
190000
Year
EBIT
less depreciation
EBT
less tax-30%
EAT
Add depreciation
EATBD
present value of EATBD = EATBD/1+r)^n r= 12%
1
58000
27170
30830
9249
21581
27170
48751
43527.68
2
58000
46550
11450
3435
8015
46550
54565
43498.88
3
58000
33250
24750
7425
17325
33250
50575
35998.29
4
29000
23750
5250
1575
3675
23750
27425
17429.08
5
29000
16910
12090
3627
8463
16910
25373
14397.32
6
29000
16910
12090
3627
8463
16910
25373
12854.75
7
29000
16910
12090
3627
8463
16910
25373
11477.46
8
29000
8550
20450
6135
14315
8550
22865
9234.79
9
29000
0
29000
8700
20300
0
20300
7320.384
10
29000
0
29000
8700
20300
0
20300
6536.057
sum of present value of EATBD
202274.7
less present value of cash outflow
190000
net present value
12274.69
Machine should be purchased as it results In positive NPV
Solution
Years
cost of machin
7 Year MACRS rate
annual depreciation = cost of machine* MACRS rate
1
190000
0.143
27170
2
190000
0.245
46550
3
190000
0.175
33250
4
190000
0.125
23750
5
190000
0.089
16910
6
190000
0.089
16910
7
190000
0.089
16910
8
190000
0.045
8550
total
1
190000
Year
EBIT
less depreciation
EBT
less tax-30%
EAT
Add depreciation
EATBD
present value of EATBD = EATBD/1+r)^n r= 12%
1
58000
27170
30830
9249
21581
27170
48751
43527.68
2
58000
46550
11450
3435
8015
46550
54565
43498.88
3
58000
33250
24750
7425
17325
33250
50575
35998.29
4
29000
23750
5250
1575
3675
23750
27425
17429.08
5
29000
16910
12090
3627
8463
16910
25373
14397.32
6
29000
16910
12090
3627
8463
16910
25373
12854.75
7
29000
16910
12090
3627
8463
16910
25373
11477.46
8
29000
8550
20450
6135
14315
8550
22865
9234.79
9
29000
0
29000
8700
20300
0
20300
7320.384
10
29000
0
29000
8700
20300
0
20300
6536.057
sum.
what this meansSolutionwhat this means.pdf
This very short document does not provide enough contextual information to generate a meaningful 3 sentence summary. It contains only a heading and subheading with no further details or body content to summarize.
we will use tables of normal distribution ( Z distribution)P( z .pdf
we will use tables of normal distribution ( Z distribution)
P( z < -1) = 0.1587
Solution
we will use tables of normal distribution ( Z distribution)
P( z < -1) = 0.1587.
The Northeast blackout of 2003 was a widespread power outage that oc.pdf
The Northeast blackout of 2003 was a widespread power outage that occurred throughout parts
of the Northeasternand Midwestern United States and the Canadian province of Ontario on
Thursday, August 14, 2003, just after 4:10 p.m. All told, 50 million people lost power for up to
two days in the biggest blackout in North American history. The event contributed to at least 11
deaths and cost an estimated $6 billion. A surge of electricity to western New York and Canada
touched off a series of power failures and enforced blackouts that left parts of at least eight states
in the Northeast and the Midwest without electricity.Transmission system operators scattered
across some 300 control centers nationwide monitor voltage and current data from SCADA
(supervisory control and data acquisition) systems placed at transformers, generators and other
critical points.The widespread failures provoked the evacuation of office buildings, stranded
thousands of commuters. As circuit breakers tripped at generating stations from New York to
Michigan and into Canada, millions of people were instantly caught up in the largest blackout in
American history. In New York City, power was shut off by officials struggling to head off a
wider blackout. Cleveland and Detroit went dark, as did Toronto and sections of New Jersey,
Pennsylvania, Connecticut and Massachusetts. Officials worked into the night to put the grid
back in operation and restore electric service. Mayor Michael R. Bloomberg said that that the
power was back on in parts of Brooklyn, the Bronx and Queens by 11 p.m. -- but not Manhattan.
The blackout began just after the stock exchanges had closed for the day, a slow summer day of
relatively light trading, as thousands of workers were about to head home. Office workers who
were still at their desks watched their computer monitors blink off without warning on a hot and
hazy afternoon. Soon hospitals and government buildings were switching on backup generators
to keep essential equipment operating, and the police were evacuating people trapped in
elevators. Airports throughout the affected states suffered serious disruptions, including the three
major airports in the New York metropolitan region, but did not close. Thousands of subway
passengers in New York City had to be evacuated from tunnels, and commuter trains also came
to a halt. Gov.George E. Pataki said that 600 trains were stranded.The office of the Canadian
prime minister, Jean Chrétien, initially said the power problems were caused by lightning in New
York State but later retracted that. Canadian officials later expressed uncertainty about the exact
cause but continued to insist the problem began on the United States side of the border. The
Nuclear Regulatory Commission said that the seven nuclear plants in New York and New Jersey
and two in the Midwest had shut down automatically when the failure occurred.Telephone
service was disrupted, especially calls to and from cellular phones.Officials said the.
The chemical reaction responsible for formation of water2H2+ O2--.pdf
The chemical reaction responsible for formation of water:
2H2+ O2-----------> 2H2O
2 molcules of hydrogen combine with 2 molecules of oxygen to give two molecules of water.
Solution
The chemical reaction responsible for formation of water:
2H2+ O2-----------> 2H2O
2 molcules of hydrogen combine with 2 molecules of oxygen to give two molecules of water..
sample mean = population mean = 2.11standard deviation = popul.pdf
sample mean = population mean = 2.11
standard deviation = population standad deviation / sqrt(sample size)
= 1.3 / sqrt(90)
= 0.137
Solution
sample mean = population mean = 2.11
standard deviation = population standad deviation / sqrt(sample size)
= 1.3 / sqrt(90)
= 0.137.
Polymerase chain reaction (PCR) is a process used to replicate DNA i.pdf
Polymerase chain reaction (PCR) is a process used to replicate DNA in vitro. DNA replication is
the biological process by which original DNA molecule replicate in to two identical replicas of
DNA. DNA polymerase III replicate the Prokaryotic DNA in 5\' to 3\' at a rate of 1 kb/s but 1-4
kb/min in a typical PCR reaction. Prokaryotic DNA is circular and replication starts at single
origin (ori C). It do not occur in PCR.
Prokaryotic DNA replication is followed by the three basic step:-
1. initiation, 2. elongation, and 3. termination.
The process of PCR is followed by the three basic step:-
1. Denaturation, 2.,Annealing, and 3. Primer Extension
1) the process of forming single stranded DNA, :-In PCR the DNA denatures at high tempreture,
at 95oC to form its single-stranded. During the prokaryotic DNA Replication, Ori C unwinds the
DNA by DNA B or helicase, and extension of ssDNA for copying during initiation step.
2) priming of the DNA polymerase,:-In PCR primer binds with the complementry strand of DNA
during the Annealing step. In prokaryotic replcation, primase form RNA primer extended by
DNA polymerase III,during the elongation step. Synthesis ofleading strand and lagging strand
takes place in this step. It do not occur in PCR.
3) the DNA polymerase itself:-In prokaryotic DNA replication DNA polymerase III synthesize
the segment of DNA for both leading and lagging strand. In PCR, DNA polymerase extends the
primer, Taq polymerase, is used in this process and extension is performed at 72oC.
4) the end product:- As a PCR two ds DNA is formed. And in prokaryotic DNA replication 2
circular DNA is formed , which remain interlinked and topoisomerase II make them seprate by
making cut over them.
Protien involed in prokaryotic DNA replication:-
Dna A protein-It initiate DNA replication by the binding with the Dna A boxes.
Dna helicase- it seprates the ds DNA.
Topoisomerase- From the replication fork, it removes positive supercoiling.
Single stranded binding protein-to prevent the reformation of ds DNA, it binds with single
stranded structure,
primase - synthesis the short RNA primer.
DNA polymerase III-In the leading and lagging strands, it synthesize the DNA.
DNA polymerase I- Removes RNA primer
DNA ligase- it joins the fragments.
Tus;-binds with ter sequence.
DNA replication different than PCR:--PCR generates large quantities of DNA from very small
amount of DNA, PCR is a INVITRO technique, while DNA replication occurs inside the iving
system, it is a invivo technique. The whole DNA is replicated, in DNA replication, while in PCR
the amplification of segment of DNA takes place.
Solution
Polymerase chain reaction (PCR) is a process used to replicate DNA in vitro. DNA replication is
the biological process by which original DNA molecule replicate in to two identical replicas of
DNA. DNA polymerase III replicate the Prokaryotic DNA in 5\' to 3\' at a rate of 1 kb/s but 1-4
kb/min in a typical PCR reaction. Prokaryotic DNA is circular and replication st.
Glutamic acid and Lycine Hydrogen bonding could .pdf
Glutamic acid and Lycine: Hydrogen bonding could occur between the NH3+
group of Lysine and the COO- group of Glutamic acid; and Van der Waals (nonpolar-nonpolar)
forces/interactions could arise between the nonpolar methylene groups of both R groups.
Phenylalanine and Isoleucine: Phenylalanine is quite hydrophobic so it works in hydrophobic
reactions. Both have nonpolar R groups so the only interactions they could undergo would be the
weakly attractive Van der Waals forces between the methylene groups of both R groups.
However, Phenylalanine is an aromatic compound with an aromatic ring, which is capable of
undergoing noncovalent Pi-system interactions but not with Isoleucine because it is not polar
enough to distort the Pi system of the aromatic ring.
Solution
Glutamic acid and Lycine: Hydrogen bonding could occur between the NH3+
group of Lysine and the COO- group of Glutamic acid; and Van der Waals (nonpolar-nonpolar)
forces/interactions could arise between the nonpolar methylene groups of both R groups.
Phenylalanine and Isoleucine: Phenylalanine is quite hydrophobic so it works in hydrophobic
reactions. Both have nonpolar R groups so the only interactions they could undergo would be the
weakly attractive Van der Waals forces between the methylene groups of both R groups.
However, Phenylalanine is an aromatic compound with an aromatic ring, which is capable of
undergoing noncovalent Pi-system interactions but not with Isoleucine because it is not polar
enough to distort the Pi system of the aromatic ring..
Need a clear pictureSolutionNeed a clear picture.pdf
This very short document discusses needing a clear picture but does not provide any details on the topic or solution. It repeats the phrase "Need a clear picture" but does not elaborate on what is needed, why it is needed, or how any solution might help provide a clear picture.
Li2SSolutionLi2S.pdf
This one sentence document contains the word "Li2S" repeated three times. It appears to be referring to lithium sulfide, a chemical compound with the formula Li2S, but provides no other context or information.
For H2SO3 pK1 = 1.8 and pK2 = 7.2 and thus K1 = .pdf
For H2SO3 pK1 = 1.8 and pK2 = 7.2 and thus K1 = 1.58E-2 and K2 = 6.31E-8
For a given pH (get to this in a minute), it can be shown [H+]=10^-pH [OH-] = Kw / [H+]
where Kw is the ion product of water (1E-14) [H2SO3] = [H+]^2 / f [HSO3-] = K1 [H+] / f
[SO3--] = K1 K2 / f where f = [H+]^2 + K1[H+] = K1K2 Given that H2SO3 is a weak acid, lets
assume that the pH is described by K1 or [H+] = sqrt(K1 C) where C is the acid conc (mol/l).
This leads to a [H+] = 0.05 and a pH = 1.07 Using the [H+] and data above yields in mol/l [H+]
8.51E-02 [OH-] 1.17E-13 [H2SO3] 3.88E-01 [HSO3-] 7.22E-02 [SO3--] 5.35E-08 Note the
above data describes the acid as 15% dissociated and the assumptiom that the pH of the solution
is best described by K1 is not necessarily a good assumption. A better way is as follows :- The
charge balance is this system is given by [H+] = [OH-] + [HSO3-] + 2*[SO3--] Set up all the
above relationships in MS Excel and Goal Seek a pH that makes the above charge balance
correct. Doing this yields a pH = 1.11 and the following concs (mol/l) [H+] 7.78E-02 [OH-]
1.28E-13 [H2SO3] 3.82E-01 [HSO3-] 7.78E-02 [SO3--] 6.31E-08 Be mindful that in all this
calculation, ionic strength impacts (ie activities) have been ignored or activity coefficients have
been taken as 1
Solution
For H2SO3 pK1 = 1.8 and pK2 = 7.2 and thus K1 = 1.58E-2 and K2 = 6.31E-8
For a given pH (get to this in a minute), it can be shown [H+]=10^-pH [OH-] = Kw / [H+]
where Kw is the ion product of water (1E-14) [H2SO3] = [H+]^2 / f [HSO3-] = K1 [H+] / f
[SO3--] = K1 K2 / f where f = [H+]^2 + K1[H+] = K1K2 Given that H2SO3 is a weak acid, lets
assume that the pH is described by K1 or [H+] = sqrt(K1 C) where C is the acid conc (mol/l).
This leads to a [H+] = 0.05 and a pH = 1.07 Using the [H+] and data above yields in mol/l [H+]
8.51E-02 [OH-] 1.17E-13 [H2SO3] 3.88E-01 [HSO3-] 7.22E-02 [SO3--] 5.35E-08 Note the
above data describes the acid as 15% dissociated and the assumptiom that the pH of the solution
is best described by K1 is not necessarily a good assumption. A better way is as follows :- The
charge balance is this system is given by [H+] = [OH-] + [HSO3-] + 2*[SO3--] Set up all the
above relationships in MS Excel and Goal Seek a pH that makes the above charge balance
correct. Doing this yields a pH = 1.11 and the following concs (mol/l) [H+] 7.78E-02 [OH-]
1.28E-13 [H2SO3] 3.82E-01 [HSO3-] 7.78E-02 [SO3--] 6.31E-08 Be mindful that in all this
calculation, ionic strength impacts (ie activities) have been ignored or activity coefficients have
been taken as 1.
find the largest electronegativity difference (mo.pdf
find the largest electronegativity difference (most ionic)
Solution
find the largest electronegativity difference (most ionic).
let an=(-1)^(2n+1)and L=-1let us ahow that L=lim anindeed if e.pdf
let an=(-1)^(2n+1)
and L=-1
let us ahow that L=lim an
indeed if e >0
we must find N such that if n>=N
then |an+1|< e
that is examine the quantity |an+1|
find the natural number N
|an+1|=|(-1)2n+1+1|
=|(-1)2n+(-1)1+1|
=|1-1+1|=1=n
thus |an+1| <= 1/n
therefore n is an integer for which N>1/e
therefore for every e>0 there exist a natural number N such that n>=N then
|an-L|<1/n<1/N infinity [(-1)2n+1] = -1
Solution
let an=(-1)^(2n+1)
and L=-1
let us ahow that L=lim an
indeed if e >0
we must find N such that if n>=N
then |an+1|< e
that is examine the quantity |an+1|
find the natural number N
|an+1|=|(-1)2n+1+1|
=|(-1)2n+(-1)1+1|
=|1-1+1|=1=n
thus |an+1| <= 1/n
therefore n is an integer for which N>1/e
therefore for every e>0 there exist a natural number N such that n>=N then
|an-L|<1/n<1/N infinity [(-1)2n+1] = -1.
D) more, vibrational A vibration happens in any .pdf
D) more, vibrational A vibration happens in any bond connection. Since ethane has
more no. of bonds in it, it has more no.of vibrations and hence more entropy
Solution
D) more, vibrational A vibration happens in any bond connection. Since ethane has
more no. of bonds in it, it has more no.of vibrations and hence more entropy.
Import java.awt.; Import acm.program.; Import acm.graphics.;.pdf
Import java.awt.*;
Import acm.program.*;
Import acm.graphics.*;
public class Sierpinski extends GraphicsProgram
{
public void run()
{
GRect box = new GRect(20, 20, 242, 242);
box.setFilled(true);
add(box);
drawGasket(20, 20, 243);
}
private void drawFigure(int x, int y, int side) {
int sub = side / 3;
GRect box = new GRect(x + sub, y + sub, sub - 1, sub - 1);
box.setFilled(true);
box.setColor(Color.WHITE);
add(box);
if (sub >= 3) {
drawFigure(x, y, sub);
drawFigure(x + sub, y, sub);
drawFigure(x + 2 * sub, y, sub);
drawFigure(x, y + sub, sub);
drawFigure(x + 2 * sub, y + sub, sub);
drawFigure(x, y + 2 * sub, sub);
drawFigure(x + sub, y + 2 * sub, sub);
drawFigure(x + 2 * sub, y + 2 * sub, sub);
}
}
C.8 SQUARE in recursive method:-
import java.awt.*;
import java.awt.event.*;
import javax.swing.*;
public class SierpinskiCarpet extends JPanel implements ActionListener
{
public static void main(String[] args)
{
JFrame window = new JFrame(\"Sierpinski Carpet\");
window.setContentPane( new SierpinskiCarpet() );
window.pack();
window.setResizable(false);
window.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
Dimension screensize = Toolkit.getDefaultToolkit().getScreenSize();
int x = Math.max(10, (screensize.width - window.getWidth()) / 2);
int y = Math.max(10, (screensize.height - window.getHeight()) / 2);
window.setLocation(x,y);
window.setVisible(true);
}
private int level;
private JRadioButton[] levelButton;
private JCheckBox[][] regionCheckbox;
private Display display;
private class Display extends JPanel
{
Display()
{
setBackground(Color.WHITE);
setPreferredSize(new Dimension(729,729));
}
protected void paintComponent(Graphics g)
{
super.paintComponent(g);
drawCarpet(g,level,0,0,729);
}
}
public SierpinskiCarpet()
{
setLayout(new BorderLayout(3,3));
setBackground(Color.GRAY);
setBorder(BorderFactory.createLineBorder(Color.GRAY,3));
display = new Display();
add(display,BorderLayout.CENTER);
Box controls = Box.createVerticalBox();
controls.setBorder(BorderFactory.createEmptyBorder(10, 10, 10, 10));
controls.setBackground(Color.WHITE);
controls.setOpaque(true);
add(controls, BorderLayout.EAST);
controls.add(new JLabel(\"Controls\"));
controls.add(Box.createVerticalStrut(30));
levelButton = new JRadioButton[7];
ButtonGroup grp = new ButtonGroup();
for (int i = 0; i < 7; i++)
{
levelButton[i] = new JRadioButton(\"Recursion Level \" + i);
grp.add(levelButton[i]);
levelButton[i].addActionListener(this);controls.add(levelButton[i]);
}
level = 4;
levelButton[4].setSelected(true);
regionCheckbox = new JCheckBox[3][3];
JPanel checks = new JPanel();
checks.setLayout(new GridLayout(3,3,5,5));
for (int r = 0; r < 3; r++)
{
for (int c = 0; c < 3; c++)
{
regionCheckbox[r][c] = new JCheckBox();
checks.add(regionCheckbox[r][c]);
regionCheckbox[r][c].addActionListener(this);
regionCheckbox[r][c].setSelected(true);
}
}
checks.setMaximumSize(new Dimension(90,90));
checks.setAlignmentX(LEFT_ALIGNMENT);
checks.setBackground(Color.WHITE);
regionCheckbox[1][1].setSelec.
How the standard developed1) The standards are developed to this .pdf
How the standard developed?
1) The standards are developed to this stage is just because of the customers.
2) Meetings will be held how to develop or how to enhance this.
3) The solution to this problem comes by thinking from the customer\'s point of view only.
4) Though the customer is not present in the meeting, the solution comes from their point of
view.
5) So the only person that lead to development is the customer.
How are standards born?
1) The rules of the individual standards body.
2) The main principal of Internet standards body is the Internet Engineering Task Force.
3) Anyone can participate in the IETF and standards coming out of this body are freely available
to the public.
4) The standards process in a nutshell it actually means one writes an Internet Draft RFC and
sends it to the IETF for review.
5) The work can directly become a standard but more often than not, it’s developed further by an
IETF working group.
6) After a time, the working group then decides whether or not to approve the draft.
7) The broader community is then given a final opportunity to comment.
8) The Internet Engineering Steering Group makes a final judgment as to whether or not there is
at least rough consensus to publish the edited document as an Internet standard.
9) A key factor that opinion within the IETF is whether or not there is running code.
Internet Society:
1) Founded 1991 by Internet Pioneers Founded 1991 by Internet Pioneers
2) International, not-for-profit, membership org. International, not-for-profit, membership org.
a) 80+ organisation organisation members members
b) 21,000+ individual members 21,000+ individual members
c) 75+ chapters, 10+ chapters forming
3) Organisation Organisation members fund activities in members fund activities in
a) Standards Standards
b) Education Education
c) Policy
Solution
How the standard developed?
1) The standards are developed to this stage is just because of the customers.
2) Meetings will be held how to develop or how to enhance this.
3) The solution to this problem comes by thinking from the customer\'s point of view only.
4) Though the customer is not present in the meeting, the solution comes from their point of
view.
5) So the only person that lead to development is the customer.
How are standards born?
1) The rules of the individual standards body.
2) The main principal of Internet standards body is the Internet Engineering Task Force.
3) Anyone can participate in the IETF and standards coming out of this body are freely available
to the public.
4) The standards process in a nutshell it actually means one writes an Internet Draft RFC and
sends it to the IETF for review.
5) The work can directly become a standard but more often than not, it’s developed further by an
IETF working group.
6) After a time, the working group then decides whether or not to approve the draft.
7) The broader community is then given a final opportunity to comment.
8) The Internet Engineering Steering Group makes a final judg.
Hi!The chlorphylls having lower Rf values indicates that they inte.pdf
Hi!
The chlorphylls having lower Rf values indicates that they interact more strongly with the
solvent. Both chlorophylls are polar while carotenes are non-polar. If the solvent being used for
TLC is mostly non-polar, we would expect the carotenes to interact more strongly and have
lower Rf values and vice versa. Other variations in Rf values (such as those between chlorophyll
a and b) may indicate slight differences in polarity (perhaps there are nonpolar regions on the
molecule even though it is polar overall) or differences in molecular weight (the heavier, the
slower the molecule moves up the plate.)
I hope this helps! Please don\'t forget to rate : )
Solution
Hi!
The chlorphylls having lower Rf values indicates that they interact more strongly with the
solvent. Both chlorophylls are polar while carotenes are non-polar. If the solvent being used for
TLC is mostly non-polar, we would expect the carotenes to interact more strongly and have
lower Rf values and vice versa. Other variations in Rf values (such as those between chlorophyll
a and b) may indicate slight differences in polarity (perhaps there are nonpolar regions on the
molecule even though it is polar overall) or differences in molecular weight (the heavier, the
slower the molecule moves up the plate.)
I hope this helps! Please don\'t forget to rate : ).
Determining the hybridization state requires that you know how many .pdf
Determining the hybridization state requires that you know how many regions of electron density
there are. This does not mean how many bonds! For the carbon with the arrow, there are 4 bonds
(2 single bonds and 1 double bond) but only 3 regions of electron density. The double bond only
occupies 1 region of electron density. If you look at the terminal -CH3 carbon on the molecule,
you\'ll notice that it also has 4 bonds (4 single bonds). This carbon has 4 regions of electron
density since each bond occupies its own region.
Once you know how many regions, it\'s simply a matter of adding exponents. The only
hybridization states of carbon are sp, sp2, and sp3. If you add the exponents of each, you get 2, 3,
and 4, respectively. These correspond to electron density regions. If it has 3 regions of electron
density (not necessarily 3 bonds!), then it\'s sp2 hybridized (s1p2= 3). Carbon can only form 4
bonds (this rule never changes), so it can have at most 4 regions of electron density. Since there
are only single, double, and triple bonds (no quadruple bond), the minimum regions of electron
desity for carbon is 2 (i.e. triple and single bond, or double and double bond).
Just remember, count regions (not bonds!) and add exponents.
Solution
Determining the hybridization state requires that you know how many regions of electron density
there are. This does not mean how many bonds! For the carbon with the arrow, there are 4 bonds
(2 single bonds and 1 double bond) but only 3 regions of electron density. The double bond only
occupies 1 region of electron density. If you look at the terminal -CH3 carbon on the molecule,
you\'ll notice that it also has 4 bonds (4 single bonds). This carbon has 4 regions of electron
density since each bond occupies its own region.
Once you know how many regions, it\'s simply a matter of adding exponents. The only
hybridization states of carbon are sp, sp2, and sp3. If you add the exponents of each, you get 2, 3,
and 4, respectively. These correspond to electron density regions. If it has 3 regions of electron
density (not necessarily 3 bonds!), then it\'s sp2 hybridized (s1p2= 3). Carbon can only form 4
bonds (this rule never changes), so it can have at most 4 regions of electron density. Since there
are only single, double, and triple bonds (no quadruple bond), the minimum regions of electron
desity for carbon is 2 (i.e. triple and single bond, or double and double bond).
Just remember, count regions (not bonds!) and add exponents..
E) 5thSolutionE) 5th.pdf
This very short document does not contain enough contextual information to generate a meaningful 3 sentence summary. It consists of the letter "E", the number "5th", the word "Solution", and repeats the letter "E" and number "5th" again without any other details.
Command language is more beneficial than other input methods in foll.pdf
Command language is more beneficial than other input methods in following cases
Some example applications are
Solution
Command language is more beneficial than other input methods in following cases
Some example applications are.
Clastic structure1)Roundingit is the degree of smoothing due to .pdf
Clastic structure:
1)Rounding:it is the degree of smoothing due to abrasion of sedimentary particles.it is expressed
as the ratio of the average radius of curvature of the edges or corners to the radius of curvature of
the max inscribed sphere.the sphere are angular,very angular,sub angular,sub ronded,well
rounded.
2)sorting:the distribution of grain size of sediments either in consolidated deposits.sorting in
sediments very poorly sorted,poorly sorted,moderately sorted,w ell sorted.the degree of sorting
may also indicate the energy,rate,duration of deposition,as well as the transport process.it is
effected by the reworking of material after deposition.
3)compaction:the process by which asediment progresivelly loses its porosity due to the effect of
loading.when a layer of sediments is originally deposited,it contains an open fame work of
particles with the pore space being usually filled with water.
4)cementation:it involves ions carried in ground water chemical participating to from new
crystalline material between sedimentary grains.it occurred as part of the diagenesis of
sediments.cementation occured primarily below the water table regardless of sedimentar grin
size present.
non clastic texture.
1)crystalline texture:it are visible and form an interlocking network.unlike igneious crystalline
textures.sedimentary crystallind textures are typically formed one material throught the entire
rock.
2)skeletal texture:it term used to describe the habit of enhedral to subhedral crystals in igneous
rock contains crystallogically oriented hallows and gaps.crystalls can bdescribed as re-entrant.the
voids with in skeletal crystals are filled with ground mass material.
3)oolite texture:it is a sedimentary rock formed by ooids,spherical grains composed of concentric
layers.oolites consisted of ooids of diameter 0.25-2mm rocks composed of ooids larger than
2mmare called pisolites.
Solution
Clastic structure:
1)Rounding:it is the degree of smoothing due to abrasion of sedimentary particles.it is expressed
as the ratio of the average radius of curvature of the edges or corners to the radius of curvature of
the max inscribed sphere.the sphere are angular,very angular,sub angular,sub ronded,well
rounded.
2)sorting:the distribution of grain size of sediments either in consolidated deposits.sorting in
sediments very poorly sorted,poorly sorted,moderately sorted,w ell sorted.the degree of sorting
may also indicate the energy,rate,duration of deposition,as well as the transport process.it is
effected by the reworking of material after deposition.
3)compaction:the process by which asediment progresivelly loses its porosity due to the effect of
loading.when a layer of sediments is originally deposited,it contains an open fame work of
particles with the pore space being usually filled with water.
4)cementation:it involves ions carried in ground water chemical participating to from new
crystalline material between sedimentary grains.it occurred as part of the diage.
Balance in the Retained Earnings account immediately after Event 2 i.pdf
Balance in the Retained Earnings account immediately after Event 2 in 2016 = Retained Earning
at the begining of 2017 - (Increase in Retained Earning in 3-5 event in 2016 - Decrease in
Retained Earning in 3-5 event in 2016)
Balance in the Retained Earnings account immediately after Event 2 in 2016 = 13000 - (0 -
22000)
Balance in the Retained Earnings account immediately after Event 2 in 2016 = 35000
Balance in the Retained Earnings account immediately after Event 2 in 2017 =Retained Earning
at the begining of 2017 + Received $95,000 for providing services in 2017.
Balance in the Retained Earnings account immediately after Event 2 in 2017 = 13000 + 95000
Balance in the Retained Earnings account immediately after Event 2 in 2017 = 108000
Answer
Balance in the Retained Earnings account immediately after Event 2 in 2016 = 35000
Balance in the Retained Earnings account immediately after Event 2 in 2017 = 108000
Solution
Balance in the Retained Earnings account immediately after Event 2 in 2016 = Retained Earning
at the begining of 2017 - (Increase in Retained Earning in 3-5 event in 2016 - Decrease in
Retained Earning in 3-5 event in 2016)
Balance in the Retained Earnings account immediately after Event 2 in 2016 = 13000 - (0 -
22000)
Balance in the Retained Earnings account immediately after Event 2 in 2016 = 35000
Balance in the Retained Earnings account immediately after Event 2 in 2017 =Retained Earning
at the begining of 2017 + Received $95,000 for providing services in 2017.
Balance in the Retained Earnings account immediately after Event 2 in 2017 = 13000 + 95000
Balance in the Retained Earnings account immediately after Event 2 in 2017 = 108000
Answer
Balance in the Retained Earnings account immediately after Event 2 in 2016 = 35000
Balance in the Retained Earnings account immediately after Event 2 in 2017 = 108000.
Digital Artifact 1 - 10VCD Environments Unit
Digital Artifact 1 - 10VCD Environments Unit - NGV Pavilion Concept Design
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we will use tables of normal distribution ( Z distribution)P( z .pdf
we will use tables of normal distribution ( Z distribution)
P( z < -1) = 0.1587
Solution
we will use tables of normal distribution ( Z distribution)
P( z < -1) = 0.1587.
The Northeast blackout of 2003 was a widespread power outage that oc.pdf
The Northeast blackout of 2003 was a widespread power outage that occurred throughout parts
of the Northeasternand Midwestern United States and the Canadian province of Ontario on
Thursday, August 14, 2003, just after 4:10 p.m. All told, 50 million people lost power for up to
two days in the biggest blackout in North American history. The event contributed to at least 11
deaths and cost an estimated $6 billion. A surge of electricity to western New York and Canada
touched off a series of power failures and enforced blackouts that left parts of at least eight states
in the Northeast and the Midwest without electricity.Transmission system operators scattered
across some 300 control centers nationwide monitor voltage and current data from SCADA
(supervisory control and data acquisition) systems placed at transformers, generators and other
critical points.The widespread failures provoked the evacuation of office buildings, stranded
thousands of commuters. As circuit breakers tripped at generating stations from New York to
Michigan and into Canada, millions of people were instantly caught up in the largest blackout in
American history. In New York City, power was shut off by officials struggling to head off a
wider blackout. Cleveland and Detroit went dark, as did Toronto and sections of New Jersey,
Pennsylvania, Connecticut and Massachusetts. Officials worked into the night to put the grid
back in operation and restore electric service. Mayor Michael R. Bloomberg said that that the
power was back on in parts of Brooklyn, the Bronx and Queens by 11 p.m. -- but not Manhattan.
The blackout began just after the stock exchanges had closed for the day, a slow summer day of
relatively light trading, as thousands of workers were about to head home. Office workers who
were still at their desks watched their computer monitors blink off without warning on a hot and
hazy afternoon. Soon hospitals and government buildings were switching on backup generators
to keep essential equipment operating, and the police were evacuating people trapped in
elevators. Airports throughout the affected states suffered serious disruptions, including the three
major airports in the New York metropolitan region, but did not close. Thousands of subway
passengers in New York City had to be evacuated from tunnels, and commuter trains also came
to a halt. Gov.George E. Pataki said that 600 trains were stranded.The office of the Canadian
prime minister, Jean Chrétien, initially said the power problems were caused by lightning in New
York State but later retracted that. Canadian officials later expressed uncertainty about the exact
cause but continued to insist the problem began on the United States side of the border. The
Nuclear Regulatory Commission said that the seven nuclear plants in New York and New Jersey
and two in the Midwest had shut down automatically when the failure occurred.Telephone
service was disrupted, especially calls to and from cellular phones.Officials said the.
The chemical reaction responsible for formation of water2H2+ O2--.pdf
The chemical reaction responsible for formation of water:
2H2+ O2-----------> 2H2O
2 molcules of hydrogen combine with 2 molecules of oxygen to give two molecules of water.
Solution
The chemical reaction responsible for formation of water:
2H2+ O2-----------> 2H2O
2 molcules of hydrogen combine with 2 molecules of oxygen to give two molecules of water..
sample mean = population mean = 2.11standard deviation = popul.pdf
sample mean = population mean = 2.11
standard deviation = population standad deviation / sqrt(sample size)
= 1.3 / sqrt(90)
= 0.137
Solution
sample mean = population mean = 2.11
standard deviation = population standad deviation / sqrt(sample size)
= 1.3 / sqrt(90)
= 0.137.
Polymerase chain reaction (PCR) is a process used to replicate DNA i.pdf
Polymerase chain reaction (PCR) is a process used to replicate DNA in vitro. DNA replication is
the biological process by which original DNA molecule replicate in to two identical replicas of
DNA. DNA polymerase III replicate the Prokaryotic DNA in 5\' to 3\' at a rate of 1 kb/s but 1-4
kb/min in a typical PCR reaction. Prokaryotic DNA is circular and replication starts at single
origin (ori C). It do not occur in PCR.
Prokaryotic DNA replication is followed by the three basic step:-
1. initiation, 2. elongation, and 3. termination.
The process of PCR is followed by the three basic step:-
1. Denaturation, 2.,Annealing, and 3. Primer Extension
1) the process of forming single stranded DNA, :-In PCR the DNA denatures at high tempreture,
at 95oC to form its single-stranded. During the prokaryotic DNA Replication, Ori C unwinds the
DNA by DNA B or helicase, and extension of ssDNA for copying during initiation step.
2) priming of the DNA polymerase,:-In PCR primer binds with the complementry strand of DNA
during the Annealing step. In prokaryotic replcation, primase form RNA primer extended by
DNA polymerase III,during the elongation step. Synthesis ofleading strand and lagging strand
takes place in this step. It do not occur in PCR.
3) the DNA polymerase itself:-In prokaryotic DNA replication DNA polymerase III synthesize
the segment of DNA for both leading and lagging strand. In PCR, DNA polymerase extends the
primer, Taq polymerase, is used in this process and extension is performed at 72oC.
4) the end product:- As a PCR two ds DNA is formed. And in prokaryotic DNA replication 2
circular DNA is formed , which remain interlinked and topoisomerase II make them seprate by
making cut over them.
Protien involed in prokaryotic DNA replication:-
Dna A protein-It initiate DNA replication by the binding with the Dna A boxes.
Dna helicase- it seprates the ds DNA.
Topoisomerase- From the replication fork, it removes positive supercoiling.
Single stranded binding protein-to prevent the reformation of ds DNA, it binds with single
stranded structure,
primase - synthesis the short RNA primer.
DNA polymerase III-In the leading and lagging strands, it synthesize the DNA.
DNA polymerase I- Removes RNA primer
DNA ligase- it joins the fragments.
Tus;-binds with ter sequence.
DNA replication different than PCR:--PCR generates large quantities of DNA from very small
amount of DNA, PCR is a INVITRO technique, while DNA replication occurs inside the iving
system, it is a invivo technique. The whole DNA is replicated, in DNA replication, while in PCR
the amplification of segment of DNA takes place.
Solution
Polymerase chain reaction (PCR) is a process used to replicate DNA in vitro. DNA replication is
the biological process by which original DNA molecule replicate in to two identical replicas of
DNA. DNA polymerase III replicate the Prokaryotic DNA in 5\' to 3\' at a rate of 1 kb/s but 1-4
kb/min in a typical PCR reaction. Prokaryotic DNA is circular and replication st.
Glutamic acid and Lycine Hydrogen bonding could .pdf
Glutamic acid and Lycine: Hydrogen bonding could occur between the NH3+
group of Lysine and the COO- group of Glutamic acid; and Van der Waals (nonpolar-nonpolar)
forces/interactions could arise between the nonpolar methylene groups of both R groups.
Phenylalanine and Isoleucine: Phenylalanine is quite hydrophobic so it works in hydrophobic
reactions. Both have nonpolar R groups so the only interactions they could undergo would be the
weakly attractive Van der Waals forces between the methylene groups of both R groups.
However, Phenylalanine is an aromatic compound with an aromatic ring, which is capable of
undergoing noncovalent Pi-system interactions but not with Isoleucine because it is not polar
enough to distort the Pi system of the aromatic ring.
Solution
Glutamic acid and Lycine: Hydrogen bonding could occur between the NH3+
group of Lysine and the COO- group of Glutamic acid; and Van der Waals (nonpolar-nonpolar)
forces/interactions could arise between the nonpolar methylene groups of both R groups.
Phenylalanine and Isoleucine: Phenylalanine is quite hydrophobic so it works in hydrophobic
reactions. Both have nonpolar R groups so the only interactions they could undergo would be the
weakly attractive Van der Waals forces between the methylene groups of both R groups.
However, Phenylalanine is an aromatic compound with an aromatic ring, which is capable of
undergoing noncovalent Pi-system interactions but not with Isoleucine because it is not polar
enough to distort the Pi system of the aromatic ring..
Need a clear pictureSolutionNeed a clear picture.pdf
This very short document discusses needing a clear picture but does not provide any details on the topic or solution. It repeats the phrase "Need a clear picture" but does not elaborate on what is needed, why it is needed, or how any solution might help provide a clear picture.
Li2SSolutionLi2S.pdf
This one sentence document contains the word "Li2S" repeated three times. It appears to be referring to lithium sulfide, a chemical compound with the formula Li2S, but provides no other context or information.
For H2SO3 pK1 = 1.8 and pK2 = 7.2 and thus K1 = .pdf
For H2SO3 pK1 = 1.8 and pK2 = 7.2 and thus K1 = 1.58E-2 and K2 = 6.31E-8
For a given pH (get to this in a minute), it can be shown [H+]=10^-pH [OH-] = Kw / [H+]
where Kw is the ion product of water (1E-14) [H2SO3] = [H+]^2 / f [HSO3-] = K1 [H+] / f
[SO3--] = K1 K2 / f where f = [H+]^2 + K1[H+] = K1K2 Given that H2SO3 is a weak acid, lets
assume that the pH is described by K1 or [H+] = sqrt(K1 C) where C is the acid conc (mol/l).
This leads to a [H+] = 0.05 and a pH = 1.07 Using the [H+] and data above yields in mol/l [H+]
8.51E-02 [OH-] 1.17E-13 [H2SO3] 3.88E-01 [HSO3-] 7.22E-02 [SO3--] 5.35E-08 Note the
above data describes the acid as 15% dissociated and the assumptiom that the pH of the solution
is best described by K1 is not necessarily a good assumption. A better way is as follows :- The
charge balance is this system is given by [H+] = [OH-] + [HSO3-] + 2*[SO3--] Set up all the
above relationships in MS Excel and Goal Seek a pH that makes the above charge balance
correct. Doing this yields a pH = 1.11 and the following concs (mol/l) [H+] 7.78E-02 [OH-]
1.28E-13 [H2SO3] 3.82E-01 [HSO3-] 7.78E-02 [SO3--] 6.31E-08 Be mindful that in all this
calculation, ionic strength impacts (ie activities) have been ignored or activity coefficients have
been taken as 1
Solution
For H2SO3 pK1 = 1.8 and pK2 = 7.2 and thus K1 = 1.58E-2 and K2 = 6.31E-8
For a given pH (get to this in a minute), it can be shown [H+]=10^-pH [OH-] = Kw / [H+]
where Kw is the ion product of water (1E-14) [H2SO3] = [H+]^2 / f [HSO3-] = K1 [H+] / f
[SO3--] = K1 K2 / f where f = [H+]^2 + K1[H+] = K1K2 Given that H2SO3 is a weak acid, lets
assume that the pH is described by K1 or [H+] = sqrt(K1 C) where C is the acid conc (mol/l).
This leads to a [H+] = 0.05 and a pH = 1.07 Using the [H+] and data above yields in mol/l [H+]
8.51E-02 [OH-] 1.17E-13 [H2SO3] 3.88E-01 [HSO3-] 7.22E-02 [SO3--] 5.35E-08 Note the
above data describes the acid as 15% dissociated and the assumptiom that the pH of the solution
is best described by K1 is not necessarily a good assumption. A better way is as follows :- The
charge balance is this system is given by [H+] = [OH-] + [HSO3-] + 2*[SO3--] Set up all the
above relationships in MS Excel and Goal Seek a pH that makes the above charge balance
correct. Doing this yields a pH = 1.11 and the following concs (mol/l) [H+] 7.78E-02 [OH-]
1.28E-13 [H2SO3] 3.82E-01 [HSO3-] 7.78E-02 [SO3--] 6.31E-08 Be mindful that in all this
calculation, ionic strength impacts (ie activities) have been ignored or activity coefficients have
been taken as 1.
find the largest electronegativity difference (mo.pdf
find the largest electronegativity difference (most ionic)
Solution
find the largest electronegativity difference (most ionic).
let an=(-1)^(2n+1)and L=-1let us ahow that L=lim anindeed if e.pdf
let an=(-1)^(2n+1)
and L=-1
let us ahow that L=lim an
indeed if e >0
we must find N such that if n>=N
then |an+1|< e
that is examine the quantity |an+1|
find the natural number N
|an+1|=|(-1)2n+1+1|
=|(-1)2n+(-1)1+1|
=|1-1+1|=1=n
thus |an+1| <= 1/n
therefore n is an integer for which N>1/e
therefore for every e>0 there exist a natural number N such that n>=N then
|an-L|<1/n<1/N infinity [(-1)2n+1] = -1
Solution
let an=(-1)^(2n+1)
and L=-1
let us ahow that L=lim an
indeed if e >0
we must find N such that if n>=N
then |an+1|< e
that is examine the quantity |an+1|
find the natural number N
|an+1|=|(-1)2n+1+1|
=|(-1)2n+(-1)1+1|
=|1-1+1|=1=n
thus |an+1| <= 1/n
therefore n is an integer for which N>1/e
therefore for every e>0 there exist a natural number N such that n>=N then
|an-L|<1/n<1/N infinity [(-1)2n+1] = -1.
D) more, vibrational A vibration happens in any .pdf
D) more, vibrational A vibration happens in any bond connection. Since ethane has
more no. of bonds in it, it has more no.of vibrations and hence more entropy
Solution
D) more, vibrational A vibration happens in any bond connection. Since ethane has
more no. of bonds in it, it has more no.of vibrations and hence more entropy.
Import java.awt.; Import acm.program.; Import acm.graphics.;.pdf
Import java.awt.*;
Import acm.program.*;
Import acm.graphics.*;
public class Sierpinski extends GraphicsProgram
{
public void run()
{
GRect box = new GRect(20, 20, 242, 242);
box.setFilled(true);
add(box);
drawGasket(20, 20, 243);
}
private void drawFigure(int x, int y, int side) {
int sub = side / 3;
GRect box = new GRect(x + sub, y + sub, sub - 1, sub - 1);
box.setFilled(true);
box.setColor(Color.WHITE);
add(box);
if (sub >= 3) {
drawFigure(x, y, sub);
drawFigure(x + sub, y, sub);
drawFigure(x + 2 * sub, y, sub);
drawFigure(x, y + sub, sub);
drawFigure(x + 2 * sub, y + sub, sub);
drawFigure(x, y + 2 * sub, sub);
drawFigure(x + sub, y + 2 * sub, sub);
drawFigure(x + 2 * sub, y + 2 * sub, sub);
}
}
C.8 SQUARE in recursive method:-
import java.awt.*;
import java.awt.event.*;
import javax.swing.*;
public class SierpinskiCarpet extends JPanel implements ActionListener
{
public static void main(String[] args)
{
JFrame window = new JFrame(\"Sierpinski Carpet\");
window.setContentPane( new SierpinskiCarpet() );
window.pack();
window.setResizable(false);
window.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
Dimension screensize = Toolkit.getDefaultToolkit().getScreenSize();
int x = Math.max(10, (screensize.width - window.getWidth()) / 2);
int y = Math.max(10, (screensize.height - window.getHeight()) / 2);
window.setLocation(x,y);
window.setVisible(true);
}
private int level;
private JRadioButton[] levelButton;
private JCheckBox[][] regionCheckbox;
private Display display;
private class Display extends JPanel
{
Display()
{
setBackground(Color.WHITE);
setPreferredSize(new Dimension(729,729));
}
protected void paintComponent(Graphics g)
{
super.paintComponent(g);
drawCarpet(g,level,0,0,729);
}
}
public SierpinskiCarpet()
{
setLayout(new BorderLayout(3,3));
setBackground(Color.GRAY);
setBorder(BorderFactory.createLineBorder(Color.GRAY,3));
display = new Display();
add(display,BorderLayout.CENTER);
Box controls = Box.createVerticalBox();
controls.setBorder(BorderFactory.createEmptyBorder(10, 10, 10, 10));
controls.setBackground(Color.WHITE);
controls.setOpaque(true);
add(controls, BorderLayout.EAST);
controls.add(new JLabel(\"Controls\"));
controls.add(Box.createVerticalStrut(30));
levelButton = new JRadioButton[7];
ButtonGroup grp = new ButtonGroup();
for (int i = 0; i < 7; i++)
{
levelButton[i] = new JRadioButton(\"Recursion Level \" + i);
grp.add(levelButton[i]);
levelButton[i].addActionListener(this);controls.add(levelButton[i]);
}
level = 4;
levelButton[4].setSelected(true);
regionCheckbox = new JCheckBox[3][3];
JPanel checks = new JPanel();
checks.setLayout(new GridLayout(3,3,5,5));
for (int r = 0; r < 3; r++)
{
for (int c = 0; c < 3; c++)
{
regionCheckbox[r][c] = new JCheckBox();
checks.add(regionCheckbox[r][c]);
regionCheckbox[r][c].addActionListener(this);
regionCheckbox[r][c].setSelected(true);
}
}
checks.setMaximumSize(new Dimension(90,90));
checks.setAlignmentX(LEFT_ALIGNMENT);
checks.setBackground(Color.WHITE);
regionCheckbox[1][1].setSelec.
How the standard developed1) The standards are developed to this .pdf
How the standard developed?
1) The standards are developed to this stage is just because of the customers.
2) Meetings will be held how to develop or how to enhance this.
3) The solution to this problem comes by thinking from the customer\'s point of view only.
4) Though the customer is not present in the meeting, the solution comes from their point of
view.
5) So the only person that lead to development is the customer.
How are standards born?
1) The rules of the individual standards body.
2) The main principal of Internet standards body is the Internet Engineering Task Force.
3) Anyone can participate in the IETF and standards coming out of this body are freely available
to the public.
4) The standards process in a nutshell it actually means one writes an Internet Draft RFC and
sends it to the IETF for review.
5) The work can directly become a standard but more often than not, it’s developed further by an
IETF working group.
6) After a time, the working group then decides whether or not to approve the draft.
7) The broader community is then given a final opportunity to comment.
8) The Internet Engineering Steering Group makes a final judgment as to whether or not there is
at least rough consensus to publish the edited document as an Internet standard.
9) A key factor that opinion within the IETF is whether or not there is running code.
Internet Society:
1) Founded 1991 by Internet Pioneers Founded 1991 by Internet Pioneers
2) International, not-for-profit, membership org. International, not-for-profit, membership org.
a) 80+ organisation organisation members members
b) 21,000+ individual members 21,000+ individual members
c) 75+ chapters, 10+ chapters forming
3) Organisation Organisation members fund activities in members fund activities in
a) Standards Standards
b) Education Education
c) Policy
Solution
How the standard developed?
1) The standards are developed to this stage is just because of the customers.
2) Meetings will be held how to develop or how to enhance this.
3) The solution to this problem comes by thinking from the customer\'s point of view only.
4) Though the customer is not present in the meeting, the solution comes from their point of
view.
5) So the only person that lead to development is the customer.
How are standards born?
1) The rules of the individual standards body.
2) The main principal of Internet standards body is the Internet Engineering Task Force.
3) Anyone can participate in the IETF and standards coming out of this body are freely available
to the public.
4) The standards process in a nutshell it actually means one writes an Internet Draft RFC and
sends it to the IETF for review.
5) The work can directly become a standard but more often than not, it’s developed further by an
IETF working group.
6) After a time, the working group then decides whether or not to approve the draft.
7) The broader community is then given a final opportunity to comment.
8) The Internet Engineering Steering Group makes a final judg.
Hi!The chlorphylls having lower Rf values indicates that they inte.pdf
Hi!
The chlorphylls having lower Rf values indicates that they interact more strongly with the
solvent. Both chlorophylls are polar while carotenes are non-polar. If the solvent being used for
TLC is mostly non-polar, we would expect the carotenes to interact more strongly and have
lower Rf values and vice versa. Other variations in Rf values (such as those between chlorophyll
a and b) may indicate slight differences in polarity (perhaps there are nonpolar regions on the
molecule even though it is polar overall) or differences in molecular weight (the heavier, the
slower the molecule moves up the plate.)
I hope this helps! Please don\'t forget to rate : )
Solution
Hi!
The chlorphylls having lower Rf values indicates that they interact more strongly with the
solvent. Both chlorophylls are polar while carotenes are non-polar. If the solvent being used for
TLC is mostly non-polar, we would expect the carotenes to interact more strongly and have
lower Rf values and vice versa. Other variations in Rf values (such as those between chlorophyll
a and b) may indicate slight differences in polarity (perhaps there are nonpolar regions on the
molecule even though it is polar overall) or differences in molecular weight (the heavier, the
slower the molecule moves up the plate.)
I hope this helps! Please don\'t forget to rate : ).
Determining the hybridization state requires that you know how many .pdf
Determining the hybridization state requires that you know how many regions of electron density
there are. This does not mean how many bonds! For the carbon with the arrow, there are 4 bonds
(2 single bonds and 1 double bond) but only 3 regions of electron density. The double bond only
occupies 1 region of electron density. If you look at the terminal -CH3 carbon on the molecule,
you\'ll notice that it also has 4 bonds (4 single bonds). This carbon has 4 regions of electron
density since each bond occupies its own region.
Once you know how many regions, it\'s simply a matter of adding exponents. The only
hybridization states of carbon are sp, sp2, and sp3. If you add the exponents of each, you get 2, 3,
and 4, respectively. These correspond to electron density regions. If it has 3 regions of electron
density (not necessarily 3 bonds!), then it\'s sp2 hybridized (s1p2= 3). Carbon can only form 4
bonds (this rule never changes), so it can have at most 4 regions of electron density. Since there
are only single, double, and triple bonds (no quadruple bond), the minimum regions of electron
desity for carbon is 2 (i.e. triple and single bond, or double and double bond).
Just remember, count regions (not bonds!) and add exponents.
Solution
Determining the hybridization state requires that you know how many regions of electron density
there are. This does not mean how many bonds! For the carbon with the arrow, there are 4 bonds
(2 single bonds and 1 double bond) but only 3 regions of electron density. The double bond only
occupies 1 region of electron density. If you look at the terminal -CH3 carbon on the molecule,
you\'ll notice that it also has 4 bonds (4 single bonds). This carbon has 4 regions of electron
density since each bond occupies its own region.
Once you know how many regions, it\'s simply a matter of adding exponents. The only
hybridization states of carbon are sp, sp2, and sp3. If you add the exponents of each, you get 2, 3,
and 4, respectively. These correspond to electron density regions. If it has 3 regions of electron
density (not necessarily 3 bonds!), then it\'s sp2 hybridized (s1p2= 3). Carbon can only form 4
bonds (this rule never changes), so it can have at most 4 regions of electron density. Since there
are only single, double, and triple bonds (no quadruple bond), the minimum regions of electron
desity for carbon is 2 (i.e. triple and single bond, or double and double bond).
Just remember, count regions (not bonds!) and add exponents..
E) 5thSolutionE) 5th.pdf
This very short document does not contain enough contextual information to generate a meaningful 3 sentence summary. It consists of the letter "E", the number "5th", the word "Solution", and repeats the letter "E" and number "5th" again without any other details.
Command language is more beneficial than other input methods in foll.pdf
Command language is more beneficial than other input methods in following cases
Some example applications are
Solution
Command language is more beneficial than other input methods in following cases
Some example applications are.
Clastic structure1)Roundingit is the degree of smoothing due to .pdf
Clastic structure:
1)Rounding:it is the degree of smoothing due to abrasion of sedimentary particles.it is expressed
as the ratio of the average radius of curvature of the edges or corners to the radius of curvature of
the max inscribed sphere.the sphere are angular,very angular,sub angular,sub ronded,well
rounded.
2)sorting:the distribution of grain size of sediments either in consolidated deposits.sorting in
sediments very poorly sorted,poorly sorted,moderately sorted,w ell sorted.the degree of sorting
may also indicate the energy,rate,duration of deposition,as well as the transport process.it is
effected by the reworking of material after deposition.
3)compaction:the process by which asediment progresivelly loses its porosity due to the effect of
loading.when a layer of sediments is originally deposited,it contains an open fame work of
particles with the pore space being usually filled with water.
4)cementation:it involves ions carried in ground water chemical participating to from new
crystalline material between sedimentary grains.it occurred as part of the diagenesis of
sediments.cementation occured primarily below the water table regardless of sedimentar grin
size present.
non clastic texture.
1)crystalline texture:it are visible and form an interlocking network.unlike igneious crystalline
textures.sedimentary crystallind textures are typically formed one material throught the entire
rock.
2)skeletal texture:it term used to describe the habit of enhedral to subhedral crystals in igneous
rock contains crystallogically oriented hallows and gaps.crystalls can bdescribed as re-entrant.the
voids with in skeletal crystals are filled with ground mass material.
3)oolite texture:it is a sedimentary rock formed by ooids,spherical grains composed of concentric
layers.oolites consisted of ooids of diameter 0.25-2mm rocks composed of ooids larger than
2mmare called pisolites.
Solution
Clastic structure:
1)Rounding:it is the degree of smoothing due to abrasion of sedimentary particles.it is expressed
as the ratio of the average radius of curvature of the edges or corners to the radius of curvature of
the max inscribed sphere.the sphere are angular,very angular,sub angular,sub ronded,well
rounded.
2)sorting:the distribution of grain size of sediments either in consolidated deposits.sorting in
sediments very poorly sorted,poorly sorted,moderately sorted,w ell sorted.the degree of sorting
may also indicate the energy,rate,duration of deposition,as well as the transport process.it is
effected by the reworking of material after deposition.
3)compaction:the process by which asediment progresivelly loses its porosity due to the effect of
loading.when a layer of sediments is originally deposited,it contains an open fame work of
particles with the pore space being usually filled with water.
4)cementation:it involves ions carried in ground water chemical participating to from new
crystalline material between sedimentary grains.it occurred as part of the diage.
Balance in the Retained Earnings account immediately after Event 2 i.pdf
Balance in the Retained Earnings account immediately after Event 2 in 2016 = Retained Earning
at the begining of 2017 - (Increase in Retained Earning in 3-5 event in 2016 - Decrease in
Retained Earning in 3-5 event in 2016)
Balance in the Retained Earnings account immediately after Event 2 in 2016 = 13000 - (0 -
22000)
Balance in the Retained Earnings account immediately after Event 2 in 2016 = 35000
Balance in the Retained Earnings account immediately after Event 2 in 2017 =Retained Earning
at the begining of 2017 + Received $95,000 for providing services in 2017.
Balance in the Retained Earnings account immediately after Event 2 in 2017 = 13000 + 95000
Balance in the Retained Earnings account immediately after Event 2 in 2017 = 108000
Answer
Balance in the Retained Earnings account immediately after Event 2 in 2016 = 35000
Balance in the Retained Earnings account immediately after Event 2 in 2017 = 108000
Solution
Balance in the Retained Earnings account immediately after Event 2 in 2016 = Retained Earning
at the begining of 2017 - (Increase in Retained Earning in 3-5 event in 2016 - Decrease in
Retained Earning in 3-5 event in 2016)
Balance in the Retained Earnings account immediately after Event 2 in 2016 = 13000 - (0 -
22000)
Balance in the Retained Earnings account immediately after Event 2 in 2016 = 35000
Balance in the Retained Earnings account immediately after Event 2 in 2017 =Retained Earning
at the begining of 2017 + Received $95,000 for providing services in 2017.
Balance in the Retained Earnings account immediately after Event 2 in 2017 = 13000 + 95000
Balance in the Retained Earnings account immediately after Event 2 in 2017 = 108000
Answer
Balance in the Retained Earnings account immediately after Event 2 in 2016 = 35000
Balance in the Retained Earnings account immediately after Event 2 in 2017 = 108000.
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SolutionDomain name system is used to determine the IP address of.pdf
- 1. Solution : Domain name system is used to determine the IP address of the host/domain name. DNS is very useful to every user where it helps retrieve the website IP address. Without DNS we cannot retrieve the website IP address. This process is known as forward DNS resolution. DNS should be registered by some companies. If any attacker has details then the dns will be controlled by the attacker. when we are sending the e-mail the dns data will be used then it can be catches anywhere in the network. So it is very important that a DNS require reliable communications.