Addition of NH3(aq) produced a white gelatinous precipitate in a col.pdf

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Addition of NH3(aq) produced a white gelatinous precipitate in a colorless solution. note: The bismuth III ions are separated by precipitation as white gelatinous bismuth (III) hydroxide upon the addition of an aqueous solution of ammonia. Bi definitely present. The addition of the ammonia water to precipitate the bismuth hydroxide also converts any copper(II) ion present to the deep blue complex ion, Cu(NH3)4^2+. since the solution is colorless, Cu definitely absent. When solution (Z) was treated with K3Fe(CN)6, a dark blue suspension was produced. KFe[Fe(CN)6] this indicates Fe3+ definitely present Co2+,Ni2+ cannot decide if it exists or not. Solution Addition of NH3(aq) produced a white gelatinous precipitate in a colorless solution. note: The bismuth III ions are separated by precipitation as white gelatinous bismuth (III) hydroxide upon the addition of an aqueous solution of ammonia. Bi definitely present. The addition of the ammonia water to precipitate the bismuth hydroxide also converts any copper(II) ion present to the deep blue complex ion, Cu(NH3)4^2+. since the solution is colorless, Cu definitely absent. When solution (Z) was treated with K3Fe(CN)6, a dark blue suspension was produced. KFe[Fe(CN)6] this indicates Fe3+ definitely present Co2+,Ni2+ cannot decide if it exists or not..

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Alternative 1: Continuing present job PV = Pmt x (1 + i) x (1 - (1 + g) n x (1 + i)-n) / (i - g) Pmt = 60,000 Discount rate (i) = 12% Growth rate (g) = 14% No of years (n) = 10 = 60000 × (1+0.12) × (1- (1+0.14)10 × (1+0.12)-10/ (0.12-0.14) = 650583.51 (Or) Year Cashflow Present value@12% Discounted cashflow 1 60000.00 1 60000.00 2 68400.00 0.8929 61071.43 3 77976.00 0.7972 62161.99 4 88892.64 0.7118 63272.03 5 101337.61 0.6355 64401.88 6 115524.87 0.5674 65551.92 7 131698.36 0.5066 66722.49 8 150136.13 0.4523 67913.96 9 171155.19 0.4039 69126.71 10 195116.91 0.3606 70361.11 Net Present value 650583.51 Alternative2: Going for MBA Year Cash flow Present value@12% Discounted cash flow 1 -45000.00 1 -45000.00 2 -45000.00 0.8929 -40178.57 3 90000.00 0.7972 71747.45 4 99000.00 0.7118 70466.24 5 108900.00 0.6355 69207.92 6 119790.00 0.5674 67972.06 7 131769.00 0.5066 66758.28 8 144945.90 0.4523 65566.16 9 159440.49 0.4039 64395.34 10 175384.54 0.3606 63245.42 Net present value 454180.31 The NPV of most attractive choice of continuing current job is 650583.51 Year Cashflow Present value@12% Discounted cashflow 1 60000.00 1 60000.00 2 68400.00 0.8929 61071.43 3 77976.00 0.7972 62161.99 4 88892.64 0.7118 63272.03 5 101337.61 0.6355 64401.88 6 115524.87 0.5674 65551.92 7 131698.36 0.5066 66722.49 8 150136.13 0.4523 67913.96 9 171155.19 0.4039 69126.71 10 195116.91 0.3606 70361.11 Net Present value 650583.51 Solution Alternative 1: Continuing present job PV = Pmt x (1 + i) x (1 - (1 + g) n x (1 + i)-n) / (i - g) Pmt = 60,000 Discount rate (i) = 12% Growth rate (g) = 14% No of years (n) = 10 = 60000 × (1+0.12) × (1- (1+0.14)10 × (1+0.12)-10/ (0.12-0.14) = 650583.51 (Or) Year Cashflow Present value@12% Discounted cashflow 1 60000.00 1 60000.00 2 68400.00 0.8929 61071.43 3 77976.00 0.7972 62161.99 4 88892.64 0.7118 63272.03 5 101337.61 0.6355 64401.88 6 115524.87 0.5674 65551.92 7 131698.36 0.5066 66722.49 8 150136.13 0.4523 67913.96 9 171155.19 0.4039 69126.71 10 195116.91 0.3606 70361.11 Net Present value 650583.51 Alternative2: Going for MBA Year Cash flow Present value@12% Discounted cash flow 1 -45000.00 1 -45000.00 2 -45000.00 0.8929 -40178.57 3 90000.00 0.7972 71747.45 4 99000.00 0.7118 70466.24 5 108900.00 0.6355 69207.92 6 119790.00 0.5674 67972.06 7 131769.00 0.5066 66758.28 8 144945.90 0.4523 65566.16 9 159440.49 0.4039 64395.34 10 175384.54 0.3606 63245.42 Net present value 454180.31 The NPV of most attractive choice of continuing current job is 650583.51 Year Cashflow Present value@12% Discounted cashflow 1 60000.00 1 60000.00 2 68400.00 0.8929 61071.43 3 77976.00 0.7972 62161.99 4 88892.64 0.7118 63272.03 5 101337.61 0.6355 64401.88 6 115524.87 0.5674 65551.92 7 131698.36 0.5066 66722.49 8 150136.13 0.4523 67913.96 9 171155.19 0.4039 69126.71 10 195116.91 0.3606 70361.11 Net Present value 650583.51.

Alternative 1 Continuing present jobPV = Pmt x (1 + i) x (1 - (1 .pdf

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#include iostream #include fstreamusing namespace std;void.pdf

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Wilhelm Johannsen first proposed the distinction between genotype and phenotype in the study of heredity while working in Denmark in 1909. The distinction is between the hereditary dispositions of organisms (their genotypes) and the ways in which those dispositions manifest themselves in the physical characteristics of those organisms (their phenotypes). This distinction was an outgrowth of Johannsen Solution Wilhelm Johannsen first proposed the distinction between genotype and phenotype in the study of heredity while working in Denmark in 1909. The distinction is between the hereditary dispositions of organisms (their genotypes) and the ways in which those dispositions manifest themselves in the physical characteristics of those organisms (their phenotypes). This distinction was an outgrowth of Johannsen.

Wilhelm Johannsen first proposed the distinction between genotype a.pdf

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True, they are identical Thats just the bottom vi.pdf

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The NH2OH is in acidic solution as a reactant and.pdf

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The atom is a basic unit of matter consisting of central nucleus surrounded by a cloud of negatively charged electrons. The atomic nucleus contains a mix of positively charged protons and electrically neutral neutrons (except in the case of hydrogen-1, which is the only stable nuclide with no neutrons). The electrons of an atom are bound to the nucleus by the electromagnetic force. Likewise, a group of atoms can remain bound to each other, forming a molecule. An atom containing an equal number of protons and electrons is electrically neutral, otherwise it has a positive charge if there are fewer electrons (electron deficiency) or negative charge if there are more electrons (electron excess). A positively or negatively charged atom is known as an ion. An atom is classified according to the number of protons and neutrons in its nucleus: the number of protons determines the chemical element, and the number of neutrons determines the isotope of the element.[1] The name atom comes from the Greek ?t?µ?? (atomos, \"indivisible\") from ?- (a-, \"not\") and t?µ?? (temno, \"I cut\")[2], which means uncuttable, or indivisible, something that cannot be divided further.[3] The concept of an atom as an indivisible component of matter was first proposed by early Indian and Greek philosophers. In the 17th and 18th centuries, chemists provided a physical basis for this idea by showing that certain substances could not be further broken down by chemical methods. During the late 19th and early 20th centuries, physicists discovered subatomic components and structure inside the atom, thereby demonstrating that the \'atom\' was divisible. The principles of quantum mechanics were used to successfully model the atom.[4][5] Atoms are minuscule objects with proportionately tiny masses. Atoms can only be observed individually using special instruments such as the scanning tunneling microscope. Over 99.94% of an atom\'s mass is concentrated in the nucleus,[note 1] with protons and neutrons having roughly equal mass. Each element has at least one isotope with unstable nuclei that can undergo radioactive decay. This can result in a transmutation that changes the number of protons or neutrons in a nucleus.[6] Electrons that are bound to atoms possess a set of stable energy levels, or orbitals, and can undergo transitions between them by absorbing or emitting photons that match the energy differences between the levels. The electrons determine the chemical properties of an element, and strongly influence an atom\'s magnetic properties. Solution The atom is a basic unit of matter consisting of central nucleus surrounded by a cloud of negatively charged electrons. The atomic nucleus contains a mix of positively charged protons and electrically neutral neutrons (except in the case of hydrogen-1, which is the only stable nuclide with no neutrons). The electrons of an atom are bound to the nucleus by the electromagnetic force. Likewise, a group of atoms can remain bound to each .

The atom is a basic unit of matter consisting of .pdf

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see the inner one is union of events from 1-inf. .pdf

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No you wouldnt include the intermediates in the.pdf

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mole of NaOH= 1240 =0.3 so molarity = 0.31.8 = .pdf

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millimoles of HBr=50*0.04=2 millimoles of KOH=25*.107=2.675 millimoles of acid remaining=0 millimoles of base remaining=0.675 Kb=[OH-][K+]/[KOH] 1=[OH-]*2/0.675 [oh-]=0.675/2 =0.3375 pH=14+log(0.3375) =13.53 Solution millimoles of HBr=50*0.04=2 millimoles of KOH=25*.107=2.675 millimoles of acid remaining=0 millimoles of base remaining=0.675 Kb=[OH-][K+]/[KOH] 1=[OH-]*2/0.675 [oh-]=0.675/2 =0.3375 pH=14+log(0.3375) =13.53.

millimoles of HBr=500.04=2 millimoles of KOH=25.pdf

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initially 2 moles of HCl per L of water suppose x.pdf

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i cant see the question.. please post the complet.pdf

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hey pls give the reactions.... .pdf

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FT-Raman spectra of some calcium phosphates, calcium carbonate and calcium hydroxide are reported. Comparing with dispersive Raman spectra, it is shown that some strong bands in the FT-Raman spectra of these compounds which have been erroneously assigned to Ramanbands, are in fact fluorescence emission bands. The origin of these fluorescencebands have been discussed in terms of rare earth impurities in a particular structure. It is further concluded that in using FT-Raman spectroscopy of minerals, care should be taken in not confusing the fluorescencebands with Raman scattering bands. Among instrumental techniques, fluorescence spectroscopy is recognized as one of the more sensitive. In fluorescence, the intensity of the emission of the sample is measured. The reason for the high sensitivity of fluorescence techniques is that the emission signal is measured above a low background level. This is inherently more sensitive than comparing two relatively large signals as in absorption spectroscopy. The sensitivity of fluorescence techniques is as much as 1000 times more sensitive than absorption spectroscopy. A spectrofluorometer with high sensitivity is an asset for the researcher. It is obvious that an instrument with high sensitivity will be able to perform experiments with the fluorescing species at a low concentration. For analytical measurements, sensitivity determines the detection limit of a material. For research in life science, it may be important to work at concentration levels that are very low. This is almost always the case in biomedical research. A more subtle requirement is the need to measure changes in fluorescence during the course of an experiment. Even though the gross signal levels may be high, the changes in the signal may be quite small. The ability to reliably measure these small changes is directly dependent upon the sensitivity of the fluorometer. The measurement of sensitivity has been discussed in the literature [1]. Unfortunately, there is little consensus on a standard procedure for determining this important capability. This paper discusses the measurement of sensitivity and details two methods that are applicable to any instrument. The Spectrofluorometer The sensitivity of fluorescence is dependent on both the fluorophore and the instrument. The response of a fluorophore will depend on the molar absorptivity and the quantum yield. These factors are, in general, beyond the control of the analyst. The sensitivity of a spectrofluorometer depends on a number of factors. Instrumental contributions to sensitivity are described below. 1. Source intensity. In general, a brighter excitation source will result in brighter emission. The source for most fluorometers is a xenon arc, which has a high intensity between 200-900 nm, the spectral region where most fluorescence experiments are performed. While a high power arc lamp is good for highest intensity, a more important criteria is the brightness of the lamp. The brightness.

FT-Raman spectra of some calcium phosphates, calc.pdf

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H is oxidised here no O2 is not oxidised as it g.pdf

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Current Stock price = Expected dividend (Required rate of return -.pdf

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YES the news stories in the Illustrated American sympathetic or non-sympathetic to the cause of the strikers. The Homestead Strike - this was at the Homestead Steel Works, owned by Andrew Carnegie but run by Henry Clay Frick. Labor went on strike because of pay rates that were continually adjusted down as well as poor working conditions. Carnegie was in Scotland at this time and Frick was the man on the spot and in charge. A miscommunication between these two resulted in Frick using too much force to suppress the strikers. Labor was responsible for the strike, but Frick (and maybe Carnegie) were resposible for subsequent events and deaths. The Homestead strike was organized and purposeful, a harbinger of the type of strike which would mark the modern age of labor relations in the United States. The AA strike at the Homestead steel mill in 1892 was different from previous large-scale strikes in American history such as the Great railroad strike of 1877 or the Great Southwest Railroad Strike of 1886. Earlier strikes had been largely leaderless and disorganized mass uprisings of workers. The AFL refused to call for a boycott of Carnegie products in September 1892. Wholesale crossing of the picket line occurred, first among Eastern European immigrants and then among all workers. The strike had collapsed so much that the state militia pulled out on October 13, ending the 95-day occupation. The AA was nearly bankrupted by the job action. Weekly Union relief for a member averaged $6.25 but totalled a staggering $10,000 per week when including 1,600 strikers. With only 192 out of more than 3,800 strikers in attendance, the Homestead chapter of the AA voted, 101 to 91, to return to work on November 20, 1892. Solution YES the news stories in the Illustrated American sympathetic or non-sympathetic to the cause of the strikers. The Homestead Strike - this was at the Homestead Steel Works, owned by Andrew Carnegie but run by Henry Clay Frick. Labor went on strike because of pay rates that were continually adjusted down as well as poor working conditions. Carnegie was in Scotland at this time and Frick was the man on the spot and in charge. A miscommunication between these two resulted in Frick using too much force to suppress the strikers. Labor was responsible for the strike, but Frick (and maybe Carnegie) were resposible for subsequent events and deaths. The Homestead strike was organized and purposeful, a harbinger of the type of strike which would mark the modern age of labor relations in the United States. The AA strike at the Homestead steel mill in 1892 was different from previous large-scale strikes in American history such as the Great railroad strike of 1877 or the Great Southwest Railroad Strike of 1886. Earlier strikes had been largely leaderless and disorganized mass uprisings of workers. The AFL refused to call for a boycott of Carnegie products in September 1892. Wholesale crossing of the picket line occurred, first among Eastern European i.

YES the news stories in the Illustrated American sympathetic or non-.pdf

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Well its quite an easy one. The given curve is a cardioid. The limits of theta, unless otherwise mentioned is always from 0 to . The secret to finding it is that, cos=(r-4)/3 . So for any value of r in polar coordinate, (r-4)/3 will give some value. But we know cos=cos(2-), so the same value wil be given by and 2- . So if we put =2-, we get = which is maximum range for non repeatation of r values. A simpler way to find the range is: You get the minimum value of r at cos=-1, so minimum r=1. So range of r must be from 1 to 4 + 3cos. r is minimm at = and maximum of r is when cos=1 that is =0. So your can range from 0 to . However, integrating from 0 to will only give you half the area ( upper part of curve). So multiply the answer with 2 to get the full area. Solution Well its quite an easy one. The given curve is a cardioid. The limits of theta, unless otherwise mentioned is always from 0 to . The secret to finding it is that, cos=(r-4)/3 . So for any value of r in polar coordinate, (r-4)/3 will give some value. But we know cos=cos(2-), so the same value wil be given by and 2- . So if we put =2-, we get = which is maximum range for non repeatation of r values. A simpler way to find the range is: You get the minimum value of r at cos=-1, so minimum r=1. So range of r must be from 1 to 4 + 3cos. r is minimm at = and maximum of r is when cos=1 that is =0. So your can range from 0 to . However, integrating from 0 to will only give you half the area ( upper part of curve). So multiply the answer with 2 to get the full area..

Well its quite an easy one. The given curve is a cardioid. The limit.pdf

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Total ionic Equation is 2Cu2+ +4 Cl - + 4H+ +SO42- --- Cu2SiO4+ 4.pdf

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Answer: SOCS are the supressor of cytokine signaling proteins and are negative feedback regulators of JAK-STAT signaling pathway. It means the SOCS are inhibitors of activation of JAK-STAT pathway proteins. As JAK-STAT pathway plays essential roles in initiation and development of cancers, mutation in SOCS family proteins could be one of the mechanism of abnormal activation of STAT . The expression of SOCS are inversly related to the STAT-3 activation. Hence overexpression of SOCS may not be lead to cancer risk. Overexpression of SOCS may supress the cytokine signaling or inhibit cytokine signaling by (i) Blocking STAT recruitment to the cytokine receptor (ii) may target protein for degradation (iii) bind to JAK and inhibit their kinase (iv) target JAK for degradation via proteosome pathway Solution Answer: SOCS are the supressor of cytokine signaling proteins and are negative feedback regulators of JAK-STAT signaling pathway. It means the SOCS are inhibitors of activation of JAK-STAT pathway proteins. As JAK-STAT pathway plays essential roles in initiation and development of cancers, mutation in SOCS family proteins could be one of the mechanism of abnormal activation of STAT . The expression of SOCS are inversly related to the STAT-3 activation. Hence overexpression of SOCS may not be lead to cancer risk. Overexpression of SOCS may supress the cytokine signaling or inhibit cytokine signaling by (i) Blocking STAT recruitment to the cytokine receptor (ii) may target protein for degradation (iii) bind to JAK and inhibit their kinase (iv) target JAK for degradation via proteosome pathway.

AnswerSOCS are the supressor of cytokine signaling proteins and a.pdf

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True, they are identical Thats just the bottom vi.pdf

The NH2OH is in acidic solution as a reactant and.pdf

The atom is a basic unit of matter consisting of .pdf

see the inner one is union of events from 1-inf. .pdf

No you wouldnt include the intermediates in the.pdf

mole of NaOH= 1240 =0.3 so molarity = 0.31.8 = .pdf

millimoles of HBr=500.04=2 millimoles of KOH=25.pdf

initially 2 moles of HCl per L of water suppose x.pdf

i cant see the question.. please post the complet.pdf

hey pls give the reactions.... .pdf

FT-Raman spectra of some calcium phosphates, calc.pdf

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Current Stock price = Expected dividend (Required rate of return -.pdf

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Well its quite an easy one. The given curve is a cardioid. The limit.pdf

Total ionic Equation is 2Cu2+ +4 Cl - + 4H+ +SO42- --- Cu2SiO4+ 4.pdf

AnswerSOCS are the supressor of cytokine signaling proteins and a.pdf

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Addition of NH3(aq) produced a white gelatinous precipitate in a col.pdf

1. Addition of NH3(aq) produced a white gelatinous precipitate in a colorless solution. note: The bismuth III ions are separated by precipitation as white gelatinous bismuth (III) hydroxide upon the addition of an aqueous solution of ammonia. Bi definitely present. The addition of the ammonia water to precipitate the bismuth hydroxide also converts any copper(II) ion present to the deep blue complex ion, Cu(NH3)4^2+. since the solution is colorless, Cu definitely absent. When solution (Z) was treated with K3Fe(CN)6, a dark blue suspension was produced. KFe[Fe(CN)6] this indicates Fe3+ definitely present Co2+,Ni2+ cannot decide if it exists or not. Solution Addition of NH3(aq) produced a white gelatinous precipitate in a colorless solution. note: The bismuth III ions are separated by precipitation as white gelatinous bismuth (III) hydroxide upon the addition of an aqueous solution of ammonia. Bi definitely present. The addition of the ammonia water to precipitate the bismuth hydroxide also converts any copper(II) ion present to the deep blue complex ion, Cu(NH3)4^2+. since the solution is colorless, Cu definitely absent. When solution (Z) was treated with K3Fe(CN)6, a dark blue suspension was produced. KFe[Fe(CN)6] this indicates Fe3+ definitely present Co2+,Ni2+ cannot decide if it exists or not.

Addition of NH3(aq) produced a white gelatinous precipitate in a col.pdf

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