6- Microorganisms called denitrifying bacteria that grow in waterlogge (1).docx

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6. Microorganisms called denitrifying bacteria that grow in waterlogged soil convert NOs ions into N2O gas as they feed on dead plant tissue (empirical formula CH:O), converting it to carbon dioxide CO and water H:O. Balance the redox reaction in slightly acidic medium that describes this conversion: Solution NO3^- (aq) + CH2O(s) ----------> N2O(g) + CO2(g) + H2O(l) oxidation half reaction Reduction half reaction CH2O(s) ------------> CO2(g) NO3^- (aq) ------------> N2O(g) CH2O(s) +H2O------------> CO2(g) 2NO3^- (aq) ------------> N2O(g)+5H2O CH2O(s) +H2O------------> CO2(g) 2NO3^- (aq) ------------> N2O(g)+5H2O CH2O(s) +H2O------------> CO2(g)+4H^+(aq) 2NO3^- (aq)+10H^+ (aq) --------> N2O(g)+5H2O CH2O(s) +H2O------------> CO2(g)+4H^+(aq)+4e^- 2NO3^- (aq)+10H^+ (aq) +8e^- --------> N2O(g)+5H2O 2CH2O(s) +2H2O------------> 2CO2(g)+8H^+(aq)+8e^- 2NO3^- (aq)+10H^+ (aq) +8e^- --------> N2O(g)+5H2O ----------------------------------------------------------------------------- 2CH2O9s) + 2NO3^-(aq) + 2H^+ (aq) -----------> 2Co2(g) + N2O(g) + 3H2o(l) .

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Based on the information in “Atom: Clash of the Titans”, please answer the following questions: 1. Why was it so important that the scientists investigating the atom be young and unattached to existing scientific thinking? 2. What connection did Bohr discover between atoms and light? 3. Why did Einstein hate Bohr’s ideas - quantum mechanics, electrons jumping and releasing light - about the atom? Solution 1) The scientists investigating the atom should be young as was then the young scientist Albert Einstein. His harsh arrogance led him barely any employment, thereby because of his undemanding job he had a lot of time to have deep thoughts. And that was the era wherin he worked on to prove the existence of atom. Though it was Boltzman who started work on this topic but due to his health conditions and alegations on him being irrelegious materialist lead to him comitting suicide. And the scientists should be unattached to existing scientific thinking as was scientist Neil Bohr who was fearless in his thinking and was ready to abandon common sense and human intution to find an explanation. 2)Bohr\'s connection between atoms and light was as follows-quantum jumps Light is emitted when an electron jumps from a higher orbit to a lower orbit and is absorbed when jumps from lower to higher orbit. In order to overcome the limitations of Bohr model and to develop a general model for atoms.The 2 developments contibuted a lot - 3)As the bohr idea of new atom and quantum jumps were not so traditional neither they matched classical science theories thus the old school people strongly opposed it. And Albert Einstein was one of them as he wanted to save the world of order and wanted to preserve the common sense. .

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6.3. (a) In which form does Cd(II) occur mainly in a water of the following composition: pH = 7.8 Solution (a) pH=7.8=-log [H+] [H+]=10^-7.8=1.585*10^-8M Also, [H+][OH-]=1*10^-14=Ionic product of water [OH-]=(1*10^-14)/[H+]=(1*10^-14)/(1.585*10^-8)=6.309*10^-7M Alkalinity=[OH-]+[CO3 2- ]=1.3*10^-3 eq/L [CO32-]=(1.3*10^-3 eq/L)-[OH-]=(1.3*10^-3 eq/L)-(6.309*10^-7M)=(1.3*10^-3 eq/L)=1/2*(1.3*10^-3 eq/L)=0.65*10^-3mol/L [For [OH-]eq=[OH]M ,but for Carbonate, 2eq=1mol] Thus,[CO32-]>>>[OH-] Also,Cd2+ +CO32- <--->CdCO3 ,log K=4.5,K=10^-4.5 But,Ca2+ +CO32- <--->CdCO3,log K=8.5 So,CdCO3 will precipitate first and CaCO3 later, Also, Cd T =1*10^-9 M<<<Ca T =1*10^-3 M And Cd 2+ +2OH- <--->Cd(OH)2 ,log K=14.3 The value of log K is much higher, so Cd(II) must be predominant as Cd(CO3) b) yes, CdCO3 precipitates due to very high equilibrium constant for this reaction compared to that for Cd(OH)2 c)[EDTA]=10^-7 mol/L Cd 2+ +EDTA 4- <--->CdEDTA 2- , log K=16.5,K=10^-16.5 K=[CdEDTA 2- ]/[Cd 2+ ][EDTA 4- ] or,[CdEDTA 2- ]=K*[Cd 2+ ][EDTA 4- ]=10^-16.5 *(10^-7)*[Cd2+]=10^-23.5 [Cd2+] But ,Cd 2+ +CO3 2- <--->CdCO3 ,log K=4.5,K=10^-4.5 Similarly,[CdCO3]=10^-4.5 *(6.309*10^-7)*[Cd2+]=(6.309*10^-11.5) [Cd2+] where., [OH-]=6.309*10^-7M Hence ,[CdEDTA 2- ]<[CdCO3] CdCO3 remains the predominant form for Cd(II) .

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6. A solution of an acid, with a general formula HA, contains 0.400 moles of H\"\'. The acid solution is neutralized by 43.4 mL of sodium hydroxide. What is the molarity of the sodium hydroxide solution? 7. Write the balanced chemical equations for the removal of water from the following hydrated crystals. a) Copper(Il) sulfate pentahydrate b) Tin(II) chloride dehydrate c) Cobalt(II) chloride hexahydrate d) Magnesium sulfate heptahydrate oint of solution made hy dissolving 6.000 q of urea Solution 6) HA + NaOH ------------------> NaA + H2O moles of HA = moles of NaOH = 0.400 moles = 0.400 volume = 43.4 mL = 0.0434 L molarity of NaOH = moles / volume = 0.400 / 0.0434 = 9.21 M 7) a) CuSO4 .5H2O ------------------> CuSO4 + 5 H2O b) SnCl2 H2O ----------------------> SnCl2 + H2O c) CoCl2.6H2O --------------------> CoCl2 + 6H2O d) MgSO4 7H2O -------------------> MgSO4 + 7H2O .

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6. What are the differences between objects and abstract data types Solution Differences between Objects and Abstract data types: ______________________________________________ Objects: Object is an instance of a class and it has states and behaviours of class. Abstract data type: It is a data type of abstract and it hides the representation of class, and it is typically implemented as a class and each instance of Abstract data type is usually an object of that class. Objects and Abstract data types are not the same. But these are not different with each other. Objects are a kind of Abstract data types. An Abstract data type is an interface.But objects are instance of a class, with its own copy of any non-static variables. Abstract data type is a collection of methods but in case of objects it is not an interface.It is an representation of a class with new operator. A class can have one or more Abstract data types in the same way class can have one or more objects also. To understand you with clear for Example, Collection interface is Abstract data type, ArrayList is a class and you can make ArrayList object with new operator. .

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6. What do the bubbles in boiling water principally c a. air b. O2 (e and H2e c. H20 d. vacuum e. nitrogen What is the correct name for Co3(PO4)4? a. cobalt phosphate b. tricobalt tetraphosphate c. cobalt (II) phosphate d. cobalt (III) phosphate e. cobalt (IV) phosphate 7. 8. How many neutrons are contained in *He? a. 0 b. 1 c. 2 d. 3 e. 4 9. How many electrons are contained in He 2 2 a. d. 3 e. 4 Solution 6) the bubbles in boiling water contains water vapours, hence H2O(g), as the vapour pressur eof li 7) The name for anion is phosphate , there are four phosphate groups The oxidation state of cobalt is 3x + 4(-3) = 0 x = +4 So the correct name is Cobal(IV)phosphate 8) neutrons = mass number - atomic number = 4-2 = 2 9) Electrons = 0 as the number of electrons in Helium in its ground state = 2 The charge is +2, which represents removal of two electrons .

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6. Both solutions and colloids are mixtures. Differentiate between them. Solution Although solutions and colloid both are mixtures but we can differentiate solution and colloid on the basis of some parameter given below: 1) Particle size: Particle size of solution is less than 1nm or (10 -9 ) m Particles size of colloid are between 1nm to 1000 nm 2) Visibility In solutions the visibility of substances dissolved in it are not possible even with optical means. While in colloid it may be visible with ultra microscope. 3) Diffusion Diffusion is very rapid in solution while very slow in colloid 4) setting Solution doesn\'t settle while colloid may settle down under centrifuge . 5) Nature: Solution is Homogeneous in nature while colloid is heterogeneous in nature. 6) Tyndall effect Solution doesn\'t show Tyndall effect Colloid shows Tyndall effect So we can say that solution and colloid both are mixture but there are a lot of difference between them. .

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6). The Town of Elmwood applied for a competitive grant from the state government for providing an after- school recreational program for at-risk children from low-income families. On 2/1/2005, Elmwood was notified that it had been awarded a grant of $100,000 for the program. The program provides up to $1,000 per child served provided that family annual income is less than $25,000. The grant provides for reimbursement of all allowable costs of conducting the program on a twice-monthly basis. No time limit is imposed for completion of the program. Any unused grant award may be carried forward to the next fiscal year. By 12/31/2005, Elmwood?s fiscal year-end, $74,000 had been expended in providing recreational services for eligible beneficiaries. Reimbursement of $71,000 had been received from the state by this date. Required: [12 pts.] a). What are the eligibility requirements of this grant? b). As of 12/31/2005, identify the following: (1). Cash Received (2). Program Revenue (3). Expenditures Solution Total Expenditure = $ 74000 Cash Received = $ 71000 Revenue = 71000-74000 = $ 3000 The program must be carried on & also the expenditure will be for the children whose family income is less than $ 25000/-. .

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