Bronsted acididy H2Se H2S note Down the group.pdf

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Bronsted acididy: H2Se > H2S note: Down the group, the hydride become more acidic. Lewis acidity: AlH3 > BH3 Solution Bronsted acididy: H2Se > H2S note: Down the group, the hydride become more acidic. Lewis acidity: AlH3 > BH3.

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conjugate base: substance formed when an acid loses a hydrogen ion Conjugate base of HSO4- is SO42- conjugate acid: substance formed when a base gains a hydrogen ion. Conjugate acid of HSO4- is H2SO4 Solution conjugate base: substance formed when an acid loses a hydrogen ion Conjugate base of HSO4- is SO42- conjugate acid: substance formed when a base gains a hydrogen ion. Conjugate acid of HSO4- is H2SO4.

conjugate base substance formed when an acid loses a hydrogen ion.pdf

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cash flow financing activities includes proceeds from selling stock.Financing activities are the activities through which capital are raised and pays it back to investors through capital invested by them. Solution cash flow financing activities includes proceeds from selling stock.Financing activities are the activities through which capital are raised and pays it back to investors through capital invested by them..

cash flow financing activities includes proceeds from selling stock..pdf

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C) interference with biotransformation enzyme activites Reason: The size of the nanoparticle is always in nanoscale 10um to 100nm in which active drug molecules are coated on the neutral chanrged cyclodextrins such as hydroxyl propyl methyl cellulose, methylcellulose in the form of liposomoes. These liposomes can have higher affinity towards the cell membrane negative change and allow the drug molecules to reach the active target regions. These liposomes (acts as surfactant i.e. both hydrophilic & lipophillic properties) can release the drug in controlled level of sustained novel drug delivery systems. Liver biotransformation enzymes are mainly involved in detoxification of foreign materials which are incompatable to the host body therefore, when nanoparticles are going to interact with live CYP enzymes finally reduce toxicity of prodrug or liposome based -nanoparticle drug. Nanoparticles are going to induce toxicity by producing reactive oxygen species (ROS) finally result in oxidative stress due to activation of signaling pathways to cause damage to DNA 1. Pharmaceutical nanoparticle drug delivery in breast cancer chemotherapies include the making the anticancer drug molecules into the “liposome coated” to make efficient drug delivery with efficient bioavailability associated with higher penetration rate into the breast cancer cells. 2. For example “making microRNAs to treat the breast cancer” into the liposome formulation as they can form micelle composed of polymers including methylcellulose, hydroxyl propyl methylcellulose. These polymers have the capability to incorporate the drug moieties into the cancer cells in naoscale. Solution C) interference with biotransformation enzyme activites Reason: The size of the nanoparticle is always in nanoscale 10um to 100nm in which active drug molecules are coated on the neutral chanrged cyclodextrins such as hydroxyl propyl methyl cellulose, methylcellulose in the form of liposomoes. These liposomes can have higher affinity towards the cell membrane negative change and allow the drug molecules to reach the active target regions. These liposomes (acts as surfactant i.e. both hydrophilic & lipophillic properties) can release the drug in controlled level of sustained novel drug delivery systems. Liver biotransformation enzymes are mainly involved in detoxification of foreign materials which are incompatable to the host body therefore, when nanoparticles are going to interact with live CYP enzymes finally reduce toxicity of prodrug or liposome based -nanoparticle drug. Nanoparticles are going to induce toxicity by producing reactive oxygen species (ROS) finally result in oxidative stress due to activation of signaling pathways to cause damage to DNA 1. Pharmaceutical nanoparticle drug delivery in breast cancer chemotherapies include the making the anticancer drug molecules into the “liposome coated” to make efficient drug delivery with efficient bioavailability associated with higher penetration rate i.

C) interference with biotransformation enzyme activitesReason.pdf

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//c++ program to operate on binary search tree #include #include #include int tree[20],t=1,x,i,s; void insert(int,int ); void delte(int); void display(int); int search(int); int search1(int,int); int main() { int ch,y; for(i=1;i<20;i++) tree[i]=-1; while(1) { cout <<\"1.insert\ 2.delete\ 3.display\ 4.search\ 5.exit\ Enter your choice:\"; cin >> ch; switch(ch) { case 1: cout <<\"enter the element to insert\"; cin >> ch; insert(1,ch); break; case 2: cout <<\"enter the element to delete\"; cin >>x; y=search(1); if(y!=-1) delete(y); else cout<<\"no such element in tree\"; break; case 3: display(1); cout<<\"\ \"; for(int i=0;i<=32;i++) cout <> x; y=search(1); if(y == -1) cout <<\"no element found in tree\"; else cout <ch) tree[2*x]=ch; else tree[2*x+1]=ch; t++; } void delete(int x) { if( tree[2*x]==-1 && tree[2*x+1]==-1) { { tree[x]=-1; } else if(tree[2*x]==-1) { tree[x]=tree[2*x+1]; tree[2*x+1]=-1; } else if(tree[2*x+1]==-1) { tree[x]=tree[2*x]; tree[2*x]=-1; } else { tree[x]=tree[2*x]; delete(2*x); } t--; } int search(int n) { if(t==1) { cout <<\"no element in tree\"; return -1; } if(tree[n]==-1) { return tree[n]; } if(tree[n]>x) { search(2*n); } else if(tree[n] ch) { search1(2*n,ch); } else { search1(2*n+1,ch); } } Solution //c++ program to operate on binary search tree #include #include #include int tree[20],t=1,x,i,s; void insert(int,int ); void delte(int); void display(int); int search(int); int search1(int,int); int main() { int ch,y; for(i=1;i<20;i++) tree[i]=-1; while(1) { cout <<\"1.insert\ 2.delete\ 3.display\ 4.search\ 5.exit\ Enter your choice:\"; cin >> ch; switch(ch) { case 1: cout <<\"enter the element to insert\"; cin >> ch; insert(1,ch); break; case 2: cout <<\"enter the element to delete\"; cin >>x; y=search(1); if(y!=-1) delete(y); else cout<<\"no such element in tree\"; break; case 3: display(1); cout<<\"\ \"; for(int i=0;i<=32;i++) cout <> x; y=search(1); if(y == -1) cout <<\"no element found in tree\"; else cout <ch) tree[2*x]=ch; else tree[2*x+1]=ch; t++; } void delete(int x) { if( tree[2*x]==-1 && tree[2*x+1]==-1) { { tree[x]=-1; } else if(tree[2*x]==-1) { tree[x]=tree[2*x+1]; tree[2*x+1]=-1; } else if(tree[2*x+1]==-1) { tree[x]=tree[2*x]; tree[2*x]=-1; } else { tree[x]=tree[2*x]; delete(2*x); } t--; } int search(int n) { if(t==1) { cout <<\"no element in tree\"; return -1; } if(tree[n]==-1) { return tree[n]; } if(tree[n]>x) { search(2*n); } else if(tree[n] ch) { search1(2*n,ch); } else { search1(2*n+1,ch); } }.

c++ program to operate on binary search tree#includeiostream.pdf

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As per section 10.29(a) of Treasury Circular 230 a conflict of interest exists if – (i) representation of one party by the tax preparer will affect any other party adversely (ii) the tax preparer’s responsibility to another client will materially limit representation of one or more parties. However in case of conflict of interest a tax preparer can represent a party or a client provided that each affected client or party waives the conflict and gives and informed and written consent to the representation. Law requires the tax preparer to obtain the written consent within a reasonable period of time but it should be no longer than 30 days after the conflict has come to the knowledge of the tax preparer. Solution As per section 10.29(a) of Treasury Circular 230 a conflict of interest exists if – (i) representation of one party by the tax preparer will affect any other party adversely (ii) the tax preparer’s responsibility to another client will materially limit representation of one or more parties. However in case of conflict of interest a tax preparer can represent a party or a client provided that each affected client or party waives the conflict and gives and informed and written consent to the representation. Law requires the tax preparer to obtain the written consent within a reasonable period of time but it should be no longer than 30 days after the conflict has come to the knowledge of the tax preparer..

As per section 10.29(a) of Treasury Circular 230 a conflict of inter.pdf

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Answer Generally chromosome reduction occurs during formation of gametes. The gametes are haploid as they are produced from diploid parent by meiosis. If the reduction doesn\'t occur, the ploidy of organism that is produced by sexually reproduction will be increased. The ploidy may be come tetraploid or hexaploid or polyploid. These polyploid organisms are mostly sterile. Ex. The gametes produced from diploid parents are also diploid due to the absence of reduction. If the two diploid gametes fuse, it develops into tetraploid organism. It may or may not alive. Even if it is alive, the tetraploid organism will be sterile. Solution Answer Generally chromosome reduction occurs during formation of gametes. The gametes are haploid as they are produced from diploid parent by meiosis. If the reduction doesn\'t occur, the ploidy of organism that is produced by sexually reproduction will be increased. The ploidy may be come tetraploid or hexaploid or polyploid. These polyploid organisms are mostly sterile. Ex. The gametes produced from diploid parents are also diploid due to the absence of reduction. If the two diploid gametes fuse, it develops into tetraploid organism. It may or may not alive. Even if it is alive, the tetraploid organism will be sterile..

AnswerGenerally chromosome reduction occurs during formation of ga.pdf

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AnsFalseExplantionIn a communication diagram, a(n) sequence .pdf

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a. less severe immunophenotypeb. strong growth increase of strain..pdf

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a) They offer defined benefit retirement pension plans to certain employees. These benefits relate primarily to local plans outside the U.S. and, to a lesser extent, plans assumed in previous acquisitions covering U.S. employees. They also provide certain other retiree benefits, primarily health care and life insurance, for the majority of our U.S. employees who become eligible for these benefits when they meet minimum age and service requirements. Generally, the health care plans require cost sharing with retirees and pay a stated percentage of expenses, reduced by deductibles and other coverages. These benefits are primarily funded by ESOP Series B shares and certain other assets contributed by the Company. b) Total global defined contribution expense was $347, $344 and $364 in 2011, 2010 and 2009, respectively.The primary U.S. defined contribution plan comprises the majority of the balances and expense for the Company’s defined contribution plans. For the U.S. DC plan, the contribution rate is set annually. Total contributions for this plan approximated 15% of total participants’ annual wages and salaries in 2011, 2010 and 2009. c) Their investment objective for defined benefit retirement plan assets is to meet the plan\'s benefit obligations, while minimizing the potential for future required Company plan contributions. The investment strategies focus on asset class diversification, liquidity to meet benefit payments and an appropriate balance of long-term investment return and risk. Target ranges for asset allocations are determined by matching the actuarial projections of the plan\'s future liabilities and benefit payments with expected long-term rates of return on the assets, taking into account investment return volatility and correlations across asset classes. Plan assets are diversified across several investment managers and are generally invested in liquid funds that are selected to track broad market equity and bond indices. Investment risk is carefully controlled with plan assets rebalanced to target allocations on a periodic basis and continual monitoring of investment manager\'s performance relative to the investment guidelines established with each investment manager. Several factors are considered in developing the estimate for the longterm expected rate of return on plan assets. For the defi ned benefi t retirement plans, these factors include historical rates of return of broad equity and bond indices and projected long-term rates of return obtained from pension investment consultants. The expected longterm rates of return for plan assets are 8–9% for equities and 5–6% for bonds. For other retiree benefit plans, the expected long-term rate of return reflects the fact that the assets are comprised primarily of Company stock. The expected rate of return on Company stock is based on the long-term projected return of 9.5% and reflects the historical pattern of favorable returns. Solution a) They offer defined benefit retirement.

a)They offer defined benefit retirement pension plans to certain e.pdf

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61nSolution61n.pdf

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29Solution29.pdf

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Not at all. Most coloured salts contain transition metals or heavy p-block elements, but even the largest alkali metal ions give colourless salts. It depends on electron distribution in the ions, not how large they are. Solution Not at all. Most coloured salts contain transition metals or heavy p-block elements, but even the largest alkali metal ions give colourless salts. It depends on electron distribution in the ions, not how large they are..

Not at all. Most coloured salts contain transitio.pdf

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Te2- Sol.pdf

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2. A -> Anova for complete block design can be used here. 3. A is correct. 7. A For Brand A, B and C, normality condition is completely met. For brand D, there is a slight deviation in Normal plot. 8. A is fine. 12. A Mean life is significantly greater as mean life of B falls outside the CI range of brand A. Solution 2. A -> Anova for complete block design can be used here. 3. A is correct. 7. A For Brand A, B and C, normality condition is completely met. For brand D, there is a slight deviation in Normal plot. 8. A is fine. 12. A Mean life is significantly greater as mean life of B falls outside the CI range of brand A..

2. A- Anova for complete block design can be used here.3. A.pdf

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18)Research has found that Hg+ and Cu+ block the alkaline phosphat.pdf

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1.85Solution1.85.pdf

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1) Strategies and structures: In Protozoans the method of movement is determined by the type of organism and the surrounding environment. Protozoans mainly move by cell extension, flagella or pseudopodia and cilia, the movement as per the presence of structure can be classified as ciliary, flagellar and amoeboid movement. Ciliates : Ciliates form the largest group of protozoa. These organisms vary in size and often live in watery environments, including oceans, marshes, bays and streams. Ciliates move using tiny cilia, which are hair-like strands that act as sensors and tiny limbs. Flagella are longer and less numerous that cilia, they use their long tail like flagella to move. Amoebas : In these two cytoskeleton get polymerized . This creates a vacancy and cytoplasmice material flow to cover the vacancy created. When amoeba moves cytoplasm moves to the arm like extension called pseudopodium. This pseudopodium extends and enlarge and hence this push the animal body towards that respective direction. 2) A) Flagellates can live as single cells, in colonies, or as parasites. Commonly live in niche\'s of water. They conduct photosynthesis and have a cell wall. They contain flagella for propulsion or to create a current to bring in food. They can inhabit the reproductive tract, alimentary canal, tissue sites and also the blood stream, lymph vessels and cerebrospinal canal. B) Pseudopods : Also called as false feet , are projections that can appear and disappear from the organism\'s body. These are used for movement and to engulf prey and digest them using enzymes. C) Apicomplexa : Unicellular and spore forming, most of them possess a unique form of organelle that comprises a type of plastid called an apicoplast, and an apical complex structure. They have apicoplast(non photosynthetic plastid) , mitochondria and nuclear genomes. Lack of cilia, sexual reproduction, use micropores for feeding, and the production of oocysts containing sporozoites as the infective form. They have unique gliding capability which enables them to cross through tissues and enter and leave their host cells. This gliding ability is made possible by the use of adhesions and small static myosin motors. 3) Key characteristics of fungi : Fungi are unicellular or multicellular. Most of the fungi grow as tubular filaments called hyphae They are haploid. Fungus are heterotrophs (they can obtain nutrients by absorption) . They absorb food and secrete enzymes to digest complex molecules Propogate by spores Asexual or sexual reproduction They can be multinucleated Fungi are achlorophyllous (lack of cholorphyll pigment) Both Fungi and protists belong to same kingdom but fungi is different from protist, protists are able to live in an anaerobic environment without oxygen but fungi need aerobic respiration to survive. Protists are unicellular but fungi are multicellular. Protists are autotrophic (make their own energy) and heterotrophic (rely on outside source to get energy), but fungi a.

1) Strategies and structuresIn Protozoans the method of movement .pdf

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1). Purple bacteria are the photosynthetic bacteria that contains the bacteriochlorophyll a or b pigments. The presence of these bacteria makes them to appear in different colors ranging from purple to orange. These bacteria are of two types namely, purple sulfur bacteria and purple non- sulfur bacteria. Metabolically, purple sulfur bacteria are anoxygenic (does not produce oxygen) because they don’t use water as reducing agent during the photosynthesis. Instead, they use elemental sulfur or sulfide as reducing agents and convert them into less toxic forms. In case of purple non-sulfur bacteria, they typically use hydrogen ions as other compounds in place of water. The metal reducing microorganisms use the oxidation-reduction reaction to reduce the metal compounds and convert them into less toxic forms. This also includes the nitrogen cycle. The toxic metals are converted into the polar compounds, which are immobilized and accumulate in the food web. For example, the nitrogen fixation, which encourages plat growth. Solution 1). Purple bacteria are the photosynthetic bacteria that contains the bacteriochlorophyll a or b pigments. The presence of these bacteria makes them to appear in different colors ranging from purple to orange. These bacteria are of two types namely, purple sulfur bacteria and purple non- sulfur bacteria. Metabolically, purple sulfur bacteria are anoxygenic (does not produce oxygen) because they don’t use water as reducing agent during the photosynthesis. Instead, they use elemental sulfur or sulfide as reducing agents and convert them into less toxic forms. In case of purple non-sulfur bacteria, they typically use hydrogen ions as other compounds in place of water. The metal reducing microorganisms use the oxidation-reduction reaction to reduce the metal compounds and convert them into less toxic forms. This also includes the nitrogen cycle. The toxic metals are converted into the polar compounds, which are immobilized and accumulate in the food web. For example, the nitrogen fixation, which encourages plat growth..

1). Purple bacteria are the photosynthetic bacteria that contains th.pdf

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Think about intermolecular theory the stronger th.pdf

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The way you would solve this problem is to look at the polarity of the molecules-- which has to deal with the electronegativity of the atoms themselves, and the atomic size and mass. Polarity affects intermolecular forces and causes molecules and atoms to deviate from the ideal behavior. H2 would be expected to behave most ideally because it has a dipole moment of 0 (non-polar). Also, it has smaller mass, it is not as easily polarizable. F2 may behave ideally, because it has a dipole moment of 0 also. But, it\'s mass is bigger and it is highly electronegative HF would be expected to deviate because it has a higher dipole moment, and it\'s polar. That dipole moment creates an intermolecular force that influecnes other polar molecules. Source(s): General Chemistry 4th edition (Hill, Petrucci, McCreary, Perry) Solution The way you would solve this problem is to look at the polarity of the molecules-- which has to deal with the electronegativity of the atoms themselves, and the atomic size and mass. Polarity affects intermolecular forces and causes molecules and atoms to deviate from the ideal behavior. H2 would be expected to behave most ideally because it has a dipole moment of 0 (non-polar). Also, it has smaller mass, it is not as easily polarizable. F2 may behave ideally, because it has a dipole moment of 0 also. But, it\'s mass is bigger and it is highly electronegative HF would be expected to deviate because it has a higher dipole moment, and it\'s polar. That dipole moment creates an intermolecular force that influecnes other polar molecules. Source(s): General Chemistry 4th edition (Hill, Petrucci, McCreary, Perry).

The way you would solve this problem is to look a.pdf

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