This is a bit complex to answer as we have HCl and NaOH present, the.pdf

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This is a bit complex to answer as we have HCl and NaOH present, they will undergo neutralization reaction to form NaCl and water. so inevitably water is also in system. acetyl chloride and water, they will react to form HCl and Acetic acid which is will react with NaOH to form sodium acetate. Considering all these cases. I think the answer should be benzene in CH2Cl2 layer, as benzene only acetyl chloride in NaOH layer as CH3COO-Na+ AlCl3 will be NaOH layer in the form of AlCl4- ion, reason being NaOH and HCl have reacted to give Na+ and Cl- ions and AlCl3 in presence of chloride ions gives AlCl4- HCl will be in NaOH in form of NaCl ... dissociated into Na+ and Cl- ions ultimately Solution This is a bit complex to answer as we have HCl and NaOH present, they will undergo neutralization reaction to form NaCl and water. so inevitably water is also in system. acetyl chloride and water, they will react to form HCl and Acetic acid which is will react with NaOH to form sodium acetate. Considering all these cases. I think the answer should be benzene in CH2Cl2 layer, as benzene only acetyl chloride in NaOH layer as CH3COO-Na+ AlCl3 will be NaOH layer in the form of AlCl4- ion, reason being NaOH and HCl have reacted to give Na+ and Cl- ions and AlCl3 in presence of chloride ions gives AlCl4- HCl will be in NaOH in form of NaCl ... dissociated into Na+ and Cl- ions ultimately.

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You must be familiar with the fact that melting point is essentially the same thing as freezing point. What is meant by this is that the same temperature will melt something (if we\'re considering an endothermic reaction) as it will freeze something (exothermic reaction). After you\'ve reached this clarity, simply observe all of the compounds and check for which ones have low melting points and then work your way up. Now the question is, how can we identify compounds with low melting points? Remember, if a compound is polar (charged, or is a dipole, like water), it will have a high boiling point (it will be very tough to boil). So compounds that have a high boiling point will most likely have high melting points as well (generally, not universally). For NaBr, Na+ and Br- and the constituents and they certainly have an electronegativity difference (uneven charge distribution in the electron density implies a dipole). We have to look at calcium nitrate next before determining the priority of NaBr. Ca(NO3)2 is obviously going to have charged constituents. The Ca will be positively charged and the nitrate groups which are oxyANIONs will be negatively charged. Since there are MORE oxygens in calcium nitrate, this compound will be more electronegative and will form a stronger dipole compared to NaBr (Br is NOT as electronegative as oxygen). Greater electronegativity implies stronger dipole in a compound and this means higher boiling point, which means higher melting point, which means higher freezing point compared to NaBr. HCl is an acid and although there are H atoms, HCl CANNOT form Hydrogen bonds (h-bonding occurs only for N-H, O-H, and F- H. Cl is obviously NOT one of the constituents that can form hydrogen bonding and therefore will have a LOW boiling point, meaning low freezing point as well. HCl falls at the very bottom of our list. Ethanol is polar (because of the free electrons from O--H from the sp3 orbital). Hydrogen bonding IMMEDIATELY implies high boiling point, so EtOH will have a relatively large melting point, and thus freezing point. Now that I\'ve laid out the information as thoroughly as possible. Simply arrange them to form your freezing point trend list given the four compounds. Should be rather straightforward. Solution You must be familiar with the fact that melting point is essentially the same thing as freezing point. What is meant by this is that the same temperature will melt something (if we\'re considering an endothermic reaction) as it will freeze something (exothermic reaction). After you\'ve reached this clarity, simply observe all of the compounds and check for which ones have low melting points and then work your way up. Now the question is, how can we identify compounds with low melting points? Remember, if a compound is polar (charged, or is a dipole, like water), it will have a high boiling point (it will be very tough to boil). So compounds that have a high boiling point will most likely have high melting poin.

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It is a molecular process involving the exchange of bonds between the two reacting chemical species, which results in the creation of products with similar or identical bonding affiliations. This is represented by the general reaction scheme: AX + BY ? BX + AY These chemical species can either be ionic or covalent. When referring to precipitation reactions between solutions of ions in inorganic chemistry, these were formerly referred to as double displacement or double replacement reactions, and these terms are still encouraged. It seems that the term double decomposition is more specifically used when at least one of the substances does not dissolve in the liquor, as the ligand or ion exchange takes place in the solid state of the reactant, i.e. AX(aq) + BY(s) ? AY(aq) + BX(s). Types: Neutralization A neutralization reaction is a specific type of double displacement reaction. Neutralization occurs when an acid reacts with an equal amount of a base. A neutralization reaction creates a solution of a salt and water. For example, hydrochloric acid reacts with sodium hydroxide to produce salt and water: HCl (aq) + NaOH (aq) ? NaCl (aq) + H2O (l) Aqueous metathesis (precipitation) Metathesis reactions can occur between two inorganic salts when one product is insoluble in water, driving the reaction forward. For example, the precipitation of silver chloride from a mixture of silver nitrate and sodium chloride causes sodium nitrate to be left in solution: AgNO3 (aq) + NaCl (aq) ? AgCl (s) + NaNO3 (aq) The formation of an insoluble gas that bubbles out of the solution, or a molecular compound such as water, also drives the reaction to completion. Therefore, a solubility chart (or general knowledge of solubility rules) can be used to predict whether two aqueous solutions will react. HSAB theory can also be used to predict the products of a metathesis reaction. Aqueous metathesis (double decomposition) The reactants need not to be dissolved for metathesis reactions to take place. An example of this is the formation of barium thiocyanate when boiling a mixture of copper(I)thiocyanate and barium hydroxide in water: Ba(OH)2 (s) + 2 CuCNS (s) ? Ba(CNS)2 (aq) + 2CuOH (s) Acid and carbonates A subcategory of aqueous metathesis reactions is the reaction of an acid with a carbonate or bicarbonate. Such a reaction always yields carbonic acid as a product, which spontaneously decomposes into carbon dioxide and water. The release of carbon dioxide gas from the reaction mixture drives the reaction to completion. For example, a common, science-fair \"volcano\" reaction involves the reaction of acetic acid with sodium bicarbonate: HCH3COO (aq) + NaHCO3 (s) ? NaCH3COO (aq) + CO2 (g) + H2O (l) Solution It is a molecular process involving the exchange of bonds between the two reacting chemical species, which results in the creation of products with similar or identical bonding affiliations. This is represented by the general reaction scheme: AX + BY ? BX + AY These che.

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When recovering an acid from its salt dissolved in water, we typically add HCl until the pH of the solution is strongly acidic (pH ~ 2). Why donâ€™t we stop when the pH is neutral? When recovering basic compounds from their aqueous salts we again skip neutral and adjust the pH until the solution is strongly basic. Why? (hint: Henderson-Hasselbalch) Solution The extraction is done with bicarbonate first and then with NaOH. But if you are doing it in reverse order, then extraction will not be proper. The three compounds mentions in the text are 2-methoxy naphthalene, 2-naphthol and naphthoic acid. Naphthol is a weak acid, naphthoic acid is strong acidic and naphthalene is neutral. NaOH is a strong base and bicarbonate is a weak base. Naphthol being a weak acid will only solubilise in strong base. Naphthoic acid is a strong acid, so it will be soluble in both strong and weak base, i.e. NaOH and NaHCO3. If you add aqueous NaOH solution to the mixture, then both naphthoic acid and naphthol will go in water layer and the naphthalene will go in organic layer, which is ether here. If you separate these layers and add NaHCO3 solution to the ether layer, you will not be left with anything which will be soluble in NaHCO3 solution. So, extraction will be difficult. On the other side, If you add NaHCO3 initially to the mixture of compounds, then naphthoic acid will go in aqueous layer and naphthalene will go in ether layer. Then when NaOH will be added, naphthol will go in this layer. Thus, all the compounds are in different layers, and easy to recover. If concentrated HCl is used instead of dilute HCl, even if the pH is kept same to recover the compounds, then there are more chances of side reactions during recovery, which is undesirable. And by adding concentrated HCl, there is increased evolution of heat. When NaHCO3 is neutralized with HCl, CO2 is evolved. If HCl is concentrated, then there will be lot of CO2 evolved and we have to handle the system with a lot of care during the recovery. These problems are not there while using dilute HCl. .

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Here is the code for youimport java.util.Scanner; import java.u.pdf

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Following are the changes mentioned in bold in order to obtain the required result and stop scrolling in the background. #include \"SDL/SDL.h\" #include //The attributes of the screen can be defined as follows const int SCN_WIDTH = 640; const int SCN_HEIGHT = 480; const int SCN_BPP = 32; //BPP defines bits per pixel SDL_Surface* Background = NULL; SDL_Surface* SpriteImage = NULL; SDL_Surface* Backbuffer = NULL; int SpriteFrame = 0; int FrameCounter = 0; const int MaxSpriteFrame = 12; const int FrameDelay = 2; int BackgroundX = 0; SDL_Surface* LoadImage(char* fileName); bool LoadFiles(); void FreeFiles(); void DrawImage(SDL_Surface* image, SDL_Surface* destSurface, int x, int y); void DrawImageFrame(SDL_Surface* image, SDL_Surface* destSurface, int x, int y, int width, int height, int frame); bool ProgramIsRunning(); int main(int argc, char* args[]) { if(SDL_Init(SDL_INIT_EVERYTHING) < 0) { printf(\"Failed to initialize SDL!\ \"); return 0; } Backbuffer = SDL_SetVideoMode(800, 600, 32, SDL_SWSURFACE); SDL_WM_SetCaption(\"Image Animation\", NULL); if(!LoadFiles()) { printf(\"Failed to load all files!\ \"); FreeFiles(); SDL_Quit(); return 0; } while(ProgramIsRunning()) { //Update\'s the sprites frame FrameCounter++; if(FrameCounter > FrameDelay) { FrameCounter = 0; SpriteFrame++; } if(SpriteFrame > MaxSpriteFrame) SpriteFrame = 0; //Background scrolling can be removed from this position //Render the scene DrawImage(Background,Backbuffer, BackgroundX, 0); DrawImage(Background,Backbuffer, BackgroundX+800, 0); DrawImageFrame(SpriteImage, Backbuffer, 350,250, 150, 120, SpriteFrame); SDL_Delay(20); SDL_Flip(Backbuffer); } FreeFiles(); SDL_Quit(); return 0; } SDL_Surface* LoadImage(char* fileName) { SDL_Surface* imageLoaded = NULL; SDL_Surface* processedImage = NULL; imageLoaded = SDL_LoadBMP(fileName); if(imageLoaded != NULL) { processedImage = SDL_DisplayFormat(imageLoaded); SDL_FreeSurface(imageLoaded); if(processedImage != NULL) { //Here we map the color key Uint32 colorKey = SDL_MapRGB(processedImage->format, 0, 0xFF, 0xFF); //Now, set all the pixels of color R 0,G 0*FF,B 0*FF to be transparent SDL_SetColorKey(processedImage, SDL_SRCCOLORKEY, colorKey); } } return processedImage; } bool LoadFiles() { Background = LoadImage(\"graphics/background.bmp\"); if(Background == NULL) return false; SpriteImage = LoadImage(\"graphics/bat.bmp\"); //The file should be preloaded and linked with the required libraries in SDL if(SpriteImage == NULL) return false; else return true; } void FreeFiles() { SDL_FreeSurface(Background); SDL_FreeSurface(SpriteImage); } void DrawImage(SDL_Surface* image, SDL_Surface* destSurface, int x, int y) //A temporary rectangle is used to hold the offsets { SDL_Rect destRect; //Giving the offsets to the rectangle destRect.x = x; destRect.y = y; //Blit the surface SDL_BlitSurface(image, NULL, destSurface, &destRect); } //Here, we need to start the main function: int main(int argc,char** args) //Now, initialize all SDL subsystems if .

Following are the changes mentioned in bold in order to obtain the r.pdf

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During meiosis, each member of a pair of genes tends to be randomly distributed into gametes (receive alleles) independently of how other chromosomes are distributed. Genes that are having their loci nearer to each other are not generally separated during chromosomal crossover and are inherited together to the offspring. These genes are known as linked genes (two different chromosomes), and they always have multiple alleles. Incompletely linked genes undergo crossing over, and the frequency of crossing over depends upon their distance. In contrast, the completely linked genes do not undergo crossing over at all. Means, they do not produce recombinant gametes. If two linked genes are separated by 10 cM (centi morgans or map units), the percent recombinants produced by this cross is always equal to the 10% and the remaining 90% will be identical to the parental genotype. And, in case of cross between the linked genes, the highest percent of offspring indicates (always, equal to or more than 50%) the paretnal and the remaining are recombinants. Solution During meiosis, each member of a pair of genes tends to be randomly distributed into gametes (receive alleles) independently of how other chromosomes are distributed. Genes that are having their loci nearer to each other are not generally separated during chromosomal crossover and are inherited together to the offspring. These genes are known as linked genes (two different chromosomes), and they always have multiple alleles. Incompletely linked genes undergo crossing over, and the frequency of crossing over depends upon their distance. In contrast, the completely linked genes do not undergo crossing over at all. Means, they do not produce recombinant gametes. If two linked genes are separated by 10 cM (centi morgans or map units), the percent recombinants produced by this cross is always equal to the 10% and the remaining 90% will be identical to the parental genotype. And, in case of cross between the linked genes, the highest percent of offspring indicates (always, equal to or more than 50%) the paretnal and the remaining are recombinants..

During meiosis, each member of a pair of genes tends to be randomly .pdf

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B parents marital statusSolutionB parents marital status.pdf

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ANSWERS 12. B collecting ducts 13. B efferent arteriol 15. juxtaglomerular cells 16. A can be active or passive 17. C loop of Henle 18. D proximal convoluted tubule 19. .C increase the surface area of the mucosa of the small intestine 20. D submucosa of the duodenum 21. E myenteric plexus 22. D secretion - gall bladder epithelial cells ( I AM NOT SURE ABOUT QUESTION NO .14 , ) Solution ANSWERS 12. B collecting ducts 13. B efferent arteriol 15. juxtaglomerular cells 16. A can be active or passive 17. C loop of Henle 18. D proximal convoluted tubule 19. .C increase the surface area of the mucosa of the small intestine 20. D submucosa of the duodenum 21. E myenteric plexus 22. D secretion - gall bladder epithelial cells ( I AM NOT SURE ABOUT QUESTION NO .14 , ).

ANSWERS12. B collecting ducts13. B efferent arteriol15. juxtag.pdf

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Array:- Arrays is a collection of data items with same data type and access using a common name. Linked List:-A linked list is a linear data structure where each element is a separate object. Each element(node) of a list is comprising of two items - the data and a reference to the next node. Both Arrays and Linked Lists having advantages and disadvantages. Solution Array:- Arrays is a collection of data items with same data type and access using a common name. Linked List:-A linked list is a linear data structure where each element is a separate object. Each element(node) of a list is comprising of two items - the data and a reference to the next node. Both Arrays and Linked Lists having advantages and disadvantages..

Array- Arrays is a collection of data items with same data type and.pdf

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This is a bit complex to answer as we have HCl and NaOH present, the.pdf

1. This is a bit complex to answer as we have HCl and NaOH present, they will undergo neutralization reaction to form NaCl and water. so inevitably water is also in system. acetyl chloride and water, they will react to form HCl and Acetic acid which is will react with NaOH to form sodium acetate. Considering all these cases. I think the answer should be benzene in CH2Cl2 layer, as benzene only acetyl chloride in NaOH layer as CH3COO-Na+ AlCl3 will be NaOH layer in the form of AlCl4- ion, reason being NaOH and HCl have reacted to give Na+ and Cl- ions and AlCl3 in presence of chloride ions gives AlCl4- HCl will be in NaOH in form of NaCl ... dissociated into Na+ and Cl- ions ultimately Solution This is a bit complex to answer as we have HCl and NaOH present, they will undergo neutralization reaction to form NaCl and water. so inevitably water is also in system. acetyl chloride and water, they will react to form HCl and Acetic acid which is will react with NaOH to form sodium acetate. Considering all these cases. I think the answer should be benzene in CH2Cl2 layer, as benzene only acetyl chloride in NaOH layer as CH3COO-Na+ AlCl3 will be NaOH layer in the form of AlCl4- ion, reason being NaOH and HCl have reacted to give Na+ and Cl- ions and AlCl3 in presence of chloride ions gives AlCl4- HCl will be in NaOH in form of NaCl ... dissociated into Na+ and Cl- ions ultimately

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