First HBr was added to remove the double bond and add a bromine. E.pdf

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First HBr was added to remove the double bond and add a bromine. Ether was then added so Br will attach to Mg. H3CH=O was then added so a oxygen will form. Then H3O+ was added so an H will be placed on the Oxygen. H2SO4 was then added so the hydroxyl would leave and cause a double bond to be formed. BR NBS cause two br to be put. Ether was added again so that Mg will attach to the two Br. 2H3 CH=O was added so two oxygen will be added. H3O+ was then added to put H on the two oxygen. H2SO4 was then added to remove the two OH and replace it wil double bonds. follow this link if you have confusion: (there is an explanation with diagrams) http://goo.gl/tK9gi Solution First HBr was added to remove the double bond and add a bromine. Ether was then added so Br will attach to Mg. H3CH=O was then added so a oxygen will form. Then H3O+ was added so an H will be placed on the Oxygen. H2SO4 was then added so the hydroxyl would leave and cause a double bond to be formed. BR NBS cause two br to be put. Ether was added again so that Mg will attach to the two Br. 2H3 CH=O was added so two oxygen will be added. H3O+ was then added to put H on the two oxygen. H2SO4 was then added to remove the two OH and replace it wil double bonds. follow this link if you have confusion: (there is an explanation with diagrams) http://goo.gl/tK9gi.

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Redox reactions are analogous in that they are electron transfer reactions. Certain chemicals are oxidizers, like O2, which take electrons. While others are reducers, which are oxidizing agents, and give electrons. The better the reducing agent a chemical it is, the worse oxidizing agent it is. Solution Redox reactions are analogous in that they are electron transfer reactions. Certain chemicals are oxidizers, like O2, which take electrons. While others are reducers, which are oxidizing agents, and give electrons. The better the reducing agent a chemical it is, the worse oxidizing agent it is..

Redox reactions are analogous in that they are el.pdf

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$mole fraction of He = mole of He total mole now.pdf$ $mole fraction of He = mole of He total mole now.pdf$

mole fraction of He = mole of He total mole now.pdf

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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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From the LCAO method: The overlap of the 1s AO on one H atom with the is AO on a second gives one bonding MO (s) at lower energy to the constituent AOs and one antibonding MO (s*) at higher energy. Each MO can hold 2 e-. H2^- has 3e- and so the MO \"scheme\" is s(2e-) s*(1e-) Bond Order = ½[S (bonding e-) - S (antibonding e-)] = ½[2-1] = 0.5 so H2^2- is predicted to be bonding. When irradiated with UV light an e- in the s MO is promoted to the s*: s(2e-) s*(1e-) ? h? ? s(1e-) s*(2e-) ? dissociation The bond order of the excited state is formally ½[1-2] = -0.5 which is repulsive: the species flies apart into H^- and H. (H atom not a period). You have to look at an MO diagram for O2; if you do that, focusing on the p-based orbitals, you\'ll see that there are 6 bonding electrons (2 in the p-sigma orbital, and 4 in the p-pi orbitals), and two electrons in the p-pi* (one in each). Thus there are six bonding electrons and two antibonding electrons, so the net is four bonding electrons. Since it takes two bonding electrons to make a bond, the bond order is 2 for O2. If you take away one of the antibonding electrons, now you have a net of five bonding electrons, which - when divided by 2 - gives you a bond order of 2.5. Because you\'ve taken away one electron from neutral O2, the charge will be +1. Solution From the LCAO method: The overlap of the 1s AO on one H atom with the is AO on a second gives one bonding MO (s) at lower energy to the constituent AOs and one antibonding MO (s*) at higher energy. Each MO can hold 2 e-. H2^- has 3e- and so the MO \"scheme\" is s(2e-) s*(1e-) Bond Order = ½[S (bonding e-) - S (antibonding e-)] = ½[2-1] = 0.5 so H2^2- is predicted to be bonding. When irradiated with UV light an e- in the s MO is promoted to the s*: s(2e-) s*(1e-) ? h? ? s(1e-) s*(2e-) ? dissociation The bond order of the excited state is formally ½[1-2] = -0.5 which is repulsive: the species flies apart into H^- and H. (H atom not a period). You have to look at an MO diagram for O2; if you do that, focusing on the p-based orbitals, you\'ll see that there are 6 bonding electrons (2 in the p-sigma orbital, and 4 in the p-pi orbitals), and two electrons in the p-pi* (one in each). Thus there are six bonding electrons and two antibonding electrons, so the net is four bonding electrons. Since it takes two bonding electrons to make a bond, the bond order is 2 for O2. If you take away one of the antibonding electrons, now you have a net of five bonding electrons, which - when divided by 2 - gives you a bond order of 2.5. Because you\'ve taken away one electron from neutral O2, the charge will be +1..

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Cis-trans isomers occur because it is impossible to rotate the double bond without breaking it. Thus, isomers that are related by \"rotation\" about the double bond are actually separate compounds - cis/trans isomers. In the example you have drawn, keep the -CH2CH3 group and the right-hand H in place and reverse the position of the -CH2CH2CH3 group and the other H (so that the H\'s are on \"opposite\" sides of the double bond). This is the trans isomer of the compound you drew (you have the cis isomer). Now, imagine that instead of the CH2CH2CH3 group you had another H. In that case, switching the position of the two H\'s would result in the same compound, and cis-trans isomers would not be possible. In general, cis/trans isomers are possible only if each carbon that is part of the double bond is bonded to two DIFFERENT groups. This is only true for I and III. In II, the right-hand carbon of the double bond is bonded to two H\'s. And in IV, the left hand carbon of the double bond is bonded to two - CH3 groups. Thus neither of these has cis-trans isomers. So the correct answer is (b), I and III. Solution Cis-trans isomers occur because it is impossible to rotate the double bond without breaking it. Thus, isomers that are related by \"rotation\" about the double bond are actually separate compounds - cis/trans isomers. In the example you have drawn, keep the -CH2CH3 group and the right-hand H in place and reverse the position of the -CH2CH2CH3 group and the other H (so that the H\'s are on \"opposite\" sides of the double bond). This is the trans isomer of the compound you drew (you have the cis isomer). Now, imagine that instead of the CH2CH2CH3 group you had another H. In that case, switching the position of the two H\'s would result in the same compound, and cis-trans isomers would not be possible. In general, cis/trans isomers are possible only if each carbon that is part of the double bond is bonded to two DIFFERENT groups. This is only true for I and III. In II, the right-hand carbon of the double bond is bonded to two H\'s. And in IV, the left hand carbon of the double bond is bonded to two - CH3 groups. Thus neither of these has cis-trans isomers. So the correct answer is (b), I and III..

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Astatine is a halogen. Column VII(7) on the periodic table. Therefore, the elements that would have similar properties to it are the other halogens: Iodine(I), Bromine(Br), Chlorine(Cl), Fluorine(F) Solution Astatine is a halogen. Column VII(7) on the periodic table. Therefore, the elements that would have similar properties to it are the other halogens: Iodine(I), Bromine(Br), Chlorine(Cl), Fluorine(F).

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RainFall.java public class RainFall { private double months[]; String monthStr[] = {\"January\", \"Fabruary\", \"March\", \"April\", \"May\", \"June\", \"July\", \"August\", \"September\",\"October\",\"November\",\"December\"}; public RainFall(double d[]){ months = d; } public void displayTotalRainFall(){ double total = 0; for(int i=0; i months[i]){ min = months[i]; minIndex = i; } } System.out.println(\"The minimum rainfall is :\"+min); System.out.println(\"The minimum rainfall month is :\"+monthStr[minIndex]); } } RainFallTest.java import java.util.Scanner; public class RainFallTest { static String monthStr[] = {\"January\", \"Fabruary\", \"March\", \"April\", \"May\", \"June\", \"July\", \"August\", \"September\",\"October\",\"November\",\"December\"}; public static void main(String[] args) { double month[] = new double[12]; Scanner scan = new Scanner(System.in); for(int i=0; i Solution RainFall.java public class RainFall { private double months[]; String monthStr[] = {\"January\", \"Fabruary\", \"March\", \"April\", \"May\", \"June\", \"July\", \"August\", \"September\",\"October\",\"November\",\"December\"}; public RainFall(double d[]){ months = d; } public void displayTotalRainFall(){ double total = 0; for(int i=0; i months[i]){ min = months[i]; minIndex = i; } } System.out.println(\"The minimum rainfall is :\"+min); System.out.println(\"The minimum rainfall month is :\"+monthStr[minIndex]); } } RainFallTest.java import java.util.Scanner; public class RainFallTest { static String monthStr[] = {\"January\", \"Fabruary\", \"March\", \"April\", \"May\", \"June\", \"July\", \"August\", \"September\",\"October\",\"November\",\"December\"}; public static void main(String[] args) { double month[] = new double[12]; Scanner scan = new Scanner(System.in); for(int i=0; i.

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Proximal convoluted tubule performs absorption of majority of water and solutes in the kidney. It is lined by cuboidal epithelium. Urethral orifice is the region which is the external opening of the urethra. Solution Proximal convoluted tubule performs absorption of majority of water and solutes in the kidney. It is lined by cuboidal epithelium. Urethral orifice is the region which is the external opening of the urethra..

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Please find my implementation: public KeyedItem remove() { //DO THIS //remove the smallest item // since Linked List is in sorted order, so head is the samllest element if(head == null) return null; // return head element and move head one step forward KeyedItem item = head.getItem(); head = head.getNext(); return item; } private Node locateNodeAdd(KeyedItem item) { //DO THIS //find the insertion location (remember FIFO for duplicates) Node curr = head; Node prev = null; while(curr != null){ if((current.getItem().getKey()).compareTo(item.getKey) >= 0){ return prev; } prev = curr; curr = curr.getNext(); } return prev; } Solution Please find my implementation: public KeyedItem remove() { //DO THIS //remove the smallest item // since Linked List is in sorted order, so head is the samllest element if(head == null) return null; // return head element and move head one step forward KeyedItem item = head.getItem(); head = head.getNext(); return item; } private Node locateNodeAdd(KeyedItem item) { //DO THIS //find the insertion location (remember FIFO for duplicates) Node curr = head; Node prev = null; while(curr != null){ if((current.getItem().getKey()).compareTo(item.getKey) >= 0){ return prev; } prev = curr; curr = curr.getNext(); } return prev; }.

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Oxidation means loss of electrons. Metals are electropositive and lose electrons easily to form positively charged ions with stable octet structures, thus they tend to be oxidized. Reduction means gain of electrons. Non-metals are electronegative and gain electrons easily to form negatively charged ions with stable octet structures, thus they tend to be reduced. Solution Oxidation means loss of electrons. Metals are electropositive and lose electrons easily to form positively charged ions with stable octet structures, thus they tend to be oxidized. Reduction means gain of electrons. Non-metals are electronegative and gain electrons easily to form negatively charged ions with stable octet structures, thus they tend to be reduced..

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Mutualism : 1. Two organisms live together and both gain from that association Chemotrophs : 2. Use chemicals to produce their own food Commensalism : 3. One organism benefits from the association and the other one is unaffected Autotrophs : 4. Produce their own food Photoautotrophs : 5. Use light to produce their own food Heterotrophs : 6. Feed on others Parasitism : 7. Affects others negatively Solution Mutualism : 1. Two organisms live together and both gain from that association Chemotrophs : 2. Use chemicals to produce their own food Commensalism : 3. One organism benefits from the association and the other one is unaffected Autotrophs : 4. Produce their own food Photoautotrophs : 5. Use light to produce their own food Heterotrophs : 6. Feed on others Parasitism : 7. Affects others negatively.

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a) The area of a disk of radius 1 is Pi. We need the integral over the unit disk of f(x,y) to be 1 so int_(x^2+y^2<=1) f(x,y) dy dx = c*Pi = 1 So c=1/Pi b) Fr(r)=P(R <= r) = P(X^2+Y^2 <= r^2) = int_(x^2+y^2<=r^2) 1/Pi dydx = Pir^2/pi ( since the disk of radius r^2 has area pir^2 ) = r^2 c) since x^2+y^2<=1 then -sqrt(1-x^2) <= y <= sqrt(1-x^2) fX(x) = int(-sqrt(1-x^2)<=y<=sqrt(1+x^2)) f(x,y) dy = 2/Pi*sqrt(1-x^2) for -1<=x<=1 Since we have a clear symmetry of variables here then fY(y) =2/pi*sqrt(1-y^2) Solution a) The area of a disk of radius 1 is Pi. We need the integral over the unit disk of f(x,y) to be 1 so int_(x^2+y^2<=1) f(x,y) dy dx = c*Pi = 1 So c=1/Pi b) Fr(r)=P(R <= r) = P(X^2+Y^2 <= r^2) = int_(x^2+y^2<=r^2) 1/Pi dydx = Pir^2/pi ( since the disk of radius r^2 has area pir^2 ) = r^2 c) since x^2+y^2<=1 then -sqrt(1-x^2) <= y <= sqrt(1-x^2) fX(x) = int(-sqrt(1-x^2)<=y<=sqrt(1+x^2)) f(x,y) dy = 2/Pi*sqrt(1-x^2) for -1<=x<=1 Since we have a clear symmetry of variables here then fY(y) =2/pi*sqrt(1-y^2).

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