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Balancing ANY chemical equation is done exactly the same. Easiest way is to start with a primary element such as Carbon, then just work your way back and forth between products and reactants until everything is balanced. Be sure to check your equation once you\'re done to make sure everything is balanced completely. Types of Chemical reactions: 1.) Combustion: A combustion reaction is when oxygen combines with another compound to form water and carbon dioxide. These reactions are exothermic, meaning they produce heat. An example of this kind of reaction is the burning of propane: C3H8 + 5O2 --> 3CO2 + 4H2O *Hint: Combustion reactions always have O2 in the reactants and CO2 and H2O somewhere in the products. 2.) Synthesis: A synthesis reaction is when two or more simple compounds combine to form a more complicated one. These reactions come in the general form of: A + B -- -> AB One example of a synthesis reaction is the combination of iron and sulfur to form iron (II) sulfide: 8 Fe + S8 ---> 8 FeS 3.) Decomposition: A decomposition reaction is the opposite of a synthesis reaction - a complex molecule breaks down to make simpler ones. These reactions come in the general form: AB ---> A + B One example of a decomposition reaction is the electrolysis of water to make oxygen and hydrogen gas: 2 H2O ---> 2 H2 + O2 4.) Single displacement: This is when one element trades places with another element in a compound. These reactions come in the general form of: A + BC ---> AC + B One example of a single displacement reaction is when magnesium replaces hydrogen in water to make magnesium hydroxide and hydrogen gas: Mg + 2 H2O ---> Mg(OH)2 + H2 5) Double displacement: This is when the anions and cations of two different molecules switch places, forming two entirely different compounds. These reactions are in the general form: AB + CD ---> AD + CB One example of a double displacement reaction is the reaction of lead (II) nitrate with potassium iodide to form lead (II) iodide and potassium nitrate: Pb(NO3)2 + 2 KI ---> PbI2 + 2 KNO3 6) Acid-base: This is a special kind of double displacement reaction that takes place when an acid and base react with each other. The H+ ion in the acid reacts with the OH- ion in the base, causing the formation of water. Generally, the product of this reaction is some ionic salt and water: HA + BOH ---> H2O + BA One example of an acid-base reaction is the reaction of hydrobromic acid (HBr) with sodium hydroxide: HBr + NaOH ---> NaBr + H2O *Hint: An acid plus a base always produces a salt and H2O as a product. Solution Balancing ANY chemical equation is done exactly the same. Easiest way is to start with a primary element such as Carbon, then just work your way back and forth between products and reactants until everything is balanced. Be sure to check your equation once you\'re done to make sure everything is balanced completely. Types of Chemical reactions: 1.) Combustion: A combustion reaction is when oxygen combines with an.
Balancing ANY chemical equation is done exactly t.pdf
Balancing ANY chemical equation is done exactly t.pdf
deepua8
Since the heat of reaction ..is negative,.... it loses heat to.. the surroundings and therefore.. is exothermic. 2 NO(g) + O2(g) ---> 2 NO2(g) change in H= -114.1kJ molecular weight of NO2=80 so no of moles= 113.8/80=1.4225 so amount of heat produced =- 114.1/(2*1.14225) =40 kJ Solution Since the heat of reaction ..is negative,.... it loses heat to.. the surroundings and therefore.. is exothermic. 2 NO(g) + O2(g) ---> 2 NO2(g) change in H= -114.1kJ molecular weight of NO2=80 so no of moles= 113.8/80=1.4225 so amount of heat produced =- 114.1/(2*1.14225) =40 kJ.
36.Kovacs reagent is used in Indole test. Kovacs reagent is 4 (p)-.pdf
36.Kovacs reagent is used in Indole test. Kovacs reagent is 4 (p)-.pdf
deepua8
pH = pKa + log ( [ HEPES] / [HEPES protonated] 7.8 = 7.47 + log ( [HEPES] / [HEPES protonated]) 7.8-7.47 = log ( [HEPES] / [HEPES protonated]) [HEPES] / [HEPES protonated] = 10 ^ 0.33 = 2.138 [HEPES] = 2.138 * [HEPES protonated] Given [HEPES] = 0.025M [HEPES protonated] = 0.025/2.138 =0.0117M For preparation of solution amount of HEPES = 0.0117+0.025 = 0.0367 M of HEPES hydrochloride and 0.025 M of Na OH is required 0.0367 M of HEPES hydrochloride and 0.025 M of Na OH is required has to be used Solution pH = pKa + log ( [ HEPES] / [HEPES protonated] 7.8 = 7.47 + log ( [HEPES] / [HEPES protonated]) 7.8-7.47 = log ( [HEPES] / [HEPES protonated]) [HEPES] / [HEPES protonated] = 10 ^ 0.33 = 2.138 [HEPES] = 2.138 * [HEPES protonated] Given [HEPES] = 0.025M [HEPES protonated] = 0.025/2.138 =0.0117M For preparation of solution amount of HEPES = 0.0117+0.025 = 0.0367 M of HEPES hydrochloride and 0.025 M of Na OH is required 0.0367 M of HEPES hydrochloride and 0.025 M of Na OH is required has to be used.
10.Real number1.Baseb. Consists of a set and rule for combining2.Bin.pdf
10.Real number1.Baseb. Consists of a set and rule for combining2.Bin.pdf
deepua8
O-N=O is a resonance structure. In a resonance structure, the atoms themselves do not move, but the bonds move. In O-O, the elements cannot change to a different element. In N=C, another atom cannot be added. And in O=N-F, the atoms cannot be moved around. Solution O-N=O is a resonance structure. In a resonance structure, the atoms themselves do not move, but the bonds move. In O-O, the elements cannot change to a different element. In N=C, another atom cannot be added. And in O=N-F, the atoms cannot be moved around..
1. Yeasts grow by budding. The cell buds and separates into 2 cells..pdf
1. Yeasts grow by budding. The cell buds and separates into 2 cells..pdf
deepua8
Most carbon compounds are poor conductors of electricity as we have seen . From the data on the boiling and melting points of the above compounds, we can conclude that the forces of attraction between these molecules are not very strong. Since these compounds are largely nonconductors of electricity, we can conclude that the bonding in these compounds does not give rise to any ions. Solution Most carbon compounds are poor conductors of electricity as we have seen . From the data on the boiling and melting points of the above compounds, we can conclude that the forces of attraction between these molecules are not very strong. Since these compounds are largely nonconductors of electricity, we can conclude that the bonding in these compounds does not give rise to any ions..
1. 252.125Solution1. 252.125.pdf
1. 252.125Solution1. 252.125.pdf
deepua8
It\'s due to the reverse of the formation reaction of CdS CdCl2 + H2S(g) <--> CdS + 2HCl(aq) if we make the HCl strong enough we can force the equilibrium the other way, and CdCl2 is soluble. Solution It\'s due to the reverse of the formation reaction of CdS CdCl2 + H2S(g) <--> CdS + 2HCl(aq) if we make the HCl strong enough we can force the equilibrium the other way, and CdCl2 is soluble..
(B) 0.815Solution(B) 0.815.pdf
(B) 0.815Solution(B) 0.815.pdf
deepua8
Has Bond in the product that are weaker than the reactants Solution Has Bond in the product that are weaker than the reactants.
Suppose AFnSolution Suppose AFn.pdf
Suppose AFnSolution Suppose AFn.pdf
deepua8
For the ethanol water graph the simple distillation should be a steady smooth increase, the fractional distillation should look like an s, that sort of takes a straight up jump in the middle. I am unsure about the other mixture though sorry, hope that helps Solution For the ethanol water graph the simple distillation should be a steady smooth increase, the fractional distillation should look like an s, that sort of takes a straight up jump in the middle. I am unsure about the other mixture though sorry, hope that helps.
C code on linked list #include stdio.h #include stdlib.h.pdf
C code on linked list #include stdio.h #include stdlib.h.pdf
deepua8
Recommended
Balancing ANY chemical equation is done exactly the same. Easiest way is to start with a primary element such as Carbon, then just work your way back and forth between products and reactants until everything is balanced. Be sure to check your equation once you\'re done to make sure everything is balanced completely. Types of Chemical reactions: 1.) Combustion: A combustion reaction is when oxygen combines with another compound to form water and carbon dioxide. These reactions are exothermic, meaning they produce heat. An example of this kind of reaction is the burning of propane: C3H8 + 5O2 --> 3CO2 + 4H2O *Hint: Combustion reactions always have O2 in the reactants and CO2 and H2O somewhere in the products. 2.) Synthesis: A synthesis reaction is when two or more simple compounds combine to form a more complicated one. These reactions come in the general form of: A + B -- -> AB One example of a synthesis reaction is the combination of iron and sulfur to form iron (II) sulfide: 8 Fe + S8 ---> 8 FeS 3.) Decomposition: A decomposition reaction is the opposite of a synthesis reaction - a complex molecule breaks down to make simpler ones. These reactions come in the general form: AB ---> A + B One example of a decomposition reaction is the electrolysis of water to make oxygen and hydrogen gas: 2 H2O ---> 2 H2 + O2 4.) Single displacement: This is when one element trades places with another element in a compound. These reactions come in the general form of: A + BC ---> AC + B One example of a single displacement reaction is when magnesium replaces hydrogen in water to make magnesium hydroxide and hydrogen gas: Mg + 2 H2O ---> Mg(OH)2 + H2 5) Double displacement: This is when the anions and cations of two different molecules switch places, forming two entirely different compounds. These reactions are in the general form: AB + CD ---> AD + CB One example of a double displacement reaction is the reaction of lead (II) nitrate with potassium iodide to form lead (II) iodide and potassium nitrate: Pb(NO3)2 + 2 KI ---> PbI2 + 2 KNO3 6) Acid-base: This is a special kind of double displacement reaction that takes place when an acid and base react with each other. The H+ ion in the acid reacts with the OH- ion in the base, causing the formation of water. Generally, the product of this reaction is some ionic salt and water: HA + BOH ---> H2O + BA One example of an acid-base reaction is the reaction of hydrobromic acid (HBr) with sodium hydroxide: HBr + NaOH ---> NaBr + H2O *Hint: An acid plus a base always produces a salt and H2O as a product. Solution Balancing ANY chemical equation is done exactly the same. Easiest way is to start with a primary element such as Carbon, then just work your way back and forth between products and reactants until everything is balanced. Be sure to check your equation once you\'re done to make sure everything is balanced completely. Types of Chemical reactions: 1.) Combustion: A combustion reaction is when oxygen combines with an.
Balancing ANY chemical equation is done exactly t.pdf
Balancing ANY chemical equation is done exactly t.pdf
deepua8
Since the heat of reaction ..is negative,.... it loses heat to.. the surroundings and therefore.. is exothermic. 2 NO(g) + O2(g) ---> 2 NO2(g) change in H= -114.1kJ molecular weight of NO2=80 so no of moles= 113.8/80=1.4225 so amount of heat produced =- 114.1/(2*1.14225) =40 kJ Solution Since the heat of reaction ..is negative,.... it loses heat to.. the surroundings and therefore.. is exothermic. 2 NO(g) + O2(g) ---> 2 NO2(g) change in H= -114.1kJ molecular weight of NO2=80 so no of moles= 113.8/80=1.4225 so amount of heat produced =- 114.1/(2*1.14225) =40 kJ.
36.Kovacs reagent is used in Indole test. Kovacs reagent is 4 (p)-.pdf
36.Kovacs reagent is used in Indole test. Kovacs reagent is 4 (p)-.pdf
deepua8
pH = pKa + log ( [ HEPES] / [HEPES protonated] 7.8 = 7.47 + log ( [HEPES] / [HEPES protonated]) 7.8-7.47 = log ( [HEPES] / [HEPES protonated]) [HEPES] / [HEPES protonated] = 10 ^ 0.33 = 2.138 [HEPES] = 2.138 * [HEPES protonated] Given [HEPES] = 0.025M [HEPES protonated] = 0.025/2.138 =0.0117M For preparation of solution amount of HEPES = 0.0117+0.025 = 0.0367 M of HEPES hydrochloride and 0.025 M of Na OH is required 0.0367 M of HEPES hydrochloride and 0.025 M of Na OH is required has to be used Solution pH = pKa + log ( [ HEPES] / [HEPES protonated] 7.8 = 7.47 + log ( [HEPES] / [HEPES protonated]) 7.8-7.47 = log ( [HEPES] / [HEPES protonated]) [HEPES] / [HEPES protonated] = 10 ^ 0.33 = 2.138 [HEPES] = 2.138 * [HEPES protonated] Given [HEPES] = 0.025M [HEPES protonated] = 0.025/2.138 =0.0117M For preparation of solution amount of HEPES = 0.0117+0.025 = 0.0367 M of HEPES hydrochloride and 0.025 M of Na OH is required 0.0367 M of HEPES hydrochloride and 0.025 M of Na OH is required has to be used.
10.Real number1.Baseb. Consists of a set and rule for combining2.Bin.pdf
10.Real number1.Baseb. Consists of a set and rule for combining2.Bin.pdf
deepua8
O-N=O is a resonance structure. In a resonance structure, the atoms themselves do not move, but the bonds move. In O-O, the elements cannot change to a different element. In N=C, another atom cannot be added. And in O=N-F, the atoms cannot be moved around. Solution O-N=O is a resonance structure. In a resonance structure, the atoms themselves do not move, but the bonds move. In O-O, the elements cannot change to a different element. In N=C, another atom cannot be added. And in O=N-F, the atoms cannot be moved around..
1. Yeasts grow by budding. The cell buds and separates into 2 cells..pdf
1. Yeasts grow by budding. The cell buds and separates into 2 cells..pdf
deepua8
Most carbon compounds are poor conductors of electricity as we have seen . From the data on the boiling and melting points of the above compounds, we can conclude that the forces of attraction between these molecules are not very strong. Since these compounds are largely nonconductors of electricity, we can conclude that the bonding in these compounds does not give rise to any ions. Solution Most carbon compounds are poor conductors of electricity as we have seen . From the data on the boiling and melting points of the above compounds, we can conclude that the forces of attraction between these molecules are not very strong. Since these compounds are largely nonconductors of electricity, we can conclude that the bonding in these compounds does not give rise to any ions..
1. 252.125Solution1. 252.125.pdf
1. 252.125Solution1. 252.125.pdf
deepua8
It\'s due to the reverse of the formation reaction of CdS CdCl2 + H2S(g) <--> CdS + 2HCl(aq) if we make the HCl strong enough we can force the equilibrium the other way, and CdCl2 is soluble. Solution It\'s due to the reverse of the formation reaction of CdS CdCl2 + H2S(g) <--> CdS + 2HCl(aq) if we make the HCl strong enough we can force the equilibrium the other way, and CdCl2 is soluble..
(B) 0.815Solution(B) 0.815.pdf
(B) 0.815Solution(B) 0.815.pdf
deepua8
Has Bond in the product that are weaker than the reactants Solution Has Bond in the product that are weaker than the reactants.
Suppose AFnSolution Suppose AFn.pdf
Suppose AFnSolution Suppose AFn.pdf
deepua8
For the ethanol water graph the simple distillation should be a steady smooth increase, the fractional distillation should look like an s, that sort of takes a straight up jump in the middle. I am unsure about the other mixture though sorry, hope that helps Solution For the ethanol water graph the simple distillation should be a steady smooth increase, the fractional distillation should look like an s, that sort of takes a straight up jump in the middle. I am unsure about the other mixture though sorry, hope that helps.
C code on linked list #include stdio.h #include stdlib.h.pdf
C code on linked list #include stdio.h #include stdlib.h.pdf
deepua8
e =mc^2. m is the defected mass, or mass that isn\'t there and is instead used as binding energy. For Mg, you have 25 nucleons: 12 protons and 13 neutrons. So your expected mass would be 1.007825x12 + 1.008665x13 = 25.206545. The defected mass is the supposed mass minus the actual mass. so its 25.206545-24.985839= .220706 amu. Then you have to convert this into kilograms. so, divide 220706 amu by 1 kg = 6.022 x 10^26 and you get: 3.664996*10^-28 Solution e =mc^2. m is the defected mass, or mass that isn\'t there and is instead used as binding energy. For Mg, you have 25 nucleons: 12 protons and 13 neutrons. So your expected mass would be 1.007825x12 + 1.008665x13 = 25.206545. The defected mass is the supposed mass minus the actual mass. so its 25.206545-24.985839= .220706 amu. Then you have to convert this into kilograms. so, divide 220706 amu by 1 kg = 6.022 x 10^26 and you get: 3.664996*10^-28.
Definition of Log-Normal DistributionA statistical distr.pdf
Definition of Log-Normal DistributionA statistical distr.pdf
deepua8
calculate the mol of NaOH so mol= concentration * volume n=C x V n= 0.175 x 0.0206 = 0.0036 then you get the ratio of acid to base from the equation above which is 1:1 so the number of mol of base is equal to the number of mol of acid n(acid)=0.0036 the concentration of that which is C=n/V C= 0.0036/0.025 C=0.144M Solution calculate the mol of NaOH so mol= concentration * volume n=C x V n= 0.175 x 0.0206 = 0.0036 then you get the ratio of acid to base from the equation above which is 1:1 so the number of mol of base is equal to the number of mol of acid n(acid)=0.0036 the concentration of that which is C=n/V C= 0.0036/0.025 C=0.144M.
Step1 NaOH (aq) ----- Na(+)(aq) + Cl(-)(aq) Ste.pdf
Step1 NaOH (aq) ----- Na(+)(aq) + Cl(-)(aq) Ste.pdf
deepua8
cl- br- Solution cl- br-.
The compounds with low oxidation states (O.S.) be.pdf
The compounds with low oxidation states (O.S.) be.pdf
deepua8
A) ionic compounds generally formed between the combination of 1st and 2nd group elements with 7th group elements . ex: NaCl, MgF, CaCl2 and so on B) Molecular compounds means vanderwall forces acting btw the molecule these may be crystals or liguids .ex Iodine(I2) is molecular crystal ,Quartz, phosphorous(P4), sulphur(S8) C) hydrocarbons in general contain carbon and hydrogen in some cases hetero atoms may present in the compound . ex CH4 , C2H6, C2H2, C6H5N02, C6H12O6 (glucose) etc D) in general 5th, 6th, 7th groups oxides , oxy acids behaves as acid . ex: N20, N205, F20,HNO3, HClO4, HCl, HF, H2SO4, H2S208 etc along with these metal with lower oxidation state can also act as acidic compound ex: MnO MnO2 MnO3 in the above three oxides oxidation states of metals in orderly +2,+4,+6. so first compound can act as an acid last one is a base Solution A) ionic compounds generally formed between the combination of 1st and 2nd group elements with 7th group elements . ex: NaCl, MgF, CaCl2 and so on B) Molecular compounds means vanderwall forces acting btw the molecule these may be crystals or liguids .ex Iodine(I2) is molecular crystal ,Quartz, phosphorous(P4), sulphur(S8) C) hydrocarbons in general contain carbon and hydrogen in some cases hetero atoms may present in the compound . ex CH4 , C2H6, C2H2, C6H5N02, C6H12O6 (glucose) etc D) in general 5th, 6th, 7th groups oxides , oxy acids behaves as acid . ex: N20, N205, F20,HNO3, HClO4, HCl, HF, H2SO4, H2S208 etc along with these metal with lower oxidation state can also act as acidic compound ex: MnO MnO2 MnO3 in the above three oxides oxidation states of metals in orderly +2,+4,+6. so first compound can act as an acid last one is a base.
A) ionic compounds generally formed between the c.pdf
A) ionic compounds generally formed between the c.pdf
deepua8
A ethers ethers contain R-O-R linkage not carbonyl Solution A ethers ethers contain R-O-R linkage not carbonyl.
A ethers ethers contain R-O-R linkage not carbony.pdf
A ethers ethers contain R-O-R linkage not carbony.pdf
deepua8
The fact that a molecule vibrates does not in itself insure that the molecule will exhibit an IR spectrum. For a particular vibrational mode to absorb infrared radiation, the vibrational motion associated with that mode must produce a change in the dipole moment of the molecule. HCl, for example, with a center of positive charge at the H atom and a center of negative charge at the Cl atom, has a dipole moment. The magnitude of the dipole moment changes as the HCl bond stretches, so this vibration absorbs IR radiation. We say that the vibration is IR active. The N2 molecule, on the other hand, has no dipole moment. Further, stretching the N-N bond does not produce a change in dipole moment, so the vibration is infrared inactive (i.e., cannot directly absorb IR radiation). It is important to realize that there are many molecules that, although possessing no permanent dipole moment, still undergo vibrations that cause changes in the value of the dipole moment from 0 to some non-zero value. An example is CO2, which has no permanent dipole moment because the individual bond dipoles exactly cancel. However, when CO2 undergoes a bending vibration, its dipole moment changes from zero to some non-zero value. This vibration produces a change in dipole moment and is IR active. These vibrational modes are responsible for the \"greenhouse\" effect in which heat radiated from the earth is absorbed (trapped) by CO2 molecules in the atmosphere. The arrows indicate the directions of motion. Vibrations labeled A and B represent the stretching of the chemical bonds, one in a symmetric (A) fashion, in which both C=O bonds lengthen and contract together (in-phase), and the other in an asymmetric (B) fashion, in which one bond shortens while the other lengthens. The asymmetric stretch (B) is infrared active because there is a change in the molecular dipole moment during this vibration. Solution The fact that a molecule vibrates does not in itself insure that the molecule will exhibit an IR spectrum. For a particular vibrational mode to absorb infrared radiation, the vibrational motion associated with that mode must produce a change in the dipole moment of the molecule. HCl, for example, with a center of positive charge at the H atom and a center of negative charge at the Cl atom, has a dipole moment. The magnitude of the dipole moment changes as the HCl bond stretches, so this vibration absorbs IR radiation. We say that the vibration is IR active. The N2 molecule, on the other hand, has no dipole moment. Further, stretching the N-N bond does not produce a change in dipole moment, so the vibration is infrared inactive (i.e., cannot directly absorb IR radiation). It is important to realize that there are many molecules that, although possessing no permanent dipole moment, still undergo vibrations that cause changes in the value of the dipole moment from 0 to some non-zero value. An example is CO2, which has no permanent dipole moment because the individual bond dipo.
The fact that a molecule vibrates does not in its.pdf
The fact that a molecule vibrates does not in its.pdf
deepua8
The compounds of interest are Na2S and H2SO4. The molecular equation then is: Na2S + H2SO4 ---> Na2SO4 + H2S. Just exchange one anion for the other. Now, we need to know a little bit about the properties of these compounds. Almost all sodium salts are soluble, so both Na2S and Na2SO4 are fully dissociated: 2 Na+ + S2- + 2 H+ + SO42- ---> 2 Na+ + SO42- + H2S. Crossing out terms that appear on both sides, we obtain the net ionic equation: 2 H+ + S2- ----> H2S. Conveniently, H2S is a gas, and escapes from solution. Thus, it selectively removes H+ and S2- ions from the equilibrium, driving the reaction to completion (favoring Na2SO4) Solution The compounds of interest are Na2S and H2SO4. The molecular equation then is: Na2S + H2SO4 ---> Na2SO4 + H2S. Just exchange one anion for the other. Now, we need to know a little bit about the properties of these compounds. Almost all sodium salts are soluble, so both Na2S and Na2SO4 are fully dissociated: 2 Na+ + S2- + 2 H+ + SO42- ---> 2 Na+ + SO42- + H2S. Crossing out terms that appear on both sides, we obtain the net ionic equation: 2 H+ + S2- ----> H2S. Conveniently, H2S is a gas, and escapes from solution. Thus, it selectively removes H+ and S2- ions from the equilibrium, driving the reaction to completion (favoring Na2SO4).
The compounds of interest are Na2S and H2SO4. Th.pdf
The compounds of interest are Na2S and H2SO4. Th.pdf
deepua8
sucrose has a formula of C12H22O11while ammonia has a formula of NH3. NH3 is polar and has anelectronegative N which can conduct charge. Solution sucrose has a formula of C12H22O11while ammonia has a formula of NH3. NH3 is polar and has anelectronegative N which can conduct charge..
sucrose has a formula of C12H22O11while ammonia h.pdf
sucrose has a formula of C12H22O11while ammonia h.pdf
deepua8
reduction strength: Fe > Pb > As Solution reduction strength: Fe > Pb > As.
reduction strength Fe Pb As .pdf
reduction strength Fe Pb As .pdf
deepua8
sp3d since it has 1s 3p and 1d orbitals available for hybridization Solution sp3d since it has 1s 3p and 1d orbitals available for hybridization.
sp3d since it has 1s 3p and 1d orbitals available.pdf
sp3d since it has 1s 3p and 1d orbitals available.pdf
deepua8
No standard potential data given. Solution No standard potential data given..
No standard potential data given. .pdf
No standard potential data given. .pdf
deepua8
x2 + 4x Solution x2 + 4x.
x2 + 4xSolutionx2 + 4x.pdf
x2 + 4xSolutionx2 + 4x.pdf
deepua8
moles of Al = 12/27 Since chlorine is in excess , Moles of Alcl3 = 12/27 * 1 = 4/9 Solution moles of Al = 12/27 Since chlorine is in excess , Moles of Alcl3 = 12/27 * 1 = 4/9.
moles of Al = 1227 Since chlorine is in excess ,.pdf
moles of Al = 1227 Since chlorine is in excess ,.pdf
deepua8
We can make a chart for the above data in a following way: Solution We can make a chart for the above data in a following way:.
We can make a chart for the above data in a following waySoluti.pdf
We can make a chart for the above data in a following waySoluti.pdf
deepua8
There is nothing much wrong in the code. Error is due to the the entity name. You may have a another entity with same name in same library. You may defined the specified entity in the specified files. However, all entities are being compiled to the same library. Compile the specified entities in different libraries. That is compile entity prob_c5_1 in different library. or you can change the entity name and compile entity in same library. Find the code with different entity name as below: and compile it in same library. library IEEE; USE ieee.std_logic_1164.all; ENTITY prob_c5_differ IS PORT( V,F,H : In std_logic; original,reduced : OUT std_logic ); END prob_c5_differ; Architecture arc OF prob_c5_differ IS begin original <= (H and F) OR (H and V); reduced <= (F OR V) AND (H); End arc; If it still give some errors than compile the entity in different library other than work. Solution There is nothing much wrong in the code. Error is due to the the entity name. You may have a another entity with same name in same library. You may defined the specified entity in the specified files. However, all entities are being compiled to the same library. Compile the specified entities in different libraries. That is compile entity prob_c5_1 in different library. or you can change the entity name and compile entity in same library. Find the code with different entity name as below: and compile it in same library. library IEEE; USE ieee.std_logic_1164.all; ENTITY prob_c5_differ IS PORT( V,F,H : In std_logic; original,reduced : OUT std_logic ); END prob_c5_differ; Architecture arc OF prob_c5_differ IS begin original <= (H and F) OR (H and V); reduced <= (F OR V) AND (H); End arc; If it still give some errors than compile the entity in different library other than work..
There is nothing much wrong in the code.Error is due to the the en.pdf
There is nothing much wrong in the code.Error is due to the the en.pdf
deepua8
The term business cycle is used in several ways in business. This article defines business cycle in context with related terms including economic cycle, recession, and depression. The first and primary meaning of business cycle refers to fluctuations in economic output in a country or countries, characterized by well-known phases of a business cycle such as recession, depression, recovery, and expansion. The business cycle or economic cycle in this sense may be accompanied by changes in stock market prices, known as the stock market cycle. For more on these cycles and their phases, see the sections below. The second meaning of business cycle sometimes refers to stages in the life span of a single company. In this regard, important phases in a company\'s life may include: birth (or start up), growth, maturity, decline, and demise. Progress through these cycles may be impacted heavily by the economic business cycle. To accountants, companies are viewed as ongoing entities that will continue in business indefinitely. In reality, the vast majority of business startups move through these stages and cease business within a few years or within the founder\'s life time at most. The third meaning of business cycle also refers to phases in the life of an ongoing business covering a year or several years, whereby the company takes in revenues from normal operations for a year or more, re-evaluates business performance and growth prospects, adjusts or changes the business model (especially competitive strategy, marketing strategy, and pricing and margin models), and then resumes business under the new model for a period before re-evaluating again. The business cycle is the pattern of expansion, contraction and recovery in the economy. Generally speaking, the business cycle is measured and tracked in terms of GDP and unemployment – GDP rises and unemployment shrinks during expansion phases, while reversing in periods of recession. Wherever one starts in the cycle, the economy is observed to go through four periods – expansion, peak, contraction and trough. Recession is typically used to mean a downturn in economic activity, but most economists use a specific definition of \"two consecutive quarters of declining real GDP\" for recession. By comparison, there is no formal definition of depression. While recessions have averaged around 10 months in length since the 1950s, the recovery/expansion phases have a much wider range of lengths, though around three years is relatively common. The movement of the economy through business cycles also highlights certain economic relationships. While growth will rise and fall with cycles, there is a long-term trend line for growth; when economic growth is above the trend line, unemployment usually falls. One expression of this relationship is Okun\'s Law, an equation that holds that every 1% of GDP above trend equates to 0.5% less unemployment. While the business cycle is a relatively simple concept, there is great debat.
The term business cycle is used in several ways in business. This ar.pdf
The term business cycle is used in several ways in business. This ar.pdf
deepua8
The Renaissance was a time of great intellectual growth and rebirth.The word Renaissance means \"rebirth\" in French. Renaissance changed the world for the better.The renaissance brought the world out of dark ages into the light.The Renaissance taught us many great ideas. The main factors for change in Renaissance are: .The invention of printing press. .The revival of classic learning. .The church was critised for the first time. .Voyages of exploration .Turning point in the history of medicine Many other achivements were accomplished during this time.In general the World became a new place.But still the ideas or beliefs stayed the same. Solution The Renaissance was a time of great intellectual growth and rebirth.The word Renaissance means \"rebirth\" in French. Renaissance changed the world for the better.The renaissance brought the world out of dark ages into the light.The Renaissance taught us many great ideas. The main factors for change in Renaissance are: .The invention of printing press. .The revival of classic learning. .The church was critised for the first time. .Voyages of exploration .Turning point in the history of medicine Many other achivements were accomplished during this time.In general the World became a new place.But still the ideas or beliefs stayed the same..
The Renaissance was a time of great intellectual growth and rebirth..pdf
The Renaissance was a time of great intellectual growth and rebirth..pdf
deepua8
The mixture of microorganism regularly found at any anatomical site is called 33 normal microflorae. These microorganisms often offer some protection from transient microbes, but can become 34 virulent and cause disease under certain circumstance. When they do so, they are termed 35 opportunistic pathogens. Individuals with lowered resistance are called 36 immunocompromised and they are vulnerable to such infection. The normal microflora/ flora are the groups of microbes normally present at specified anatomical sites (GIT, oral cavity, nose, skin, etc.). They provide protection by not letting other harmful bacteria/microbes grow at these sites. In immunocompromised individuals (natural immune system becomes less effective, like in patients suffering from AIDS, Tuberculosis, etc.), the normal microflora may become virulent/ pathogenic and cause disease in the host. Solution The mixture of microorganism regularly found at any anatomical site is called 33 normal microflorae. These microorganisms often offer some protection from transient microbes, but can become 34 virulent and cause disease under certain circumstance. When they do so, they are termed 35 opportunistic pathogens. Individuals with lowered resistance are called 36 immunocompromised and they are vulnerable to such infection. The normal microflora/ flora are the groups of microbes normally present at specified anatomical sites (GIT, oral cavity, nose, skin, etc.). They provide protection by not letting other harmful bacteria/microbes grow at these sites. In immunocompromised individuals (natural immune system becomes less effective, like in patients suffering from AIDS, Tuberculosis, etc.), the normal microflora may become virulent/ pathogenic and cause disease in the host..
The mixture of microorganism regularly found at any anatomical site .pdf
The mixture of microorganism regularly found at any anatomical site .pdf
deepua8
the double bond in these compound is highly sensitive to the (Ozone).it craeates a radical,it breaks the all double in the polymer and weakens and the tire loses it.elasticity.polyurethane and polyisobutylene are not affect by Ozone. Solution the double bond in these compound is highly sensitive to the (Ozone).it craeates a radical,it breaks the all double in the polymer and weakens and the tire loses it.elasticity.polyurethane and polyisobutylene are not affect by Ozone..
the double bond in these compound is highly sensitive to the (Ozone).pdf
the double bond in these compound is highly sensitive to the (Ozone).pdf
deepua8
Please find the answers and explanations below: Answer 1: According to the arterial blood gas data, it can be seen the PO2 and PCO2 of Charles are within normal ranges but PCO and PN2 are very high. This clearly suggests toxic accumulation of these gases in his blood probably due to partial combustion of kerosene oil. Thus option d seems most correct. Answer 8: Option c is correct. Toxicity of CO and N2 has not changed the PO2 and PCO2 values of Charles but PCO and PN2 are exceedingly high. Since CO and N2 have higher affinitiy for hemoglobin binding than O2, they will prevent transport of oxygen from the blood to cells and the exchange of CO2 for O2. Thus, ventilation will be affected in later stages but fundamentally, the transport of gases will be highly and directly impaired. Answer 9: Option c is correct. Emphysema is a respiratory disease where toxic chemicals are entrapped in the alveolar/arterial blood gas and prevent further transport of gases to different cells of the body. It primarily occurs due to tobacco smoke and lead to chronic obstructive pulmonary disorder. The condition of Charles is very much similar to it. Answer 10: Option d is correct: Increasing the blood pCO2 levels will further dampen the respiratory system and impair the medical condition of Charles. He should be given general alkalosis of blood along with a gaseous mixture of high oxygen content to sustain life. Solution Please find the answers and explanations below: Answer 1: According to the arterial blood gas data, it can be seen the PO2 and PCO2 of Charles are within normal ranges but PCO and PN2 are very high. This clearly suggests toxic accumulation of these gases in his blood probably due to partial combustion of kerosene oil. Thus option d seems most correct. Answer 8: Option c is correct. Toxicity of CO and N2 has not changed the PO2 and PCO2 values of Charles but PCO and PN2 are exceedingly high. Since CO and N2 have higher affinitiy for hemoglobin binding than O2, they will prevent transport of oxygen from the blood to cells and the exchange of CO2 for O2. Thus, ventilation will be affected in later stages but fundamentally, the transport of gases will be highly and directly impaired. Answer 9: Option c is correct. Emphysema is a respiratory disease where toxic chemicals are entrapped in the alveolar/arterial blood gas and prevent further transport of gases to different cells of the body. It primarily occurs due to tobacco smoke and lead to chronic obstructive pulmonary disorder. The condition of Charles is very much similar to it. Answer 10: Option d is correct: Increasing the blood pCO2 levels will further dampen the respiratory system and impair the medical condition of Charles. He should be given general alkalosis of blood along with a gaseous mixture of high oxygen content to sustain life..
Please find the answers and explanations belowAnswer 1 According.pdf
Please find the answers and explanations belowAnswer 1 According.pdf
deepua8
Slides from the launch of the final report from our QAA Collaborative Enhancement project, 'When Quality Assurance Meets Innovation in Higher Education' 14 May 2024
When Quality Assurance Meets Innovation in Higher Education - Report launch w...
When Quality Assurance Meets Innovation in Higher Education - Report launch w...
Gary Wood
EADTU-EU Summit 2024
e-Sealing at EADTU by Kamakshi Rajagopal
e-Sealing at EADTU by Kamakshi Rajagopal
EADTU
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More from deepua8
e =mc^2. m is the defected mass, or mass that isn\'t there and is instead used as binding energy. For Mg, you have 25 nucleons: 12 protons and 13 neutrons. So your expected mass would be 1.007825x12 + 1.008665x13 = 25.206545. The defected mass is the supposed mass minus the actual mass. so its 25.206545-24.985839= .220706 amu. Then you have to convert this into kilograms. so, divide 220706 amu by 1 kg = 6.022 x 10^26 and you get: 3.664996*10^-28 Solution e =mc^2. m is the defected mass, or mass that isn\'t there and is instead used as binding energy. For Mg, you have 25 nucleons: 12 protons and 13 neutrons. So your expected mass would be 1.007825x12 + 1.008665x13 = 25.206545. The defected mass is the supposed mass minus the actual mass. so its 25.206545-24.985839= .220706 amu. Then you have to convert this into kilograms. so, divide 220706 amu by 1 kg = 6.022 x 10^26 and you get: 3.664996*10^-28.
Definition of Log-Normal DistributionA statistical distr.pdf
Definition of Log-Normal DistributionA statistical distr.pdf
deepua8
calculate the mol of NaOH so mol= concentration * volume n=C x V n= 0.175 x 0.0206 = 0.0036 then you get the ratio of acid to base from the equation above which is 1:1 so the number of mol of base is equal to the number of mol of acid n(acid)=0.0036 the concentration of that which is C=n/V C= 0.0036/0.025 C=0.144M Solution calculate the mol of NaOH so mol= concentration * volume n=C x V n= 0.175 x 0.0206 = 0.0036 then you get the ratio of acid to base from the equation above which is 1:1 so the number of mol of base is equal to the number of mol of acid n(acid)=0.0036 the concentration of that which is C=n/V C= 0.0036/0.025 C=0.144M.
Step1 NaOH (aq) ----- Na(+)(aq) + Cl(-)(aq) Ste.pdf
Step1 NaOH (aq) ----- Na(+)(aq) + Cl(-)(aq) Ste.pdf
deepua8
cl- br- Solution cl- br-.
The compounds with low oxidation states (O.S.) be.pdf
The compounds with low oxidation states (O.S.) be.pdf
deepua8
A) ionic compounds generally formed between the combination of 1st and 2nd group elements with 7th group elements . ex: NaCl, MgF, CaCl2 and so on B) Molecular compounds means vanderwall forces acting btw the molecule these may be crystals or liguids .ex Iodine(I2) is molecular crystal ,Quartz, phosphorous(P4), sulphur(S8) C) hydrocarbons in general contain carbon and hydrogen in some cases hetero atoms may present in the compound . ex CH4 , C2H6, C2H2, C6H5N02, C6H12O6 (glucose) etc D) in general 5th, 6th, 7th groups oxides , oxy acids behaves as acid . ex: N20, N205, F20,HNO3, HClO4, HCl, HF, H2SO4, H2S208 etc along with these metal with lower oxidation state can also act as acidic compound ex: MnO MnO2 MnO3 in the above three oxides oxidation states of metals in orderly +2,+4,+6. so first compound can act as an acid last one is a base Solution A) ionic compounds generally formed between the combination of 1st and 2nd group elements with 7th group elements . ex: NaCl, MgF, CaCl2 and so on B) Molecular compounds means vanderwall forces acting btw the molecule these may be crystals or liguids .ex Iodine(I2) is molecular crystal ,Quartz, phosphorous(P4), sulphur(S8) C) hydrocarbons in general contain carbon and hydrogen in some cases hetero atoms may present in the compound . ex CH4 , C2H6, C2H2, C6H5N02, C6H12O6 (glucose) etc D) in general 5th, 6th, 7th groups oxides , oxy acids behaves as acid . ex: N20, N205, F20,HNO3, HClO4, HCl, HF, H2SO4, H2S208 etc along with these metal with lower oxidation state can also act as acidic compound ex: MnO MnO2 MnO3 in the above three oxides oxidation states of metals in orderly +2,+4,+6. so first compound can act as an acid last one is a base.
A) ionic compounds generally formed between the c.pdf
A) ionic compounds generally formed between the c.pdf
deepua8
A ethers ethers contain R-O-R linkage not carbonyl Solution A ethers ethers contain R-O-R linkage not carbonyl.
A ethers ethers contain R-O-R linkage not carbony.pdf
A ethers ethers contain R-O-R linkage not carbony.pdf
deepua8
The fact that a molecule vibrates does not in itself insure that the molecule will exhibit an IR spectrum. For a particular vibrational mode to absorb infrared radiation, the vibrational motion associated with that mode must produce a change in the dipole moment of the molecule. HCl, for example, with a center of positive charge at the H atom and a center of negative charge at the Cl atom, has a dipole moment. The magnitude of the dipole moment changes as the HCl bond stretches, so this vibration absorbs IR radiation. We say that the vibration is IR active. The N2 molecule, on the other hand, has no dipole moment. Further, stretching the N-N bond does not produce a change in dipole moment, so the vibration is infrared inactive (i.e., cannot directly absorb IR radiation). It is important to realize that there are many molecules that, although possessing no permanent dipole moment, still undergo vibrations that cause changes in the value of the dipole moment from 0 to some non-zero value. An example is CO2, which has no permanent dipole moment because the individual bond dipoles exactly cancel. However, when CO2 undergoes a bending vibration, its dipole moment changes from zero to some non-zero value. This vibration produces a change in dipole moment and is IR active. These vibrational modes are responsible for the \"greenhouse\" effect in which heat radiated from the earth is absorbed (trapped) by CO2 molecules in the atmosphere. The arrows indicate the directions of motion. Vibrations labeled A and B represent the stretching of the chemical bonds, one in a symmetric (A) fashion, in which both C=O bonds lengthen and contract together (in-phase), and the other in an asymmetric (B) fashion, in which one bond shortens while the other lengthens. The asymmetric stretch (B) is infrared active because there is a change in the molecular dipole moment during this vibration. Solution The fact that a molecule vibrates does not in itself insure that the molecule will exhibit an IR spectrum. For a particular vibrational mode to absorb infrared radiation, the vibrational motion associated with that mode must produce a change in the dipole moment of the molecule. HCl, for example, with a center of positive charge at the H atom and a center of negative charge at the Cl atom, has a dipole moment. The magnitude of the dipole moment changes as the HCl bond stretches, so this vibration absorbs IR radiation. We say that the vibration is IR active. The N2 molecule, on the other hand, has no dipole moment. Further, stretching the N-N bond does not produce a change in dipole moment, so the vibration is infrared inactive (i.e., cannot directly absorb IR radiation). It is important to realize that there are many molecules that, although possessing no permanent dipole moment, still undergo vibrations that cause changes in the value of the dipole moment from 0 to some non-zero value. An example is CO2, which has no permanent dipole moment because the individual bond dipo.
The fact that a molecule vibrates does not in its.pdf
The fact that a molecule vibrates does not in its.pdf
deepua8
The compounds of interest are Na2S and H2SO4. The molecular equation then is: Na2S + H2SO4 ---> Na2SO4 + H2S. Just exchange one anion for the other. Now, we need to know a little bit about the properties of these compounds. Almost all sodium salts are soluble, so both Na2S and Na2SO4 are fully dissociated: 2 Na+ + S2- + 2 H+ + SO42- ---> 2 Na+ + SO42- + H2S. Crossing out terms that appear on both sides, we obtain the net ionic equation: 2 H+ + S2- ----> H2S. Conveniently, H2S is a gas, and escapes from solution. Thus, it selectively removes H+ and S2- ions from the equilibrium, driving the reaction to completion (favoring Na2SO4) Solution The compounds of interest are Na2S and H2SO4. The molecular equation then is: Na2S + H2SO4 ---> Na2SO4 + H2S. Just exchange one anion for the other. Now, we need to know a little bit about the properties of these compounds. Almost all sodium salts are soluble, so both Na2S and Na2SO4 are fully dissociated: 2 Na+ + S2- + 2 H+ + SO42- ---> 2 Na+ + SO42- + H2S. Crossing out terms that appear on both sides, we obtain the net ionic equation: 2 H+ + S2- ----> H2S. Conveniently, H2S is a gas, and escapes from solution. Thus, it selectively removes H+ and S2- ions from the equilibrium, driving the reaction to completion (favoring Na2SO4).
The compounds of interest are Na2S and H2SO4. Th.pdf
The compounds of interest are Na2S and H2SO4. Th.pdf
deepua8
sucrose has a formula of C12H22O11while ammonia has a formula of NH3. NH3 is polar and has anelectronegative N which can conduct charge. Solution sucrose has a formula of C12H22O11while ammonia has a formula of NH3. NH3 is polar and has anelectronegative N which can conduct charge..
sucrose has a formula of C12H22O11while ammonia h.pdf
sucrose has a formula of C12H22O11while ammonia h.pdf
deepua8
reduction strength: Fe > Pb > As Solution reduction strength: Fe > Pb > As.
reduction strength Fe Pb As .pdf
reduction strength Fe Pb As .pdf
deepua8
sp3d since it has 1s 3p and 1d orbitals available for hybridization Solution sp3d since it has 1s 3p and 1d orbitals available for hybridization.
sp3d since it has 1s 3p and 1d orbitals available.pdf
sp3d since it has 1s 3p and 1d orbitals available.pdf
deepua8
No standard potential data given. Solution No standard potential data given..
No standard potential data given. .pdf
No standard potential data given. .pdf
deepua8
x2 + 4x Solution x2 + 4x.
x2 + 4xSolutionx2 + 4x.pdf
x2 + 4xSolutionx2 + 4x.pdf
deepua8
moles of Al = 12/27 Since chlorine is in excess , Moles of Alcl3 = 12/27 * 1 = 4/9 Solution moles of Al = 12/27 Since chlorine is in excess , Moles of Alcl3 = 12/27 * 1 = 4/9.
moles of Al = 1227 Since chlorine is in excess ,.pdf
moles of Al = 1227 Since chlorine is in excess ,.pdf
deepua8
We can make a chart for the above data in a following way: Solution We can make a chart for the above data in a following way:.
We can make a chart for the above data in a following waySoluti.pdf
We can make a chart for the above data in a following waySoluti.pdf
deepua8
There is nothing much wrong in the code. Error is due to the the entity name. You may have a another entity with same name in same library. You may defined the specified entity in the specified files. However, all entities are being compiled to the same library. Compile the specified entities in different libraries. That is compile entity prob_c5_1 in different library. or you can change the entity name and compile entity in same library. Find the code with different entity name as below: and compile it in same library. library IEEE; USE ieee.std_logic_1164.all; ENTITY prob_c5_differ IS PORT( V,F,H : In std_logic; original,reduced : OUT std_logic ); END prob_c5_differ; Architecture arc OF prob_c5_differ IS begin original <= (H and F) OR (H and V); reduced <= (F OR V) AND (H); End arc; If it still give some errors than compile the entity in different library other than work. Solution There is nothing much wrong in the code. Error is due to the the entity name. You may have a another entity with same name in same library. You may defined the specified entity in the specified files. However, all entities are being compiled to the same library. Compile the specified entities in different libraries. That is compile entity prob_c5_1 in different library. or you can change the entity name and compile entity in same library. Find the code with different entity name as below: and compile it in same library. library IEEE; USE ieee.std_logic_1164.all; ENTITY prob_c5_differ IS PORT( V,F,H : In std_logic; original,reduced : OUT std_logic ); END prob_c5_differ; Architecture arc OF prob_c5_differ IS begin original <= (H and F) OR (H and V); reduced <= (F OR V) AND (H); End arc; If it still give some errors than compile the entity in different library other than work..
There is nothing much wrong in the code.Error is due to the the en.pdf
There is nothing much wrong in the code.Error is due to the the en.pdf
deepua8
The term business cycle is used in several ways in business. This article defines business cycle in context with related terms including economic cycle, recession, and depression. The first and primary meaning of business cycle refers to fluctuations in economic output in a country or countries, characterized by well-known phases of a business cycle such as recession, depression, recovery, and expansion. The business cycle or economic cycle in this sense may be accompanied by changes in stock market prices, known as the stock market cycle. For more on these cycles and their phases, see the sections below. The second meaning of business cycle sometimes refers to stages in the life span of a single company. In this regard, important phases in a company\'s life may include: birth (or start up), growth, maturity, decline, and demise. Progress through these cycles may be impacted heavily by the economic business cycle. To accountants, companies are viewed as ongoing entities that will continue in business indefinitely. In reality, the vast majority of business startups move through these stages and cease business within a few years or within the founder\'s life time at most. The third meaning of business cycle also refers to phases in the life of an ongoing business covering a year or several years, whereby the company takes in revenues from normal operations for a year or more, re-evaluates business performance and growth prospects, adjusts or changes the business model (especially competitive strategy, marketing strategy, and pricing and margin models), and then resumes business under the new model for a period before re-evaluating again. The business cycle is the pattern of expansion, contraction and recovery in the economy. Generally speaking, the business cycle is measured and tracked in terms of GDP and unemployment – GDP rises and unemployment shrinks during expansion phases, while reversing in periods of recession. Wherever one starts in the cycle, the economy is observed to go through four periods – expansion, peak, contraction and trough. Recession is typically used to mean a downturn in economic activity, but most economists use a specific definition of \"two consecutive quarters of declining real GDP\" for recession. By comparison, there is no formal definition of depression. While recessions have averaged around 10 months in length since the 1950s, the recovery/expansion phases have a much wider range of lengths, though around three years is relatively common. The movement of the economy through business cycles also highlights certain economic relationships. While growth will rise and fall with cycles, there is a long-term trend line for growth; when economic growth is above the trend line, unemployment usually falls. One expression of this relationship is Okun\'s Law, an equation that holds that every 1% of GDP above trend equates to 0.5% less unemployment. While the business cycle is a relatively simple concept, there is great debat.
The term business cycle is used in several ways in business. This ar.pdf
The term business cycle is used in several ways in business. This ar.pdf
deepua8
The Renaissance was a time of great intellectual growth and rebirth.The word Renaissance means \"rebirth\" in French. Renaissance changed the world for the better.The renaissance brought the world out of dark ages into the light.The Renaissance taught us many great ideas. The main factors for change in Renaissance are: .The invention of printing press. .The revival of classic learning. .The church was critised for the first time. .Voyages of exploration .Turning point in the history of medicine Many other achivements were accomplished during this time.In general the World became a new place.But still the ideas or beliefs stayed the same. Solution The Renaissance was a time of great intellectual growth and rebirth.The word Renaissance means \"rebirth\" in French. Renaissance changed the world for the better.The renaissance brought the world out of dark ages into the light.The Renaissance taught us many great ideas. The main factors for change in Renaissance are: .The invention of printing press. .The revival of classic learning. .The church was critised for the first time. .Voyages of exploration .Turning point in the history of medicine Many other achivements were accomplished during this time.In general the World became a new place.But still the ideas or beliefs stayed the same..
The Renaissance was a time of great intellectual growth and rebirth..pdf
The Renaissance was a time of great intellectual growth and rebirth..pdf
deepua8
The mixture of microorganism regularly found at any anatomical site is called 33 normal microflorae. These microorganisms often offer some protection from transient microbes, but can become 34 virulent and cause disease under certain circumstance. When they do so, they are termed 35 opportunistic pathogens. Individuals with lowered resistance are called 36 immunocompromised and they are vulnerable to such infection. The normal microflora/ flora are the groups of microbes normally present at specified anatomical sites (GIT, oral cavity, nose, skin, etc.). They provide protection by not letting other harmful bacteria/microbes grow at these sites. In immunocompromised individuals (natural immune system becomes less effective, like in patients suffering from AIDS, Tuberculosis, etc.), the normal microflora may become virulent/ pathogenic and cause disease in the host. Solution The mixture of microorganism regularly found at any anatomical site is called 33 normal microflorae. These microorganisms often offer some protection from transient microbes, but can become 34 virulent and cause disease under certain circumstance. When they do so, they are termed 35 opportunistic pathogens. Individuals with lowered resistance are called 36 immunocompromised and they are vulnerable to such infection. The normal microflora/ flora are the groups of microbes normally present at specified anatomical sites (GIT, oral cavity, nose, skin, etc.). They provide protection by not letting other harmful bacteria/microbes grow at these sites. In immunocompromised individuals (natural immune system becomes less effective, like in patients suffering from AIDS, Tuberculosis, etc.), the normal microflora may become virulent/ pathogenic and cause disease in the host..
The mixture of microorganism regularly found at any anatomical site .pdf
The mixture of microorganism regularly found at any anatomical site .pdf
deepua8
the double bond in these compound is highly sensitive to the (Ozone).it craeates a radical,it breaks the all double in the polymer and weakens and the tire loses it.elasticity.polyurethane and polyisobutylene are not affect by Ozone. Solution the double bond in these compound is highly sensitive to the (Ozone).it craeates a radical,it breaks the all double in the polymer and weakens and the tire loses it.elasticity.polyurethane and polyisobutylene are not affect by Ozone..
the double bond in these compound is highly sensitive to the (Ozone).pdf
the double bond in these compound is highly sensitive to the (Ozone).pdf
deepua8
Please find the answers and explanations below: Answer 1: According to the arterial blood gas data, it can be seen the PO2 and PCO2 of Charles are within normal ranges but PCO and PN2 are very high. This clearly suggests toxic accumulation of these gases in his blood probably due to partial combustion of kerosene oil. Thus option d seems most correct. Answer 8: Option c is correct. Toxicity of CO and N2 has not changed the PO2 and PCO2 values of Charles but PCO and PN2 are exceedingly high. Since CO and N2 have higher affinitiy for hemoglobin binding than O2, they will prevent transport of oxygen from the blood to cells and the exchange of CO2 for O2. Thus, ventilation will be affected in later stages but fundamentally, the transport of gases will be highly and directly impaired. Answer 9: Option c is correct. Emphysema is a respiratory disease where toxic chemicals are entrapped in the alveolar/arterial blood gas and prevent further transport of gases to different cells of the body. It primarily occurs due to tobacco smoke and lead to chronic obstructive pulmonary disorder. The condition of Charles is very much similar to it. Answer 10: Option d is correct: Increasing the blood pCO2 levels will further dampen the respiratory system and impair the medical condition of Charles. He should be given general alkalosis of blood along with a gaseous mixture of high oxygen content to sustain life. Solution Please find the answers and explanations below: Answer 1: According to the arterial blood gas data, it can be seen the PO2 and PCO2 of Charles are within normal ranges but PCO and PN2 are very high. This clearly suggests toxic accumulation of these gases in his blood probably due to partial combustion of kerosene oil. Thus option d seems most correct. Answer 8: Option c is correct. Toxicity of CO and N2 has not changed the PO2 and PCO2 values of Charles but PCO and PN2 are exceedingly high. Since CO and N2 have higher affinitiy for hemoglobin binding than O2, they will prevent transport of oxygen from the blood to cells and the exchange of CO2 for O2. Thus, ventilation will be affected in later stages but fundamentally, the transport of gases will be highly and directly impaired. Answer 9: Option c is correct. Emphysema is a respiratory disease where toxic chemicals are entrapped in the alveolar/arterial blood gas and prevent further transport of gases to different cells of the body. It primarily occurs due to tobacco smoke and lead to chronic obstructive pulmonary disorder. The condition of Charles is very much similar to it. Answer 10: Option d is correct: Increasing the blood pCO2 levels will further dampen the respiratory system and impair the medical condition of Charles. He should be given general alkalosis of blood along with a gaseous mixture of high oxygen content to sustain life..
Please find the answers and explanations belowAnswer 1 According.pdf
Please find the answers and explanations belowAnswer 1 According.pdf
deepua8
More from deepua8
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Definition of Log-Normal DistributionA statistical distr.pdf
Definition of Log-Normal DistributionA statistical distr.pdf
Step1 NaOH (aq) ----- Na(+)(aq) + Cl(-)(aq) Ste.pdf
Step1 NaOH (aq) ----- Na(+)(aq) + Cl(-)(aq) Ste.pdf
The compounds with low oxidation states (O.S.) be.pdf
The compounds with low oxidation states (O.S.) be.pdf
A) ionic compounds generally formed between the c.pdf
A) ionic compounds generally formed between the c.pdf
A ethers ethers contain R-O-R linkage not carbony.pdf
A ethers ethers contain R-O-R linkage not carbony.pdf
The fact that a molecule vibrates does not in its.pdf
The fact that a molecule vibrates does not in its.pdf
The compounds of interest are Na2S and H2SO4. Th.pdf
The compounds of interest are Na2S and H2SO4. Th.pdf
sucrose has a formula of C12H22O11while ammonia h.pdf
sucrose has a formula of C12H22O11while ammonia h.pdf
reduction strength Fe Pb As .pdf
reduction strength Fe Pb As .pdf
sp3d since it has 1s 3p and 1d orbitals available.pdf
sp3d since it has 1s 3p and 1d orbitals available.pdf
No standard potential data given. .pdf
No standard potential data given. .pdf
x2 + 4xSolutionx2 + 4x.pdf
x2 + 4xSolutionx2 + 4x.pdf
moles of Al = 1227 Since chlorine is in excess ,.pdf
moles of Al = 1227 Since chlorine is in excess ,.pdf
We can make a chart for the above data in a following waySoluti.pdf
We can make a chart for the above data in a following waySoluti.pdf
There is nothing much wrong in the code.Error is due to the the en.pdf
There is nothing much wrong in the code.Error is due to the the en.pdf
The term business cycle is used in several ways in business. This ar.pdf
The term business cycle is used in several ways in business. This ar.pdf
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The Renaissance was a time of great intellectual growth and rebirth..pdf
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The mixture of microorganism regularly found at any anatomical site .pdf
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the double bond in these compound is highly sensitive to the (Ozone).pdf
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Please find the answers and explanations belowAnswer 1 According.pdf
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Slides from the launch of the final report from our QAA Collaborative Enhancement project, 'When Quality Assurance Meets Innovation in Higher Education' 14 May 2024
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When Quality Assurance Meets Innovation in Higher Education - Report launch w...
Gary Wood
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PSYPACT- Practicing Over State Lines May 2024.pptx
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VAMOS CUIDAR DO NOSSO PLANETA! .
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Transport (British English) or Transportation (American English) ransportation has developed along three basic Mode (Media):- 1. Land Transportation (way)– (a) Road Transportation (b) Rail Transportation 2. Water Transportation 3. Air Transportation Tramway Inland water transport Ocean transport These may be classified as under: (a). Liners (b). Tramps Liners Vs Tramps Figure- Layout airport runway design TRAFFIC SIGNS
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Nguyen Thanh Tu Collection
Program code examples (known also as worked examples) play a crucial role in learning how to program. Instructors use examples extensively to demonstrate the semantics of the programming language being taught and to highlight the fundamental coding patterns. Programming textbooks allocate considerable space to present and explain code examples. To make the process of studying code examples more interactive, CS education researchers developed a range of tools to engage students in the study of code examples. These tools include codecasts (codemotion,codecast,elicasts), interactive example explorers (WebEx, PCEX), and tutoring systems (DeepTutor). An important component in all types of worked examples is code explanations associated with specific code lines or code chunks of an example. The explanations connect examples with general programming knowledge explaining the role and function of code fragments or their behavior. In textbooks, these explanations are usually presented as comments in the code or as explanations on the margins. The example explorer tools allow students to examine these explanations interactively. Tutoring systems, which engage students in explaining the code, use these model explanations to check student responses and provide scaffolding. In all these cases, to make a worked example re-usable beyond its presentation in a lecture, the explanations have to be authored by instructors or domain experts i.e., produced and integrated into a specific system. As the experience of the last 10 years demonstrated, these explanations are hard to obtain. Those already collected are usually “locked” in a specific example-focused system and can’t be reused. The purpose of this working group is to support broader re-used of worked examples augmented with explanations. Our current plan is to develop а standard approach to represent explained examples. This approach will enable an example created for any of the existing systems to be explored in a standard format and imported into any other example-focused system. We plan to follow a successful experience of the PEML working group focused on re-using programming exercises.
SPLICE Working Group:Reusable Code Examples
SPLICE Working Group:Reusable Code Examples
Peter Brusilovsky
Dhaka Textiles Ltd, a leading Bangladeshi textile manufacturer, faced a communication crisis when rumors spread among employees about possible benefit cuts. This crisis not only disrupted operations but also damaged the company’s reputation.
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Dabee Kamal
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Andreas Schleicher presents at the launch of ‘What does child empowerment mean today? Implications for education and well-being’ on the 15 May 2024. The report was launched by Mathias Cormann, OECD Secretary-General and can be found here: https://www.oecd-ilibrary.org/education/what-does-child-empowerment-mean-today_8f80ce38-en
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EduSkills OECD
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8 Tips for Effective Working Capital Management.pdf" This comprehensive guide, "8 Tips for Effective Working Capital Management", serves as a valuable resource for businesses aiming to optimize their financial operations. 💼 In this PDF, you will find practical tips and strategies to effectively manage your working capital, ensuring a healthy and sustainable financial position for your organization. for more information visit now https://www.mbaassignmentexperts.com/management-assignment-help
8 Tips for Effective Working Capital Management
8 Tips for Effective Working Capital Management
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PSYPACT- Practicing Over State Lines May 2024.pptx
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Analyzing and resolving a communication crisis in Dhaka textiles LTD.pptx
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Stl Algorithms in C++ jjjjjjjjjjjjjjjjjj
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