Once HCl is added in solution, the ions would dis.pdf
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Once HCl is added in solution, the ions would dissociate into their component parts (H+ and Cl-). As per the second equation, adding more H+ would initially increase the quantity of products, thusly shifting the reaction towards the reactants by LeChatlier\'s Principle. If NaOH was added, the dissociation would create OH- anions that would soak up H+ cations. By removing products, the equlibrium is shifted towards the products by LeChatlier\'s Principle. Best of wishes! (: Solution Once HCl is added in solution, the ions would dissociate into their component parts (H+ and Cl-). As per the second equation, adding more H+ would initially increase the quantity of products, thusly shifting the reaction towards the reactants by LeChatlier\'s Principle. If NaOH was added, the dissociation would create OH- anions that would soak up H+ cations. By removing products, the equlibrium is shifted towards the products by LeChatlier\'s Principle. Best of wishes! (:.
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Hydrogenation (1)
I hope You all like it. I hope It is very beneficial for you all. I really thought that you all get enough knowledge from this presentation. This presentation is about materials and their classifications. After you read this presentation you knowledge is not as before.
946 LTE Labs Le.chateliers-lab.pdf Before beginning this experiment.pdf
9:46 LTE Labs Le.chateliers-lab.pdf Before beginning this experiment in the laboratory you
should be alieto answer the following questions PRE LAB QUESTIONS 1. Beiefly stabe
LeChidelier\'s Principle. 2. Consider the following equilibrium In which direction will the
equilibrium shit if a. HSO,ibs added? Why? b. Buicl, is added? Why C NaCl is added? Why d.
Heat is added? Why? Consider the folewing equilibrium foe nitous acid, HNOs,a weak acid In
which direction will the equilibrium shift if a NaOH is added? b. NaNO, is added HCl is added?
d. The acid solution is made more dilute? Complete and balance the following equations and
then wrae balanced net ionic equations. 4· 3. On the busis of LeChitelier\'s Principle, explain
why Ag-C0 disos when HNO, is added Previous Next Dashboard Calendar To DoNotifications
Inbox
Solution
Answer:
1. Le Chtelier\'s principle states that if a dynamic equilibrium is disturbed by changing the
conditions, the position of equilibrium shifts to counteract the change to re-establish a new
equilibrium
2.a) H2SO4 is added:
the equilibrium shifts to the left because as there are more sulfate ions are added.
b)BaCl2 is added
The equilibrium shifts to the left because as there are more Ba^+2 are added.
c) NaCl is added
There is no shift in the equilibrium. As there are no Na^+1 and Cl^-1 are present in the
equilibrium mixture
d) Heat is added
The equilibrium will shift to the right. As the forward reaction is endothermic, when the heat is
added, the reaction will favours the product
3. a) NaOH is added
The equilibrium will shift to the left because as there are more OH^-1 ions added, it will react
with H+ present in the right side and forms more water.
b)NaNO2 is added
The equilibrium will shift to the left because the more NO2^-1 ions added, it forms more HNO2
by reacts with H3O^+1.
c) HCl is added
The equilibrium will shift to the right because more H+ is added in the equilibrium mixture
d) The acid solution is made more dilute
The equilibrium will shift to the right as there are more water is added.
4 a) AgNO3 (aq) + HCl (aq) <<>> AgCl (s) + HNO3 (aq)
Net ionic equation: Ag^+1 (aq) + Cl^-1 (aq) >> AgCl (s)
b) NH3(aq) + HCl (aq) <<<>> NH4Cl (s)
Net equation: NH3 (aq) + H^+1 (aq) + Cl^-1 (aq) > NH4Cl(s)
c) Na2CO3 (aq) + 2HNO3 (aq) <<>> H2O(l) + 2NaNO3 (aq) + CO2 (g)
Net ionic equation: CO3 ^-2 (aq) + 2H+ (aq) >> H2O (l) + CO2 (g)
5. Ag2CO3 (s) + 2HNO3 (aq) <<<>>> 2AgNO3 (aq) + H2O (l) + CO2(g)
Ag2CO3 is a solid . When adding nitric acid to silver carbonate, it forms soluble silver nitrate..
Recommended
Hydrogenation (1)
I hope You all like it. I hope It is very beneficial for you all. I really thought that you all get enough knowledge from this presentation. This presentation is about materials and their classifications. After you read this presentation you knowledge is not as before.
946 LTE Labs Le.chateliers-lab.pdf Before beginning this experiment.pdf
9:46 LTE Labs Le.chateliers-lab.pdf Before beginning this experiment in the laboratory you
should be alieto answer the following questions PRE LAB QUESTIONS 1. Beiefly stabe
LeChidelier\'s Principle. 2. Consider the following equilibrium In which direction will the
equilibrium shit if a. HSO,ibs added? Why? b. Buicl, is added? Why C NaCl is added? Why d.
Heat is added? Why? Consider the folewing equilibrium foe nitous acid, HNOs,a weak acid In
which direction will the equilibrium shift if a NaOH is added? b. NaNO, is added HCl is added?
d. The acid solution is made more dilute? Complete and balance the following equations and
then wrae balanced net ionic equations. 4· 3. On the busis of LeChitelier\'s Principle, explain
why Ag-C0 disos when HNO, is added Previous Next Dashboard Calendar To DoNotifications
Inbox
Solution
Answer:
1. Le Chtelier\'s principle states that if a dynamic equilibrium is disturbed by changing the
conditions, the position of equilibrium shifts to counteract the change to re-establish a new
equilibrium
2.a) H2SO4 is added:
the equilibrium shifts to the left because as there are more sulfate ions are added.
b)BaCl2 is added
The equilibrium shifts to the left because as there are more Ba^+2 are added.
c) NaCl is added
There is no shift in the equilibrium. As there are no Na^+1 and Cl^-1 are present in the
equilibrium mixture
d) Heat is added
The equilibrium will shift to the right. As the forward reaction is endothermic, when the heat is
added, the reaction will favours the product
3. a) NaOH is added
The equilibrium will shift to the left because as there are more OH^-1 ions added, it will react
with H+ present in the right side and forms more water.
b)NaNO2 is added
The equilibrium will shift to the left because the more NO2^-1 ions added, it forms more HNO2
by reacts with H3O^+1.
c) HCl is added
The equilibrium will shift to the right because more H+ is added in the equilibrium mixture
d) The acid solution is made more dilute
The equilibrium will shift to the right as there are more water is added.
4 a) AgNO3 (aq) + HCl (aq) <<>> AgCl (s) + HNO3 (aq)
Net ionic equation: Ag^+1 (aq) + Cl^-1 (aq) >> AgCl (s)
b) NH3(aq) + HCl (aq) <<<>> NH4Cl (s)
Net equation: NH3 (aq) + H^+1 (aq) + Cl^-1 (aq) > NH4Cl(s)
c) Na2CO3 (aq) + 2HNO3 (aq) <<>> H2O(l) + 2NaNO3 (aq) + CO2 (g)
Net ionic equation: CO3 ^-2 (aq) + 2H+ (aq) >> H2O (l) + CO2 (g)
5. Ag2CO3 (s) + 2HNO3 (aq) <<<>>> 2AgNO3 (aq) + H2O (l) + CO2(g)
Ag2CO3 is a solid . When adding nitric acid to silver carbonate, it forms soluble silver nitrate..
Acids, Bases And Buffers Pharmaceutical Inorganic chemistry UNIT-II (Part-I)
Acids, Bases And Buffers Pharmaceutical Inorganic chemistry UNIT-II (Part-I)
Acids, Bases are defined by Four main theories,
1.Traditional theory / concept
2.Arrhenius theory
3.Bronsted and Lowry theory
4.Lewis theory
Importance of acids and bases in pharmacy
Buffers: Buffer action
Buffer capacity Buffers system
Types of Buffers : Generally buffers are of two types:
1. Acidic buffers
2. Basic buffers
There are some other buffer system:
3. Two salts acts as acid-base pair. Ex- Potassium hydrogen phosphate and potassium dihydrogen phosphate.
4. Amphoteric electrolyte. Ex- Solution of glycine.
5. Solution of strong acid and solution of strong base. Ex- Strong HCl with KCl Mechanism of Buffer action: Mechanism of Action of acidic buffers: Buffer equation-Henderson-Hasselbalch equation:
Standard Buffer Solutions Preparation of Buffer Solutions: Buffers in pharmaceutical systems or Application of buffer: Stability of buffers Buffered isotonic solution Types of Buffer Isotonic solution
1. Isotonic Solutions:
2. Hypertonic Solutions:
3. Hypotonic Solution:
Measurement of Tonicity: 1. Hemolytic method: 2. Cryoscopic method or depression of freezing point:
Methods of adjusting the tonicity:
Class I methods:
In this type, sodium chloride or other substances are added to the solution in sufficient quantity to make it isotonic. Then the preparation is brought to its final volume withan isotonic or a buffered isotonic diluting solution.
These methods are of two types:
Cryoscopic method
Sodium chloride equivalent method.
Class II methods:
In this type, water is added in sufficient quantity make the preparation isotonic. Then the preparation is brought to its volume with an isotonic or a buffered isotonic diluting solution.
These methods are of two types:
White-Vincent method
Sprowls method.
When recovering an acid from its salt dissolved in water- we typically.docx
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 don’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.
.
Acid base eqm and solubility equilibrium
This chapter talks about:
Acid –base equilibria
solubility equilibria
Buffer solution
Acid-base titration
Molar solubility and solubility
pH and Solubility
For more materials subscribe to channel:
https://www.youtube.com/channel/UCzXxV4xER9NIWt316gfeO1w
You need to know the solubility rules in order to.pdf
You need to know the solubility rules in order to answer this question. When HCl is
added and forms a precipitate, we\'re looking for something that is insoluble when combined
with chlorine atoms. That would be Hg+. So we know that is in the starting solution for sure. The
first precipitate would be HgCl. Next, H2SO4 is added and a precipitate is formed so we\'re
looking for a cation that is insoluble when combined with sulfate (SO4^2-) This would be Ca2+.
(Hg+ would work too except we removed all of those ions when we removed the first
precipitate.) So, the second precipitate would be CaSO4 Next, NaOH is added and according to
the solubility rules, we know that hydroxides are generally all insoluble except when they are
combined with ammonium ion or any Group I ions. So since no precipitate is formed, none of
the other cations can be in the starting solution because if they were, then a precipitate would
have been formed with OH. So, the only cations that were present in the original solution were
Ca2+ and Hg+.
Solution
You need to know the solubility rules in order to answer this question. When HCl is
added and forms a precipitate, we\'re looking for something that is insoluble when combined
with chlorine atoms. That would be Hg+. So we know that is in the starting solution for sure. The
first precipitate would be HgCl. Next, H2SO4 is added and a precipitate is formed so we\'re
looking for a cation that is insoluble when combined with sulfate (SO4^2-) This would be Ca2+.
(Hg+ would work too except we removed all of those ions when we removed the first
precipitate.) So, the second precipitate would be CaSO4 Next, NaOH is added and according to
the solubility rules, we know that hydroxides are generally all insoluble except when they are
combined with ammonium ion or any Group I ions. So since no precipitate is formed, none of
the other cations can be in the starting solution because if they were, then a precipitate would
have been formed with OH. So, the only cations that were present in the original solution were
Ca2+ and Hg+..
Acid and bases (PCI Syllabus, B.Pharm)
Arrhenius concept of acids and bases, bronsted-lowry theory of acids and bases,amphoteric nature of water , characteristics of strong acids , characteristics of weak acids , characteristics of strong bases, characteristics of weak bases, conjugate acids, conjugate base,introduction on buffers , preparation of buffers, types of buffer, acidic buffer, basic buffer, how do buffers act, why doesn't the ph of buffers doesn't change , Handerson-Hasselbach equation, buffer capacity, pharmaceutical buffers, why maintainance of body ph is important, osmolarity of blood, isotonic, hypertonic and hypotonic solution, pharmaceutical buffer system,phosphate-buffered saline, methods to measure tonicity, hemolytic method, methods to adjust tonicity, cryoscopic method, NaCl equivalent method
When two data sets are pooled together the pooled mean will alway.pdf
When two data sets are pooled together the pooled mean will always be more accurate, between
original mean values and more precise.
Ans: (e)
Solution
When two data sets are pooled together the pooled mean will always be more accurate, between
original mean values and more precise.
Ans: (e).
DELTOID flexes and medially rotates arm; prime mover of arm abduct.pdf
DELTOID : flexes and medially rotates arm; prime mover of arm abduction; extends and
laterally rotates arm. LATISSIMUS DORI : prime mover of arm extension; adducts and medially
rotates arm (big muscle). PECTORALIS MAJOR : prime mover of arm flexion, adducts and
medially rotates arm ( big muscle). TERES MAJOR : Extends, adducts and medially rotates arm.
Solution
DELTOID : flexes and medially rotates arm; prime mover of arm abduction; extends and
laterally rotates arm. LATISSIMUS DORI : prime mover of arm extension; adducts and medially
rotates arm (big muscle). PECTORALIS MAJOR : prime mover of arm flexion, adducts and
medially rotates arm ( big muscle). TERES MAJOR : Extends, adducts and medially rotates arm..
More Related Content
Similar to Once HCl is added in solution, the ions would dis.pdf
Acids, Bases And Buffers Pharmaceutical Inorganic chemistry UNIT-II (Part-I)
Acids, Bases And Buffers Pharmaceutical Inorganic chemistry UNIT-II (Part-I)
Acids, Bases are defined by Four main theories,
1.Traditional theory / concept
2.Arrhenius theory
3.Bronsted and Lowry theory
4.Lewis theory
Importance of acids and bases in pharmacy
Buffers: Buffer action
Buffer capacity Buffers system
Types of Buffers : Generally buffers are of two types:
1. Acidic buffers
2. Basic buffers
There are some other buffer system:
3. Two salts acts as acid-base pair. Ex- Potassium hydrogen phosphate and potassium dihydrogen phosphate.
4. Amphoteric electrolyte. Ex- Solution of glycine.
5. Solution of strong acid and solution of strong base. Ex- Strong HCl with KCl Mechanism of Buffer action: Mechanism of Action of acidic buffers: Buffer equation-Henderson-Hasselbalch equation:
Standard Buffer Solutions Preparation of Buffer Solutions: Buffers in pharmaceutical systems or Application of buffer: Stability of buffers Buffered isotonic solution Types of Buffer Isotonic solution
1. Isotonic Solutions:
2. Hypertonic Solutions:
3. Hypotonic Solution:
Measurement of Tonicity: 1. Hemolytic method: 2. Cryoscopic method or depression of freezing point:
Methods of adjusting the tonicity:
Class I methods:
In this type, sodium chloride or other substances are added to the solution in sufficient quantity to make it isotonic. Then the preparation is brought to its final volume withan isotonic or a buffered isotonic diluting solution.
These methods are of two types:
Cryoscopic method
Sodium chloride equivalent method.
Class II methods:
In this type, water is added in sufficient quantity make the preparation isotonic. Then the preparation is brought to its volume with an isotonic or a buffered isotonic diluting solution.
These methods are of two types:
White-Vincent method
Sprowls method.
When recovering an acid from its salt dissolved in water- we typically.docx
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 don’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.
.
Acid base eqm and solubility equilibrium
This chapter talks about:
Acid –base equilibria
solubility equilibria
Buffer solution
Acid-base titration
Molar solubility and solubility
pH and Solubility
For more materials subscribe to channel:
https://www.youtube.com/channel/UCzXxV4xER9NIWt316gfeO1w
You need to know the solubility rules in order to.pdf
You need to know the solubility rules in order to answer this question. When HCl is
added and forms a precipitate, we\'re looking for something that is insoluble when combined
with chlorine atoms. That would be Hg+. So we know that is in the starting solution for sure. The
first precipitate would be HgCl. Next, H2SO4 is added and a precipitate is formed so we\'re
looking for a cation that is insoluble when combined with sulfate (SO4^2-) This would be Ca2+.
(Hg+ would work too except we removed all of those ions when we removed the first
precipitate.) So, the second precipitate would be CaSO4 Next, NaOH is added and according to
the solubility rules, we know that hydroxides are generally all insoluble except when they are
combined with ammonium ion or any Group I ions. So since no precipitate is formed, none of
the other cations can be in the starting solution because if they were, then a precipitate would
have been formed with OH. So, the only cations that were present in the original solution were
Ca2+ and Hg+.
Solution
You need to know the solubility rules in order to answer this question. When HCl is
added and forms a precipitate, we\'re looking for something that is insoluble when combined
with chlorine atoms. That would be Hg+. So we know that is in the starting solution for sure. The
first precipitate would be HgCl. Next, H2SO4 is added and a precipitate is formed so we\'re
looking for a cation that is insoluble when combined with sulfate (SO4^2-) This would be Ca2+.
(Hg+ would work too except we removed all of those ions when we removed the first
precipitate.) So, the second precipitate would be CaSO4 Next, NaOH is added and according to
the solubility rules, we know that hydroxides are generally all insoluble except when they are
combined with ammonium ion or any Group I ions. So since no precipitate is formed, none of
the other cations can be in the starting solution because if they were, then a precipitate would
have been formed with OH. So, the only cations that were present in the original solution were
Ca2+ and Hg+..
Acid and bases (PCI Syllabus, B.Pharm)
Arrhenius concept of acids and bases, bronsted-lowry theory of acids and bases,amphoteric nature of water , characteristics of strong acids , characteristics of weak acids , characteristics of strong bases, characteristics of weak bases, conjugate acids, conjugate base,introduction on buffers , preparation of buffers, types of buffer, acidic buffer, basic buffer, how do buffers act, why doesn't the ph of buffers doesn't change , Handerson-Hasselbach equation, buffer capacity, pharmaceutical buffers, why maintainance of body ph is important, osmolarity of blood, isotonic, hypertonic and hypotonic solution, pharmaceutical buffer system,phosphate-buffered saline, methods to measure tonicity, hemolytic method, methods to adjust tonicity, cryoscopic method, NaCl equivalent method
Similar to Once HCl is added in solution, the ions would dis.pdf (20)
Acids, Bases And Buffers Pharmaceutical Inorganic chemistry UNIT-II (Part-I)
Acids, Bases And Buffers Pharmaceutical Inorganic chemistry UNIT-II (Part-I)
Balancing_Chemical_Equations_using_the_Law_of_Mass.ppt
Balancing_Chemical_Equations_using_the_Law_of_Mass.ppt
Balancing_Chemical_Equations_using_the_Law_of_Mass.ppt
Balancing_Chemical_Equations_using_the_Law_of_Mass.ppt
When recovering an acid from its salt dissolved in water- we typically.docx
When recovering an acid from its salt dissolved in water- we typically.docx
You need to know the solubility rules in order to.pdf
You need to know the solubility rules in order to.pdf
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When two data sets are pooled together the pooled mean will alway.pdf
When two data sets are pooled together the pooled mean will always be more accurate, between
original mean values and more precise.
Ans: (e)
Solution
When two data sets are pooled together the pooled mean will always be more accurate, between
original mean values and more precise.
Ans: (e).
DELTOID flexes and medially rotates arm; prime mover of arm abduct.pdf
DELTOID : flexes and medially rotates arm; prime mover of arm abduction; extends and
laterally rotates arm. LATISSIMUS DORI : prime mover of arm extension; adducts and medially
rotates arm (big muscle). PECTORALIS MAJOR : prime mover of arm flexion, adducts and
medially rotates arm ( big muscle). TERES MAJOR : Extends, adducts and medially rotates arm.
Solution
DELTOID : flexes and medially rotates arm; prime mover of arm abduction; extends and
laterally rotates arm. LATISSIMUS DORI : prime mover of arm extension; adducts and medially
rotates arm (big muscle). PECTORALIS MAJOR : prime mover of arm flexion, adducts and
medially rotates arm ( big muscle). TERES MAJOR : Extends, adducts and medially rotates arm..
Event Cash Flow Assets Claims Income Statement OAIAFA Cash.pdf
Event Cash Flow Assets Claims Income Statement OA/IA/FA Cash + Supplies stock +
Fixed asset = Note payable + Bank loan short term + Accounts payable + Common Stock +
Retained Earnings Revenue - Expense = Net Income a FA 75,000 + + = + + + 75,000
+ - = b OA (5,000) + 5,000 + = + + + + - = c IA (10,000) + + 80,000 =
70,000 + + + + - = d OA 27,000 + + = + 27,000 + + + - = e FA
(3,500) + + = + + + + (3,500) - = f OA 17,000 + + = + + + + 17,000 - =
17,000 g OA (4,000) + + = + + + + - 4,000 = (4,000) h OA + 1,000 + = + +
1,000 + + - = + + = + + + + - = + + = + + + + - = + + = + + + + - =
+ + = + + + + - = + + = + + + + - = 96,500 + 6,000 + 80,000 =
70,000 + 27,000 + 1,000 + 75,000 + (3,500) 17,000 - 4,000 = 13,000
Solution
Event Cash Flow Assets Claims Income Statement OA/IA/FA Cash + Supplies stock +
Fixed asset = Note payable + Bank loan short term + Accounts payable + Common Stock +
Retained Earnings Revenue - Expense = Net Income a FA 75,000 + + = + + + 75,000
+ - = b OA (5,000) + 5,000 + = + + + + - = c IA (10,000) + + 80,000 =
70,000 + + + + - = d OA 27,000 + + = + 27,000 + + + - = e FA
(3,500) + + = + + + + (3,500) - = f OA 17,000 + + = + + + + 17,000 - =
17,000 g OA (4,000) + + = + + + + - 4,000 = (4,000) h OA + 1,000 + = + +
1,000 + + - = + + = + + + + - = + + = + + + + - = + + = + + + + - =
+ + = + + + + - = + + = + + + + - = 96,500 + 6,000 + 80,000 =
70,000 + 27,000 + 1,000 + 75,000 + (3,500) 17,000 - 4,000 = 13,000.
We know that Molarity = No . of moles Volume of.pdf
We know that Molarity = No . of moles / Volume of solution in L 0.5 M = n / 1.0 L
---> n = 0.5 mol Li2SO4 ---> 2Li+ + SO4 2- So 1 mole of Li2SO4 produces 1 mole of SO4 2- &
2 moles of Li+ So No . of moles of SO4 2- ions present in 0.5 mol of Li2SO4 is 1 * 0.5 = 0.5
mol
Solution
We know that Molarity = No . of moles / Volume of solution in L 0.5 M = n / 1.0 L
---> n = 0.5 mol Li2SO4 ---> 2Li+ + SO4 2- So 1 mole of Li2SO4 produces 1 mole of SO4 2- &
2 moles of Li+ So No . of moles of SO4 2- ions present in 0.5 mol of Li2SO4 is 1 * 0.5 = 0.5
mol.
there are no molecules given..pls repost and post.pdf
there are no molecules given..pls repost and post on my profile
Solution
there are no molecules given..pls repost and post on my profile.
The false statement is Oxygen forms binary compou.pdf
The false statement is Oxygen forms binary compounds with nonmetals called acid
anhydrides.
Solution
The false statement is Oxygen forms binary compounds with nonmetals called acid
anhydrides..
The area is how many Hydrogens are on the graph. .pdf
The area is how many Hydrogens are on the graph. You would need a correlation
table to determine what groups are on the figure. First, calculate the degrees of saturation. (18-
10)/2 = 4 degrees this could be a ring wth 3 pi bonds maybe benxene
Solution
The area is how many Hydrogens are on the graph. You would need a correlation
table to determine what groups are on the figure. First, calculate the degrees of saturation. (18-
10)/2 = 4 degrees this could be a ring wth 3 pi bonds maybe benxene.
SO4{2-} (aq) + Sn{2+} (aq) + 4 H{+} (aq)=========.pdf
SO4{2-} (aq) + Sn{2+} (aq) + 4 H{+} (aq)============> H2SO3(aq) + Sn{4+}
(aq)+ H2O(l)
Solution
SO4{2-} (aq) + Sn{2+} (aq) + 4 H{+} (aq)============> H2SO3(aq) + Sn{4+}
(aq)+ H2O(l).
its e1 OMe attacks at 3-position by transfer of a.pdf
its e1 OMe attacks at 3-position by transfer of anion to stable form
Solution
its e1 OMe attacks at 3-position by transfer of anion to stable form.
H and F are identical, you just need to rotate H .pdf
H and F are identical, you just need to rotate H until it matches F, I is also identical
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imposable on their mirror image) to each other.
Solution
H and F are identical, you just need to rotate H until it matches F, I is also identical
to G. H and F are enantiomers of I and G, this also means that they are Chiral (non super
imposable on their mirror image) to each other..
From weakest to strongest A, B, C Butanal is an .pdf
From weakest to strongest: A, B, C Butanal is an aldehyde which means it has the
C=O so it cannot hydrogen bond, only dipole-dipole interactions. However, butanol can
hydrogen-bond because of the OH group so it would have the strongest intermolecular force.
Pentane would have weakest because it has no electronegative atoms.
Solution
From weakest to strongest: A, B, C Butanal is an aldehyde which means it has the
C=O so it cannot hydrogen bond, only dipole-dipole interactions. However, butanol can
hydrogen-bond because of the OH group so it would have the strongest intermolecular force.
Pentane would have weakest because it has no electronegative atoms..
Effusion is inversely proportional to the mass. .pdf
Effusion is inversely proportional to the mass. Hence, Methane > HF > Propane >
Hydrogen sulfide
Solution
Effusion is inversely proportional to the mass. Hence, Methane > HF > Propane >
Hydrogen sulfide.
Benzoin and methanol both have an alcohol group, .pdf
Benzoin and methanol both have an alcohol group, but benzoin has a ketone group
and phenyl groups within it also. The interesting part about benzoin is that it is a conjugated pi
system, meaning that electrons from the phenyl groups can shift around the double bonds
throughout the whole molecule without effecting anything; this also impacts the ketone group.
This is also known as resonance, you can see an example here
http://alevelchem.com/drop/peptide_angles.gif Because of this effect, the whole rest of the
benzoin moiety (besides the alcohol group) can be considered one giant add-on (Let\'s call it R)
If methanol looks like this... H3C-OH Then our benzoin because of its delocalized resonant
electrons will, in comparison, just look like... R-OH When placed side by side this is how both
molecules will act with each other. Therefore the polarities of both molecules are similar because
electronically both molecules function the same way (Methanol with a CH3, and Benzoin with a
conjugated R group).
Solution
Benzoin and methanol both have an alcohol group, but benzoin has a ketone group
and phenyl groups within it also. The interesting part about benzoin is that it is a conjugated pi
system, meaning that electrons from the phenyl groups can shift around the double bonds
throughout the whole molecule without effecting anything; this also impacts the ketone group.
This is also known as resonance, you can see an example here
http://alevelchem.com/drop/peptide_angles.gif Because of this effect, the whole rest of the
benzoin moiety (besides the alcohol group) can be considered one giant add-on (Let\'s call it R)
If methanol looks like this... H3C-OH Then our benzoin because of its delocalized resonant
electrons will, in comparison, just look like... R-OH When placed side by side this is how both
molecules will act with each other. Therefore the polarities of both molecules are similar because
electronically both molecules function the same way (Methanol with a CH3, and Benzoin with a
conjugated R group)..
When acid and base react together they form salt and water.This reac.pdf
When acid and base react together they form salt and water.This reaction is known as
Neutralization Reaction. ForExample
HCl+NaOH ---------> NaCl + H2O
HCl is acid while NaOH is a base, when they react with eachother they form NaCl, that is a salt
(table salt)
These reactions are also reversible reactions.
Solution
When acid and base react together they form salt and water.This reaction is known as
Neutralization Reaction. ForExample
HCl+NaOH ---------> NaCl + H2O
HCl is acid while NaOH is a base, when they react with eachother they form NaCl, that is a salt
(table salt)
These reactions are also reversible reactions..
There are 37 classes in java.lang package as per Java SE7 The java.pdf
There are 37 classes in java.lang package as per Java SE7
The java.lang package contains classes that are fundamental to the design of the Java language.
1) Object, Class, ClassLoader, System and Compiler classes
2) Wrapper classes:- Boolean, Character, Byte, Short, Integer, Long, Float, and Double
3) Math, StrictMath classes
4) String, StringBuffer and StringBuilder classes
5) Process, ProcessBuilder, Runtime, RuntimePermission classes
6) Enum>, Throwable, Void classes
----------------------------------------------
1) Object, Class, ClassLoader, System and Compiler classes
----------------------------------------------
Object: The Object class is the root of the class hierarchy. Every class has Object as a
superclass. All objects, including arrays, implement the methods of this class.
Class: The Class class instance represent classes and interfaces in a running Java application.It
has no public constructor.
ClassLoader: The ClassLoader class is an object that is responsible for loading classes. This class
is an abstract class. It may be used by security managers to indicate security domains.
System: The System class contains several useful class fields and methods. It cannot be
instantiated.
Some of the facilities provided by System class are standard input, output, error output streams
and access to externally defined properties and environment variables. This class has a utility
method for quickly copying a portion of an array. Also it has methods for loading files and
libraries
Compiler: The Compiler class is provided to support Java-to-native-code compilers and related
services. By design, it serves as a placeholder for a JIT compiler implementation.
SecurityManager: The SecurityManager class allows applications to implement a security policy.
It allows an application to determine, before performing a possibly unsafe or sensitive operation,
what the operation is and whether it is being attempted in a security context that allows the
operation to be performed. The application can allow or disallow the operation.
----------------------------------------------
2) Wrapper classes:- Boolean, Character, Byte, Short, Integer, Long, Float, and Double
----------------------------------------------
Frequently it is necessary to represent a value of primitive type as if it were an object. The
wrapper classes serve this purpose.
An object of type Double, for example, contains a field whose type is double, representing that
value in such a way that a reference to it can be stored in a variable of reference type. These
classes also provide a number of methods for converting among primitive values, as well as
supporting such standard methods as equals and hashCode.
Other classes in support to wrapper classes:
----------------------------------------------
Number: The Number class is the superclass of classes BigDecimal, BigInteger, Byte, Double,
Float, Integer, Long, and Short.The Subclasses of Number must provide methods to convert the
represented num.
The solution is as belowEmployeeDemo.javaimport java.util.Scann.pdf
The solution is as below:
EmployeeDemo.java
import java.util.Scanner;
public class EmployeeDemo
{
static Scanner scan = new Scanner(System.in);
public static void main(String[] args)
{
Company clerk = new Company( );
Employee employee = new Employee();// one student
int count,dependent;
String fname,lname;
float hourly_rate,no_of_hours,l_tax,f_tax,s_tax;
System.out.println(\"Enter number of employees:\");
int numberOfEmployees = scan.nextInt();
for (count = 0; count < numberOfEmployees; count++)
{
System.out.println(\"Enter data for employee number \" + count);
System.out.println(\"First Name \");
fname = scan.next();
System.out.println(\"Last Name \");
lname = scan.next();
System.out.println(\"Number of dependents \");
dependent = scan.nextInt();
System.out.println(\"Hourly rate \");
hourly_rate = scan.nextFloat();
System.out.println(\"Number of hours worked \");
no_of_hours = scan.nextFloat();
System.out.println(\"Local tax withheld to date \");
l_tax = scan.nextFloat();
System.out.println(\"Federal tax withheld to date \");
f_tax = scan.nextFloat();
System.out.println(\"State tax withheld to date \");
s_tax = scan.nextFloat();
employee.readInput(fname,lname,dependent,hourly_rate,no_of_hours,l_tax,f_tax,s_tax);
employee.calculateData(count);
employee.writeOutput(count);
clerk.colectDataForCompanyReport(employee,count);
}
clerk.printDataForCompanyReport();
}
}
Employee.java
import java.util.ArrayList;
public class Employee {
private ArrayList fname,lname;
private ArrayList dependent;
private ArrayList
hourly_rate,no_of_hours,l_tax,f_tax,s_tax,g_wages,cf_tax,cl_tax,cs_tax,t_tax,ct_tax,net_pay;
public Employee(){
fname = new ArrayList();
lname = new ArrayList();
dependent = new ArrayList();
hourly_rate = new ArrayList();
no_of_hours = new ArrayList();
l_tax = new ArrayList();
f_tax = new ArrayList();
s_tax = new ArrayList();
g_wages = new ArrayList();
cl_tax = new ArrayList();
cf_tax = new ArrayList();
cs_tax = new ArrayList();
t_tax = new ArrayList();
ct_tax = new ArrayList();
net_pay = new ArrayList();
}
public ArrayList getFname() {
return fname;
}
public void setFname(ArrayList fname) {
this.fname = fname;
}
public ArrayList getLname() {
return lname;
}
public void setLname(ArrayList lname) {
this.lname = lname;
}
public ArrayList getDependent() {
return dependent;
}
public void setDependent(ArrayList dependent) {
this.dependent = dependent;
}
public ArrayList getHourly_rate() {
return hourly_rate;
}
public void setHourly_rate(ArrayList hourly_rate) {
this.hourly_rate = hourly_rate;
}
public ArrayList getNo_of_hours() {
return no_of_hours;
}
public void setNo_of_hours(ArrayList no_of_hours) {
this.no_of_hours = no_of_hours;
}
public ArrayList getL_tax() {
return l_tax;
}
public void setL_tax(ArrayList l_tax) {
this.l_tax = l_tax;
}
public ArrayList getF_tax() {
return f_tax;
}
public void setF_tax(ArrayList f_tax) {
this.f_tax = f_tax;
}
public ArrayList getS_tax() {
return s_tax;
}
public void setS_tax(ArrayList s_tax) {
this.s_tax = s_ta.
The precations to be given to the workers areSolutionThe prec.pdf
The precations to be given to the workers are:
Solution
The precations to be given to the workers are:.
At 1700 cm-1 a strong signal in an IR spectum is .pdf
At 1700 cm-1 a strong signal in an IR spectum is the result of a ketone (C=O). (It
can shift if the compound is conjugated).
Solution
At 1700 cm-1 a strong signal in an IR spectum is the result of a ketone (C=O). (It
can shift if the compound is conjugated)..
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Once HCl is added in solution, the ions would dis.pdf
- 1. Once HCl is added in solution, the ions would dissociate into their component parts (H+ and Cl-). As per the second equation, adding more H+ would initially increase the quantity of products, thusly shifting the reaction towards the reactants by LeChatlier's Principle. If NaOH was added, the dissociation would create OH- anions that would soak up H+ cations. By removing products, the equlibrium is shifted towards the products by LeChatlier's Principle. Best of wishes! (: Solution Once HCl is added in solution, the ions would dissociate into their component parts (H+ and Cl-). As per the second equation, adding more H+ would initially increase the quantity of products, thusly shifting the reaction towards the reactants by LeChatlier's Principle. If NaOH was added, the dissociation would create OH- anions that would soak up H+ cations. By removing products, the equlibrium is shifted towards the products by LeChatlier's Principle. Best of wishes! (: