polysaccharides are long chains containing thousands of monosaccharide.docx
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polysaccharides are long chains containing thousands of monosaccharides most of all contain glucose as monosaccharide unit but differ in three important ways. list them. Solution A few differences between polysacharradies and monosacharadies are listed below. Monosaccharide: I) single sugar molecules. II) sweet in taste III) soluble in water IV) simple unbranched structures V) often reducing sugars Polysaccharides: I) monosacharadies are building blocks for polysaccharides II) not sweet tasting III) insoluble in water IV) complex branced structures V) non-reducing carbohydrate molecules .
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Please Type Answer! Explain how the Brnsted-Lowry definition of bases.docx
Please Type Answer!
Explain how the Brønsted-Lowry definition of bases differs from the Arrhenius definition. Give an example of a compound that is not a base under the Arrhenius definition but is under the Brønsted-Lowry definition.
Solution
The Bronsted-Lowry Theory of acids and bases states that
An acid is a proton (hydrogen ion) donor.
A base is a proton (hydrogen ion) acceptor.
Whereas according to Arrhenius theory
Acids are substances which produce hydrogen ions in solution.
Bases are substances which produce hydroxide ions in solution.
So we see that the definition of acid is same in both theories but the base definition is different.
NH3 ( ammonia) is not a base under arhenius theory as it cannot give OH- but it can accept a H+ that\'s why it is a bronsted- base.
.
Please show work- Thank you!Solutiona) First law of thermodynamics- q.docx
Please show work. Thank you!
Solution
a)
First law of thermodynamics:
q - w = du
where work done w = p*dv = p*(v2 - v1) = 0 since for constant volume process, v1 = v2.
Therefore, q = du
For ideal gas, du = Cv*dT
so q = Cv*dT
For liquids, du = v dp
so, q = v*dp
b)
First law of thermodynamics:
q - w = du
where work done w = p*dv = p*(v2 - v1)
Therefore, q = du + p*dv
q = (u2 - u1) + p*(v2 - v1)
q = (u2 + p*v2) - (u1 + p*v1)
But enthalpy h = u + pv
So, q = h2 - h1
q = dh
For ideal gas, dh = Cp*dT
so q = Cp*dT
For liquids, dh = Cp*(T2 - T1)
.
please show workSolution1- From the conservation of angular momentum t.docx
please show work
Solution
1.
From the conservation of angular momentum the angular momentum of the system remains same if the lady bug moves towards the ege from the center
2. moment of inertia increases
I = m*r2
I = Idisk + I_ladybug
previously I_ladybug = 0
3. Angular velocity is decreases because the moment of inertia increases to make the angular momentum remains same
4. the rotational kinetic energy is
k.e = 0.5*I*W^2
As the I increased , W will decreased so that the rotational k.e is decreased
.
Please respond in detail- Thank you! Discuss some of the manifestation.docx
Please respond in detail, Thank you!
Discuss some of the manifestations of the globalization of world capital markets.
Solution
Some of The Manifestations of Globalization of World capital markets are as follows:
1) Increasing globalization of financial intermediaries such as banks and other institutions.
2) Interconnectedness of financial markets such as stock markets etc (one can observe a high degree of correlation between these).
3) increased discussion of foreign financial markets in the US press
4) Growth in markets for international financial instruments such as Euro dollar & Euro Bonds.
5) Also disruption in a major foreign market can cause disruption in the US and vice versa. etc.,
.
please no cursive handwritingSolution10- -HCN- - 0-150 M HCN ---- H+ +.docx
please no cursive handwriting
Solution
10.
[HCN] = 0.150 M
HCN <==> H+ + CN-
let x amount dissociated
Ka = [CN-][H+]/[HCN]
4.9 x 10^-10 = x^2/0.150
x = [H+] = 8.6 x 10^-6 M
pH = -log[H+] = 5.07
pH before addition of base,
A) 5.07
--
11.
N3H + NaOH --> N3Na + H2O
moles N3H = 0.150 M x 25 ml = 3.75 mmol
moles NaOH added = 0.150 M x 26 ml = 3.9 mmol
excess [OH-] = 0.15 mmol/51 ml = 0.003 moles
[H+] = 1 x 10^-14/0.003 = 3.4 x 10^-12 M
pH = -log[H+] = 11.47
pH after adding 26 ml base,
B) 11.47
--
12. moles HOCl = 0.0998 M x 50 ml = 4.99 mmol
moles NaOH needed = 4.99 mmol
Volume NaOH needed = 4.99 mmol/0.120 M = 41.6 ml
Volume needed,
C) 41.6 ml
.
Problem 10-132 -LO2- Mr- Earl Pearl- accountant for Margie K.docx
Problem 10-132 [LO2]
Solution
a) Lost contribution margin -13000 Less fixed expenses avoided: Rent 1950 Utilities 920 Supervisors\' salaries 3,900 Maintenance 1,200 Administrative expenses 3,090 11060 Disadvantage of dropping Product C -1940 B) Since there is net disvantage of (1940) to dropping Product C , it should not be dropped.
.
Problem 1- Determine whether the following molecules are identical- co.docx
Problem 1: Determine whether the following molecules are identical, constitutional isomers, enantiomers or diastereomers ?? b. a. OH C. OH ?? e. ?? OH CI Cl h. Br m. m.
Solution
constitutional isomers differ by the connectivity of the bond, enantiomers are mirror images of each other and diastereomers has the same connectivity of bond but are different in spatial position by rotating the bond they can be made identical.
a) diastereomers
b)diastereomers
c)enantiomers
d)diastereomers
e)diastereomers
f)diastereomers
g)diastereomers
h)diastereomers
i) constitutional isomers
j)constitutional isomers
k)diastereomers
l)diastereomers
m) enantiomers
n) diastereomers
.
Please provide answer to the question below and why- Question 10- P.docx
Please provide answer to the question below and why?
Solution
b) Significant Influence
Reason: If the Parent Company acquired shares in company between 20% - 50%, it is known as Significant Influence.
.
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Please Type Answer! Explain how the Brnsted-Lowry definition of bases.docx
Please Type Answer!
Explain how the Brønsted-Lowry definition of bases differs from the Arrhenius definition. Give an example of a compound that is not a base under the Arrhenius definition but is under the Brønsted-Lowry definition.
Solution
The Bronsted-Lowry Theory of acids and bases states that
An acid is a proton (hydrogen ion) donor.
A base is a proton (hydrogen ion) acceptor.
Whereas according to Arrhenius theory
Acids are substances which produce hydrogen ions in solution.
Bases are substances which produce hydroxide ions in solution.
So we see that the definition of acid is same in both theories but the base definition is different.
NH3 ( ammonia) is not a base under arhenius theory as it cannot give OH- but it can accept a H+ that\'s why it is a bronsted- base.
.
Please show work- Thank you!Solutiona) First law of thermodynamics- q.docx
Please show work. Thank you!
Solution
a)
First law of thermodynamics:
q - w = du
where work done w = p*dv = p*(v2 - v1) = 0 since for constant volume process, v1 = v2.
Therefore, q = du
For ideal gas, du = Cv*dT
so q = Cv*dT
For liquids, du = v dp
so, q = v*dp
b)
First law of thermodynamics:
q - w = du
where work done w = p*dv = p*(v2 - v1)
Therefore, q = du + p*dv
q = (u2 - u1) + p*(v2 - v1)
q = (u2 + p*v2) - (u1 + p*v1)
But enthalpy h = u + pv
So, q = h2 - h1
q = dh
For ideal gas, dh = Cp*dT
so q = Cp*dT
For liquids, dh = Cp*(T2 - T1)
.
please show workSolution1- From the conservation of angular momentum t.docx
please show work
Solution
1.
From the conservation of angular momentum the angular momentum of the system remains same if the lady bug moves towards the ege from the center
2. moment of inertia increases
I = m*r2
I = Idisk + I_ladybug
previously I_ladybug = 0
3. Angular velocity is decreases because the moment of inertia increases to make the angular momentum remains same
4. the rotational kinetic energy is
k.e = 0.5*I*W^2
As the I increased , W will decreased so that the rotational k.e is decreased
.
Please respond in detail- Thank you! Discuss some of the manifestation.docx
Please respond in detail, Thank you!
Discuss some of the manifestations of the globalization of world capital markets.
Solution
Some of The Manifestations of Globalization of World capital markets are as follows:
1) Increasing globalization of financial intermediaries such as banks and other institutions.
2) Interconnectedness of financial markets such as stock markets etc (one can observe a high degree of correlation between these).
3) increased discussion of foreign financial markets in the US press
4) Growth in markets for international financial instruments such as Euro dollar & Euro Bonds.
5) Also disruption in a major foreign market can cause disruption in the US and vice versa. etc.,
.
please no cursive handwritingSolution10- -HCN- - 0-150 M HCN ---- H+ +.docx
please no cursive handwriting
Solution
10.
[HCN] = 0.150 M
HCN <==> H+ + CN-
let x amount dissociated
Ka = [CN-][H+]/[HCN]
4.9 x 10^-10 = x^2/0.150
x = [H+] = 8.6 x 10^-6 M
pH = -log[H+] = 5.07
pH before addition of base,
A) 5.07
--
11.
N3H + NaOH --> N3Na + H2O
moles N3H = 0.150 M x 25 ml = 3.75 mmol
moles NaOH added = 0.150 M x 26 ml = 3.9 mmol
excess [OH-] = 0.15 mmol/51 ml = 0.003 moles
[H+] = 1 x 10^-14/0.003 = 3.4 x 10^-12 M
pH = -log[H+] = 11.47
pH after adding 26 ml base,
B) 11.47
--
12. moles HOCl = 0.0998 M x 50 ml = 4.99 mmol
moles NaOH needed = 4.99 mmol
Volume NaOH needed = 4.99 mmol/0.120 M = 41.6 ml
Volume needed,
C) 41.6 ml
.
Problem 10-132 -LO2- Mr- Earl Pearl- accountant for Margie K.docx
Problem 10-132 [LO2]
Solution
a) Lost contribution margin -13000 Less fixed expenses avoided: Rent 1950 Utilities 920 Supervisors\' salaries 3,900 Maintenance 1,200 Administrative expenses 3,090 11060 Disadvantage of dropping Product C -1940 B) Since there is net disvantage of (1940) to dropping Product C , it should not be dropped.
.
Problem 1- Determine whether the following molecules are identical- co.docx
Problem 1: Determine whether the following molecules are identical, constitutional isomers, enantiomers or diastereomers ?? b. a. OH C. OH ?? e. ?? OH CI Cl h. Br m. m.
Solution
constitutional isomers differ by the connectivity of the bond, enantiomers are mirror images of each other and diastereomers has the same connectivity of bond but are different in spatial position by rotating the bond they can be made identical.
a) diastereomers
b)diastereomers
c)enantiomers
d)diastereomers
e)diastereomers
f)diastereomers
g)diastereomers
h)diastereomers
i) constitutional isomers
j)constitutional isomers
k)diastereomers
l)diastereomers
m) enantiomers
n) diastereomers
.
Please provide answer to the question below and why- Question 10- P.docx
Please provide answer to the question below and why?
Solution
b) Significant Influence
Reason: If the Parent Company acquired shares in company between 20% - 50%, it is known as Significant Influence.
.
Prepare the financial ratios for J-C- Penney Company- Inc- (JCP) for t.docx
Prepare the financial ratios for J.C. Penney Company, Inc. (JCP) for the years indicated below. Use the JCP financial documents (Balance Sheets and Income Statements) for Calendar Years 2011, 2012, 2013 to calculate the required Financial Ratios and reference in your assessment analysis. Including the following:
Prepare the financial ratios for 2011, 2012, 2013.
Comment on the differences between J.C. Penney\'s ratios and the industry averages.
Determine the trends evident in the financial statements.
Write a 3-page memo divided into the following sections:
Liquidity.
Asset management.
Long-term debt-paying ability.
Profitability.
I have the documents that were given to me (balance sheet and income statement) but I dont know how to attach them for your review. These are made up numbers not the real statements online.
Solution
Take a snap and attach the image.
.
Please name the following net ionic c THAT HAVE AN ARROW NEXT TO THEM.docx
Please name the following net ionic c
THAT HAVE AN ARROW NEXT TO THEM
4. Naming Ionic Compounds Write the correct name for each of the following ionic compounds Formula Name Nslac0 2 Pb(HCO) 3. Al:S uummum Sulud ? 10. Na CrO Sadumshromate Potnssiun Chroanat 12. ?CuP 13. MgiN 15 NH MnO 17 BaSO 8. SnlNO, 19. AgCIO 20. Cu(HSO;2 um Sullate Silver chiias
Solution
2)
Pb(HCO3)2 ----------------> lead bicarbonate
6)
CoF3 ----------------> cobalt (III) fluoride
7)
Fe2(HPO4)3----------------> Iron (III) monohydrogen phosphate
9)
SnO2 -------------------->Tin (IV) oxide
12)
Cu3P --------------------> copper(I)phosphide
15)
NH4 MnO4 ------------------------> ammonium permanganate
.
Preparation of cyclohexene from cyclohexanol Preparation of cyclohex.docx
Preparation of cyclohexene from cyclohexanol
Preparation of cyclohexene from cyclohexanol
Solution
cyclohexanol ................> cyclohexene + H2O
this is the dehydration reaction.
here in this reaction water is formed. this is exothermic reaction. so temperature of reaction mixture increases.
so there is possibility of reverse reaction in presence of water molecules and acid.
so that again cyclohexanol is produced.
2) sulfuric acid is used as it is dehydration reaction.
HCl take part in reaction and pruduce addition product chloro cyclohexane.
.
Potassiumiodide(KI)exhibitspredominantly ionic bonding- The K-- and I-.docx
Potassiumiodide(KI)exhibitspredominantly ionic bonding. The K ?? and I ?? ions have elec- tron structures that are identical to which two inert gases?
Solution
Potassium (K) ion has an electron configuration identical to the inert gas Argon
The iodine ion has an electron configuration identical to the inert gas Xenon.
.
Portfolio Beta Your investment club has only two stocks in its portfol.docx
Portfolio Beta Your investment club has only two stocks in its portfolio; $25,000 is invested in a stock with a beta of 0.4, and $75,000 is invested in a stock with a beta of 2.3. What is the portfolio\'s beta? Round your answer to two decimal places.
Solution
Portfolio Value = $100,000
Investment in first Stock= $25,000
Investment second Stock= $75,000
Weight of first Stock: $25,000 / $100,000 = 0.25
Weight of second Stock: $75,000 / $100,000 = 0.75
Beta of first Stock= 0.4
Beta of second Stock= 2.3
Beta (portfolio) = [Weight (Stock A) X Beta (Stock A)] + [Weight (Stock B) X Beta (Stock B)]
=[0.25 x 0.4]+ [0.75 x 2.3]
=0.1+1.725=1.83
.
Polyvinyl alcohol (PVA) is generally obtained by hydrolysis from polyv.docx
Polyvinyl alcohol (PVA) is generally
obtained by hydrolysis from polyvinyl
acetate, The solubility of PVA in water is
increased up to 87% of the extent of
hydrolysis but the solubility is rather
reduced It is showing, Explain why
Solution
This may be due to the hydrogen bonding interactions.
In the case of fully hydrolysed polyvinyl alcohols, the inner molecuar hydrogen bonding will be very high.That causes insolubility. But in the case of partially hydrolysed polyvinyl alcohols,the hydrogen bonding is balancing with with the desired aqueous solubility. The residual acetate groups reduce the combination by reducing the regularity of the chain structure.
Hence the answer.
.
678020731-Sumas-y-Restas-Para-Colorear.pdf
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Take a snap and attach the image.
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Please name the following net ionic c
THAT HAVE AN ARROW NEXT TO THEM
4. Naming Ionic Compounds Write the correct name for each of the following ionic compounds Formula Name Nslac0 2 Pb(HCO) 3. Al:S uummum Sulud ? 10. Na CrO Sadumshromate Potnssiun Chroanat 12. ?CuP 13. MgiN 15 NH MnO 17 BaSO 8. SnlNO, 19. AgCIO 20. Cu(HSO;2 um Sullate Silver chiias
Solution
2)
Pb(HCO3)2 ----------------> lead bicarbonate
6)
CoF3 ----------------> cobalt (III) fluoride
7)
Fe2(HPO4)3----------------> Iron (III) monohydrogen phosphate
9)
SnO2 -------------------->Tin (IV) oxide
12)
Cu3P --------------------> copper(I)phosphide
15)
NH4 MnO4 ------------------------> ammonium permanganate
.
Preparation of cyclohexene from cyclohexanol Preparation of cyclohex.docx
Preparation of cyclohexene from cyclohexanol
Preparation of cyclohexene from cyclohexanol
Solution
cyclohexanol ................> cyclohexene + H2O
this is the dehydration reaction.
here in this reaction water is formed. this is exothermic reaction. so temperature of reaction mixture increases.
so there is possibility of reverse reaction in presence of water molecules and acid.
so that again cyclohexanol is produced.
2) sulfuric acid is used as it is dehydration reaction.
HCl take part in reaction and pruduce addition product chloro cyclohexane.
.
Potassiumiodide(KI)exhibitspredominantly ionic bonding- The K-- and I-.docx
Potassiumiodide(KI)exhibitspredominantly ionic bonding. The K ?? and I ?? ions have elec- tron structures that are identical to which two inert gases?
Solution
Potassium (K) ion has an electron configuration identical to the inert gas Argon
The iodine ion has an electron configuration identical to the inert gas Xenon.
.
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Portfolio Beta Your investment club has only two stocks in its portfolio; $25,000 is invested in a stock with a beta of 0.4, and $75,000 is invested in a stock with a beta of 2.3. What is the portfolio\'s beta? Round your answer to two decimal places.
Solution
Portfolio Value = $100,000
Investment in first Stock= $25,000
Investment second Stock= $75,000
Weight of first Stock: $25,000 / $100,000 = 0.25
Weight of second Stock: $75,000 / $100,000 = 0.75
Beta of first Stock= 0.4
Beta of second Stock= 2.3
Beta (portfolio) = [Weight (Stock A) X Beta (Stock A)] + [Weight (Stock B) X Beta (Stock B)]
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.
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Polyvinyl alcohol (PVA) is generally
obtained by hydrolysis from polyvinyl
acetate, The solubility of PVA in water is
increased up to 87% of the extent of
hydrolysis but the solubility is rather
reduced It is showing, Explain why
Solution
This may be due to the hydrogen bonding interactions.
In the case of fully hydrolysed polyvinyl alcohols, the inner molecuar hydrogen bonding will be very high.That causes insolubility. But in the case of partially hydrolysed polyvinyl alcohols,the hydrogen bonding is balancing with with the desired aqueous solubility. The residual acetate groups reduce the combination by reducing the regularity of the chain structure.
Hence the answer.
.
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- 1. polysaccharides are long chains containing thousands of monosaccharides most of all contain glucose as monosaccharide unit but differ in three important ways. list them. Solution A few differences between polysacharradies and monosacharadies are listed below. Monosaccharide: I) single sugar molecules. II) sweet in taste III) soluble in water IV) simple unbranched structures V) often reducing sugars Polysaccharides: I) monosacharadies are building blocks for polysaccharides II) not sweet tasting III) insoluble in water IV) complex branced structures V) non-reducing carbohydrate molecules