Consider an electrolytic cell with a platinum anode and a silver cathode in a 1.0 M AgNO3(aq) solution.
a) (3 pts) What species can be reduced in this solution? Which species is preferentially reduced? Write the reduction half- reaction. (Note that oxyanions like nitrate are not commonly reduced in aqueous electrolysis due to kinetic reasons.)
b) (2 pts) Which species is oxidized during the electrolysis? Write the oxidation half-reaction. Note that acid will form as a byproduct of the oxidation.
c) (2 pts) Write the overall electrolysis reaction in net ionic and molecular forms.
d) (2 pts) Determine the external electric potential needed for the electrolysis under standard conditions.
e) (5 pts) If the electrolysis is carried out for 1.00 hour using 1.00 A current, how many grams of metal will be deposited at the cathode and how many liters of gas will form at the anode at 1.00 atm pressure and 25°C?
Answers
Answer:
Check the attached image below
Explanation:
The step by step solution to the question above can be seen in the attached image below.
Related Questions
While estimating the Caloric content of the chemical constituents present in the food item, the carbohydrate amount is discounted to some extent compared to the average value measured using bomb calorimetry. This is because the fiber content __________. Select the correct answer below: is not actually considered a carbohydrate is not digestible cannot be burned in the calorimeter is classified as a carbohydrate but stores energy like a fat.
Answers
Answer:
Fiber is "Not Digestible"
Explanation:
Carbohydrates that contain fiber cannot be completely digested. the indigestible components of fiber are measured in the calorimeter, but they are not accessible for energy in the human body.
While estimating the Caloric content of the chemical constituents present in the food item, the carbohydrate amount is discounted to some extent compared to the average value measured using bomb calorimetry. This is because the fiber content is not digestible.
How is calorie content measured?
At its most fundamental, a calorie is a measure of electricity. One Calorie (same as one kilocalorie, or 1,000 energy) is the quantity of power that is required to warm one kilogram of water 1 degree Celsius at sea stage. The strength content of meals become traditionally measured by the usage of bomb calorimetry.
As an alternative, the entire caloric price is calculated by using including the calories supplied by using the strength-containing nutrients: protein, carbohydrate, fats, and alcohol.
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A student wishes to calculate the experimental value of Ksp for AgI. S/he follows the procedure in Part 3 and finds Ecell to be 0.839 V. S/he does not perform Part 2; instead, s/he uses the theoretical slope of the Nernst plot for an Ag/Ag+ concentration cell, as instructed by her/his TA; this value is −0.0591 V. (a) What is the concentration of Ag+(aq) in the saturated solution of AgI (i.e., [Ag+ ]dilute)? ([Ag+ ]conc = 1.0 ✕ 10−1 M.) M (b) Using [I− ] = 0.20 M, calculate the experimental Ksp. (c) Suppose s/he mistakenly uses 1.039 V as Ecell. How does this affect [Ag+ ]dilute? Will it be too high, too low, or unaffected?
Answers
Answer:
a) [Ag+]dilute = 6.363 × 10⁻¹⁶ M
b) 1.273 × 10⁻¹⁶
c) 2.629×10⁻¹⁹ M Thus; the value for [Ag+ ]dilute will be too low
Explanation:
In an Ag | Ag+ concentration cell ,
The anode reaction can be written as :
Ag ----> Ag+(dilute) + e- &:
The cathode reaction can be written as:
Ag+(concentrated) + e- ----> Ag
The Overall Reaction : is
Ag+(concentrated) -----> Ag+(dilute)
However, the Standard Reduction potential of cell = E°cell = 0
( since both cathode and anode have same Ag+║Ag )
Also , given that the theoretical slope is - 0.0591 V
Therefore; the reduction potential of cell ; i.e
Ecell = E°cell - 0.0591 V × log ( [Ag+]dilute / [Ag+]concentrated )
0.839 V = 0 - 0.0591 V × log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) )
log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) ) = - 14.1963
[Ag+]dilute = [tex]\mathbf{10^{-14.1963} }[/tex] × 1.0 × 10⁻¹ M
[Ag+]dilute = 6.363 × 10⁻¹⁶ M
b)
AgI ----> Ag + (dilute) + I⁻
So , Solubility product = Ksp = [Ag⁺]dilute × [I⁻]
= 6.363 × 10⁻¹⁶ M × 0.20 M
= 1.273 × 10⁻¹⁶
c) If s/he mistakenly uses 1.039 V as Ecell; then the value for [Ag+]dilute will be :
Ecell = E°cell - 0.0591 V × log ( [Ag+]dilute / [Ag+]concentrated )
1.039 V = 0 - 0.0591 V × log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) )
log ( [Ag+]dilute / ( 1.0 × 10⁻¹ M ) ) = - 17.5804
[Ag+]dilute = [tex]\mathbf{10^{-17.5804} }[/tex] × 1.0 × 10⁻¹ M
[Ag+]dilute = 2.629×10⁻¹⁹ M
Thus, the value for [Ag+ ]dilute will be too low
Complete the following sentences. Match the words in the left-hand column with the appropriate blank in the sentences in the right-hand column. View Available Hint(s) ResetHelp 1. A monosaccharide is a(n) glucose if the carbonyl group is on the end of the carbon chain. 2. If a carbohydrate, like xylulose, has five carbon atoms and a carbonyl group on the second carbon, it is called a(n) aldotriose. 3. Any carbohydrate with the carbonyl group on the second carbon is a(n) ketopentose. 4. With the carbonyl group on the end of a six-carbon chain, the carbohydrate would be classified as a(n) aldose. 5. The most common carbohydrate, ketose, has six carbon atoms. 6. Glyceraldehyde is an example of a(n) aldohexose, because it has three carbon atoms.
Answers
1. If a carbohydrate, like xylulose, has five carbon atoms and a carbonyl group on the second carbon, it is called a(n) ketopentose.
2. Any carbohydrate with the carbonyl group on the second carbon is a(n) ketose.
3. The most common carbohydrate, glucose, has six carbon atoms.
4. A monosaccharide is a(n) aldose if the carbonyl group is on the end of the carbon chain.
5. Glyceraldehyde is an example of a(n) triose, because it has three carbon atoms.
6. With the carbonyl group on the end of a six-carbon chain, the carbohydrate would be classified as a(n) aldohexose.
1. The term "ketopentose" refers to a five-carbon carbohydrate with a carbonyl group on the second carbon.
2. A "ketose" is a carbohydrate with a carbonyl group (ketone group) on an internal carbon atom.
3. "Glucose" is a common hexose sugar with six carbon atoms.
4. An "aldose" is a monosaccharide with a carbonyl group at the end of the carbon chain.
5. "Triose" denotes a three-carbon sugar, and glyceraldehyde is an example.
6. An "aldohexose" is a six-carbon sugar with the carbonyl group at the end of the carbon chain.
In summary, Carbohydrates are classified based on carbon atom count and carbonyl group placement. Xylulose is a ketopentose. Glucose is a common hexose. An aldose has the carbonyl group at the chain end, while glyceraldehyde is a triose. An aldohexose has a six-carbon chain with the carbonyl group at the end.
The question probable may be:
Complete the following sentences. Match the words in the left-hand column with the appropriate blank in the sentences in the right-hand column.
aldose, ketose, ketopentose, aldohexose, triose, glucose
1. 1. If a carbohydrate, like xylulose, has five carbon atoms and a carbonyl group on the second carbon, it is called a(n)
2. 2. Any carbohydrate with the carbonyl group on the second carbon is a(n)
3. The most common carbohydrate,_________, has six carbon atoms.
4. A monosaccharide is a(n)______ if the carbonyl group is on the end of the carbon chain.
5. Glyceraldehyde is an example of a(n) _________ , because it has three carbon atoms.
6. With the carbonyl group on the end of a six-carbon chain, the carbohydrate would be classified as a(n)____________.
Enter your answer in the provided box. Lead(II) nitrate is added slowly to a solution that is 0.0400 M in Cl− ions. Calculate the concentration of Pb2+ ions (in mol / L) required to initiate the precipitation of PbCl2.
(Ksp for PbCl2 is 2.40 × 10−4.)
Answers
Answer:
the concentration of Pb2+ ions is [tex][Pb^{2+}] } = 0.15 mol/L[/tex]
Explanation:
From the question we are told that
The concentration of [tex]Cl^{-}[/tex] is [tex]C_s = 0.0400 M[/tex]
The reaction is
[tex]Pb^{2+} + 2Cl^- ----> PbCl_2[/tex]
The solubility constant for this reaction is mathematically represented as
[tex]K_{sp} = [Pb^{2+}][Cl^{-}]^2[/tex]
Substituting values
[tex]2.4 0 * 10^{-4} = 0.0400^2 * [Pb^{2+}][/tex]
[tex][Pb^{2+}] } = \frac{2.4 0 * 10^{-4} }{0.0400^2 }[/tex]
[tex][Pb^{2+}] } = 0.15 mol/L[/tex]
Which molecular formula is also an empirical formula?
A. PCl3
B. C2H4
C. H2O2
D. C6H12O6
Answers
Answer:
The molecular formula [tex]PCl_{3}[/tex] is also an empirical formula.
Explanation:
Empirical formula of a compound represents the simplest positive integer ratio of constituting atoms.
Molecular formula may or may not be equal to empirical formula.
In general, if [tex]A_{x}B_{y}[/tex] is the empirical formula then molecular formula is [tex]A_{nx}B_{ny}[/tex] where n = 1, 2, 3......
Empirical formula Molecular formula
[tex]PCl_{3}[/tex] [tex]PCl_{3}[/tex]
[tex]CH_{2}[/tex] [tex](CH_{2})_{2}[/tex] or [tex]C_{2}H_{4}[/tex]
OH [tex](OH)_{2}[/tex] or [tex]H_{2}O_{2}[/tex]
[tex]CH_{2}O[/tex] [tex](CH_{2}O)_{6}[/tex] or [tex]C_{6}H_{12}O_{6}[/tex]
Hence the molecular formula [tex]PCl_{3}[/tex] is also an empirical formula.
A molecular formula which also represents the empirical formula has its atoms in the simplest whole number ratio. For the given options, PCl3 is an example of this as its phosphorus to chlorine ratio is 1:3.
Explanation:The empirical formula represents the simplest whole number ratio of its atoms while the molecular formula represents the actual number of atoms in a molecule. A molecular formula will also represent the empirical formula if the ratio of its atoms are in their simplest form. Considering the options, we can say that A. PCl3 is also an empirical formula because the ratio of phosphorus to chlorine atoms is in its simplest form, 1:3.
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A neutralization reaction between an acid and sodium hydroxide formed water and the salt named sodium sulfate. What was the formula of the acid that was neutralized? A neutralization reaction between an acid and sodium hydroxide formed water and the salt named sodium sulfate. What was the formula of the acid that was neutralized? H2SO4 Na2SO4 H2S HCl none of the above
Answers
Answer: The formula of the acid that was neutralized is [tex]H_2SO_4[/tex]
Explanation:
Neutralization is a chemical reaction in which acid and base combine to form salt and water. Acids donate [tex]H^+[/tex] and bases donate [tex]OH^-[/tex] in water which combine to form water molecules.
As the salt formed is [tex]Na_2SO_4[/tex] , the acid must contain [tex]SO_4[/tex] and thus the acid is [tex]H_2SO_4[/tex]
The balanced chemical reaction is ;
[tex]H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O[/tex]
The formula of the acid that was neutralized is [tex]H_2SO_4[/tex]
The formula of the acid that was neutralized is H₂SO₄.
The reaction is between an acid and a base sodium hydroxide.
The reaction between an acid and a base is called a neutralization reaction. The end product of neutralization reaction is salt and water.
An acid are substance which when dissolve in water produces hydrogen ion.
The acid involve in the neutralization reaction with sodium hydroxide to produce sodium sulphate is hydrogen sulphate( H₂SO₄).
The chemical reaction can be represented as follows:
H₂SO₄ + NaOH → Na₂SO₄ + H₂O
The balanced equation will be
H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O
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Do the following calculation and express the answer using correct scientific notation. (6.00x10^23)y(3.00)/284)
Answers
To calculate the given expression, multiply 6.00x10^23 by 3.00 and divide the result by 284 to get 6.33x10^21.
Explanation:To calculate the given expression and express the answer using scientific notation, we need to perform the following steps:
Multiply 6.00x10^23 by 3.00 to get 1.80x10^24.Divide the result by 284 to get 6.33x10^21.So, the answer is 6.33x10^21.
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Matt has found a book on various marine zones. A picture illustrates a zone that shows diffused penetration of sunlight in water, but no marine plants. There are a few marine fish though. Which zone is illustrated?
Answers
Answer:
the mesopelagic, dysphotic, or twilight zone
Explanation:
Marine zones are the divisions of the ocean. The ocean is divided into two basic parts; the pelagic or open ocean, and the benthic or sea floor.
The pelagic zone is further divided into five broad zones according to how far down sunlight penetrates and they are:
1) the epipelagic, euphotic, or sunlit zone: the top layer of the ocean where enough sunlight penetrates for plants to carry on photosynthesis.
2) the mesopelagic, dysphotic, or twilight zone: a dim zone where some light penetrates, but not enough for plants to grow.
3) the bathypelagic, aphotic, or midnight zone: the deep ocean layer where no light penetrates.
4) the abyssal zone: the pitch-black bottom layer of the ocean; the water here is almost freezing and its pressure is immense.
5) the hadal zone: the waters found in the ocean's deepest trenches.
Use the observation in the first column to answer the question in the second column. observation question The enthalpy of vaporization of Substance C is bigger than that of Substance D. Which has the higher boiling point? Substance C Substance D Neither, C and D have the same boiling point. It's impossible to know without more information. At 1atm pressure, Substance E boils at 89.°C and Substance F boils at 128.°C. Which has a higher enthalpy of vaporization? Substance E Substance F Neither, E and F have the same enthalpy of vaporization. It's impossible to know without more information. At −3°C, Substance A has a vapor pressure of 147.torr and Substance B has a vapor pressure of 177.torr. Which has a higher boiling point? Substance A Substance B Neither, A and B have the same boiling point. It's impossible to know without more information.
Answers
Answer:
See explaination
Explanation:
1. Substance C
With Vapor pressure (VP) of C = 136 torr and VP of D = 186 torr. Since VP (C) < VP (D), this entaills that one needs to supply more energy to get C into the vapor phase. Hence, enthalpy of vaporization of C will be higher.
2. Substance A
B.pt (A) = 109 C
B,pt (B) = 125 C
Having A boils at a lower temperature than B, this automatically will make A to have a higher vapor pressure.
3. Substance E
Since E has a lower enthalpy of vaporization, it can be brought into the vapor phase with lower energy as compared to F. Hence, E will have a higher vapor pressure.
Substance C has a higher boiling point than Substance D, Substance F has a higher boiling point (and thereby a higher enthalpy of vaporization) than Substance E, and Substance A has a higher boiling point than Substance B due to its lower vapor pressure.
Explanation:The boiling point of a substance is directly related to its enthalpy of vaporization. In other words, a substance with a higher enthalpy of vaporization requires more heat to transform from a liquid to a gaseous state, making it have a higher boiling point.
So, based on the given observations:
Substance C has a higher enthalpy of vaporization than Substance D, therefore Substance C has a higher boiling point.Substance E boils at 89°C and Substance F boils at 128°C, therefore Substance F has a higher boiling point and consequently a higher enthalpy of vaporization.Substance A has a lower vapor pressure than Substance B at the same temperature (-3°C). Therefore, Substance A would require more energy to convert to the gaseous phase and so between Substance A and Substance B, Substance A has a higher boiling point.Learn more about Enthalpy of Vaporization here:https://brainly.com/question/32361849
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What causes the formation of ionic bonds?
Answers
When two or more atoms loss or gain electrons to produce an ion there is a formation of ionic bonds.
What do you mean by an ionic bond ?The bond is produced when an atom, specially a metal, loses an electron or electrons, and becomes a positive ion, or cation. Another atom, typically a non-metal, is able to get the electron(s) to become a negative ion, or anion.
Ionic bonds affect a cation and an anion.Ionic bonds are produce only between metals and nonmetals.
The ionic bond is produce through the transportation of electrons from the metal atoms to the non-metal atoms. The metal atoms lose their valence electrons to attain a stable noble gas electron arrangement.
Thus, The strong electrostatic forces of attraction between the oppositely-charged ions is known as ionic bond.
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One type of water softener is precipitation softening, also known as ion-exchange. Based on the reactants used (Na2CO3 and CaCl2), what ions would remain in the softened water that would be consumed by the homeowners? What could be some negative aspects of consuming these ions?
Answers
Answer:
Explanation:
In ion exchange softener , the minerals of hard water are in the form of ions of Mg⁺² and Ca⁺² . They are replaced by Na⁺ ion . The Na⁺ that resides on the bed of resin used in ion exchange softener make all heavy ions like Mg⁺² and Ca⁺² . In this way all the hard ions of hard water are removed and they are replaced by Na⁺ ion . hard ions are removed because they form insoluble compounds so they are precipitated out.
The ion that remains in soft water is Na⁺ ion.
The negative aspect of consuming this ion is that it is harmful for those suffering from heart ailment like heart pressure.
A sample of a gas at room temperature occupies a volume of 40.0 L at a pressure of 262 torr.If the pressure changes to 1310 torr,with no change in the temperature or moles of gas,what is the new volume,V2
Answers
Answer:
a) p1v1= p2v2
852 x 40.0= 4260 x v2
v2= 8 L
b) 852 x 40 = p2 x 69.0
p2= 494 torr
what is the mass of 1.25 moles of Na2O?
Answers
Answer:
Explanation:
I have the same question
what is number of unpaired
electron of Mg
Answers
Answer:
2 unpaired Electrons
Explanation:
Answer:
I was told 2 I'm not so sure, But if it's wrong I'm Sorry :( I'm having a sorta bad day...I'm trying to help everyone to make my troubles go away god bless you and have a wonderful day, ~~ Night.
What are the possible phenotypes of the offspring?
freckles with curly hair
freckles with straight hair
no freckles with curly hair
no freckles with straight hair
SHIRTS
TUNG
W
WORD
CLE
COLE
ROUTE
G
LOBREGA
Savo
WORLD
BAL
What’s the answer
Answers
Answer:freckles with straight hair
Explanation:
Answer:
The Answer is B Freckles with Straight Hair
Explanation:
Early Merry Christmas Everyone
What does STP mean? What are the values for an ideal gas? Are the values different for a real gas?
Answers
Answer: STP is a standard reference point of temperature and pressure, used when measuring gases. The universal value of STP is 1 atm (pressure) and 0o C. Note that this form specifically stated 0o C degree, not 273 Kelvin, even thought you will have to convert into Kelvin when plugging this value into the Ideal Gas equation or any of the simple gas equations. yes they are deal gases are those which follow ideal gas equation (which is PV=nRT).
Explanation:
2. If 2.50g of sodium hydroxide is being reacted with 4.30g of magnesium chloride, how many grams of magnesium hydroxide would be produced?
how should i go about solving this equation
Answers
Answer:
1.822 g of magnesium hydroxide would be produced.
Explanation:
Balanced reaction: [tex]2NaOH+MgCl_{2}\rightarrow Mg(OH)_{2}+2NaCl[/tex]
Compound Molar mass (g/mol)
NaOH 39.997
[tex]MgCl_{2}[/tex] 95.211
[tex]Mg(OH)_{2}[/tex] 58.3197
So, 2.50 g of NaOH = [tex]\frac{2.50}{39.997}[/tex] mol of NaOH = 0.0625 mol of NaOH
4.30 g of [tex]MgCl_{2}[/tex] = [tex]\frac{4.30}{95.211}[/tex] mol of [tex]MgCl_{2}[/tex] = 0.0452 mol of [tex]MgCl_{2}[/tex]
According to balanced equation-
2 mol of NaOH produce 1 mol of [tex]Mg(OH)_{2}[/tex]
So, 0.0625 mol of NaOH produce [tex](\frac{0.0625}{2})[/tex] mol of NaOH or 0.03125 mol of NaOH
1 mol of [tex]MgCl_{2}[/tex] produces 1 mol of [tex]Mg(OH)_{2}[/tex]
So, 0.0452 mol of [tex]MgCl_{2}[/tex] produce 0.0452 mol of [tex]Mg(OH)_{2}[/tex]
As least number of moles of [tex]Mg(OH)_{2}[/tex] are produced from NaOH therefore NaOH is the limiting reagent.
So, amount of [tex]Mg(OH)_{2}[/tex] would be produced = 0.03125 mol
= [tex](0.03125\times 58.3197)[/tex] g
= 1.822 g
To find the grams of magnesium hydroxide produced, you need to balance the chemical equation and calculate stoichiometry.
To find the grams of magnesium hydroxide produced, we need to balance the chemical equation and calculate the stoichiometry.
The balanced chemical equation for the reaction is:
2NaOH + MgCl2 → Mg(OH)2 + 2NaCl
Using the molar masses of sodium hydroxide (NaOH), magnesium chloride (MgCl2), and magnesium hydroxide (Mg(OH)2), we can calculate the grams of magnesium hydroxide produced:
Calculate the number of moles of sodium hydroxide and magnesium chloride using their respective masses and molar masses.Use the stoichiometry of the balanced equation to determine the moles of magnesium hydroxide produced.Convert the moles of magnesium hydroxide to grams using its molar mass.After performing these calculations, we find that 4.59g of magnesium hydroxide would be produced.
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Use molecular orbital theory to determine whether He2 2+ or He2 + is more stable. Use molecular orbital theory to determine whether He2 2+ or He2 + is more stable. The He2 + ion is more stable since it has a higher bond order (bond order = 1) than the He2 2+ ion (bond order = 1/2). The He2 + ion is more stable since it has a higher bond order (bond order = 2) than the He2 2+ ion (bond order = 1). The He2 2+ ion is more stable since it has a lower bond order (bond order = 1/2) than the He2 + ion (bond order = 1). The He2 2+ a lower bond order (bond order = 3/2) than the He2 + ion (bond order = 2). The He2 2+ ion is more stable since it has a higher bond order (bond order = 1) than the He2 + ion (bond order = 1/2).
Answers
Answer:
The He₂ 2+ ion is more stable since it has a higher bond order (bond order = 1) than the He₂ + ion (bond order = 1/2).
Explanation:
Molecular orbital of He₂⁺
[tex]1\sigma_{1s}^21\sigma(star)_{1s}^1[/tex]
There are two electrons in bonding and 1 electron in antibonding orbital
Bond order = [tex]\frac{(2-1)}{2}[/tex]
= [tex]\frac{1}{2}[/tex]
Molecular orbital of He₂⁺²
[tex]1\sigma_{1s}^21\sigma(star)_{1s}^0[/tex]
There are two electrons in bonding and 0 electron in antibonding orbital
Bond order = [tex]\frac{(2-0)}{2}[/tex]
= 1
So bond order of He₂⁺² is 1 which is more stable than He₂⁺ whose bond order is [tex]\frac{1}{2}[/tex] .
In comparing the He2 2+ and He2 + ions and viewing their stability through the lens of Molecular Orbital Theory, we see that the He2 + ion is more stable due to its higher bond order.
Explanation:Using Molecular Orbital Theory, He2 2+ and He2 + ions can be compared based on their bond order. Bond order is a measure of the stability of a bond. A higher bond order signifies greater stability. The He2 + ion has a bond order of 1, while the He2 2+ ion has a bond order of 0.5. Therefore, the major difference between He2 2+ and He2+ comes from their bond orders, and the He2+ ion is considered more stable because its bond order is higher.
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Match the terms in the first list with the characteristics in the second list.1. adsorption chromatography2. partition chromatography3. ion-exchange chromatography4. molecular exclusion chromatography5. affinity chromatography
A. Ions in mobile phase are attracted to counterions covalently attached to stationary phase
B. Solute in mobile phase is attracted to specific groups covalently attached to stationary phase
C. Solute equilibrates between mobile phase and surface of stationary phase
D. Solute equilibrates between mobile phase and film of liquid attached to stationary phase
E. Different-sized solutes penetrate pores in stationary phase to different extents. Large solutes are eluted first.
Answers
Answer: Please See answers below
Explanation:
A. Ions in mobile phase are attracted to counterions covalently attached to stationary phase -----Ion Exchange Chromatography
B. Solute in mobile phase is attracted to specific groups covalently attached to stationary phase -----Affinity Chromatography
C. Solute equilibrates between mobile phase and surface of stationary phase -----Adsorption Chromatography
D. Solute equilibrates between mobile phase and film of liquid attached to stationary phase --- Partition Chromatography
E. Different-sized solutes penetrate pores in stationary phase to different extents. Large solutes are eluted first. ---Molecular Exclusion Chromatography
Answer =
3) How many moles are there in 454 g of carbon? How many atoms is this?
1 mol =
-
8C
1 mol =
atoms C
Answer=
Answers
Answer:
37.83 moles
Explanation:
moles = grams / formula weight = 454 grams / 12g/mol = 37.83 moles
. Solve: Turn off Show summary. Use the Choose reaction drop down menu to see other
equations, and balance them. Check your answers and then write the balanced equations.
2
6
2
Alt
HCI->
AlCl3 +
Hz
NaCl →
Na +
Cl₂
Na2S +
HCI →
NaCl +
H₂S
CHA +
O₂ →
CO2 +
НО
Practice Balance the following chemical equations. (These equations are not in the Gizmo.) help plz
Answers
Balancing chemical equations involves adjusting the coefficients of the reactants and products in order to ensure adherence to the Law of Conservation of Mass. An example of a balanced chemical equation is 2NaOH(aq) + H₂(g) + Cl₂(g) which can also be written in complete ionic form and net ionic form. Always remember to first balance individual atoms, then oxygen, and lastly hydrogen.
Explanation:The subject of your question pertains to balancing chemical equations, which is a key concept in the field of Chemistry. Balancing chemical equations is the process of adjusting the coefficients of reactants and products in a chemical reaction to ensure that the law of conservation of mass is adhered to. This law states that matter cannot be created nor destroyed, meaning every atom we start with, we must end with and in the same quantity.
For example, let's look at the equation 2NaOH(aq) + H₂(g) + Cl₂(g). In its balanced form, this equation would look like this:
An important tip to remember while balancing equations is to first check individual atom balances (aside from O and H), then balance O, and finally, H.
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What is the H* concentration in a solution with a
pH of 1.25?
x 10" Mn=
Answers
Answer:
[H+] = 0.0562 M
Explanation:
Message me for more help.
H⁺ concentration in a solution with a pH of 1.25 is 0.0562.
What is pH?pH may be defined as a measure of acidity of a substance.
More precisely, pH is defined as the negative logarithm of the hydrogen ion concentration.
pH = - log [H⁺]
The range of pH extends from zero to 14.
A pH value of 7 is neutral, because pure water has a pH value of exactly 7. Values lower than 7 are acidic; values greater than 7 are basic or alkaline.
pH measurements can be made using litmus paper or pH paper known to change colors around a certain pH value.
A universal indicator is a mixture of indicator solutions intended to provide a color change over a pH range of 2 to 10.
pH measurements can also be done with pH meters that give the exact value of pH.
Given,
pH = 1.25
We know that, pH = - log [H⁺]
[H⁺] = antilog [ -pH]
[H⁺] = antilog [ -1.25]
= 0.0562 M
Therefore, H⁺ concentration in a solution with a pH of 1.25 is 0.0562.
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(a) Your TA will give you a 1H NMR spectrum of the 3-nitroaniline product. Using your NMR knowledge and the special NMR section in the lab book (especially pages 48 and 49), assign the 4 peaks in the spectrum to each of the protons in the product - you will need to use coupling pattern and chemical shift to complete the assignment. (b) Explain your assignment, specifically how you differentiated between HA and Hc
Answers
Answer: provided in the answer segment
Explanation:
Below is a step by step process to analyzing this problem
Let us begin;
From 1H-NMR singlet at 5.80 ppm show N-H peak as shown in structure.
Here, H(A) hydrogen has no neighbor hydrogen so it appears integrated singlet at 7.38 ppm.
H(C) hydrogen has the next 2 neighbor hydrogen H(B) and H(D) so it appears as a triplet at region 7.23-7.29 ppm.
H(B) hydrogen has next to one neighbor hydrogen H(C) show doublet at 6.92-6.98 ppm.
H(D) hydrogen has next to one neighbor hydrogen H(C) show doublet at 7.28-7.32 ppm.
(b). From our basic chemistry knowledge, we know that benzene molecule is planer so H(A) is more deshielding because of two substituent groups than H(C), which makes the delta value of H(A) is greater than H(C).
cheers i hope this helped!!!!!
The 1H NMR spectrum of 3-nitroaniline compounds has four different protons which can be differentiated by their chemical shift and coupling pattern. HA is usually the most deshielded due to its closeness to the nitro group. The coupling pattern helps especially in differentiating protons like HA and HC.
Explanation:In 3-nitroaniline compound, the protons, designated as HA, HB, HC, and HD all show up distinctly in the 1H NMR spectrum. The key to differentiating them comes down to their chemical shift and their coupling pattern.
The most deshielded proton (highest chemical shift) is usually HA due to its proximity to the nitro group. HB and HC are the aromatic protons on the benzene ring, and their positions relative to the aniline (amino) group aids in their differentiation. Finally, HD is usually the most shielded proton (lowest chemical shift), as it is the amino proton.
Coupling helps to differentiate between otherwise closely overlapping protons. Due to the nature of the coupling constants involved, HA and HC will generally have distinct coupling patterns that allow them to be differentiated.
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What volume (mL) of a 15% (m/v) NaOH solution contains 120 g NaOH
Answers
Answer:
8.0*10^2
Explanation:
What volume (in milliliters) of a 15% (m/v) NaOH solution contains 120 g NaOH?
For 15% (m/v), the conversion factor can be 15%/100 mL or 100 mL/15%. In this case, 100 mL/15% is used because we need mL as our final expression. 100 mL/15% * 120 g NaOH= 800 mL NaOH. In scientific notation: 8.0*10² or 8.0e² mL NaOH.
To find the volume of a 15% (m/v) NaOH solution that contains 120 grams of NaOH, use the formula V = (mass of solute x 100) / %m/v, which results in 800 mL. Therefore, the correct answer is E) 800 mL.
To find the volume of a 15% (m/v) NaOH solution that contains 120 grams of NaOH, we will use the formula for mass/volume percentage concentration:
%m/v = (mass of solute/volume of solution) x 100%
Given:
Mass of NaOH (solute) = 120 g
Concentration of NaOH solution = 15% (m/v)
We need to find the volume of the solution (V).
Rearranging the formula to solve for V:
V = (mass of solute x 100%) / %m/v
Substituting the given values:
V = (120 g x 100%) / 15%
Converting the percentage:
V = (120 g x 100) / 15 = 12,000 / 15 = 800 mL
Therefore, 800 mL of a 15% NaOH solution contains 120 g NaOH. The correct answer is E) 800 mL.
What volume (mL) of a 15% (m/v) NaOH solution contains 120 g NaOH?
A) 18 mL
B) 0.13 mL
C) 13 mL
D) 120 mL
E) 800 mL
Draw three resonance structures for CNO-. This species has its three atoms bonded sequentially in the following fashion: C-N-O. Draw your resonance structures so that the atoms in them are bonded together in this order. Select the most important resonance structure for this species based on the formal charges on the atoms of the three resonance structures you have drawn. Now select the statement from the multiple choices which is true about this most important resonance structure. In the most important resonance structure of CNO- : a) The leftmost bond (between C and N) is a double bond. b) The rightmost bond (between N and O) is a double bond. c) The formal charge on the leftmost (C) atom is -2. d) The number of nonbonding pairs (lone pairs) of electrons on the leftmost (C) atom is 2. e) The number of nonbonding (lone) pairs of electrons on the rightmost (O) atom is 3.
Answers
To calculate the charge of each atom.
Formal charge= Valence electrons-(non bonding electron+bonding electron/2)
The possible resonance of CNO- ( check attachment)
The second structure is the most stable because in this structure each atoms carry less formal charge than other.
(a) In the most stable resonance structure of CNO- leftmost bond is a triple bond.
Therefore the statement is False
(b) in the most stable resonance structure of CNO- rightmost bond is single.
Therefore the statement is true
(c) in the most stable resonance
structure of CNO- the formal charge on the leftmost is -1.
Therefore the statement is False
(d) in the most stable resonance structure of CNO-,the number of non bonding electron pairs on the leftmost carbon is 1.
Therefore the statement is False
(e) in the most stable resonance structure of CNO-the number of non bonding electron pairs on rightmost oxygen atom is 3.
Therefore the statement is False
Solutions of sodium acetate and acetic acid are combined in equal volume to produce a buffer. Identify the combination that will produce the buffer with the highest buffer capacity. Select one: a. 0.01 M CH3COOH , 0.1 M CH3COONa b. 0.1 M CH3COOH , 0.1 M CH3COONa c. 0.1 M CH3COOH , 0.01 M CH3COONa d. 0.01 M CH3COOH , 0.01 M CH3COONa
Answers
Answer:
Option b=> 0.1 M CH3COOH , 0.1 M CH3COONa.
Explanation:
In chemistry, when an acid or a base(alkali) is added to a solution and the solution pH does not change then we say that the solution is a BUFFER SOLUTION (that is a solution, upon the addition of acid or base will show resistance in pH value).
In order to be able to answer this question efficiently we have to consider the mathematical representation or Equation below;
pH = pKa + log ( [ A^-] / [HA] )--------------------------------------------------------------------(1).
What the mathematical representation or Equation (1) above is telling is is that if [A^-] or [HA] is high then there will be the production of buffer with higher buffer capacity.
Cardiac muscle causes movement in the .
Answers
Answer:
organ
Explanation:
it causes movement in the heart and the heart is an organ
What is the mole ratio of PbO2 to water?
PbO, H,0 = 1:0
Answers
Answer:
The mole ratio of PbO2 to H2O is 1 : 2.
Explanation:
The balanced reaction equation is:
Pb + PbO2 + 2H2SO4 → 2PbSO4 + 2H2O
On the reactant side, we have 1 mole of Pb, 1 mole of PbO2, 2 moles of H2SO4.
On the product side, we have 2 moles of PbSO4 and 2 moles of H2O.
This means that for ever 1 mole of PbO2 consumed, 2 moles of water is formed as product.
Hence, the mole ratio of PbO2 to H2O is 1 : 2.
What of the following is true if the pH of a solution decreases?
[H+] increases
[OH-] increases
Kw increases
Kw decreases
None
Answers
Answer:
[H⁺] increases
Explanation:
What of the following is true if the pH of a solution decreases?
[H⁺] increases. TRUE.The pH and the [H⁺] are related through the following expression.
pH = -log [H⁺]
According to this, there is an inverse relation between the pH and the [H⁺]. When the pH decreases, the [H⁺] increases.
[OH⁻] increases. FALSE.The [H⁺] and the [OH⁻] are inversely related through the following expression.
[H⁺] × [OH⁻] = Kw
As we saw, when the pH decreases, the [H⁺] increases. Then, the [OH⁻] decreases.
Kw increases. FALSE. Kw is constant at a given temperature.Kw decreases. FALSE. Kw is constant at a given temperature.(a) What is the total volume (in L) of gaseous products, measured at 350°C and 735 torr, when an automobile engine burns 100. g of C8H18 (a typical component of gasoline)? Page 253 (b) For part (a), the source of O2 is air, which is 78% N2, 21% O2, and 1.0% Ar by volume. Assuming all the O2 reacts, but no N2 or Ar does, what is the total volume (in L) of the engine’s gaseous exhaust?
Answers
Answer:
Part A
The volume of the gaseous product is [tex]V = 787L[/tex]
Part B
The volume of the the engine’s gaseous exhaust is [tex]V_e = 2178 \ L[/tex]
Explanation:
Part A
From the question we are told that
The temperature is [tex]T = 350^oC = 350 +273 =623K[/tex]
The pressure is [tex]P = 735 \ torr = \frac{735}{760} = 0.967\ atm[/tex]
The of [tex]C_8 H_{18} = 100.0g[/tex]
The chemical equation for this combustion is
[tex]2 C_8 H_{18}_{(l)} + 25O_2_{(l)} ----> 16CO_2_{(g)} + 18 H_2 O_{(g)}[/tex]
The number of moles of [tex]C_8 H_{18}[/tex] that reacted is mathematically represented as
[tex]n = \frac{mass \ of \ C_8H_{18} }{Molar \ mass \ of C_8H_{18} }[/tex]
The molar mass of [tex]C_8 H_{18}[/tex] is constant value which is
[tex]M = 114.23 \ g/mole[/tex]
So [tex]n = \frac{100 }{114.23} }[/tex]
[tex]n = 0.8754 \ moles[/tex]
The gaseous product in the reaction is [tex]CO_2_{(g)}[/tex] and water vapour
Now from the reaction
2 moles of [tex]C_8 H_{18}[/tex] will react with 25 moles of [tex]O_2[/tex] to give (16 + 18) moles of [tex]CO_2_{(g)}[/tex] and [tex]H_2 O_{(g)}[/tex]
So
1 mole of [tex]C_8 H_{18}[/tex] will react with 12.5 moles of [tex]O_2[/tex] to give 17 moles of [tex]CO_2_{(g)}[/tex] and [tex]H_2 O_{(g)}[/tex]
This implies that
0.8754 moles of [tex]C_8 H_{18}[/tex] will react with (12.5 * 0.8754 ) moles of [tex]O_2[/tex] to give (17 * 0.8754) of [tex]CO_2_{(g)}[/tex] and [tex]H_2 O_{(g)}[/tex]
So the no of moles of gaseous product is
[tex]N_g = 17 * 0.8754[/tex]
[tex]N_g = 14.88 \ moles[/tex]
From the ideal gas law
[tex]PV = N_gRT[/tex]
making V the subject
[tex]V = \frac{N_gRT}{P}[/tex]
Where R is the gas constant with a value [tex]R = 0.08206 \ L\cdot atm /K \cdot mole[/tex]
Substituting values
[tex]V = \frac{14.88* 0.08206 *623}{0.967}[/tex]
[tex]V = 787L[/tex]
Part B
From the reaction the number of moles of oxygen that reacted is
[tex]N_o = 0.8754 * 12.5[/tex]
[tex]N_o = 10.94 \ moles[/tex]
The volume is
[tex]V_o = \frac{10.94 * 0.08206 *623}{0.967}[/tex]
[tex]V_o = 579 \ L[/tex]
No this volume is the 21% oxygen that reacted the 79% of air that did not react are the engine gaseous exhaust and this can be mathematically evaluated as
[tex]V_e = V_o * \frac{0.79}{0.21}[/tex]
Substituting values
[tex]V_e = 579 * \frac{0.79}{0.21}[/tex]
[tex]V_e = 2178 \ L[/tex]
The total volume of gaseous products produced from the combustion of 100g C8H18 is approximately 382 L. By considering the composition of air and that N2 and Ar do not react, you can calculate the total volume of the gaseous exhaust.
Upon seeing this question, the first step is to write the balanced chemical equation for the combustion of C8H18, which is ubiquitous in gasoline:
C8H18 + 12.5 O2 -> 8 CO2 + 9 H2O
This looks complex, but the Ideal Gas Law simplifies everything. According to the stoichiometric ratio, 1 mole of C8H18 will produce 8 moles of CO2 and 9 moles of H2O when burnt, totalling 17 moles of gas. The molar mass of C8H18 is around 114 g/mol, so 100g is approximately 0.877 moles.
Therefore, upon combustion, 0.877 * 17 = 14.9 moles of gas are produced. Applying the Ideal Gas Law (PV=nRT), keeping in mind that 350°C is 623K and 1 atm is 760 torr, we have: (nR(623K)/735torr) = V. Plugging the numbers and making sure to keep the units consistent, we get V = 14.9 x 0.0821 x 623 / 0.967 = approximately 382 L.
Coming to part (b), considering the composition of air and assuming no N2 or Ar reacts, you find that for every L of O2 burned, 1L of air is consumed. Adding the total volume of gas from part (a) and the extra nitrogen and argon, you find the total volume of the engine's gaseous exhaust.
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