which peptide will be eluted first from an anion exchange column at ph 7.3?

Answers

Answer 1

The peptide with the highest net positive charge will be eluted first from an anion exchange column at pH 7.3.

Anion exchange chromatography separates molecules based on their net charge. At pH 7.3, some peptides may have a net positive charge, while others may have a net negative charge or no charge. The column is filled with a negatively charged resin, which will attract and bind positively charged molecules. As the elution buffer is introduced, molecules with a higher positive charge will bind more tightly to the column and will be eluted last. In contrast, molecules with a lower positive charge will be less tightly bound and will be eluted first.

To determine which peptide will be eluted first from an anion exchange column at pH 7.3, we must consider the net charge of each peptide. The net charge of a peptide is determined by the number of positive and negative charges on its amino acid residues. At pH 7.3, some amino acids are partially ionized, meaning that they have a net charge.

For example, lysine and arginine have a net positive charge at pH 7.3, while aspartic acid and glutamic acid have a net negative charge. Therefore, peptides that contain more lysine or arginine residues will have a higher net positive charge and will bind more tightly to the negatively charged resin in the column. These peptides will be eluted last.

On the other hand, peptides that contain more aspartic acid or glutamic acid residues will have a net negative charge and will not bind as tightly to the column. These peptides will be eluted first.

In conclusion, the peptide with the highest net positive charge will be eluted last from an anion exchange column at pH 7.3, while the peptide with the lowest net positive charge or a net negative charge will be eluted first.

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Related Questions

which line correctly shows the solubility vs. temperature profile for this substance?

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The correct solubility vs. temperature profile for a substance can be determined by plotting the solubility values of the substance at different temperatures.

The resulting graph should show a positive correlation between temperature and solubility if the substance is soluble in water. If the substance is insoluble, then the graph should show a flat line indicating no change in solubility with temperature.

The solubility vs. temperature profile for a substance is an important characteristic used to determine the ability of a substance to dissolve in water at different temperatures. The profile can be determined by plotting the solubility values of the substance at different temperatures. If the substance is soluble, then the graph should show a positive correlation between temperature and solubility. If the substance is insoluble, then the graph should show a flat line indicating no change in solubility with temperature. The correct solubility vs. temperature profile can be used to predict the solubility of the substance under different temperature conditions.

The solubility vs. temperature profile of a substance is an important characteristic used to determine the ability of the substance to dissolve in water at different temperatures. The profile can be determined by plotting the solubility values of the substance at different temperatures. The correct profile should show a positive correlation between temperature and solubility for soluble substances and a flat line for insoluble substances. The correct solubility vs. temperature profile can be used to predict the solubility of the substance under different temperature conditions.

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calculate the mole fraction of sodium chloride in a solution containing 129 g of sodium chloride per 2.50 mol comphor. the molar mass of sodium chloride is 58.5 g/mol.

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The mole fraction of sodium chloride in the given solution containing 129 g of sodium chloride per 2.50 mol camphor is 0.469 or 46.9%.

To calculate the mole fraction of sodium chloride in the given solution, we need to first calculate the number of moles of sodium chloride present in the solution. Using the given mass of sodium chloride and its molar mass, we can calculate the number of moles of sodium chloride as follows:
Number of moles of sodium chloride = Mass of sodium chloride / Molar mass of sodium chloride
= 129 g / 58.5 g/mol = 2.21 mol
Now, to calculate the mole fraction of sodium chloride, we need to divide the number of moles of sodium chloride by the total number of moles in the solution. We are given that the solution contains 2.50 mol of camphor. Therefore, the total number of moles in the solution is:
Total number of moles in the solution = Number of moles of sodium chloride + Number of moles of camphor
= 2.21 mol + 2.50 mol = 4.71 mol
The mole fraction of sodium chloride can now be calculated as follows:
Mole fraction of sodium chloride = Number of moles of sodium chloride / Total number of moles in the solution
= 2.21 mol / 4.71 mol = 0.469 or 46.9%

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the mineral hornblende has more than one direction of cleavage. how many directions of cleavage does it have? what are the angles of its cleavage?

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In the case of hornblende, the commonly observed and characteristic cleavage directions are at angles of approximately 56° and 124°.

The two directions of cleavage in hornblende are not at right angles to each other. They are typically inclined at approximately 56° and 124°. These angles are not precisely at right angles but can be categorized as close to perpendicular, although not exactly 90 degrees.

The cleavage planes in hornblende are often observed as smooth, shiny surfaces along which the mineral readily breaks. The cleavage in hornblende is considered prismatic or parallel to the long axis of the mineral.

It's important to note that the cleavage properties of minerals can vary to some extent depending on the specific crystallographic orientation and the quality of the crystal.

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what is the sequence of electron flow between the four prosthetic groups

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Answer:

Explanation:TC consists of four complexes: NADH dehydrogenase (Complex I), Succinate dehydrogenase (Complex II), Cytochrome b and c1 (Complex III), and Cytochrome c oxidase (Complex IV).

what is the poh of an aqueous solution at 25.0 °c that contains 3.98 ⋅ 10-9 m hydroxide ion?

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The pOH of the aqueous solution at 25.0 °c that contains 3.98 ⋅ 10^{-9} m hydroxide ion is 4.40.

The hydroxide ion concentration of the solution can be used to calculate the pOH, which is defined as the negative logarithm of the hydroxide ion concentration.

The formula for calculating pOH is:

pOH = -log[OH-]

Substituting the given hydroxide ion concentration of 3.98 × 10^{-9} M into the formula, we get:

pOH = -log(3.98 × 10^-9)

pOH = 4.40

Therefore, the pOH of the aqueous solution at 25.0 °C that contains 3.98 × 10^-9 M hydroxide ion is 4.40. Note that to find the pH of the solution, we would subtract the pOH from 14, giving us a pH of 9.60 (pH + pOH = 14).

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what kind of intermediate is formed when an alkene is exposed to a strong acid?

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when an alkene is exposed to a strong acid, an intermediate called a carbocation is formed. This is because the acid protonates the alkene, causing the loss of a leaving group and the formation of a positively charged carbon atom.

that the strong acid, such as sulfuric acid or hydrochloric acid, donates a proton to the alkene, creating a more reactive intermediate. The resulting carbocation is an unstable species and can undergo various reactions, including addition, elimination, and rearrangement. These reactions are commonly used in organic synthesis to form complex molecules.
when an alkene is exposed to a strong acid, a carbocation intermediate is formed.

When an alkene reacts with a strong acid, the acid donates a proton (H+) to the double bond of the alkene. This results in the formation of a positively charged carbocation intermediate. The carbocation is then attacked by the nucleophile (usually the conjugate base of the strong acid) to form the final product. This reaction mechanism is known as electrophilic addition.

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What evidence supports the statement that the stomata are fully open at a light intensity of 30 kilolux in a transpiration investigation

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The evidence supporting the statement that stomata are fully open at a light intensity of 30 kilolux in a transpiration investigation is an observation of increased transpiration rate and stomatal conductance.

This can be determined by measuring the rate of water loss from leaves (transpiration) and the movement of water vapor through stomata (stomatal conductance). A higher light intensity leads to increased photosynthesis, which requires carbon dioxide uptake through open stomata. Therefore, if the light intensity is at 30 kilolux and the transpiration rate and stomatal conductance are high, it suggests that the stomata are fully open to facilitate gas exchange for photosynthesis. the statement that stomata are fully open at a light intensity of 30 kilolux in a transpiration investigation is an observation of increased transpiration rate and stomatal conductance.

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The Ksp for Pb3(PO4)2(s) is 1.0×10-54. Calculate the solubility of Pb3(PO4)2(s). Ignore any potential reaction of the ions with water. Solubility for Pb3(PO4)2(s) = ____ mol/L

Answers

The solubility of Pb3(PO4)2(s) is 1.28 x 10^-18 mol/L.


To calculate the solubility of Pb3(PO4)2(s), we need to first write the balanced dissolution equation and then set up the expression for the solubility product constant (Ksp).
The balanced dissolution equation for Pb3(PO4)2(s) is:
Pb3(PO4)2(s) ⇌ 3Pb2+(aq) + 2PO4^3-(aq)
Now, let the solubility of Pb3(PO4)2 be 's' mol/L.

Then, the concentration of Pb2+ ions would be 3s, and the concentration of PO4^3- ions would be 2s.
Ksp = [Pb2+]^3 * [PO4^3-]^2
1.0 x 10^-54 = (3s)^3 * (2s)^2
Solving for 's' yields:
s = 1.28 x 10^-18 mol/L


Summary: The solubility of Pb3(PO4)2(s) is 1.28 x 10^-18 mol/L, which is the concentration at which Pb3(PO4)2(s) will dissolve in water without any additional reactions with water ions.

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When 4.28 g of a certain molecular compound X are dissolved in 60.0 g of dibenzyl ether [(C₆H₅CH₂)₂0] , the freezing point of the solution is measured to be -3.2°C . Calculate the molar mass of X.
If you need any additional information on dibenzyl ether, use only what you find in the ALEKS Data resource. Also, be sure your answer has a unit symbol, and is rounded to significant digit.

Answers

The molar mass of X is 202.9 g/mol.

To figure out this problem, we have to use the freezing point depression formula:

ΔT = i × Kf × m

ΔT is the change in freezing point,

Kf is the freezing point depression constant of the solvent (dibenzyl ether),

m = molality of the solution, and

i = van't Hoff factor, which corresponds to the number of particles into which from each one molecule of the solute X dissociates in solution.

Molality = moles of solute ÷ mass of solvent in kg

Convert the mass of dibenzyl ether to kg:

mass of dibenzyl ether

= 60.0 g = 0.0600 kg

To find the moles of X in the solution:

Moles of X = mass of X ÷ molar mass of X

The molar mass of X is M:

Moles of X = 4.28 g ÷ M

Now we can calculate the molality:

molality is equal to (4.28 g / M) / 0.0600 kg

molality is equal to 71.3 g/mol / M

To find the van't Hoff factor i.

We can assume that X does not dissociate in solution, so i = 1.

Now use the freezing point depression to find the molar mass of X:

ΔT = Kf · m · i

-3.2 °C = Kf · (71.3 g/mol / M) · 1

The freezing point depression is constant for dibenzyl ether  

= 9.80°C·kg/mol.

Put this value into the equation and solving for M:

M = (Kf · m) / ΔT

M = (9.80°C·kg/mol) · (71.3 g/mol / M) / (-3.2°C)

M = 202.9 g/mol

The molar mass of X is 202.9 g/mol.

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How many moles are in 1.25 g of calcium chloride

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Moles Are In 1.25g Of Calcium Chloride (CaCl21 2: is 1.55 X 10-2 Moles. The correct option is D.

To determine how many moles are in 1.25g of Calcium Chloride (CaCl2), we need to first find the molar mass of CaCl2. The molar mass is calculated by adding the atomic masses of Calcium (Ca) and Chlorine (Cl) in the compound. The atomic mass of Calcium is approximately 40.08 g/mol, and the atomic mass of Chlorine is approximately 35.45 g/mol. Since there are two Chlorine atoms in CaCl2, we multiply the atomic mass of Chlorine by 2:

Molar mass of CaCl2 = (40.08 g/mol) + 2 * (35.45 g/mol) = 110.98 g/mol

Now, we can use the given mass of Calcium Chloride (1.25g) and the calculated molar mass to find the number of moles:

Number of moles = (mass of CaCl2) / (molar mass of CaCl2) = (1.25 g) / (110.98 g/mol) ≈ 0.0113 mol

Comparing this value to the given options, it is closest to option D. Therefore, there are approximately 1.55 x 10^-2 moles in 1.25g of Calcium Chloride (CaCl2).

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Complete question:

How Many Moles Are In 1.25g Of Calcium Chloride (CaCl21 2.

a. 8.88 X 101 Moles

B. 1.39 X 102 Moles

C. 1.13x102 Moles

D. 1.55 X 10-2 Moles

If W, X, Y, and Z are different monodentate ligands, how many geometric isomers are there for each ion? Square planar [NIWXYZ]²+ Express your answer as an integer. ν ÎÏ

Answers

In a square planar complex, there are two possible arrangements of ligands around the central metal ion: cis and trans.

In chemistry, geometric isomerism refers to the phenomenon where compounds have the same molecular formula and connectivity but differ in the spatial arrangement of their atoms or ligands.

For the [NIWXYZ]²+ ion, there are four different monodentate ligands, which means that there are four possible ways to arrange them around the central metal ion. However, since W, X, Y, and Z are all different, there are only two possible arrangements: cis and trans.

For the complex [NIWXYZ]²+, we can arrange the ligands in cis and trans configurations as follows:

Cis: W and X are adjacent to each other, as are Y and Z.

Trans: W and Z are opposite each other, as are X and Y.

Therefore, there are two possible geometric isomers for the [NIWXYZ]²+ ion.

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which base, ammonia (nh3) or triethylamine [(ch3ch2)3n], would be a better choice for use in converting 1-chlorohexane to hex-1-ene? explain briefly.

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In order to convert 1-chlorohexane to hex-1-ene, you will need to perform an elimination reaction. The better base for this reaction would be triethylamine [(CH3CH2)3N] rather than ammonia (NH3).

The reason is that triethylamine is a bulkier, more sterically hindered base compared to ammonia. Bulkier bases favor E2 (bimolecular elimination) reactions, which result in the formation of alkenes like hex-1-ene. On the other hand, ammonia is a smaller, less hindered base, which would lead to a mixture of elimination and substitution reactions.

Using triethylamine increases the selectivity for the desired elimination product, hex-1-ene.

So, in summary, triethylamine [(CH3CH2)3N] would be a better choice for use in converting 1-chlorohexane to hex-1-ene due to its bulky structure that favors E2 elimination reactions.

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Predict the major product formed by 1,4-addition of HCI to 1,3-cyclohexadiene.

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The major product formed by the 1,4-addition of HCl to 1,3-cyclohexadiene is 3-chlorocyclohexene, which results from an electrophilic addition reaction displaying regioselectivity.


When HCl reacts with 1,3-cyclohexadiene in a 1,4-addition manner, the major product formed is 3-chlorocyclohexene. The reaction is an example of an electrophilic addition reaction, in which an electrophile, HCl in this case, reacts with a nucleophile, the double bond of 1,3-cyclohexadiene.

During this reaction, the hydrogen atom of HCl gets attached to one of the carbon atoms of the double bond, while the chlorine atom gets attached to the carbon atom four positions away. This leads to the formation of 3-chlorocyclohexene as the major product. The product exhibits regioselectivity, meaning that the reaction occurs predominantly in one particular orientation, forming a major product over other possible products.

As a result of an electrophilic addition reaction with regioselectivity, 3-chlorocyclohexene is the main byproduct of the 1,4-addition of HCl to 1,3-cyclohexadiene.

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complete and balance the following half-reaction: mn2+(aq)→mno2(s) (basic solution)

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To complete and balance the half-reaction mn2+(aq)→mno2(s) (basic solution), we first need to identify the reactant and product. In this case, the reactant is mn2+(aq) and the product is mno2(s). In summary, the balanced half-reaction for mn2+(aq)→mno2(s) (basic solution) is: mn2+(aq) + 4OH-(aq) → mno2(s) + 2H2O(l)

Next, we need to add water molecules and hydroxide ions to balance the charges. We can start by adding water molecules to balance the oxygen atoms. Since there are already two oxygen atoms on the product side, we need to add two water molecules on the reactant side. This gives us:
mn2+(aq) + 2H2O(l) → mno2(s) + 4OH-(aq)
We still need to balance the charges by adding hydroxide ions. There are currently no hydroxide ions on the reactant side, so we need to add four hydroxide ions to balance the charge. This gives us the final balanced half-reaction:
mn2+(aq) + 4OH-(aq) → mno2(s) + 2H2O(l)
This half-reaction shows the oxidation of mn2+ to mno2 in a basic solution.

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when a β- particle is emitted from an unstable nucleus, the atomic number of the nucleus:
increases by 1.
increases by 2. decreases by 1.
decreases by 2. does not change.

Answers

When a β- particle is emitted from an unstable nucleus, the atomic number of the nucleus decreases by 1. This is because β- particles are actually electrons that are emitted from the nucleus during beta decay, which occurs when a neutron is converted into a proton and an electron.

The electron is then ejected from the nucleus as a β- particle, while the proton remains in the nucleus. Since the atomic number of an element is determined by the number of protons in its nucleus, the emission of a β- particle results in a decrease in the atomic number. This process can also result in the creation of a new element, as the conversion of a neutron into a proton can change the identity of the element.
When a β- particle is emitted from an unstable nucleus, the atomic number of the nucleus increases by 1. This process occurs because a neutron inside the nucleus transforms into a proton, which results in the release of a β- particle (an electron) and a small, neutral particle called an antineutrino. The increase in the number of protons causes the atomic number to rise by 1.

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Question 1 of 10
Which of the following is an example of appropriate use of another scientist's
data?
A. Writing the other scientist a private thank-you note
B. Leaving the other scientist's data out of your published results
because it does not support your conclusions
OC. Collecting your own data and copying the other scientist's ideas
and words exactly
D. Clearly noting in your published results the source of all of the
data used to support your conclusions

Answers

The correct answer is D. Clearly noting in your published results the source of all of the data used to support your conclusions. This is an important part of scientific research and ensures that other scientists can properly credit the original source of the data.

which of these compounds is a secondary alcohol? propan-2-ol methanol 2-methoxypropane 2-methylpropan-2-ol

Answers

2-methylpropan-2-ol is a secondary alcohol, the correct answer is option (d)

The term "2-methylpropan-2-ol" refers to a specific type of organic compound that belongs to the alcohol family. The compound is also known by its IUPAC name, which is "2-methylpropan-2-ol."

The "2-methyl" part of the name indicates that the compound contains a methyl group (-CH3) attached to the second carbon atom of the parent chain, which is propane. The "-ol" suffix indicates that the compound belongs to the alcohol family, meaning it contains a hydroxyl group (-OH) attached to one of the carbon atoms in the molecule.

The term "secondary alcohol" refers to the specific type of alcohol that has a hydroxyl group (-OH) attached to a secondary carbon atom, which is a carbon atom that is bonded to two other carbon atoms. In the case of 2-methylpropan-2-ol, the hydroxyl group is attached to the second carbon atom of the parent chain, which is a secondary carbon atom, and thus, it is considered a secondary alcohol.

It is important to note that the classification of alcohols as primary, secondary, or tertiary depends on the number of carbon atoms bonded to the carbon atom bearing the hydroxyl group. Primary alcohols have one carbon atom bonded to the carbon bearing the hydroxyl group, while tertiary alcohols have three carbon atoms bonded to the carbon bearing the hydroxyl group. Therefore the correct option is d.

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find the concentration of i− in 0.010 m agno3 saturated with agi . include activity coefficients in the solubility-product expression. the sp of agi is 8.3

Answers

The concentration of I⁻ in a 0.010 M AgNO₃ solution saturated with AgI is 0.079 M.

The solubility product expression for AgI is:

Ksp = [Ag⁺][I⁻]

At equilibrium, the concentration of Ag⁺ is equal to the solubility of AgI, which we can denote as "s". The concentration of I⁻ is not initially known, but we can assume that it is x.

The balanced chemical equation for the dissolution of AgI in water is:

AgI(s) ⇌ Ag⁺(aq) + I⁻(aq)

The AgNO₃ dissociates completely into Ag⁺ and NO₃⁻. Therefore, the initial concentration of Ag⁺ is equal to the initial concentration of AgNO₃, which is 0.010 M.

The NO₃⁻ ion does not react with AgI, so we can ignore it for the purposes of this calculation.

To account for the activity coefficients, we can use the following expression for the solubility product:

Ksp = γ(Ag⁺) * γ(I⁻) * [Ag⁺] * [I⁻]

The activity coefficient for a 1:1 electrolyte like AgI can be approximated as:

γ = (1 + α * √(I))²

where α is a constant related to the size of the ions (typically assumed to be 0.5 for small ions) and I is the ionic strength of the solution, defined as:

I = (1/2) * Σmi(zᵢ)²

where mi is the molality of ion i and zᵢ is its charge.

In this case, the ionic strength is dominated by Ag⁺, since I⁻ is assumed to be much less concentrated. Therefore:

I = (1/2) * (0.010 mol/kg) * (1²) = 0.00005

Using α = 0.5, we can calculate the activity coefficients:

γ(Ag⁺) = (1 + 0.5 * √(0.00005))² = 1.026

γ(I⁻) = (1 + 0.5 * √(0.00005))² = 1.026

Substituting these values into the solubility product expression, we get:

8.3 = 1.026² * (0.010 M) * x

Solving for x, we get:

x = 0.079 M

Therefore, the concentration of I⁻ in a 0.010 M AgNO₃ solution saturated with AgI is 0.079 M.

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a compound with an oh and an or group attached to the same carbon is known as ______. ©gmu_76508

Answers

A compound with an OH group and an OR group attached to the same carbon is known as a hemiacetal because a hemiacetal is a compound with both an OH and an OR group attached to the same carbon atom.

Here's a breakdown of the terms:
Compound: A substance formed from two or more elements chemically bonded together.
Group: A specific arrangement of atoms in a molecule that imparts certain chemical properties to the molecule.
Carbon: A non-metallic element that forms the basis for organic chemistry, as it readily forms bonds with other elements, particularly hydrogen, oxygen, and nitrogen.

A hemiacetal or a hemiketal has the general formula R¹R²C(OH)OR, where R¹ or R² is hydrogen or an organic substituent. They generally result from the addition of an alcohol to an aldehyde or a ketone, although the latter are sometimes called hemiketals. Most sugars are hemiacetals.
In summary, A compound with an OH group and an OR group attached to the same carbon is known as a hemiacetal because a hemiacetal is a compound with both an OH and an OR group attached to the same carbon atom.

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"Identify each of the solutions as acidic, basic, or neutral. [oh−]=1.0×10^−7M [oh−]=1.0×10^−9M[h3o+]=0.0001M[h3o+]=7.3×10^−3M"

Answers

[oh−]=1.0×10^−7M and [h3o+]=7.3×10^−3M are acidic solutions, [oh−]=1.0×10^−9M is a basic solution, and [h3o+]=0.0001M is a neutral solution.


The explanation for this answer is that the pH scale ranges from 0 to 14, with 7 being neutral. A pH below 7 is considered acidic, while a pH above 7 is considered basic.

The [oh−] and [h3o+] concentrations can be used to calculate the pH of a solution using the formula pH=-log[H+].

In the case of these solutions, the pH values are as follows: [oh−]=1.0×10^−7M has a pH of 7.0, [oh−]=1.0×10^−9M has a pH of 9.0, [h3o+]=0.0001M has a pH of 4.0, and [h3o+]=7.3×10^−3M has a pH of 2.14.


In summary, the [oh−]=1.0×10^−7M and [h3o+]=7.3×10^−3M solutions are acidic, [oh−]=1.0×10^−9M is basic, and [h3o+]=0.0001M is neutral. The pH values were calculated using the concentrations of [oh−] and [h3o+], and the pH scale was used to determine if the solutions were acidic, basic, or neutral.

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identify the bicyclic product from the diels-alder reaction

Answers

Answer:

cyclohexene

Explanation:

The Diels-Alder reaction is a conjugate addition reaction of a conjugated diene to an alkene or alkyne (the dienophile) to produce a cyclohexene.

When SO2(g) reacts with O2(g) according to the following reaction, 98.9 kJ of energy are evolved for each mole of SO2(g) that reacts. Complete the following thermochemical equation.2SO2(g) + O2(g)2SO3(g) H = kJ

Answers

Thermochemical equation for the reaction is:

[tex]2SO_2(g) + O_2(g) - > 2SO_3(g) \Delta H = -198.0 kJ/mol.[/tex]

The given reaction releases 98.9 kJ of energy for each mole of SO2(g) that reacts. To write the thermochemical equation for this reaction, we need to include the enthalpy change (ΔH) for the reaction.

Since the reaction releases energy, the enthalpy change is negative. The enthalpy change can be calculated by subtracting the enthalpies of the reactants from the enthalpies of the products.

The enthalpy change for the reaction can be written as:

[tex]\Delta H = 2 * \Delta Hf[SO_3(g)] - 2 * \Delta Hf[SO_2(g)] - \Delta Hf[O_2(g)][/tex]

where ΔHf is the standard enthalpy of formation.

Using the standard enthalpies of formation for the compounds, we can calculate the enthalpy change for the reaction as:

ΔH = -198.0 kJ/mol

Therefore, the thermochemical equation for the reaction is:

[tex]2SO_2(g) + O_2(g) - > 2SO_3(g) \Delta H = -198.0 kJ/mol.[/tex]

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rhodium has a density of 12.41 g/cm3 and crystallizes with the face-centered cubic unit cell. calculate the radius of a rhodium atom. molar mass of rhodium is 102.9055g/mol

Answers

The radius of a rhodium atom is approximately 1.347 × 10^(-10) cm.

To calculate the radius of a rhodium atom, we can use the formula:

radius = (3 * molar mass) / (4 * density * Avogadro's number)^(1/3)

First, let's calculate the value inside the parentheses:

(4 * density * Avogadro's number) = (4 * 12.41 g/cm³ * 6.022 × 10^23 atoms/mol)

Now, let's calculate the radius:

radius = (3 * 102.9055 g/mol) / ((4 * 12.41 g/cm³ * 6.022 × 10^23 atoms/mol)^(1/3))

Performing the calculations:

radius ≈ 1.347 × 10^(-10) cm

Therefore, the radius of a rhodium atom is approximately 1.347 × 10^(-10) cm.

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Which test tube had the most concentrated acid? How can you tell from the diagram?​

Answers

The test tube that has the most concentrated acid is test tube B.

Concentration refers to the amount of a substance in a defined space. Another definition is that concentration is the ratio of solute in a solution to either solvent or total solution.

There are various methods of expressing the concentration of a solution.

Concentrations are usually expressed in terms of molarity, defined as the number of moles of solute in 1 L of solution.

Solutions of known concentration can be prepared either by dissolving a known mass of solute in a solvent and diluting to a desired final volume or by diluting the appropriate volume of a more concentrated solution (a stock solution) to the desired final volume.

Test tube B has the most concentrated acid as the bubbles released by its reaction with metals is maximum.

Thus, the ideal selection is Test tube B.

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in the diagram of the rock cycle which processes are represented by X ,Y and Z​

Answers

The processes represented by X Y and Z are; Melting, cooling, and weathering

What is the rock cycle?

Through a succession of geological occurrences and processes, rocks naturally change through time from one kind to another. This process is known as the rock cycle.

It explains the never-ending cycle of processes that alter, create, and reform rocks both on the surface and inside the earth. The rock cycle allows the three main types of rocks—igneous, sedimentary, and metamorphic—to change into one another.

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The most abundant component of the atmospheres of Uranus and Neptune is:
A: Hydrogen.
B: Helium.
C: Methane.
D: Ammonia.
E: Nitrogen.

Answers

The most abundant component of the atmospheres of Uranus and Neptune is methane, which is represented by the chemical formula CH4.

Uranus and Neptune are known as ice giants and their atmospheres are primarily composed of hydrogen and helium. However, methane is the most abundant component after hydrogen and helium. Methane makes up about 2.3% of Uranus' atmosphere and about 1.5% of Neptune's atmosphere. The presence of methane gives these planets their characteristic blue color as it absorbs red light and reflects blue light. The remaining components of the atmospheres of Uranus and Neptune are made up of trace amounts of ammonia, water vapor, and other gases.

In conclusion, the most abundant component of the atmospheres of Uranus and Neptune is methane, which is responsible for their distinctive blue color. While hydrogen and helium are the primary components of their atmospheres, methane makes up a significant percentage and is essential in understanding the composition and characteristics of these ice giants.

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how many moles of h c l are in 47.3 ml of a 1.65 m h c l solution?

Answers

There are approximately 0.078 moles of HCl in 47.3 mL of a 1.65 M HCl solution.

To determine the number of moles of HCl in a solution, we need to use the equation:moles of solute = concentration of solution x volume of solutionIn this case, we are given the concentration of the HCl solution as 1.65 M (moles per liter), and the volume of the solution as 47.3 mL (milliliters). However, the volume needs to be converted to liters for the calculation to work properly.1 mL = 0.001 L (since there are 1000 mL in 1 L)Therefore, the volume of the solution in liters is:47.3 mL x (1 L/1000 mL) = 0.0473 L.

Now we can plug the values into the equation:moles of HCl = concentration of solution x volume of solutionmoles of HCl = 1.65 M x 0.0473 Lmoles of HCl = 0.077945 molesSo there are approximately 0.078 moles of HCl in 47.3 mL of a 1.65 M HCl solution.

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what weight of solid oxalic acid dihydrate, h2c2o4 · 2h2o, could be completely neutralized with 35.0 ml of 0.240 m naoh?

Answers

Answer is: approximately 0.756 grams of solid oxalic acid dihydrate (H2C2O4 · 2H2O) could be completely neutralized with 35.0 mL of 0.240 M NaOH.

To determine the weight of solid oxalic acid dihydrate (H2C2O4 · 2H2O) that can be completely neutralized with 35.0 mL of 0.240 M NaOH:

The balanced chemical equation for the reaction between oxalic acid (H2C2O4) and sodium hydroxide (NaOH) is:

H2C2O4 + 2NaOH → Na2C2O4 + 2H2O

To calculate the amount of oxalic acid dihydrate, we first need to determine the number of moles of NaOH used in the reaction:

Moles of NaOH = Volume (L) × Concentration (mol/L)

= 0.0350 L × 0.240 mol/L

= 0.0084 mol

Since the stoichiometry of the reaction shows that one mole of oxalic acid reacts with two moles of NaOH, the number of moles of oxalic acid dihydrate (H2C2O4 · 2H2O) is also 0.0084 mol.

Now, let's calculate the molar mass of oxalic acid dihydrate:

Molar mass of H2C2O4 · 2H2O = (2 × atomic mass of H) + (2 × atomic mass of C) + (4 × atomic mass of O) + (2 × atomic mass of H) + (2 × atomic mass of O)

= (2 × 1.008 g/mol) + (2 × 12.011 g/mol) + (4 × 16.00 g/mol) + (2 × 1.008 g/mol) + (2 × 16.00 g/mol)

= 90.034 g/mol

Finally, we can calculate the weight of solid oxalic acid dihydrate:

Weight = Moles × Molar mass

= 0.0084 mol × 90.034 g/mol

= 0.756 g

Therefore, approximately 0.756 grams of solid oxalic acid dihydrate (H2C2O4 · 2H2O) could be completely neutralized with 35.0 mL of 0.240 M NaOH.

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A copper penny dropped into a solution of nitric acid produces a mixture of nitrogen oxides. The following reaction describes the formation of NO, one of the products.
3cu(s) + 8H^+(aq) + 2 NO^-_3(aq) ? 2NO(g) + 3Cu^2+(aq) + 4H_2O(l)
Using the provided standard potentials, calculate E_cell for this reaction.
Cu^2+(aq) + 2e^- Cu(s) E?=0.3419 V
4H^+ (aq) + NO_3^-(aq) + 3e^- NO(g) + 2H_2O(l) E?= 0.96 V

Answers

The standard cell potential for the given reaction is approximately 1.3019 V.

The overall cell reaction will be written as;

3Cu(s) + 8H⁺(aq) + 2NO₃⁻(aq) → 2NO(g) + 3Cu²⁺(aq) + 4H₂O(l)

The standard cell potential, E°cell, can be calculated using the standard reduction potentials of the half-reactions involved in the cell reaction;

E°cell =E°(reduction at cathode) - E°(reduction at anode)

The reduction half-reactions are;

Cu²⁺(aq) + 2e⁻ → Cu(s) E°(Cu²⁺/Cu) = 0.3419 V

NO₃⁻(aq) + 4H⁺(aq) + 3e⁻ → NO(g) + 2H₂O(l) E°(NO₃⁻/NO) = 0.96V

Since copper is reduced (gains electrons) at the cathode, its reduction potential is used with a positive sign. NO₃⁻ is oxidized (loses electrons) at the anode, so its reduction potential is used with a negative sign.

E°cell =E°(reduction at cathode) - E°(reduction at anode)

E°cell = E°(Cu²⁺/Cu) - E°(NO₃⁻/NO)

E°cell = 0.3419 V - (-0.96 V)

E°cell = 1.3019 V

Therefore, the E_cell for this reaction is 1.3019 V.

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Balance the following six overall redox reactions. For 1 - 4 indicated whether the reaction willgenerate acid. 1. CusFeS, (bornite) + 02 (g) + Cu?* + Fe(OH)3 (s) + SO,?-2.ZnS + + Zn^2 + SO.?.3. FesSa ++ Fe(OH)36) + SO,?-4. FeS2 Pyrite + NO; + N2 (g) + FeOOH Gosthite + SO,?. 5. MnOs + CI + * Mn?* + Cl,6.As2S; + NO; +> HAsO, + S + NO

Answers

The balanced equations for the given redox reaction is Cu₅FeS₄ + 18O₂ → 5Cu²⁺ + 2Fe(OH)₃ + 12SO₄²⁻

Here are the balanced equations for the given redox reactions:

Cu₅FeS₄ + 18O₂ → 5Cu²⁺ + 2Fe(OH)₃ + 12SO₄²⁻ (acidic)

2ZnS + 3O₂ → 2Zn²⁺ + 2SO₄²⁻ (acidic)

4Fe₃S₄ + 15O₂ → 2Fe₂O₃ + 12SO₄²⁻ + 8H₂O (acidic)

4FeS₂ + 15NO₃⁻ + 7H₂O → 3N₂ + 4FeOOH + 21H⁺ + 12SO₄²⁻ (acidic)

The reactions 1-3 are marked as "acidic" because they produce protons (H⁺) in their balanced equations. In other words, these reactions occur in acidic media. Reaction 4 is also marked as "acidic" because it requires acid (H⁺) for the reaction to proceed.

The correct question is :

Balance the following six overall redox reactions. Indicate whether the reaction will generate acid.

1. Cu₅FeS₄ (bornite) + O₂(g) → Cu²⁺ +  Fe(OH)₃ (s) + SO₄²⁻

2. ZnS + O₂(g) → Zn₂+ + SO₄²⁻

3. Fe₃S₄ + O₂ (g) → Fe(OH)₃ (s) + SO₄²⁻

4. FeS₂ Pyrite + NO₃⁻ → N₂ (g) + FeOOH Goethite + SO₄²⁻

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