A strawberry nutritional drink used for a liquid diet is flavored with methyl butanoate. Draw the structure of methyl butanoate.

Answers

Answer 1

Answer:

See explanation and image attached

Explanation:

Methyl butanoate is an ester. Esters have the general molecular formula, RCOOR where the two Rs may represent the same or different alkyl groups.

Methyl butanoate is has a fruity odor, smelling like apples or pineapples fragrance. It is also called methyl butyrate.

The structure of the compound is shown in the image attached to this answer.

A Strawberry Nutritional Drink Used For A Liquid Diet Is Flavored With Methyl Butanoate. Draw The Structure

Related Questions

For the following acids of varying concentrations, which are titrated with 0.50 M NaOH, rank the acids in order of least to most volume of base needed to completely neutralize the acid.

a. 0.2M H2C6H5O7
b. 0.2M H2C2O4

Answers

Answer:

0.2M H2C6H5O7 < 0.2M H2C2O4

Explanation:

A weak acid/base ionizes to a very small extent in water. Hence, if we say that a substance is a weak acid/base, its percentage of ionization in solution is very little.

More volume of a very weak acid is required to neutralize a strong base. Since NaOH is a strong base, the weaker acid among the duo will require more volume for neutralization.

Since H2C6H5O7 is a weaker acid than H2C2O4, equal concentration of the both acids will require less volume of H2C2O4 than H2C6H5O7 to neutralize 0.50 M NaOH.

H₂C₆H₅O₇ is a weaker acid than H₂C₂O₄, and will require the least volume of 0.50 M NaOH to be neutralized.

H₂C₆H₅O₇ < H₂C₂O₄

The strength of an acid is related to the value of its dissociation constant, Ka or its pKa (negative logarithm of Ka)

Strong acids have high Ka values or low pKa value, whereas weak acids have low Ka values and high pKa values.

Between two acids, the acid with a higher Ka or lower pKa values is the stronger acid.

Acids are classified as either strong or weak depending on how well it ionizes in solution to produce hydrogen ions.

Strong acids ionizes completely to produce hydrogen ions.

Weak acid ionizes partially to a varying degrees in water to produce hydrogen ions.

In neutralization reactions between acids and bases, stronger acids will require the most volume of base or alkali in order to be neutralized.

H₂C₂O₄ has a Ka value of 5.9 x 10⁻² and a pKa value of 1.23

H₂C₆H₅O₇ has a Ka value of 8.4 x 10⁻⁴ and a pKa value of 3.08

Hence H₂C₂O₄ is a stronger acid than H₂C₆H₅O₇

For equal molar concentrations of the two acids, H₂C₂O₄ will produce more hydrogen ions than H₂C₆H₅O₇, and thus, will require more volume base (0.50 M NaOH) to be neutralized.

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which of the following is indicated by the ph value of a solution?
a- it's hydrogen ion concentration
b- its ammonium ion concentration
c- ability to undergo chemical reaction
d- its ratio of solute amount to solvent volume

Answers

Answer:

c- ability to undergo chemical reaction

Consider the following reaction:
Cr(NO3)3 (aq) + 2NaF (aq) --> 3NaNO3 (aq) + CrF3 (s)
If 21.0 grams of NaF are needed to precipitate all of the Cr+3 ions present in 0.125L of a solution of Cr(NO3)3, what is the molarity of the Cr(NO3)3 solution?

Your answer should be to 2 decimal places.

Answers

Answer:

2.01

Explanation:

First, let's convert grams to moles

(Na) 22.99 + (F) 18.998 = 41.988

Every mole of NaF is  41.988 grams

21/41.988 = 0.500143 moles of NaF

For every Cr+3, we will need 2 NaF, so Cr+3 will be half of NaF

0.500143/2 = 0.250071

molarity = moles/liters

0.250071/0.125 = 2.0057 M

Which phenomenon explained below is an example of deposition?
Select the correct answer below:

A) Hail is formed from water droplets lifted by air currents to an altitude where they turn into pellets of ice.

B) Frost forms when cold evening temperatures convert the humidity in the air to thin layers of ice on the ground.

C) In the winter, the top few inches of a pond turn to ice.

D) The visible cloud arising from a boiling tea kettle is not actually steam, but droplets of liquid water that form as the
steam cools in the air.

Answers

Answer:

b

Explanation:

deposition is when water turns from gas to solid. b is the only one that fits

Deposition is frost forms when cold evening temperatures convert the humidity in the air to thin layers of ice on the ground.

What is deposition?

Deposition is a process that involves collection of large mass or when mean distance between molecules are reduced. It can also be explained as gathering of substances together to form a larger mass.

Therefore, the phenomenon explained in the given example about deposition is frost forms when cold evening temperatures convert the humidity in the air to thin layers of ice on the ground.

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the nutrition label on rice lists the amounts of protein, carbohydrates and fats in one serving. these substances are important for human nutrition

Answers

Answer:

Carbohydrates, proteins, and fats are biological macromolecules that are made up of chemical elements which are inherent to chemistry.

Chemistry explain how these macromolecules are bonded together at the molecular level and give an explanation for their behavior.

Explanation:

The decrease in the water table due to overuse of water.

Answers

Answer:  Groundwater and surface water are connected. When groundwater is overused, the lakes, streams, and rivers connected to groundwater can also have their supply diminished. Land subsidence occurs when there is a loss of support below ground. This is most often caused by human activities, mainly from the overuse of groundwater, when the soil collapses, compacts, and drops.

Hope it Helps. :)

How many grams of magnesium chloride can be produced from 2.30 moles of chlorine gas reacting w excess magnesium Mg(s)+Cl2(g)->MgCl2(s)

Answers

The mass of magnesium chloride produced from 2.30 moles of chlorine gas is 218.99 grams.

How to calculate moles in stoichiometry?

Stoichiometry refers to the study and calculation of quantitative (measurable) relationships of the reactants and products in chemical reactions.

According to this question, magnesium reacts with chlorine gas to form magnesium chloride as follows:

Mg + Cl₂ → MgCl₂

Based on the above chemical equation, 1 mole of chlorine gas forms 1 mole of magnesium chloride.

This means that 2.30 moles of chlorine gas will 2.30 moles of magnesium chloride.

Next, we convert moles of magnesium chloride to mass as follows:

molar mass of magnesium chloride = 95.211g/mol

mass of magnesium chloride = 95.211 × 2.30 = 218.99 grams.

Therefore, 218.99 grams of magnesium chloride will be formed.

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11 Explain how you would obtain solid lead carbonate from a mixture of lead carbonate and sodium chloride

Answers

Explanation:

Add water, Na2CO3 dissolves, filter, PbCO3 stays in the paper and dissolved Na2CO3 goes through as the solution. Dry the PbCO3 and you have the dry solid.

OR

Add water to dissolve then filter to obtain PbCo3 as you're residue and Na2Co3 as the filtrate. Dry the insoluble PbCo3 between filter papers and you obtain solid PbCo3

Van der Waals forces hold molecules together by: A. moving electrons from one molecule to another. B. attracting a lone pair of electrons to the positive charge of a hydrogen. C. inducing temporary dipoles that attract each other. D. sharing electrons between atoms.

Answers

Van der Waals forces hold molecules together by inducing temporary dipoles that attract each other. That is option C

Van Der Waals forces are example of those intermolecular forces which are weaker than ionic and covalent bonds that exists between molecules.

Van Der Waals forces was postulated by a Dutch physicist known as Van Der Waals. He postulated the existence of weak, short-range forces of attraction, which are independent of normal bonding forces, between non-polar molecules. He came to this conclusion after studying the  of real gases at low temperatures and high pressures that:

electrons in a non-polar molecule such as hydrogen are close to one nucleus as to the other, although momentary concentration at one end of the molecule may occur, this momentary concentration of electron cloud on one side create a temporary dipole in the hydrogen molecule, that is, one side of the molecule acquires a partial negative charge while the other side acquires a partial positive charge of equal magnitude, the temporary dipole induces a similar dipole in an adjacent behavior molecule, this results in a temporary dipole-induced dipole attraction between the positive and negative ends of the adjacent molecules.

This is how weak Van Der Waals forces are set up. Therefore, option C is CORRECT

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how many mols of nh4cl are present in 279.0ml of a 0.975 M nh4cl soution?

Answers

Answer:

how many mols of nh4cl are present in 279.0ml of a 0.975 M nh4cl soution?

state function and non state function ​

Answers

Answer:

State functions represent quantities or properties of a thermodynamic system, while non-state functions represent a process during which the state functions change. For example, the state function PV is proportional to the internal energy of an ideal gas, but the work W is the amount of energy transferred as the system performs work.

Explanation:

cesium-131 has a half life of 9.7 days. what percent of a cesium-131 sample remains after 60 days?

Answers

1.37% of cesium–131 will remain after 60 days

Explanation:

From the question given above, the following data were obtained:

Half-life (t½) = 9.7 days

Time (t) = 60 days

Percentage remaining after 60 days =?

Next, we shall determine the number of half-lives that has elapsed. This can be obtained as follow:

Half-life (t½) = 9.7 days

Time (t) = 60 days

Number of half-lives (n) =?

n = t / t½

n = 60 / 9.7

Finally, we shall determine the percentage remaining. This can be obtained as follow:

Let the original amount be N₀

Let the amount remaining be N

Number of half-lives (n) = 60 / 9.7

N = N₀ / 2ⁿ

Divide both side by N₀

N/N₀ = 1/2ⁿ

N/N₀ = 1 / 2⁽⁶⁰÷⁹•⁷⁾

N/N₀ = 0.0137

Multiply by 100 to express in percentage

N/N₀ = 0.0137 × 100

N/N₀ = 1.37%

Therefore, the percentage remaining after 60 days is 1.37%

NOTE; N/N₀ is the fraction remaining.

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Determine the value of the equilibrium constant for the following reaction, where the following amounts of each species are present at equilibrium in a 5.00 L container: 1.34 mol HCl, 4.30 mol O2, 30 g H2O, and 2.42 mol Cl2.
4 HCl(g) O2(g) ----> 2 H2O(l) 2 Cl2(g)

Answers

Explanation:

here's the answer to your question about

Which one of the following is not matches the organelle with its function

Answers

Answer:

rip there isnt a photo

Explanation:

i do know a lot about cells tho lol

1. A
a stiff structure that surrounds and
protects a coll: found in plant, fungus, and some bacteria cells.
2.
is living things consisting of many cells.
3.
a green pigment that traps energy
from the sun.
the process in which plants and
some other organisms use the energy in sunlight to make food.
5. A
found in the nucleus of a cell,
a long nucleic acid molecule containing the genetic instructions
for the development and functioning of all living organisms.

Answers

Answer:

1 cell wall

2 yes

3 chloroplast

4 photosynthesis

5 Deoxyribonucleic acid (I believe)

hope this helped a little and pls mark brainiest if it did :)

Explanation:

The cell wall is a rigid layer that is found outside the cell membrane and surrounds the cell, providing structural support and protection.

Assuming a mixture of equal volumes of o xylene and cyclohexane,which of these will distill off first?

Answers

cyclohexane will distill off first as it will have lower boiling point compared to ortho xylene which has higher molecular mass

What is the molarity of a solution that contains 0.75 mol Naci in 3.0 L of solution? Select one: O a. 4.0 M O b. 2.3 M O d. 3.8 M O d. 0.25 M Clear my choice​

Answers

Answer:

[tex]\boxed {\boxed {\sf D. \ 0.25 \ M}}[/tex]

Explanation:

Molarity is a measure of concentration in moles per liter.

[tex]molarity= \frac{moles \ of \ solute}{ liters \ of \ solution}[/tex]

The solution contains 0.75 moles of sodium chloride and has a volume of 3.0 liters.

moles of solute = 0.75 mol NaCl liters of solution = 3.0 L

Substitute these values into the formula.

[tex]molarity= \frac{ 0.75 \ mol \ NaCl}{3.0 \ L}[/tex]

[tex]molarity= 0.25 \ mol \ NaCl/L[/tex]

Molarity has the molar (M) as its unit. 1 molar is equal to 1 mole per liter.

[tex]molarity= 0.25 \ M \[/tex]

The molarity of the solution is 0.25 Molar and Choice D is correct.

For an atoms electrons, how many energy sublevels are present in the principal energy level n = 4?
A. 4
B. 9
C. 10
D. 16
E. 32

Answers

Answer:

by the own's formula energy sublevels are 2 the power of n or principal quantum number this means 2 the power of 4 equal to 32

Starting from (R)-3-methylhex-1-yne as the substrate at the center of your page, draw a reaction map showing the regiochemical and stereochemical outcome or outcomes for each of the following series of reagents. Name each of your products, including stereochemical designations for any chirality centers that are generated.

a. HgSO4, H2SO4, H2O
b. 1. 9-BBN; 2. H2O2, NaOH
c. Br2, CCl4
d. HBr

Answers

Solution :

A substrate is defined as the chemical species that are being observed in the chemical reaction where the substrate reacts with a reagent and forms a product. It can also be referred to the surface where some other chemical reactions are performed.

Stereochemistry is defined as the study of relative spatial arrangement of the atoms which forms the structure of the molecules and their respective manipulations.

In the context, the products including the stereochemical designations for any chirality centers starting from the  (R)-3-methylhex-1-yne as the substrate are attached below.  

Which subshells are found in each of the following shells
electron subshell - M shell

Answers

Answer:

3

Explanation:

The electron shells are labelled as K,L,M,N,O,P, and Q or 1,2,3,4,5,6, and 7.

As we go from innermost shell outwards, this number denotes the number of subshell in the shell. Electrons in outer shells have higher average energy and travel farther from the nucleus than those in inner shells.

Hence, M shell contains s,p and d subshells.

Which redox reaction would most likely occur if silver and copper metal were added to a solution that contained silver and copper ions?
A. Cu + Agt Cu2+ + 2Ag
B. Cu2+ + 2Ag* → Cu + 2Ag
C. Cu2+ + 2Ag → Cu + 2Ag+
D. Cu + 2Ag Cu²+ + 2Ag+​


give the wrong answer and I'm reporting ​

Answers

Answer:

B

Explanation:

b/c copper is readuction agent

The most likely redox reaction that would occur if silver and copper metal were added to a solution that contained silver and copper ions is   [tex]\rm Cu^{2+} + 2Ag \rightarrow Cu + 2Ag^+[/tex]. The correct answer is option C.

Redox reaction is a reaction in which reduction and oxidation takes place simultaneously.

In this reaction:

[tex]\rm Cu^{2+} + 2Ag \rightarrow Cu + 2Ag^+[/tex]

Copper metal has a higher reduction potential than silver metal, which means that it will be oxidized to [tex]\rm Cu^{2+}[/tex] ions before silver metal is oxidized to  [tex]\rm Ag^+[/tex] ions.

The [tex]\rm Cu^{2+}[/tex] ions in the solution will then react with the silver metal to form [tex]\rm Ag^+[/tex] ions and Copper metal. This reaction is an example of a displacement reaction, where a more reactive metal removes a less reactive metal from its compound.

Therefore, option C. [tex]\rm Cu^{2+} + 2Ag \rightarrow Cu + 2Ag^+[/tex] is the correct answer.

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Sally has constructed a concentration cell to measure Ksp for MCln. She constructs the cell by adding 2 mL of 0.05 M M(NO3)n to one compartment of the microwell plate. She then makes a solution of MCln by adding KCl to M(NO3)n. She adds 7.903 mL of the resulting mixture to a second compartment of the microwell plate. Sally knows n = +2. She has already calculated [Mn+] in the prepared MCln solution using the Nernst equation. [Mn+] = 8.279 M

Required:
How many moles of [Cl-] must be dissolved in that compartment?

Answers

Answer:

0.1309 mol

Explanation:

From the given information:

The metal ion, two ions of [tex]M^{+}[/tex] reacted with Cl⁻ to form [tex]MCl_n[/tex] i.e. the compound formed is [tex]MCl_2[/tex].

The concentration of the metal ion formed [tex][M^+][/tex] = 8.279 M

The concentration of the chlorine ion formed [tex][Cl^-][/tex] = 2 × 8.279 M

= 16.558 M

We know that:

[tex]\mathsf{Molarity = \dfrac{no \ of \ moles }{volume (mL)}}[/tex]

The number of moles of [tex][Cl^-][/tex] = [tex]16.558 \ mol.L^{-1} \times 7.903 \ mL \times \dfrac{1 \ L}{1000 \ mL}[/tex]

= 0.1309 mol

Which statement describes the 3d, 4s, and 4p orbitals of Arsenic (As) based on its electronic configuration and position in the periodic table?
The 3d and 4s orbitals are completely filled, and the 4p orbital is partially filled.
The 3d orbital is completely filled, and the 4s and 4p orbitals are partially filled.
The 3d, 4s, and 4p orbitals are completely filled.
The 3d, 4s, and 4p orbitals are partially filled.

Answers

Answer:

The 3d and 4s orbitals are completely filled, and the 4p orbital is partially filled.

Explanation:

The correct answer is: The 3d and 4s orbitals are completely filled, and the 4p orbital is partially filled.

The d orbital contains 10 electrons, the s orbital takes 2 electrons and the p orbital takes six electrons.

The orbital in chemistry is defined as a region in space where there is a high probability of finding an electron. There are s, p, d, f orbitals in chemistry which correspond to sharp, principal, diffuse and fundamental.

The electronic configuration of arsenic is 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3.

From this electronic configuration, we can see that the 4s and 3d orbitals are half filled while the 4p orbital is half filled.

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Determine whether each of the examples represents a colligative property or a non-colligative property. boiling point elevation Choose... color Choose... freezing point depression Choose... vapor pressure lowering Choose... density Choose...

Answers

Answer:

boiling point elevation - colligative property

color - non-colligative property

freezing point depression - colligative property

vapor pressure lowering - colligative property

density - non-colligative property

Explanation:

A colligative property is a property that depends on the number of particles present in the system.

Freezing point depression, boiling point elevation and vapour pressure lowering are all colligative properties of solutions.

Colour and density do not depend on the number of particles present hence they are not colligative properties.

The boiling point elevation, freezing point depression, and vapor pressure lowering are colligative properties. And color and density are non-colligative properties.

Explanation:

The colligative properties are the properties depending upon the number of particles of solute not on the nature of the solute.Example of colligative properties:Vapor pressure loweringElevation boiling pointDepression in freezing pointOsmotic pressureThe non-colligative properties are the properties depending upon the nature of solute and solvent.Example of non-colligative properties :ViscositySurface tensionDensitySolubility

So, from this, we can conclude that boiling point elevation, freezing point depression, and vapor pressure lowering are colligative properties. And color and density are non-colligative properties.

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Calculate the molarity of a solution consisting of 65.5 g of K2S0 4 in 5.00 L of solution. ​

Answers

Answer:

Molarity is 0.075 M.

Explanation:

Moles:

[tex]{ \tt{ = \frac{65.5}{RFM} }}[/tex]

RFM of potassium sulphate :

[tex]{ \tt{ = (39 \times 2) + 32 + (16 \times 4)}} \\ = 174 \: g[/tex]

substitute:

[tex]{ \tt{moles = \frac{65.5}{174} = 0.376 \: moles}}[/tex]

In volume of 5.00 l:

[tex]{ \tt{5.00 \: l = 0.376 \: moles}} \\ { \tt{1 \: l = ( \frac{0.376}{5.00} ) \: moles}} \\ { \tt{molarity = 0.075 \: mol \: l {}^{ - 1} }}[/tex]

Determine the effect each given mutation would have on the rate of glycolysis in muscle cells.

a. loss of binding site for fructose 1 ,6-bisphophate in pyruvate kinase.
b. loss of allosteric binding site for ATP in pyruvate kinase.
c. loss of allosteric binding site for AMP in phosphofructokinase.
d. loss of regulatory binding site for ATP in phosphofructokinase.

1. Increase
2. decrease
3. No effect

Answers

Answer:

a. Decrease

b. Increase

c. Increase

d. No effect

Explanation:

Glycolysis is present in muscle cells which converts glucose to pyruvate, water and NADH. It produces two molecules of ATP. Cellular respiration produces more molecules of ATP from pyruvate in mitochondria. Glycolysis increases in pyruvate kinase.

a. Loss of binding site for fructose 1,6-bisphosphate in pyruvate kinase: Decrease

b. Loss of allosteric binding site for ATP in pyruvate kinase: No effect

c. Loss of allosteric binding site for AMP in phosphofructokinase: Increase

d. Loss of regulatory binding site for ATP in phosphofructokinase: Increase

A. An important substrate in the glycolysis pathway is fructose 1,6-bisphosphate. It stimulates pyruvate kinase, an essential enzyme in glycolysis. The amount of pyruvate kinase that is activated will decrease if the fructose 1,6-bisphosphate binding site in pyruvate kinase is eliminated. As a result the rate of glycolysis in the muscle cells will probably decrease.

B. The allosteric ATP binding site of pyruvate kinase controls how active the enzyme is. However, pyruvate kinase is not significantly regulated by ATP in muscle cells. Therefore, it is unlikely that deletion of the ATP-binding allosteric site in pyruvate kinase would have no effect on the rate of glycolysis in muscle cells.

C. The rate-limiting enzyme in glycolysis, phosphofructokinase, is activated from all forms by AMP. It increases the rate of glycolysis by stimulating the activity of phosphofructokinase. If the allosteric binding site for AMP is eliminated, phosphofructokinase activation will be reduced. As a result, the rate of glycolysis in muscle cells will decrease.

D. Phosphofructokinase is inhibited allosterically by ATP. It regulates the rate of glycolysis by a feedback mechanism. High ATP concentrations cause phosphofructokinase to bind to its regulatory site, limiting its activity and delaying glycolysis. If the regulatory binding site for ATP is eliminated, the inhibitory action of ATP on phosphofructokinase would be lost. As a result, muscle cells will glycolysis at a faster rate.

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1. Draw the condensed structural formula of sodium benzoate showing all charges, atoms including any lone pairs in the side chain functional group, and all sigma and pi bonds.
2. Draw the condensed structural formula of benzoic acid showing all atoms including any lone pairs in the side chain functional group, and all sigma and pi bonds. Indicate the acidic hydrogen.
3. Draw the condensed structural formula of tetrahydrofuran (THF) showing all heteroatoms plus their lone pairs and all sigma and pi bonds.

Answers

The structures are shown in the image attached.

A structural formula is the representation of the molecule in which all atoms and bonds in the molecule are shown.

Since the question requires that all the lone pairs, formal charges and sigma and pi bonds should be shown, then the simple condensed structural formula becomes insufficient in this case.

I have attached images of the structural formula of sodium benzoate (image 1), benzoic acid (image 2)  and tetrahydrofuran (image 3).

All the formal charges, lone pairs as well as sigma and pi bonds are fully shown.

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When comparing Be2 and H2:

I. Be2 is more stable because it contains both bonding and antibonding valence electrons.
II. H2 has a higher bond order than Be2.
III. H2 is more stable because it only contains 1s electrons.
IV. H2 is more stable because it is diamagnetic, whereas Be2 is paramagnetic

a. II,III,IV
b.II,III
c.III only
d.I,II
e.III,IV.

Answers

Answer:

The answer is "Option b".

Explanation:

H2 does have bond energy of 1, while Be2 has a covalent bond of zero. Be2 has eight electrons, each of which dwells in a distinct orbital. As just a result, four of them are linked molecular orbitals and two are antibonding molecular orbitals, respectively. As just a result, this molecule is unstable. This chemical orbital, with a bond order of 1, has just two electrons. As a result, it is a very solid substance. H2's bond length is higher than Be2's. Since it only has one electron, H2 is more stable than that of other compounds.

calculate the volume of 20.5g of oxygen occupied at standard temperature and pressure.what the volume​

Answers

Answer :

volume of a gas = weight * 22.4 l / gram molecular weight

volume of o2 = ?

weight given = 20.5 g

gram molecular weight of oxygen = 32 (because of 2 oxygen atoms )

volume of oxygen = 20.5 * 22.4 / 32

volume of oxygen = 14.35 liters  

Explanation:

hope this helps you

if wrong just correct me

Another method for creating a buffer, in situ, is to add an appropriate amount of a strong base, e.g., NaOH, to a weak acid OR add an appropriate amount of a strong acid, e.g., HNO3, to a weak base. As an example, mixing 1.0 mol of acetic acid with 0.5 mol of NaOH will result in a buffer solution with 0.5 mol of acetic acid and 0.5 mol of acetate. The acetate is created by the reaction of acetic acid and the strong base, hydroxide. Given this information, which of the following, when mixed with the appropriate amount of HCl, would create a buffer solution?

a. HNO3
b. HClO2
c. LiCl
d. NH3

Answers

Answer:

As an example, mixing 1.0 mol of acetic acid with 0.5 mol of NaOH will result in a buffer solution with 0.5 mol of acetic acid and 0.5 mol of acetate. The acetate is created by the reaction of acetic acid and the strong base, hydroxide.

When HClO2 is mixed with the appropriate amount of HCl it would create a buffer solution. That is option B.

Methods used to form buffer solution

A buffer solution is the solution that resists a change in pH of a solution when acid or base is added because it is made up of weak acid and the conjugate base or weak base and the conjugate acid.

The methods that can be used to form a buffer solution include:

Adding a strong base to a weak acid: For example, mixing 1.0 mol of acetic acid with 0.5 mol of NaOH will result in a buffer solution with 0.5 mol of acetic acid and 0.5 mol of acetate.

Adding a weak acid to a conjugate base: For example HCl is a strong acid which will react with a conjugate base such as HClO2.

Although HCl is a strong acid, it can be converted to a weak acid through dilution with water. It is in this context that it can be used to form a buffer solution.

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