What volume of H2O(g) is produced when 8.00 mol of C2H4(g) reacts at STP?

C2H4 + 3O2 >>>>> 2CO2 + 2H2O

Answer is in Word document below.

On the drawing of the lecithin:

1) alcohol glycerol is purple colored.

2) choline and phosphate group are orange colored.

3) stearic acid (saturated fatty acid) is red colored.

4) oleic acid (monounsaturated fatty acid) is green colored.

Lecithin is a yellow brownish fatty substance in animal and plant tissues.

As the balloon’s altitude increases, its volume also increases.

At high altitude, the atmospheric pressure on the outside of the balloon is less that it is at Earth’s surface.

As the pressure on the outside of the balloon decreases, the balloon’s volume increases because the pressure inside the balloon pushes the balloon outward.

As the balloon’s volume increases, the pressure inside the balloon decreases until it is equal to the pressure on the outside of the balloon.

are answers for edg

A weather balloon's volume increases as it rises due to decreased atmospheric pressure in accordance to Boyle's Law. The ideal gas law better explains this behavior considering the changing temperatures. Eventually, the balloon will burst when its material's limit is reached.

As a weather balloon rises into the atmosphere, the atmospheric pressure decreases. This occurs because the atmosphere becomes thinner with elevation, resulting in fewer air molecules to exert force on any given surface area. According to Boyle's Law, which states that the pressure of a gas is inversely proportional to its volume when temperature is held constant, the volume of the gas inside the balloon will increase as the balloon ascends and the outside pressure decreases. However, since the temperature is not constant during ascent and typically drops, the ideal gas law, which combines Boyle's Law and Charles's Law, is a better descriptor of the balloon's behavior. This law indicates that the relationship between the temperature, volume, and pressure of the gas will dictate the expansion of the balloon. Notably, as the volume expands and the balloon material reaches its stretching limit, it will eventually burst.

Answer: 0.0857 L [tex]BaCl_2[/tex]

Explanation: It's a stoichiometry problem. Balanced equation is given from which there is 1:1 mol ratio between [tex]BaCl_2[/tex] and [tex]BaSO_4[/tex] .

Chemist wants to produce 12.00 grams of barium sulfate by reacting a 0.6000 M barium chloride solution with excess sulfuric acid.

We know that molarity is moles of solute per liter of solution. Molarity of barium chloride is given. If we know its moles then its volume could easily be calculated.

From given grams of barium sulfate, we calculate its moles and then using mol ratio we calculate the moles of barium chloride.

Molar mass of barium sulfate is 233.38 gram per mol.

The complete set up is shown below using dimensional analysis:

[tex]12.00gBaSO_4(\frac{1molBaSO_4}{233.38gBaSO_4})(\frac{1molBaCl_2}{1molBaSO_4})(\frac{1LBaCl_2}{0.6000molBaCl_2})[/tex]

= 0.0857 L [tex]BaCl_2[/tex]

So, 0.0857 L or 85.7 mL of [tex]BaCl_2[/tex] should be used.

Answer:

See explanation

Explanation:

First, in order to know this it's neccesary to remember how a SN2 reaction takes place. A Sn2 reaction is a bimolecular concerted reaction where all bonds are broken and making in only one step.

For this to occur, we need a strong nucleophyle (such a strong base) and a substract with a great outgoing group (The halides are great leaving groups).

The nucleophyle attacks on the back side of the molecule with the bromine, and the result is a molecule with inverted configuration and the bromine is replaced by the nucleophyle.

However, this step is fast and concerted, and in order to do this faster, the reactant must be (Ideally) with no substituent, because if the molecule is bulky, the nucleophyle's attack to the back side is hard. This doesn't mean that it will not undergo, but it will be harder and slower.

Because of this reason, we can see that from all the alkyl bromides there, the least bulky is the 1-bromopentane, so this will be the more reactive in Sn2, followed by 1-bromo-methylbutane, then the 1-bromo-2-methylbutane and finally the 2 - bromo - 2 - methylpentane.

In the picture, you have the structures of these molecules, so you can see how the steric hindrance affects this.

Orbital shell notation of fluorine is 2. 7 while that of oxygen s 2. 6. This means that these elements (that follow each other in the periodic table) will have high electronegativity in molecules due to their high atomic number (which causes them to strongly attract electron orbital shell closer to their nucleus). NB: Atomic number of a peroid increased from left to right of the periodic table.

Therefore, in the first molecule, the negative dipole would most likely be located between the F atoms In the second molecule the negative molecule would be most likely located in the between the O and F atoms.

In the given question, the oxygen atom serves as the negative pole in both [tex]\rm CF_2O \ and\ CHFO[/tex].

Molecules are the smallest units of a compound that retain all of the chemical properties of that compound. They consist of two or more atoms that are bonded together chemically.

Therefore, the negative pole in both [tex]\rm CF_2O \ and\ CHFO[/tex] is located on the oxygen atom.

Learn more about molecules here:

https://brainly.com/question/32298217

#SPJ6

The given question is not complete. The complete question is:

Where, approximately, is the negative pole on each of these molecules [tex]\rm CF_2O \ and\ CHFO[/tex] ?

Answer:

Steam forces the high-pressure turbine to turn.

Explanation:

Hello,

Producing electricity via nuclear power stations involves energy conversion from one form to another until energy in the form of electricity is obtained. Typically, the splitting of uranium atoms through the process of nuclear fission is performed to produce heat energy that is used to boil water into high-pressure steam which turns or spins the fans in the turbines that powers the generator to obtain the electricity.

Best regards.

Answer : The number of moles of carbon in 2.0 mole of caffeine are, 16 moles

Solution :

The formula of caffeine is, [tex]C_8H_{10}N_4O_2[/tex]

As we know that, 1 mole of caffeine contains 8 moles of carbon atom, 10 moles of hydrogen atoms, 4 moles of nitrogen atoms and 2 moles of oxygen atoms.

According to the question,

As, 1 mole of caffeine contains 8 moles of carbon atoms

So, 2 moles of caffeine contains [tex]8\times 2=16[/tex] moles of carbon atoms

Therefore, the number of moles of carbon in 2.0 mole of caffeine are, 16 moles

The balanced equation in basic solution is:

[tex]\[ \text{NO}_2^{-}(\text{aq}) + \text{Al}(\text{s}) + \text{H}_2\text{O}(\text{l}) + 2\text{OH}^{-}(\text{aq}) \rightarrow \text{NH}_4^{+}(\text{aq}) + \text{AlO}_2^{-}(\text{aq}) + \text{OH}^{-}(\text{aq}) + 2\text{e}^{-} \][/tex]

This equation ensures conservation of mass and charge.

To balance the given equation:

[tex]\[ \text{NO}_2^{-}(\text{aq}) + \text{Al}(\text{s}) \rightarrow \text{NH}_4^{+}(\text{aq}) + \text{AlO}_2^{-}(\text{aq}) \][/tex]

We first need to identify the elements present on both sides of the equation:

- Nitrogen (N)

- Oxygen (O)

- Aluminum (Al)

The unbalanced equation shows:

- 1 Nitrogen atom on each side

- 2 Oxygen atoms on the left and 2 Oxygen atoms on the right

- 1 Aluminum atom on each side

To balance the equation, we can start by balancing the atoms that appear only once on each side. In this case, Aluminum (Al) is already balanced.

Next, let's balance the Oxygen atoms by adding water (H₂O) molecules to the side lacking Oxygen atoms.

We will then balance the Hydrogen (H) atoms by adding Hydroxide ions (OH⁻) to the other side. Since the solution is basic, we need to add OH⁻ ions to balance the Hydrogen atoms.

[tex]\[ \text{NO}_2^{-}(\text{aq}) + \text{Al}(\text{s}) + \text{H}_2\text{O}(\text{l}) \rightarrow \text{NH}_4^{+}(\text{aq}) + \text{AlO}_2^{-}(\text{aq}) + \text{OH}^{-}(\text{aq}) \][/tex]

Now, the equation is balanced in terms of atoms.

The balanced equation in basic solution is:

[tex]\[ \text{NO}_2^{-}(\text{aq}) + \text{Al}(\text{s}) + \text{H}_2\text{O}(\text{l}) + 2\text{OH}^{-}(\text{aq}) \rightarrow \text{NH}_4^{+}(\text{aq}) + \text{AlO}_2^{-}(\text{aq}) + \text{OH}^{-}(\text{aq}) + 2\text{e}^{-} \][/tex]

The number of particles in 1.43 g or a molecular compound is 3.69 × 10²¹.  

• The mass of the compound given is 1.43 grams and the molecular mass of the compound is 233 g/mol.  

• The mole of a compound can be determined by using the formula,  

n(number of moles) = Weight/Molecular mass = 1.43/233 = 0.00613 moles

• The number of particles in 1 mole is 6.022 × 10²³ particles

The number of particles in 0.00613 moles is,  

= 0.0063 × 6.022 × 10²³

= 3.69 × 10²¹ particles

Thus, the number of particles in the given case is 3.69 × 10²¹ particles.

To know more about:

https://brainly.com/question/15109723

Answer:

C

Explanation:

The resistance of Antarctic fish to freezing, due to antifreeze proteins in their blood. This is the only answer choice where natural selection is working, and not artificial selection by human means.

Answer: [tex]3.5\times 10^{-4}[/tex]

Explanation:

To calculate the moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}[/tex]

Given mass of theobromine [tex](C_7H_8N_4O_2)[/tex] = 0.064 g

Molar mass of theobromine [tex](C_7H_8N_4O_2)=12\times 7+1\times 8+14\times 4+16\times 2=180 g[/tex]

[tex]\text{Number of moles}=\frac{0.064g}{180g/mol}=3.5\times 10^{-4}moles[/tex]

Final answer:

To calculate the moles of theobromine in a chocolate bar, divide the mass of theobromine by its molar mass. There are approximately 0.000355 moles of theobromine in a 43g chocolate bar containing 0.064g of theobromine.

Explanation:

To find out how many moles of theobromine are in a 43g chocolate bar that contains 0.064g of theobromine, we first need to know the molar mass of theobromine.

The molar mass of theobromine (C₇H₈N₄O₂) is 180.16 g/mol. Using the formula:

Moles = mass (g) / Molar Mass (g/mol)

We can calculate the number of moles:

Moles of theobromine = 0.064g / 180.16 g/mol = 0.000355 moles (approximately).

Therefore, there are approximately 0.000355 moles of theobromine in a 43g chocolate bar.

The number of carbon atoms that are present in a mole of 12 C is 6.022 x 10²³

The mole is a SI unit of measurement that is used to calculate the quantity of any substance.

Mole = Number of 12C atoms in exactly 12 g of 12 C.

6.022 x 10²³ is the Avogadro number. The quantity of moles is equal to the exact number of atoms in grams.

This is the theoretical atomic mass it contains, 6 protons and 6 neutrons. The carbon 12 of isotopes. One mole of carbon will contain exact 6.022 x 10²³ moles.

In 12 g of 12 C there are 6.022 x 10²³ atoms 12 C.

1 mole = 6.022 x 10²³

12 g of 12C contains 6.022 x 10²³  carbon atoms or 1 mol of carbon atoms.

Thus, the number of carbon atoms in a mole of 12 C is 6.022 x 10²³.

To learn more about moles, refer to the link:

https://brainly.com/question/26416088

#SPJ6

A mole of any substance contains exactly 6.02 x 10^23 entities, which could be atoms, molecules, ions, etc. Therefore, a mole of 12C contains 6.02 x 10^23 carbon atoms.

The term mole indicates the amount of a substance that contains the same number of specific entities, such as atoms, molecules, ions, etc., as exactly 12 grams of 12C carbon contains. This quantity is known as Avogadro's number, which is 6.02 x 10^23 entities per mole. Therefore, in one mole of 12C, there are 6.02 x 10^23 carbon atoms.

https://brainly.com/question/24175158

#SPJ3

Final answer:

The question about how many kilograms of CO₂ are used per year to produce Arm and Hammer baking soda requires specific data from the company, but chemical principles and reactions illustrate how CO₂ is used in similar processes. Examples like the reaction of KOH with CO₂ to produce potassium carbonate, and the CO₂ emissions from gasoline, offer an educational understanding of CO₂ use and release in the environment.

Explanation:

Quantifying CO₂ Emissions in Industry and Chemical Reactions

The original question regarding the amount of carbon dioxide (CO₂) used per year to produce Arm and Hammer baking soda cannot be directly answered without specific industrial data from the company. However, we can discuss the chemical processes and general principles that may be involved in the production and the use of CO₂ in related reactions.

For example, in the production of sodium bicarbonate (baking soda), a reaction occurs between sodium carbonate and carbon dioxide to form sodium bicarbonate.

Similarly, other reactions outlined for educational purposes, like the reaction of potassium hydroxide with carbon dioxide producing potassium carbonate and water, demonstrate how CO₂ is utilized in chemical processes.

Specifically, in the reaction between 224.4 grams of KOH and 88.0 grams of CO₂, 138.4 grams of potassium carbonate and 36.0 grams of water are formed.

When comparing CO₂ emissions from gasoline consumption, it's noted that a 40-liter tank of gasoline, with a density of 0.75 kg/L, would release a certain mass of CO₂ upon combustion.

This amount of CO₂ can be compared to typical human mass for perspective. For instance, the combusted gasoline might release more CO₂ than the average weight of a person.

Furthermore, general environmental data is presented to understand the mass concentration of CO₂ in the atmosphere resulting from oil combustion, leading to an increase in atmospheric CO₂ concentration measured in parts per million by mass (ppmm) and volume (ppmv).

Answer:

Because it helps to remove water from the system.

Explanation:

The saturated sodium chloride solution has a strong affinity for water molecules and there is the possibility of changing the saturated solution to a dilute solution in the presence of pure water. Because of these reasons, the saturated sodium chloride solution removes water molecules from the system to become a diluted solution. That is the reason why the saturated solution was used instead of pure water.

Final answer:

Saturated sodium chloride solution is used over pure water to transfer crude product in an experiment due to its higher density and the ability to maintain phase separation, aiding in a more efficient and precise transfer process.

Explanation:

In the context of a chemical experiment, saturated sodium chloride solution is used instead of pure water to transfer the crude product after an initial distillation due to its unique properties. A saturated NaCl solution has a higher density compared to water, which is around 1.2 g/mL. This increased density is significant because it helps to separate the crude product from solvents that may have similar densities to water.

When dealing with a sparing soluble hydrocarbon (HC), phase separation occurs after the saturation point. The presence of a high concentration of NaCl in the transfer medium helps ensure that the HC and water remain separate, thus making the transfer of the HC more efficient. In essence, the use of saturated sodium chloride creates a denser medium which can assist in the separation due to the difference in solubility and density between the aqueous layer and organic compounds.

Furthermore, the saturated solution is at a state of solution equilibrium, where the rate of dissolution equals the rate of recrystallization. This ensures that adding the hydrocarbon or further NaCl will not significantly change the composition of the transfer medium.

To calculate the time passed for a 700 g sample of iodine to decay to 43.75 g, we determine the number of half-lives that have elapsed, which is 4, as 43.75 g is approximately 1/16th of 700 g. Since the half-life for iodine is 8 days, the total time passed is 4 half-lives times 8 days, equalling 32 days.

The half-life of iodine is 8 days. If a 700.00 g sample decays to 43.75 g, to find out how much time has passed, we use the concept of half-lives. A half-life is the time taken for half of a radioactive substance to decay into another element. Knowing the original amount and the final amount of the substance, we can calculate the number of half-lives that have elapsed.

To determine the number of half-lives, we divide the final amount by the initial amount and keep halving until we reach a value less than or equal to 1. In this case, we divide 43.75 g by 700 g and get approximately 0.0625. This corresponds to 4 half-lives since (1/2)^4 = 1/16, and when multiplied by the initial amount (700 g), gives us approximately 43.75 g. Since each half-life is 8 days, 4 half-lives would be 32 days.

Therefore, the time that has passed since the 700 g of iodine began to decay is 32 days.

Answer:

The reduction reaction is: Cl₂ + 2 e⁻ ⇄ 2 Cl⁻

The oxidation reaction is: Cs ⇄ Cs⁺ + 1 e⁻

Explanation:

In a redox reaction, two half-reactions happen simultaneously. The oxidation refers to a species losing electrons. The reduction refers to a species gaining electrons.

Let's consider the following reaction.

Cl₂ + 2 Cs ⇄ 2CsCl

We can write the ionic equation.

Cl₂ + 2 Cs ⇄ 2 Cs⁺ + 2 Cl⁻

The reduction reaction is: Cl₂ + 2 e⁻ ⇄ 2 Cl⁻

We added 2 electrons to the left (electrons gained) to balance the half-reaction electrically.

The oxidation reaction is: Cs ⇄ Cs⁺ + 1 e⁻

We added 1 electron to the right (electrons lost) to balance the half-reaction electrically.

The redox reaction involves the electron transfer between the species. In reduction, the electrons are gained, while in oxidation the electrons are lost.

In reduction reactions, the chemical species gains electrons, whereas, in oxidation reactions, the electrons are donated or lost by the chemical species.

The overall balanced reaction is given as,

[tex]\rm Cl_{2} + 2 Cs \leftrightarrow 2CsCl[/tex]

The ionic reaction for the above reaction is given as,

[tex]\rm Cl_{2} + 2 Cs \leftrightarrow 2 Cs^{+} + 2 Cl^{-}[/tex]

The reduction half of the reaction is given by adding electrons on the left side of the reaction as:

[tex]\rm Cl_{2} + 2 e^{-} \leftrightarrow 2 Cl^{-}[/tex]

The oxidation half of the reaction is given by adding electrons on the right side of the reaction as:

[tex]\rm Cs \leftrightarrow Cs^{+} + 1 e^{-}[/tex]

Therefore, electrons are lost in the oxidation reaction and are gained in the reduction reaction.

Learn more about redox reactions here:

https://brainly.com/question/17218392

Answer:

Explanation:

Hello,

Saponification chemical reactions are defined as such reactions between either potassium hydroxide or sodium hydroxide and a fatty acid to break  hydroxiles in order to attach sodium or potassium to the organic chain. They are a type of esterification chemical reaction.

In this case, lactic acid reacts with sodium hydroxide as shown below:

[tex]CH_3CHOHCOOH+NaOH-->CH_3CHOHCOONa+H_2O[/tex]

As it is seen, sodium lactate is produced rather than lactic acid due to the easiness that the hydroxile has to break and to subsequently attract the ionized sodium to form the lactate, even do, free sodium cations easily break the hydroxile and form sodium lactate.

Best regards.

Answer: The least penetrating of the given radioactive emissions will be alpha particles.

Explanation:

There are 3 radioactive particles which are emitted during radioactive processes:

1. Alpha particles: These particles are emitted when a nuclei undergoes alpha decay process. These particles have low energy associated with them.

2. Beta particles: These particles are emitted when a nuclei undergoes beta decay process. These particles have higher energy than alpha particles.

3. Gamma radiations: These radiations are emitted when an unstable nuclei undergoes gamma ray emission process and gives an excess energy through a spontaneous electromagnetic process. These radiations have the highest energy of all the radioactive particles.

Penetrating power of the particles is directly proportional to the energy of the particles, therefore:

[tex]\text{Penetrating power}\propto \text{Energy of the particles}[/tex]

More the energy of the particles, more will be the penetrating power and vice-versa.

Increasing order of penetrating power will be:

[tex]\text{Alpha particles}<\text{Beta particles}<\text{Gamma radiations}[/tex]

Hence, alpha particles is the least penetrating among the following radioactive particles.

The volume would be occupied by a 33.3 g sample of gold is 1723.6cm³. Density is the mass of a specific material per unit volume.

Density is the mass of a specific material per unit volume. d = M/V, in which d is density, M is weight, and V is volume, is the formula for density. Grams per cubic centimeter are a typical unit of measurement for density. For instance, whereas Earth has a density of 5.51 grams, water has a density of 1 grams.

Another way to state density is in kilograms per cubic meter (in metre-kilogram-second or SI units). For instance, air weighs 1.2 pounds per cubic metre. In textbooks and manuals, the densities of typical solids, liquids, as well as gases are stated.

Density = mass / volume

0.01932 kg/cm³=  33.3 g/ volume

volume =33.3/ 0.01932=1723.6cm³

Therefore, the volume would be occupied by a 33.3 g sample of gold is 1723.6cm³.

To learn ore about density, here:

https://brainly.com/question/29775886

#SPJ6

Final answer:

The volume of carbon dioxide produced at STP from the reaction of 0.900 moles of ethyl alcohol with 1.42 moles of oxygen is approximately 40.3452 liters, determined by the stoichiometric ratio of the reactants and the molar volume of a gas at STP.

Explanation:

To determine the volume of carbon dioxide produced from the reaction of oxygen with ethyl alcohol, we first need the balanced chemical equation for the reaction. The combustion of ethyl alcohol (C₂H₅OH) typically occurs according to the following equation:

C₂H₅OH (l) + 3 O₂ (g) → 2 CO₂ (g) + 3 H₂O (g)

According to the stoichiometry of the reaction, one mole of ethyl alcohol reacts with three moles of oxygen to produce two moles of carbon dioxide. Given that we have 1.42 moles of oxygen and 0.900 moles of ethyl alcohol, we need to find the limiting reactant, which is the reactant that will run out first and dictates how much product can be formed.

We can calculate the moles of carbon dioxide produced based on the stoichiometry and the limiting reactant. At STP, one mole of a gas occupies 22.414 liters. The reaction tells us that for every 1 mole of ethyl alcohol, 2 moles of carbon dioxide are produced. Since the question provides more moles of oxygen than ethyl alcohol (and the molar ratio is 1:3), ethyl alcohol is the limiting reactant.

Therefore, from 0.900 moles of ethyl alcohol, we will get 1.8 moles of carbon dioxide (0.900 moles × 2). Using the molar volume of a gas at STP (22.414 L/mol), we can calculate the volume of carbon dioxide:

Volume of CO₂ = 1.8 moles × 22.414 L/mol = 40.3452 L

The volume of carbon dioxide produced at STP by the reaction would be approximately 40.3452 liters.

The reaction of [tex]1.42[/tex]mol of oxygen with [tex]0.900[/tex] mol of ethyl alcohol produces [tex]21.22[/tex] at STP, in liters of carbon dioxide.

To determine the volume of carbon dioxide formed at STP when [tex]1.42[/tex] mol of oxygen reacts with [tex]0.900[/tex] mol of ethyl alcohol ([tex]\text{C}_2\text{H}_5\text{OH}[/tex] ), the balanced chemical equation must come first:

According to the equation, 1 mole of ethyl alcohol reacts with 3 moles of oxygen to produce 2 moles of carbon dioxide.

Moles of [tex]O_2[/tex]: Given [tex]1.42[/tex] mol [tex]O_2[/tex]

Moles of [tex]\text{C}_2\text{H}_5\text{OH}[/tex]: Given [tex]0.900[/tex]  mol [tex]\text{C}_2\text{H}_5\text{OH}[/tex]

Since 1 mole of [tex]\text{C}_2\text{H}_5\text{OH}[/tex] requires 3 moles of [tex]O_2[/tex], for [tex]0.900[/tex] mol of [tex]\text{C}_2\text{H}_5\text{OH}[/tex], it requires [tex]2.7[/tex] mol of [tex]O_2[/tex]. As only [tex]1.42[/tex] moles of [tex]O_2[/tex] are available, [tex]O_2[/tex] is the limiting reactant.

Moles of [tex]CO_2[/tex] = [tex]\frac{1.42 \, \text{mol O}_2 \times 2 \, \text{mol CO}_2}{3 \, \text{mol O}_2} = 0.947 \, \text{mol CO}_2[/tex]

Volume of [tex]CO_2[/tex] = [tex]0.947 \text{ mol} \times 22.414 \, \text{L/mol} = 21.22 \, \text{L}[/tex]

The given structures represents a soap is E. more than one of the compounds is a soap.  

To understand Soaps are typically the sodium or potassium salts of long-chain fatty acids, also known as carboxylic acids. These long-chain fatty acids contain a hydrophobic (nonpolar) hydrocarbon tail and a hydrophilic (polar) carboxylate head, which allows them to interact with both water and oils.

Now, let’s analyze the options provided:

a. [tex]CH_3CO_2^- K^+[/tex]- This compound does not have a long hydrocarbon chain. It is the acetate ion, which does not classify as a soap.

b. [tex]CH_3(CH_2)_{14}CO_2^- Na^+[/tex] - This is sodium hexadecanoate, which is indeed a soap, as it features a long hydrocarbon chain and a carboxylate group.

c. [tex]CH_3(CH_2)_{12}COOH[/tex] - This is lauric acid, which is not a salt and thus is not a soap.

d. [tex]CH_3(CH_2)_7CO_2(CH_2)_7 Na[/tex] - This represents a soap because it consists of a long hydrocarbon chain (from both ends) and includes a sodium carboxylate.

e. More than one of the compounds is a soap. - From our analysis, options b and d are soaps.

Complete question

Which of these structures represents a soap?

a. [tex]CH_3CO_2^- K^+[/tex]

b.[tex]CH_3(CH_2)_{14}CO_2^- Na^+[/tex]

c.[tex]CH_3(CH_2)_{12}COOH[/tex]

d. [tex]CH_3(CH_2)_7CO_2(CH_2)_7 Na[/tex]

e. More than one of the compounds is a soap.