Lecithins are glycerophospolipids (phosphoglycerides) containing choline, (CH3)3N CH2CH2OH. Draw a lecithin with stearic acid, CH3(CH2)16COOH, at carbon 1 and oleic acid, CH3(CH2)7CH=CH(CH2)7COOH, at carbon 2.

The energy required to split a spherical drop of water into three smaller ones of the same size can be calculated considering the change in surface area of the drops and the surface tension of water. The energy is equivalent to the work done against the surface tension.

To solve this question, we need to understand that when a spherical drop of water splits into multiple smaller drops, energy is required. This energy is equivalent to the work done against surface tension. The energy required, also known as the surface energy, is derived from the change in the surface area of the water drops. Surface tension (γ) is the energy required per unit increase in area.

Let's consider the initial drop of water has a diameter D and the smaller drops each having diameter d. The initial surface area of the spherical drop, using the formula for the surface area of a sphere 4πr² (where r is the radius), is 4π(D/2)² and the total surface area of the 3 smaller drops is 3 * 4π(d/2)². The change in surface area ΔA = 3 * 4π(d/2)² - 4π(D/2)².

The energy required which is derived from the surface tension formula is ΔE = γ ΔA. By substituting the ΔA in this equation, we can calculate the energy required. This problem might require additional details like the ratio between the diameters D and d.

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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.

You should revise a hypothesis when the experimental results do not support the original hypothesis (Option D), based on the process of the scientific method.

In the process of scientific investigation, a hypothesis is essentially a predicted answer to a research question, which is then tested through experiments. You would revise a hypothesis in the instance where experimental results do not support the original hypothesis (Option D). This is based on the scientific method, where the hypothesis is accepted or revised based on the evidence collected.

For example, if you hypothesize that plants grow faster when exposed to classical music and you conduct an experiment that shows no significant difference in growth rates between plants exposed to classical music and those that weren't, you would then revise your hypothesis, potentially considering variables you hadn't initially accounted for or a different predicted outcome.

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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.

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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.

The molecular formula for [tex]CH_{2}[/tex] carrying a molar mass of 112 g/mol would be:

- [tex]C_{8} H_{16}[/tex]

Given that,

Empirical Formula of the compound [tex]=[/tex] [tex]CH_{2}[/tex]

The molar mass of the compound [tex]= 112 g/mol[/tex]

As we know,

Empirical Mass  = [tex]12 + 2[/tex]

[tex]= 14[/tex]

It is known that,

Molar mass [tex]= n[/tex] × [tex]empirical mass[/tex]

∵ [tex]n = molar mass/empirical mass[/tex]

[tex]= 112/14[/tex]

[tex]= 8[/tex]

∵ The Molecular Formula of the compound would be determined by multiplying n into the empirical formula,

so,

Molecular Formula = [tex]n[/tex] × [tex]emprical formula[/tex]

[tex]=[/tex]  ([tex]CH_{2}[/tex]) × [tex]8[/tex]

= [tex]C_{8} H_{16}[/tex]

Thus,  [tex]C_{8} H_{16}[/tex] is the correct answer.

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Explanation:

As density is the amount of mass divided by volume of the substance.

Mathematically,     Density = [tex]\frac{mass}{volume}[/tex]

It is given that volume is 70.0 ml and density is 1.00 g/ml. Therefore, mass of the given substance will be as follows.

                Density = [tex]\frac{mass}{volume}[/tex]

                 1.00 g/ml = [tex]\frac{mass}{70.0 ml}[/tex]

                        mass = 70.0 g

As we known that heat of vaporization of water is 2260 J/g for 1 g of a substance. Therefore, heat of vaporization of water for 70.0 g will be as follows.

                     [tex]70.0 g \times 2260 J/g[/tex]

                       = 158200 J

or,                     = 158.2 kJ                (as 1 kJ = 1000 J)

Thus, we can conclude that 158.2 kJ heat is required to vaporize given sample of water.

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.

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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.

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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] ?

A solution in chemistry is a homogeneous mixture of two or more substances, where one substance (the solute) is dissolved in another substance (the solvent). It might involve physical or chemical changes.

In chemistry, a solution is a homogeneous mixture composed of two or more substances. In such a mixture, a solute is a substance dissolved in another substance, known as the solvent. The process of dissolving can involve physical changes, like if you dissolve granulated sugar in water, you see it disappear since it changes to the microscopic level. However, it can also involve chemical changes, for instance, when you dissolve baking soda in vinegar. You observe a chemical reaction producing carbon dioxide gas, among other things, which indicates a new substance is created.

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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.

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.

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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.

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.

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The concentration of hydronium ions (H3O+) in a solution with a pH of 8.7 is approximately 2.0 × 10^-9 M. This is determined using the formula [H3O+] = 10^{-pH}.

The student is asking about the concentration of hydronium ions, or H3O+, in a solution with a pH of 8.7. The pH scale is used to determine the acidity or basicity of a solution and is calculated as the negative logarithm of the hydronium ion concentration. To find the H3O+ concentration from pH, we use the formula:

[H3O+] = 10^{-pH}

So, for a solution with a pH of 8.7:

[H3O+] = 10-8.7

On a calculator, you would take the antilog, or the "inverse" log, of -8.7 to find the H3O+ concentration:

[H3O+] = antilog (-8.7)

Or simply calculate:

[H3O+] = 10-8.7

This results in the hydronium ion concentration of approximately 2.0 × 10-9 M.

The [H₃O⁺] of a solution with a pH of 8.7 is approximately 2 × 10^-9 M.

The [H₃O⁺] of a solution with pH = 8.7 can be calculated using the formula:

[H₃O⁺] = 10-pH.

For pH = 8.7, the calculation is [H₃O⁺] = 10-8.7.

Using a calculator, you find [H₃O⁺] ≈ 2.015 × 10-9.

Comparing this result with the given options, the closest one is 2 × 10-9 M.

Therefore, the correct answer is [H₃O⁺] ≈ 2 × 10-9 M.

The complete question is:

The [H_3 O^+ ]of a solution with pH=8.7 is

5.3M

8.7×10^(-1) M

8.7M

2×10^(-9) M

5×10^(-6) M

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).

Final answer:

In ionic compounds such as NaCl, CaS, BaO, KBr, and LiF, 1 or 2 electrons are transferred to form the ionic bond, depending on the charges of the ions involved, to achieve electrical neutrality.

Explanation:

The number of electrons transferred between the cation and the anion to form an ionic bond in one formula unit of each compound can be determined by considering the charges of the ions involved.

Each compound aims for overall electric neutrality by balancing the total positive and negative charges.

Accordingly, NaCl transfers 1 electron, both CaS and BaO transfer 2 electrons each, and KBr and LiF also transfer 1 electron each.

Chloric acid, HClO3, contains 1.19% percent hydrogen by mass

Chloric acid is a colorless liquid. It will accelerate the burning of combustible materials and can ignite most on contact. It is corrosive to metals and tissue. It is used as a reagent in chemical analysis and to make other chemicals.

Chloric acid is not found in nature. It is prepared. Physical properties: Chloric acid is a colorless liquid. Its density is 1 g mL-1.

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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.

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.

The reaction orders for CH3Cl and H2O are determined to be second order for both reactants as evidenced by the quadrupled rate increase when their concentrations are doubled. So the correcct option is D.

To determine the reaction orders for CH3Cl and H2O in the given reaction, we analyze the provided experimental data by comparing initial concentrations and initial rates.

Based on these observations, the answer is D. CH3Cl: second order and H2O: second order. Both the concentrations of CH3Cl and H2O have a squared relationship to the rate, characteristic of second-order reactions.

To convert 450 atmospheres to millimeters of mercury, multiply 450 atm by the conversion factor of 760 mmHg per 1 atm, resulting in 342,000 mmHg.

To convert the pressure of a gas from atmospheres to millimeters of mercury (mmHg), we use the standard conversion factor between these two units of pressure. The relationship is 1 atmosphere (atm) is equivalent to 760 millimeters of mercury (mmHg).

Given that the gas pressure is 450 atmospheres (atm), the conversion to mmHg would be as follows:
(450 atm)
(760 mmHg / 1 atm) = 342,000 mmHg

Step-by-Step Calculation:

Therefore, a pressure of 450 atm is equal to 342,000 mmHg.