Question 31
Algae development will cause all conditions except:
a. Reduced water clarity
b. Increased chlorine consumption
c. Rapid fall in pH
d. Slimy growth

Compound 1 is a stronger acid than Compound 2 because the anion of Compound 1 is better stabilized by Option A. resonance effect. This allows for the distribution of the negative charge over a larger area, making the anion more stable and the acid stronger.

This means that the negative charge on the anion of Compound 1 is spread out over multiple atoms, making it more stable and less likely to react with other molecules. In contrast, Compound 2 does not have this stabilization effect, making it a weaker acid. Dehydration, inductive effects, and hydrogen bonding do not play significant roles in determining the acidity of these compounds. Hence, the correct answer is A. Compound 1 is a stronger acid because the anion of Compound 1 is better stabilized by the resonance effect.

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B) 2.77 × 10-3 molecules. To calculate the number of aspirin molecules in a 500-milligram tablet, we need to convert the mass to moles using the molar mass of aspirin.

The molar mass of aspirin (C9H8O4) is: 9 x 12.01 g/mol (for carbon) + 8 x 1.01 g/mol (for hydrogen) + 4 x 16.00 g/mol (for oxygen) = 180.16 g/mol So, the number of moles of aspirin in a 500-milligram tablet is: 500 mg ÷ 1000 mg/g ÷ 180.16 g/mol = 2.77 x 10^-3 moles Finally, we can convert moles to molecules by multiplying by Avogadro's number (6.02 x 10^23 molecules/mol): 2.77 x 10^-3 moles x 6.02 x 10^23 molecules/mol = 1.67 x 10^21 molecules Therefore, there are approximately 1.67 x 10^21 aspirin molecules in one 500-milligram tablet.

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The word equation for the reaction between silver nitrate (AgNO3) and hydrochloric acid (HCl) to produce solid silver chloride (AgCl) and an aqueous solution of nitric acid (HNO3) is:

AgNO3(aq) + HCl(aq) → AgCl(s) + HNO3(aq)

In this reaction, the silver ion (Ag+) from the silver nitrate solution (AgNO3) combines with the chloride ion (Cl-) from the hydrochloric acid solution (HCl) to form solid silver chloride (AgCl), which precipitates out of the solution. The nitrate ion (NO3-) from the silver nitrate solution combines with the hydrogen ion (H+) from the hydrochloric acid solution to form nitric acid (HNO3), which remains in solution.

The backwash rate for both conventional, rapid and high rate sand filters is typically 10 gpm/ft2.

This rate is used to remove accumulated particles and debris from the filter bed during the backwashing process. Backwashing is a critical process in the operation of sand filters as it helps to maintain the filter bed's efficiency and prolongs the life of the filter. During backwashing, water is forced through the filter bed in the opposite direction to the flow of water during filtration. This flow reversal dislodges and flushes out trapped particles and debris from the filter bed, which is then carried away by the backwash water. The backwash rate of 10 gpm/ft2 is the industry standard and ensures effective cleaning of the filter bed while preventing damage to the filter media.

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Yes, there is a way to convert primary and secondary alkyl halides into carboxylic acids through a multi-step process known as the Grignard reaction.

First, perform a nucleophilic substitution reaction on the alkyl halide. This involves treating the primary or secondary alkyl halide with a nucleophile, such as hydroxide ions (OH-), to replace the halogen atom. This results in the formation of an alcohol.

Next, oxidize the alcohol to form a carboxylic acid. For primary alcohols, you can use a strong oxidizing agent like potassium permanganate (KMnO4) or chromic acid (H[tex]^{2}[/tex]CrO[tex]^{4}[/tex]). The reaction will convert the primary alcohol into a carboxylic acid. For secondary alcohols, first oxidize them to ketones using an oxidizing agent like potassium dichromate (K[tex]^{2}[/tex]Cr[tex]^{2}[/tex]O[tex]^{7}[/tex]).

Conversion of Ketone to Carboxylic Acid
After obtaining a ketone from the secondary alcohol, perform a reaction called the Baeyer-Villiger Oxidation to convert the ketone into an ester. This involves using an oxidizing agent, such as peroxyacids or hydrogen peroxide, to insert an oxygen atom between the carbonyl group and one of the alkyl groups. Finally, hydrolyze the ester using a base or an acid to form a carboxylic acid.

By following these steps, you can successfully convert primary and secondary alkyl halides into carboxylic acids.

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To prepare a supersaturated solution, the following steps should be arranged in the correct order:

Dissolve the solute in a suitable solvent: The solute, typically a solid, is added to the solvent and stirred or heated to facilitate dissolution.

Heat the solution: Applying heat increases the solubility of the solute in the solvent, allowing more solute to dissolve.

Filter the solution: This step involves removing any insoluble impurities or undissolved particles from the solution by passing it through a filter paper or other filtration medium.

Cool the solution slowly: The supersaturation is achieved by cooling the solution slowly, allowing the excess solute to remain dissolved even though it would normally exceed the solubility limit at lower temperatures.

Seed the solution: Introducing a small crystal or seed of the solute into the cooled solution provides a starting point for crystal growth and encourages the formation of the desired end product.

Hence, by following these steps in the correct order (1, 2, 3, 4, 5), a supersaturated solution can be prepared for the recrystallization process.

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The header for a member function that overloads the = operator for the Bird class would be:
Bird& operator=(const Bird& other);

Explanation -  Here's the header for a member function that overloads the = operator for the class Bird:
```cpp
Bird& operator=(const Bird& other);```
This header declares a member function that takes a reference to a constant Bird object named 'other' and returns a reference to a Bird object. The purpose of this function is to define how the assignment operator (=) should work when used with objects of the Bird class.

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4) One way to check if the bore of a spotting capillary is small enough to ensure the ability to spot small enough spots of sample on the spotting line is to use a visualization reagent such as iodine or ninhydrin.

5) Water is not commonly used as an eluent in column chromatography for several reasons. One reason is that water is a highly polar solvent, which can lead to poor resolution of nonpolar compounds.

4. A small amount of the visualization reagent is applied to the spotting line, and the spotting capillary is then used to spot a small amount of the sample solution onto the same line. If the bore of the capillary is small enough, a clearly visible, small spot will form on the line. If the bore of the capillary is too large, the spot will be too large and diffuse.

5. In addition, water is a poor eluent for some types of stationary phases, such as reverse-phase chromatography, where hydrophobic interactions between the sample and the stationary phase are important for separation.

Water can also cause irreversible damage to some types of stationary phases, such as silica gel, by hydrolyzing the surface siloxane groups and altering the surface chemistry of the column. Finally, water can interfere with certain detection methods, such as UV spectroscopy, by causing high background absorbance.

As a result, organic solvents such as methanol, acetonitrile, or a mixture of them are commonly used as eluents in column chromatography.

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The overall order for this reaction is 3.

The overall order of a chemical reaction is determined by adding up the individual orders of each reactant in the rate law equation. In this case, the rate law is given as rate = k [A]0[B]1[C]2, where the exponents represent the orders of each reactant.

Since the order of A is 0, it does not affect the rate of the reaction. The order of B is 1, which means that the rate is directly proportional to the concentration of B.

Finally, the order of C is 2, which means that the rate is proportional to the square of the concentration of C. Adding up the orders of all the reactants gives an overall order of 3 for this reaction.

0 + 1 + 2 = 3

Therefore, the reaction is third order overall.

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The near-surface zone that is formed when water soaks into the ground and is held by molecular attraction as a surface film on soil particles is called the zone of saturation.

This zone is characterized by high soil moisture content and a high concentration of dissolved minerals, which are held in solution by the water molecules. The water in the zone of saturation is also important for sustaining plant growth and providing a habitat for a variety of microorganisms. The near-surface zone where water soaks into the ground and is held by molecular attraction as a surface film on soil particles is called the "zone of capillarity" or "capillary fringe." This zone occurs just above the water table, and the water is retained by the soil due to molecular forces and surface tension.

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The balanced chemical equation for this reaction would be 2 Na(s) + 2 H₂O(l) → 2 NaOH(aq) + H₂(g)

Place the reactants and products in a word equation in step 1 after identifying them. Put the chemical names into chemical formulae in step two. Write the state symbols and arrange them according to the chemical equation. Balance the chemical equation in step three.

A chemical reaction is stated using words rather than chemical formulae in a word equation. The word equation for the reaction's reactants (starting materials), products (outcomes), and direction should be written in a way that it may be translated into a chemical equation.

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The given waste treatment methods, namely Venturi scrubbers, spray towels, and packed towers, are examples of absorption methods.

Absorption is a process in which one substance is dissolved or taken up by another substance. In the context of waste treatment, absorption involves the transfer of pollutants from a gas stream into a liquid stream.

Venturi scrubbers use a high-velocity liquid stream to capture and absorb pollutants from the gas stream. The liquid droplets produced by the scrubber collide with the pollutants, causing them to dissolve and become trapped in the liquid. Spray towers work in a similar way, but use a fine mist of liquid droplets to capture pollutants. Packed towers, on the other hand, contain a packing material that provides a large surface area for the liquid to contact the gas stream, promoting absorption.

In contrast, adsorption involves the attachment of pollutants to a surface, while dialysis involves the separation of substances using a semipermeable membrane, and filtration involves the physical separation of solids from liquids or gases. The given waste treatment methods are examples of absorption methods, specifically using liquids to absorb pollutants from gas streams.

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The element with the greatest second ionization energy among the given options is (B) Na (Sodium).

To understand this, let's first define the terms:
1. Element: A substance consisting of atoms with the same atomic number (number of protons in the nucleus).
2. Greatest: Refers to the largest value among the given options.
3. Second Ionization Energy: The energy required to remove a second electron from an atom after one electron has already been removed.
In the periodic table, ionization energy increases from left to right across a period and decreases from top to bottom within a group. Sodium (Na) is in Group 1 and has a higher second ionization energy than the other elements listed, which are in Groups 2 and 13-15. This is because removing a second electron from sodium requires breaking into a more stable, lower energy level (core electrons), while the other elements still have valence electrons to lose before reaching that point.

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Scientists determine the concentrations of chemicals that harm organisms through a process called toxicity testing. This involves exposing organisms, such as fish or algae, to different concentrations of a chemical and monitoring their response. The response can include changes in behavior, growth, or mortality.

By analyzing the data collected from the toxicity tests, scientists can determine the concentration at which a chemical begins to cause harm to the organisms. This information is used to establish regulatory limits for the use of chemicals to protect both the environment and human health. Toxicology testing is a procedure used to identify the concentration of substances that are harmful to living things. Toxicology testing is done to find out how much of a chemical, at what concentration or dose, harms a particular organism or set of species. The test usually entails exposing the organisms to various chemical concentrations and tracking their reactions over time. Changes in behaviour, growth rates, reproductive success, and survival are only a few examples of the reactions.

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The concentrations of lactate and lactic acid at equilibrium, in terms of the initial concentration of lactic acid and the pH of the solution after the addition of NaOH.

To answer this question, we need to use the Henderson-Hasselbalch equation, which relates the pH of a solution to the ratio of the concentrations of a weak acid and its conjugate base. The equation is:

pH = pKa + log([A-]/[HA])

where pH is the solution's pH, pKa is the acid dissociation constant of the weak acid, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the weak acid.

In this case, the weak acid is lactic acid (HC3H5O3) and its conjugate base is lactate (C3H5O3-). The pKa of lactic acid is 3.86, which is also the pH of the solution after the addition of concentrated NaOH.

Therefore, we can rearrange the Henderson-Hasselbalch equation to solve for the concentration of lactate and lactic acid at equilibrium:

[A-]/[HA] = 10^(pH - pKa)

[A-]/[HA] = 10^(3.86 - 3.86)

[A-]/[HA] = 1

This means that at equilibrium, the concentration of lactate is equal to the concentration of lactic acid. However, we still need to know the total concentration of lactate and lactic acid in the solution in order to calculate their individual concentrations.

We can use the fact that lactic acid is a monoprotic acid (meaning it donates one proton in its reaction with water) to set up an equilibrium expression for its dissociation:

HC3H5O3 ⇌ C3H5O3- + H+

The equilibrium constant for this reaction is Ka = [C3H5O3-][H+]/[HC3H5O3]. At equilibrium, the total concentration of lactate and lactic acid is equal to the initial concentration of lactic acid, since the addition of NaOH does not affect the total number of moles of the weak acid.

Let's call the total concentration of lactate and lactic acid [HA]total. Then we have:

Ka = [C3H5O3-][H+]/[HC3H5O3] = x^2/([HA]total - x)

where x is the concentration of H+ (which is also equal to the concentration of lactate). We can assume that x is small compared to [HA]total, since lactic acid is a weak acid with a low dissociation constant. Therefore, we can approximate [HA]total - x as [HA]total.

Solving for x, we get:

x = sqrt(Ka[HA]total)

Plugging in the values, we get:

x = sqrt(1.38e-4 M * [HC3H5O3]initial)

where [HC3H5O3]initial is the initial concentration of lactic acid before the addition of NaOH. Note that we need to know [HC3H5O3]initial in order to calculate x, since we are assuming that the total concentration of lactate and lactic acid is equal to [HC3H5O3]initial.

Finally, we can calculate the concentrations of lactate and lactic acid at equilibrium:

[C3H5O3-] = x = sqrt(1.38e-4 M * [HC3H5O3]initial)

[HC3H5O3] = [HA]total - x = [HC3H5O3]initial - sqrt(1.38e-4 M * [HC3H5O3]initial)

These expressions give the concentrations of lactate and lactic acid at equilibrium, in terms of the initial concentration of lactic acid and the pH of the solution after the addition of NaOH.

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If ΔH is greater than zero, and ΔS is less than zero, the process is always non-spontaneous at low temperatures

This is because ΔG = ΔH - TΔS, and if ΔH is positive and ΔS is negative, then ΔG will always be positive at low temperatures, indicating a non-spontaneous process. However, at high temperatures, the positive ΔS term will become more significant and may overcome the positive ΔH term, resulting in a spontaneous process.
Hi! If ΔH is greater than zero and ΔS is less than zero, the process is always non-spontaneous .A nonspontaneous reaction is a reaction that does not favor the formation of products at the given set of conditions. In order for a reaction to be nonspontaneous, it must be endothermic, accompanied by a decrease in entropy, or both.

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At low temperatures, the process is always non-spontaneous if H is more than zero and S is less than zero.

This is because ΔG = ΔH - TΔS, and if ΔH is positive and ΔS is negative, then ΔG will always be positive at low temperatures, indicating a non-spontaneous process. However, at high temperatures, the positive ΔS term will become more significant and may overcome the positive ΔH term, resulting in a spontaneous process.

Hi! If ΔH is greater than zero and ΔS is less than zero, the process is always non-spontaneous .A nonspontaneous reaction is a reaction that does not favor the formation of products at the given set of conditions. In order for a reaction to be nonspontaneous, it must be endothermic, accompanied by a decrease in entropy, or both.

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An excess of a beginning substance, such as butanol or acetic acid, is not employed in the production of butyl acetate because it would be challenging to separate the result from the excess starting substance using just simple distillation.

This is because it is difficult to separate butanol and acetic acid effectively using simple distillation because both substances have boiling values that are close to those of butyl acetate.

A liquid combination is heated to vaporize the more volatile component, which is subsequently condensed to collect the purified component. This is known as simple distillation. However, if there is too much starting material, the product and starting material's boiling temperatures may coincide, causing them to co-distill and making it challenging to get a pure product.

In the case of butyl acetate synthesis, using too much starting material would cause butanol and acetic acid to co-distill with butyl acetate. This would result in a mixture of products that would need additional purification steps, like fractional distillation or additional chemical treatments, to separate and obtain pure butyl acetate.

It is feasible to obtain a larger yield of pure butyl acetate with fewer purification steps by carefully managing the reaction's stoichiometry and utilizing only the necessary amount of starting material, which makes the synthesis procedure more effective and affordable.

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False. While both low explosives and high explosives release energy upon detonation, high explosives generally create a crater due to their faster and more violent energy release. Low explosives, on the other hand, burn at a slower rate and typically do not produce a crater.

An explosive event that occurs at, immediately above, or below the surface can cause material to be ejected from the ground's surface, creating an explosion crater.

An explosive event causes material from the ground to be displaced and ejected, creating a crater. Usually, it has a bowl-like shape. Three processes—high-pressure gas, shock waves—are in charge of making the crater:

-Ground deformation caused by plasticity.

-Ejecta, or material that is thrown up by an explosion, from the ground.

-The ground's surface spalling.

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Yes, both low explosives and high explosives have the ability to create a crater in the spot where the bomb detonates. This is because when explosives are detonated, they rapidly release large amounts of energy, which generates a high-pressure shock wave.

This shock wave then travels outward from the point of detonation, causing the surrounding ground to be displaced and ejected, resulting in a crater. However, the size and depth of the crater will depend on the type and amount of explosives used, as well as the nature of the surrounding soil and terrain. High explosives, which typically contain a higher percentage of energetic materials, are generally more powerful and capable of creating larger and deeper craters compared to low explosives.

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To find the number of moles of copper in the brass sample, we need to use the given mass of copper and the atomic mass of copper:

moles of copper = mass of copper / atomic mass of copper

Substituting the values given in the question, we get:

moles of copper = 3.56 g / 63.546 g/mol
moles of copper = 0.056 moles

Therefore, the answer is option b) 0.0560 moles.

To determine the number of moles of copper in the sample, use the given mass and atomic mass of copper.

Given mass of copper = 3.56 g
Atomic mass of copper = 63.546 g/mol

Moles of copper = (Given mass of copper) / (Atomic mass of copper)
Moles of copper = 3.56 g / 63.546 g/mol

Moles of copper ≈ 0.0560 moles

Your answer: b) 0.0560 moles

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To find the number of moles of copper in the sample, we will use the mass of copper and its atomic mass. Your question is: A sample of brass contains 3.56 g of copper, Cu. How many moles of copper does the sample contain? The atomic mass of copper is 63.546 g/mol.

Step 1: Write down the given information.
Mass of copper (Cu) = 3.56 g
Atomic mass of copper (Cu) = 63.546 g/mol

Step 2: Calculate the moles of copper.
To find the moles, divide the mass of copper by its atomic mass:
moles of Cu = mass of Cu / atomic mass of Cu

Step 3: Plug in the values and solve.
moles of Cu = 3.56 g / 63.546 g/mol ≈ 0.0560 moles

Your answer is (b) 0.0560 moles.

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The role of cytotoxic T cells is to attack body cells that have been infected with specific viruses and bacteria, in order to prevent the spread of infection.

These T cells are able to recognize and target cells that display fragments of the virus or bacteria on their surface, and then release toxic substances that destroy the infected cells. This helps to limit the damage caused by the invading pathogen and also triggers the production of antibodies to help clear the infection.

The role of cytotoxic T cells is to attack body cells that have been infected by specific viruses and bacteria. They play a crucial role in the immune response by eliminating harmful pathogens and protecting the body.

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Answer:

sound can not travel through stone

It seems that you are referring to a substance with high concentrations of sodium (Na) and potassium (K) that is beneficial for the house. This substance could be a type of water softener.

Water softeners contain high concentrations of Na and K ions, which help to reduce the hardness of water by exchanging hard water minerals like calcium and magnesium with sodium or potassium ions. Using a water softener can benefit your house by preventing scale build-up in pipes and appliances, and improving the efficiency of soap and detergent use.By eliminating these minerals, water softeners can help to prevent scale buildup in pipes, fixtures, and appliances, which can make them more efficient and last longer.

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The magnitude of the urban heat island (UHI) effect on summer temperatures can vary depending on factors such as the size of the urban area, the surrounding landscape, and local weather conditions. However, studies have shown that the UHI effect can lead to temperatures in urban areas that are 1-3°C (1.8-5.4°F) warmer on average during the summer compared to nearby rural areas.

In some cases, the temperature difference between urban and rural areas can be as much as 10°C (18°F) during heatwaves.

The urban heat island effect is a phenomenon that occurs in built-up areas where there is a high concentration of buildings, roads, and other structures made of materials that absorb and re-radiate heat.

During the day, the sun's rays heat up these surfaces, which in turn release heat into the surrounding air. This causes urban areas to be warmer on average than surrounding rural areas.

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In terms of the energy change in the reaction, the negative value indicates that the reaction is exothermic as the reaction releases 1560.74 kJ of energy for every mole of C2H6 that reacts with 7/2 moles of O2.

The model of chemical energy changes is given below:

Balanced chemical equation:

C2H6 + 7/2 O2 → 2CO2 + 3H2O

Now, to calculate the energy change in the reaction, we will use a table of enthalpy values. The enthalpy change for each of the reactants and products is given in the table below:

Reactants:

C2H6: -84.68 kJ/mol

O2: 0 kJ/mol

Products:

CO2: -393.51 kJ/mol

H2O: -285.83 kJ/mol

The energy change in the reaction can be calculated using the formula:

ΔH = ∑(products) - ∑(reactants)

ΔH = [2(-393.51 kJ/mol) + 3(-285.83 kJ/mol)] - [-84.68 kJ/mol + 7/2(0 kJ/mol)]

ΔH = -1560.74 kJ/mol

Therefore, the energy change in the reaction is -1560.74 kJ/mol.

To create a model of the energy change in the reaction, we can use an energy level diagram. In this diagram, the energy of the reactants is shown on the left, the energy of the products is shown on the right, and the activation energy is shown as a barrier between them.

The energy level diagram for this reaction is shown below:

Reactants (C2H6 + 7/2 O2)

                |

                |

       Activation energy

                |

                |

Products (2CO2 + 3H2O)

As shown in the diagram, the reactants have a higher energy level than the products, and the activation energy is required to get the reaction started.

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In azeotropic distillation, an azeotrope is formed between two or more components with similar boiling points. An azeotrope is a mixture of components that exhibits a constant boiling point and vapor-liquid composition, making it challenging to separate the components using traditional distillation techniques.

The specific components of an azeotrope depend on the particular mixture you are working with. Commonly studied azeotropes include water-ethanol, water-isopropanol, and water-hydrochloric acid. In the water-ethanol azeotrope, for example, the components are water and ethanol, with an azeotropic composition of approximately 95% ethanol and 5% water by volume.

Azeotropic distillation is used to overcome the limitation of traditional distillation methods. By adding a third component, called an entrainer, the azeotrope can be broken, allowing for the separation of the original components. The entrainer's choice is crucial, as it must selectively form an azeotrope with one of the original components without forming a new azeotrope with the other component.

The specific components of an azeotrope vary based on the mixture being studied, but they typically consist of two or more substances that form a constant boiling mixture. Azeotropic distillation helps to separate these components by adding an entrainer to break the azeotrope.

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The rotating element in a centrifugal pump is commonly called the impeller.

An impeller is a rotating component of a centrifugal pump that helps to increase the velocity and pressure of a fluid as it passes through the pump. It consists of a series of curved blades that are arranged in a circular pattern around a central shaft.

When the impeller rotates, the blades create a centrifugal force that causes the fluid to move outward from the center of the impeller. This increased velocity and pressure of the fluid allow it to be pumped to a higher elevation or over a longer distance.

Impellers come in a variety of designs, including closed, semi-open, and open. Closed impellers are used for fluids with low levels of impurities, while open impellers are better suited for fluids with higher levels of impurities.

Impellers are commonly used in various industries such as oil and gas, water treatment, and chemical processing, to pump fluids in large quantities.

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If the pressure remains constant and the composition of the gas in the balloon is not known, the balloon is suspended over a Bunsen burner.

However, in general, when a balloon is exposed to heat, the gas molecules inside the balloon will begin to move faster and collide more frequently, causing the pressure inside the balloon to increase. This increase in pressure can cause the balloon to expand or burst if the material is not strong enough to withstand the pressure.

Therefore, it is possible that the balloon may increase significantly in size until it bursts if it is made of a material that cannot withstand the increased pressure caused by the heat.

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Explanation and Answer:

We need to determine the limiting reagent to find out how many moles of H3PO4 can form during the reaction.

From the given information, we know that 2.0 mol of P4O10 can form 8.0 mol of H3PO4. This means that the molar ratio of P4O10 to H3PO4 is 2:8, or 1:4.

Similarly, we know that 8.0 mol of H2O can form 5.3 mol of H3PO4. This means that the molar ratio of H2O to H3PO4 is 8:5.3, or approximately 1.51:1.

To determine the limiting reagent, we need to compare the amount of H3PO4 that can be produced from each reactant.

For P4O10:

Molar ratio of P4O10 to H3PO4 = 1:4

Therefore, 2.0 mol P4O10 can produce 8.0 mol H3PO4

For H2O:

Molar ratio of H2O to H3PO4 = 1.51:1

Therefore, (8.0 mol H2O) x (1 mol H3PO4/1.51 mol H2O) = 5.3 mol H3PO4 can be produced from 8.0 mol H2O

Since we can produce less H3PO4 from H2O than from P4O10, H2O is the limiting reagent.

Therefore, the maximum amount of H3PO4 that can be produced is 5.3 mol.

Answer:C

Explanation:it is C.