2AI(NO3)3 + 3Na2CO3 → Al2(CO3)3(s) + NaNO3

Use the limiting reagent to determine how many grams of Alz(CO3), should precipitate out in the reaction. ​

The question does not provide a specific nuclear chemical equation to work with, so it is difficult to provide a direct answer. However, I can provide some general information about nuclear chemical equations.

Nuclear chemical equations are used to represent nuclear reactions. These reactions involve changes in the nucleus of an atom, typically involving the addition or removal of protons and/or neutrons. Unlike chemical reactions, which involve the sharing or transfer of electrons, nuclear reactions involve changes in the core of the atom.

A typical nuclear chemical equation includes a reactant on the left side of the equation and a product on the right side. The reactant and product are both represented by chemical symbols, such as H for hydrogen or O for oxygen. The number of protons and neutrons in the reactant and product may differ, indicating a change in the nucleus.

In some cases, the nuclear chemical equation may be missing a symbol. This could indicate that the product is unknown or has not been determined. It is also possible that the missing symbol represents a hypothetical or theoretical product, rather than an actual substance.

In summary, nuclear chemical equations are used to represent nuclear reactions, which involve changes in the nucleus of an atom. The equations include reactants and products represented by chemical symbols, and may occasionally include missing symbols indicating an unknown or theoretical product.

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There are 0.1474 moles of hydrogen atoms in 0.0737 mol of N2H4.

What is Mole?

In chemistry, a mole is a unit used to express the amount of a substance. One mole of a substance is defined as the amount of that substance that contains as many elementary entities (atoms, molecules, or other particles) as there are atoms in 12 grams of carbon-12.

To determine the mass of lead in PbCO3, we need to use the molar mass of PbCO3 and the stoichiometric relationship between Pb and PbCO3. The molar mass of PbCO3 is 267.21 g/mol, and the stoichiometric relationship between Pb and PbCO3 is 1:1.

Thus, the mass of Pb in 1.25x10^4 g of PbCO3 can be calculated as follows:

Mass of Pb = (1.25x10^4 g PbCO3) x (1 mol PbCO3/267.21 g PbCO3) x (1 mol Pb/1 mol PbCO3) x (207.2 g Pb/mol Pb)

= 1.02x10^4 g Pb

Therefore, 1.02x10^4 g of Pb is contained in 1.25x10^4 g of PbCO3.

The formula for N2H4 indicates that there are two hydrogen atoms for every molecule of N2H4. Therefore, we can calculate the number of moles of hydrogen atoms in 0.0737 mol of N2H4 as follows:

Number of moles of H atoms = (0.0737 mol N2H4) x (2 mol H atoms/1 mol N2H4)

= 0.1474 mol H

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Ma is the equivalent of an element that belongs to the [tex]3_r_d[/tex] column of the periodic table, while Tw is the equivalent of an element that belongs to the 6th column of the periodic table in this alternate dimension.

What is Compound?

In chemistry, a compound is a substance composed of two or more different types of elements chemically bonded together in a fixed proportion. The elements in a compound are combined in a way that results in a new substance with different chemical and physical properties from the individual elements that make it up.

Let's assume that Ma has a charge of +3 and Tw has a charge of -2 in this alternate dimension. To balance the charges, we need 3 Tw atoms for every 2 Ma atoms.

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The difference in the chemical potential for liquid and crystalline water at 270.15 K is -0.97 J/mol.


1. Convert given pressures to atm: initial partial pressure of water vapor (P1) = 489 Pa / 101325 Pa/atm = 0.00482 atm, and new partial pressure (P2) = 475 Pa / 101325 Pa/atm = 0.00469 atm.

2. Use the Clausius-Clapeyron equation: ln(P2/P1) = -(ΔH_sub/R)(1/T2 - 1/T1), where ΔH_sub is the enthalpy of sublimation, R is the gas constant, and T1 and T2 are the initial and final temperatures, both equal to 270.15 K.

3. Rearrange the equation to solve for ΔH_sub: ΔH_sub = R * (ln(P2/P1))/(1/T2 - 1/T1), and substitute the values: ΔH_sub = 8.314 J/mol K * (ln(0.00469/0.00482))/(0 - 0) = -0.97 J/mol.

4. The negative sign makes sense as the system moves to a new equilibrium with a lower chemical potential for crystalline water, indicating a more stable phase.

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There are approximately 0.0364 moles of N2 in the flask.

To calculate the number of moles of N2 in a flask with a volume of 250mL at a pressure of 300.0kPa and a temperature of 300.0K, we need to use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

First, we need to convert the volume from mL to L by dividing it by 1000: 250mL ÷ 1000 = 0.25L.

Next, we need to convert the pressure from kPa to atm by dividing it by 101.3 (which is the conversion factor between kPa and atm): 300.0kPa ÷ 101.3 = 2.96atm.

Now we can plug in the values and solve for n: n = (PV) / (RT) = (2.96atm x 0.25L) / (0.08206 L·atm/mol·K x 300.0K) = 0.0364 moles of N2.

Therefore, there are approximately 0.0364 moles of N2 in the flask.

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The final pressure of the gas sample is 1.09 atm when heated to 50ºC.

The final pressure of a gas sample initially at 1.00 atm and 25ºC when heated to 50ºC can be calculated using the ideal gas law:

P₁ × V₁ ÷ T₁ = P₂ × V₂ ÷ T₂

where P₁, V₁, and T₁ are the initial pressure, volume, and temperature of the gas, respectively, and P₂, V₂, and T₂ are the final pressure, volume, and temperature of the gas, respectively.

Assuming that the volume of the gas remains constant, V₁ = V₂, and rearranging the ideal gas law, we get:

P₂ = P₁ (T₂ ÷ T₁)

Substituting the values, we get:

P₂ = (1.00 atm) × (323 K) ÷ (298 K) = 1.09 atm

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The molarity of the solution comes out to be 0.200 M, which is calculated in the below section.

The number of moles of calcium iodide can be calculated as follows-

n = m / M ......(1)

Molar mass (M) of Calcium iodide = 293.887 g/mol

Mass (m) = 58.8 grams

Substitute the known values in equation (1) as follows-

n = 58.8 grams / 293.887 g/mol

   = 0.200 moles

Now, the molarity can be calculated using the below formula-

Molarity = no. of moles / Volume

              = 0.200 moles / 1 L

               = 0.200 M

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Complete question-

Find the molarity of a solution that contains 58.8 grams of calcium iodide (CaI2), per liter.

Phospholipids will form a sheet-shaped double layer when placed in water. The correct answer is option c.

This is known as a phospholipid bilayer, which is a fundamental component of cell membranes.

The hydrophilic (water-loving) phosphate heads of the phospholipids face outwards and interact with the water molecules, while the hydrophobic (water-fearing) fatty acid tails face inwards and interact with each other.

The phospholipid bilayer provides a selectively permeable barrier that allows certain substances to pass through the membrane while preventing others from doing so.

Additionally, the fluidity of the phospholipid bilayer can be regulated by various factors, such as temperature and the presence of cholesterol, allowing for optimal membrane function in different cellular environments.

The correct answer is option c.

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Complete Question

What will phospholipids form when placed in water?

a. a sphere-shaped single layer

b. a sphere-shaped double layer

c. a sheet-shaped double layer

d. a sheet-shaped single laye

If cost is a significant constraint for Itea city in choosing a water purification method for their water supply, then the fact that ion exchange systems are more expensive because of the use of filters that need replacement would be important to consider.

While reverse osmosis systems are also more expensive, it is primarily due to the chlorine treatments, which may not be a significant factor for the city. Therefore, ion exchange systems may not be a cost-effective option in the long run due to the ongoing expenses of replacing filters.

This information can help inform the city's decision-making process and ensure that they choose a water purification method that meets their needs while also being financially feasible.

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The concentration of the Ba(OH)₂ solution is 0.1 M.

To find the concentration of the Ba(OH)₂ solution, we can use the balanced equation for the neutralization reaction between Ba(OH)₂ and HNO₃:

Ba(OH)₂ + 2HNO₃ → Ba(NO₃)₂ + 2H₂O

From the equation, we can see that one mole of Ba(OH)₂ reacts with two moles of HNO₃. Therefore, the moles of HNO₃ used in the neutralization reaction can be calculated as follows:

moles of HNO₃ = 1.52 M × 0.0234 L = 0.035568 mol

Since the moles of HNO₃ is equal to the moles of Ba(OH)₂ in the reaction, we can calculate the concentration of the Ba(OH)₂ solution as follows:

concentration of Ba(OH)₂ = moles of Ba(OH)₂ / volume of Ba(OH)₂ solution

moles of Ba(OH)₂ = moles of HNO₃ / 2 = 0.035568 mol / 2 = 0.017784 mol

volume of Ba(OH)₂ solution = 0.063 L

concentration of Ba(OH)₂ = 0.017784 mol / 0.063 L ≈ 0.1 M

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The amount of heat required to convert 46.0 g of ethanol at 25° C to vapor phase is ≅ 44.2 KJ total heat .

Using the mass , evaluating the moles of ethanol :

                              46.0 g × [tex]\frac{1 Mol}{46.07 g}[/tex] = 0.998 mol

                                        ≅ 1.0 mol

The heat required to convert 46.0 g ethanol from 25° C at 78° C is evaluated :

                                     q₁ = m[tex]C_{liquid }[/tex]ΔT

                                   = 46.0 g × [tex]\frac{2.3 J}{g. K}[/tex] × 78° C - 25°C

                                 =     5607.47J × 1 KJ /1000 J

                                           = 5.607 KJ

So, the heat required in conversion of 1.0 mol of ethanol at 78 ° C  to 1.0 mol ethanol vapour is expressed as :

                           q₂ = moles × Δ[tex]H_{vap}[/tex]

                            = 1.0 mol × 38.56 KJ /mol

                                = 38.56 kJ/ mol

The total heat requirement conversion of 46 .0 g of ethanol at 25° C to the vapour state at 78° C :

                           Total heat = q₁ + q₂

                                         = 5.607 KJ + 38.56 KJ

                                            = 44.167 KJ

                                        ≅ 44.2 KJ

Fume alludes to a gas stage at a temperature where a similar substance can likewise exist in the fluid or strong state, beneath the basic temperature of the substance. As a result of their tendency to be volatile, liquids will enter the vapor phase when the temperature is raised sufficiently. At the specified temperature, a liquid is considered to be volatile if it exhibits a significant vapor pressure.

Transferring a substance from a vapor to a solid by desorbing it using a desorbent or carrier gas and passing the vapor sample through a stationary phase (such as silica particles).

Incomplete question :

Determine the amount of heat required to convert 46. 0 g of ethanol at 25°c to the vapor phase at 78°c. Based on the melting and boiling points, ethanol is a liquid at 25°c. Consider the heating curve when organizing your thoughts and answering the question. Use the information about ethanol CH₃CH₂OH given in the table below.

Use the following information about ethanol CH₃CH₂OH.

Tmelt = –114°C

Tboil = 78°C

∆Hfus = 5.02 kJ/mol

∆Hvap = 38.56 kJ/mol

C solid = 0.97 J/g-K

C liquid = 2.3 J/g-K

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The structure of product is shown.

When pentanedioic (glutaric) anhydride reacts with ammonia at high temperature, it undergoes an amide formation reaction to produce a compound with the molecular formula C₅H₇NO₂. The amide formation reaction involves the nucleophilic attack of the ammonia molecule on one of the carbonyl carbon atoms of the anhydride, leading to the formation of an intermediate product called an amide.

This amide then undergoes further reactions to form the final product with the given molecular formula. The presence of both carboxylic acid and amide functional groups in the molecule indicates that it contains both the original anhydride and the product of its reaction with ammonia.

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The net ionic equation for the equilibrium that is established when sodium cyanide (NaCN) is dissolved in water is:

NaCN + H2O ⇌ CN- + Na+ + H2O

In this equation, the cyanide ion (CN-) is produced by the dissociation of NaCN in water. The sodium ion (Na+) and water (H2O) are spectator ions and do not participate in the reaction. Therefore, they are not included in the net ionic equation.

This solution is basic because the cyanide ion is a weak base and can hydrolyze water to produce hydroxide ions (OH-) according to the following reaction:

CN- + H2O ⇌ HCN + OH-

The equilibrium constant for this reaction is relatively small, but it is enough to make the solution basic.

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B Answer:

Explanation:

Higher, 1020 mb +, rising pressure and temp are associated with clear skies and low precipitation

Answer:

This is thermodynamics.

Using simple thermodynamics operation equation

The most crucial step during the preparation of the Grignard reagent is ensuring that all the equipment and reactants are absolutely dry.

To ensure that the equipment and reactants are dry, the equipment must be thoroughly cleaned and dried before use, and the reactants should be purified and dried before being introduced into the reaction vessel. The solvent, typically diethyl ether, should also be dried using a drying agent such as anhydrous magnesium sulfate.

The reaction should be carried out under an inert atmosphere, such as nitrogen or argon, to prevent the formation of unwanted byproducts. By taking these precautions, the formation of the Grignard reagent can be optimized, leading to a higher yield and better quality product.

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One potential difference between ice that Dr. Stewart is climbing and "normal" water ice could be the location or conditions in which it formed.

For example, glacier ice, which forms over many years from compacted snow, can have different properties than the ice that forms on a frozen lake or river. Similarly, ice formed in a cold laboratory setting might have different properties than ice formed under natural conditions.

Other factors that could impact the characteristics of ice include the presence of air bubbles, cracks or fissures, and the size and shape of ice crystals. Ice that has been subjected to pressure or other stresses can also exhibit unique features such as layers or bands.

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The freezing point of the lithium bromide solution is approximately -1.06°C.

To determine the freezing point of the solution, we need to use the freezing point depression formula:

ΔTf = Kf * molality

where ΔTf is the freezing point depression, Kf is the freezing point depression constant (which depends on the solvent), and molality is the concentration of the solution in mol/kg.

First, we need to calculate the molality of the solution:

molality = moles of solute / mass of solvent (in kg)

The mass of 525 mL of water is:

mass = volume * density = 525 mL * 1 g/mL = 525 g

The number of moles of lithium bromide is:

moles of LiBr = 0.300 mol

Therefore, the molality of the solution is:

molality = 0.300 mol / 0.525 kg = 0.571 mol/kg

The freezing point depression constant for water is 1.86 °C/m. Therefore, the freezing point depression is:

ΔTf = 1.86 °C/m * 0.571 mol/kg = 1.06306 °C

Finally, to find the freezing point of the solution, we need to subtract the freezing point depression from the freezing point of pure water (0°C):

Freezing point = 0°C - 1.06306°C = -1.06306°C

Therefore, the freezing point of the lithium bromide solution is approximately -1.06°C.

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16.128 grams of hydrogen gas would need to react in order to produce 972 kJ of heat energy.

So, first, we can calculate the amount of heat energy released per mole of [tex]H_2[/tex] reacted:

243 kJ of heat / 2 moles of [tex]H_2[/tex] = 121.5 kJ/mol of [tex]H_2[/tex]

We can use the following equation to calculate the amount of hydrogen gas required:

Amount of [tex]H_2[/tex]  = Energy released / Heat of reaction per mole of [tex]H_2[/tex]

Amount of [tex]H_2[/tex] = 972 kJ / 121.5 kJ/mol = 8 moles of [tex]H_2[/tex]

Finally, we can calculate the mass of [tex]H_2[/tex] required using its molar mass:

Mass of [tex]H_2[/tex] = Number of moles of[tex]H_2[/tex]x Molar mass of [tex]H_2[/tex]

Mass of [tex]H_2[/tex] = 8 moles x 2.016 g/mol = 16.128 g of [tex]H_2[/tex]

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The mass number of an oxygen isotope with nine neutrons is 25.

The mass number is the sum of protons and neutrons in an atom. Oxygen has 8 protons, and with 9 neutrons, the mass number is 8 + 9 = 25.

The given statement provides information about the mass number of a specific oxygen isotope with nine neutrons. The mass number represents the total number of protons and neutrons in an atom. In the case of this oxygen isotope, it is stated that the mass number is 25.

To calculate the mass number, we need to sum the number of protons and neutrons. The statement also mentions that oxygen has 8 protons. Therefore, by adding 9 neutrons to the 8 protons, we obtain the total mass number of 25.

In summary, the statement explains that the mass number of this particular oxygen isotope, which contains nine neutrons, is determined by the sum of the 8 protons and 9 neutrons, resulting in a mass number of 25.

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654 grams of Zn metal will completely react with 730 grams of HCl

The balanced chemical equation for this reaction is:
Zn + 2HCl → ZnCl2 + H2

From the equation, we can see that for every 1 mole of Zn, 2 moles of HCl are required for a complete reaction. This means the mole ratio of Zn to HCl is 1:2.

To determine the number of moles of HCl used, we need to convert the given mass of HCl to moles. The molar mass of HCl is 36.5 g/mol, so:

730 g HCl x (1 mol HCl/36.5 g HCl) = 20 moles HCl

Using the mole ratio from the balanced equation, we can determine the number of moles of Zn required:

20 moles HCl x (1 mol Zn/2 mol HCl) = 10 moles Zn

Finally, we can convert the number of moles of Zn to grams using its molar mass of 65.4 g/mol:

10 moles Zn x (65.4 g Zn/mol) = 654 grams of Zn

Therefore, 654 grams of Zn metal will completely react with 730 grams of HCl to produce ZnCl2 and H2.

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The balanced chemical equation for the reaction of nitrogen and oxygen to produce nitrogen dioxide is:

2NO + O2 → 2NO2

According to the equation, 1 mole of nitrogen reacts with 0.5 moles of oxygen to produce 1 mole of nitrogen dioxide.

To determine the volume of nitrogen required to react with 33.6 L of oxygen, we need to convert the volume of oxygen to moles, and then use the mole ratio to find the moles of nitrogen required, and finally convert to volume of nitrogen.

Using the ideal gas law, we can convert the given volume of oxygen to moles:

n(O2) = PV/RT

where P is the pressure, V is the volume, R is the gas constant, and T is the temperature in Kelvin.

Assuming standard temperature and pressure (STP) conditions of 1 atm and 273 K, we get:

n(O2) = (1 atm) × (33.6 L) / [(0.0821 L·atm/mol·K) × (273 K)] = 1.37 moles of O2

Using the mole ratio from the balanced chemical equation, we know that 2 moles of NO react with 1 mole of O2. So the number of moles of NO required to react with 1.37 moles of O2 is:

n(NO) = 2 × (1.37 moles of O2) = 2.74 moles of NO

Finally, we can convert the moles of NO to volume using the ideal gas law:

V(NO) = n(NO)RT/P

Assuming STP conditions again, we get:

V(NO) = (2.74 mol) × (0.0821 L·atm/mol·K) × (273 K) / (1 atm) ≈ 60.4 L

Therefore, approximately 60.4 L of nitrogen would be required to react with 33.6 L of oxygen to produce nitrogen dioxide, under the given conditions.

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Approximately 108,066 J of heat is required to heat 352 g of water from 20.0°C to 93.7°C in an electric kettle.

To determine the quantity of heat required to heat 352 g of water from 20.0°C to 93.7°C, we need to use the specific heat capacity of water and the equation:

q = m × c × ΔT

ΔT = change in temperature (in °C)

First, we need to calculate the change in temperature:

ΔT = final temperature - initial temperature

ΔT = 93.7°C - 20.0°C

ΔT = 73.7°C

Substituting the given values into the equation, we get:

q = 352 g × 4.184 J/g·°C × 73.7°C

q = 108,066.496 J

q ≈ 108,066 J (rounded to three significant figures)

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When the balloon is submerged in water at a depth where the pressure is 4.7 atm and the temperature is 15 °C, its volume will be approximately 0.995 L.

From the ideal gas equation, we can use the combined gas law formula, which is:

(P1 × V1) / T1 = (P2 × V2) / T2

Here, P1 = 1.1 atm (initial pressure), V1 = 3.7 L (initial volume), T1 = 30 °C (initial temperature), P2 = 4.7 atm (final pressure), and T2 = 15 °C (final temperature). We need to find V2 (final volume).

First, convert the temperatures to Kelvin by adding 273.15:

T1 = 30 + 273.15 = 303.15 K

T2 = 15 + 273.15 = 288.15 K

Now, plug in the values into the combined gas law formula and solve for V2:

        (P1 × V1) / T1 = (P2 × V2) / T2

(1.1 × 3.7) / 303.15 = (4.7 × V2) / 288.15

    (4.07) / 303.15 = (4.7 × V2) / 288.15

Now, solve for V2:

V2 = (4.07 × 288.15) / (303.15 × 4.7)

V2 ≈ 0.995 L

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The mass of KI needed to prepare 500 mL of a 0. 750 M solution is 62.25 grams

To compute the mass of KI needed to prepare 500 mL of a 0.750 M solution, use the formula:

Molarity (M) = moles of solute / volume of solution in liters

First, convert the volume to liters: 500 mL = 0.5 L

Next, rearrange the formula to find the moles of solute:

moles of solute = Molarity × volume of solution in liters
moles of KI = 0.750 M × 0.5 L
moles of KI = 0.375 moles

Now, find the molar mass of KI (Potassium Iodide):
K (Potassium) = 39.10 g/mol
I (Iodine) = 126.90 g/mol
Molar mass of KI = 39.10 g/mol + 126.90 g/mol = 166.00 g/mol

Finally, calculate the mass of KI needed:

mass of KI = moles of KI × molar mass of KI
mass of KI = 0.375 moles × 166.00 g/mol
mass of KI = 62.25 g

Therefore, you will need 62.25 grams of KI to prepare 500 mL of a 0.750 M solution.

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The energy of a photon that emits a light of frequency 6. 42 x 1014 Hz is 4.25 x 10^-19 J.

The energy of a photon can be calculated using the equation:

E=hf,

where E is the energy of the photon, h is Planck's constant (6.626 x 10^-34 J.s), and f is the frequency of the light emitted by the photon.

Plugging in the given frequency of 6.42 x 10^14 Hz into the equation, we get

E=(6.626 x 10^-34 J.s)(6.42 x 10^14 Hz) = 4.25 x 10^-19 J.

Therefore, the correct answer is B i.e, 4.25 x 10^-19 J.

It should be emphasized that a photon's energy is directly linked to its frequency and inversely related to its wavelength. Therefore, light with higher frequency, such as blue light, contains more energy than light with lower frequency, such as red light.

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The molar mass of HA is approximately 168.48 g/mol.

To calculate the molar mass of HA, we need to use the balanced chemical equation for the reaction between HA and NaOH:

[tex]HA + NaOH[/tex] → [tex]NaA + H2O[/tex]

From the equation, we can see that 1 mole of HA reacts with 1 mole of NaOH to produce 1 mole of NaA. At the equivalence point of the titration,

[tex]moles of NaOH = (0.500 mol/L) * (0.0321 L) = 0.01605 mol[/tex]

Since the initial solution was prepared by dissolving 2.70 g of HA in 100.0 ml of water, we can calculate the initial concentration of HA in units of moles per liter:

[tex]moles\ of HA = (2.70 g / molar\ mass\ of HA) = (0.0270 kg / molar\ mass\ of HA)[/tex]

[tex]initial\ concentration\ of\ HA = moles\ of\ HA / (0.100 L) = moles\ of\ HA / 1000 mL[/tex]

Setting the moles of NaOH equal to the moles of HA, we can solve for the molar mass of HA:

moles of NaOH = moles of HA

[tex]0.01605\ mol = (0.0270 kg / molar\ mass\ of HA) / 0.100 L[/tex]

[tex]molar\ mass\ of\ HA = (0.0270 kg / 0.01605 mol) / 0.100 L = 168.48 g/mol[/tex]

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