V
A student dissolves 11.S g of sodium hydroxide (NaOH) in 250. g of water in a well-insulated open cup. He then observes the temperature of the water rise
from 20.0 °C to 31.3 °C over the course of 6.7 minutes.
Use this data, and any information you need from the ALEKS Data resource, to answer the questions below about this reaction:
NaOH(s) -. Na (ag) + OH (ag)
You can make any reasonable assumptions about the physical properties of the solution. Be sure answers you calculate using measured data are rounded to 3
significant digits.
Note for advanced students: it's possible the student did not do the experiment carefully, and the values you calculate may not be the same as the known and
published values for this reaction.
is this reaction exothermic, endothermic, or neither?
Oexothermic
O endothermic
O neither
0.°
If you said the reaction was exothermic or endothermic, calculate the amount of
heat that was released or absorbed by the reaction in this case.
Calculate the reaction enthalpy AH.
nen per mole of NaOH.
kJ

Several factors can dictate entropy in an equation.

These include:

1. Phase changes

⇒ When a solid turns to a liquid, the entropy increases as the particles have more freedom to move around and thus have a greater ability for 'disorder'. Same goes for a liquid turning to a gas. In a gas, the intermolecular forces are much weaker than that of a solid or liquid, allowing the particles more freedom.

So, going from a solid to liquid to gas increases entropy, and going the other way, from gas to liquid to solid, decreases entropy.

Example:

H₂O(l) -> H₂O(g)

This will have a positive entropy change, as the water molecules are becoming gaseous and thus have more freedom.

2. Dissolution

⇒ Similarly, breaking up particles of a solute when dissolving in a solvent will increase entropy as the particles are no longer bound together.

So, dissolving a solute will increase entropy.

Example:

NaCl(s) -> NaCl(aq)

This will have a positive entropy change, as the NaCl particles are more free after being separated.

3. Number of products and reactants

⇒ Generally, if you have more moles of products than reactants, if they are the same phase then entropy will increase. Note this is not necessarily true if you form a gas from two non-gas reactants, as the gas will still have more entropy.

4. Temperature

⇒ Increasing temperature will increase entropy as the particles have more kinetic energy and are then moving faster.

-------------------------------------------

l. 2N2(g) + O2(g) → 2N2O(g)

3 moles of gas are forming 2 moles of gas. The phase of products and reactants are the same, so since we have less moles of product than reactant, entropy will be negative.

II. CaCO3(s) → CaO(s) + CO2(g)

1 mole of solid is forming 1 mole of solid and 1 mole of gas. There is a phase change from solid to gas, and there are more moles of product than reactant, entropy will be positive.

III. Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)​

While 3 moles of reactant are forming only 2 moles product, we are forming a gas from non-gaseous reactants, so entropy will be positive regardless.

Answer: , the work required to separate the two ions to an infinite distance is 4.63 x 10^-19 J.

Explanation: One can determine the potential energy of two point charges by utilizing the specified equation:

The value of U is directly proportional to the product of q1 and q2, and inversely proportional to the distance between them (r), where k is a constant factor.

The potential energy denoted by U is determined by the Coulomb constant, k, which has a value of 9 x 10^9 Nm^2/C^2. The calculation of U involves the charges of two particles, q1 and q2, as well as the distance between them, denoted by r.

We have an instance where a sodium ion (Na+) has a charge of 1.6x10^-19 C, and a chloride ion (Cl-) has a charge of -1.6x10^-19 C, positioned 4.95 Nm apart from each other.

Once we insert the given numbers, the result obtained is:

The expression for U can be obtained by utilizing the equation U = (kQq)/r, with k being the Coulomb's constant equal to 9 x 10^9 Nm^2/C^2, Q and q representing the electric charges of -1.6x10^-19 C and -1.6x10^-19 C, respectively, and r being the distance between the charges of 4.95 Nm.

The value of U is negative 4.63 times 10 to the power of negative 19 Joules.

It is important to observe that the negative symbol signifies the negativity of the potential energy, implying that effort must be exerted in order to disassociate the two ions.

The polar ice caps are large areas of ice that cover the Earth's polar regions, specifically the North Pole (Arctic) and the South Pole (Antarctica). These ice caps consist mainly of ice sheets, glaciers, icebergs, and sea ice, which have significant impacts on the global climate, ecosystems, and sea levels. Here's how they help the world:

1. Climate regulation: The polar ice caps play a crucial role in maintaining the Earth's temperature by reflecting sunlight back into space. The bright, reflective surface of the ice (called albedo) helps to cool the planet and counteract the greenhouse effect. When the ice caps melt, the darker ocean or land underneath absorbs more sunlight, leading to higher temperatures and further ice melting – a process known as the ice-albedo feedback loop.

2. Ocean circulation: The polar ice caps influence global ocean circulation patterns. Cold, dense water sinks near the polar regions, driving the thermohaline circulation (also known as the global ocean conveyor belt). This circulation helps to distribute heat and nutrients around the world, impacting weather patterns and supporting marine ecosystems.

3. Sea level regulation: The ice caps store a vast amount of freshwater in the form of ice. If all the ice in Antarctica and Greenland were to melt, global sea levels would rise by around 65 meters (213 feet), with devastating consequences for coastal communities and ecosystems. By keeping this water locked in ice, the polar ice caps help to regulate sea levels and protect coastal areas from flooding.

4. Ecosystem support: The polar ice caps support unique and fragile ecosystems that are home to a wide variety of plants and animals. Some of these species, such as Arctic foxes, polar bears, and emperor penguins, are specifically adapted to life in the harsh polar environments. The ice caps are also critical to the survival of many marine species, including fish, seals, and whales, which rely on the ice for breeding grounds, shelter, and hunting.

5. Scientific research: The polar regions offer valuable opportunities for scientific research, particularly in the fields of climate science, glaciology, and paleoclimatology. Ice cores taken from the ice caps provide a historical record of Earth's climate, allowing scientists to understand past climate changes and make more informed predictions about the future.

Oxygen has oxidation state of -2 in most compounds, but it can have different oxidation states in some compounds. In peroxide (O₂²⁻), oxygen has an oxidation state of -1

Nomenclature is the system of naming chemical compounds according to a set of rules.

In peroxide (O₂²⁻), oxygen has an oxidation state of -1. This is because the two oxygen atoms share the two electrons that make up the covalent bond equally, resulting in an oxidation state of -1 for each oxygen atom.

So, while oxygen in peroxide has a charge of 2-, it does not have a 2- charge in general.

As for nomenclature, when naming compounds containing oxygen, the oxidation state of the oxygen atom is usually indicated by using a suffix. For example, the suffix "-ite" indicates a lower oxidation state, while the suffix "-ate" indicates a higher oxidation state.

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The first quarter is the point in the lunar cycle where the moon is one (1) week after the new moon has risen. At first quarter, the moon will also be in its "waning gibbous" phase.

The moon is called a "new moon" when it lies between the sun and the Earth, as opposed to a "full moon," which occurs when the moon is directly above. Furthermore, because we cannot even see the new moon from Earth, unlike full moons, it appears as though it doesn't exist at all.

A half (1/2) or 50% of the moon is lighted during the waning gibbous phase, which occurs once per new moon.

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It took Approximately 52 minutes and 40 seconds for an electromagnetic signal to travel to Earth from Pioneer 11 when it was near Jupiter.

The speed of light, which is the speed at which an electromagnetic signal travels in a vacuum, is approximately 299,792,458 meters per second.

To find out how long it takes for an electromagnetic signal to travel from Pioneer 11 to Earth when it was near Jupiter, we can use the formula:

The distance between Pioneer 11 and Earth when it was near Jupiter is approximately 998 million km, or 998,000,000 km.

Converting this distance to meters, we get:

Dividing the distance by the speed of light, we get:

Converting this time to minutes and seconds, we get:

Rounding to the nearest minute and second, we get:

Therefore, it took approximately 52 minutes and 40 seconds for an electromagnetic signal to travel from Pioneer 11 to Earth when it was near Jupiter.

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Answer: the mass of 7.55 x 10^21 molecules of water is 0.226 g, which corresponds to option (c).

Explanation: The atomic weight of water (H2O) is:

2(1.008 g/mol) + 15.999 g/mol = 18.015 g/mol

Avogadro's number (NA) is:

6.022 x 10^23 molecules/mol

To calculate the mass of 7.55 x 10^21 particles of water, able to utilize the taking after equation:

mass = (number of atoms / Avogadro's number) x atomic weight

mass = (7.55 x 10^21 / 6.022 x 10^23) x 18.015 g/mol

mass = 0.226 g

Given the following standard enthalpies of formation for the following substances, H2O (g) = -241.8 kJ/mol KOH (aq)=-482.4 kJ/mol KOH(s) =-425.8 kJ/mol The enthalpy of the reaction is -390.2 kJ/mol.

The standard enthalpy of formation, also known as the standard heat of formation, is the change in enthalpy that occurs during the synthesis of one mole of a substance from its component parts in their reference states, with all substances in their standard states. The IUPAC recommends using the standard pressure value of p = 105 Pa (= 100 kPa = 1 bar), while previous to 1982, the value of 1.00 atm (101.325 kPa) was used.A standard temperature doesn't exist. Its designation is fH. This symbol's superscript Plimsoll denotes that the process has been carried out under typical circumstances at the given temperature (typically 25 °C or 298.15 K).

The enthalpy of the reaction can be calculated using Hess's Law and the standard enthalpies of formation for the reactants and products. First, we need to determine the enthalpy of the reaction, ∆Hrxn. This is the sum of the standard enthalpies of formation for the products minus the sum of the standard enthalpies of formation for the reactants.

∆Hrxn = [2(-482.4) kJ/mol + (-241.8) kJ/mol] - [2(-425.8) kJ/mol + (-285.8) kJ/mol]

∆Hrxn = -390.2 kJ/mol

Therefore, the enthalpy of the reaction is -390.2 kJ/mol.

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0.785 g of lithium hydroxide (LiOH) are produced when 0.38 g of lithium nitride (Li₃N) reacts with an excess of water.

we can see that 3 moles of lithium hydroxide (LiOH) are produced for every 1 mole of lithium nitride (Li₃N) that reacts.

To determine the mass of LiOH produced from 0.38 g of Li₃N, we need to first calculate the number of moles of Li₃N present:

molar mass of Li₃N = 3 x atomic mass of Li + 1 x atomic mass of N

= 3 x 6.94 g/mol + 1 x 14.01 g/mol

= 34.83 g/mol

moles of Li₃N = mass / molar mass

= 0.38 g / 34.83 g/mol

= 0.01093 mol

Since the balanced equation shows that 1 mole of Li₃N produces 3 moles of LiOH, we can calculate the number of moles of LiOH produced:

moles of LiOH = 3 x moles of Li₃N

= 3 x 0.01093 mol

= 0.03279 mol

Finally, we can use the molar mass of LiOH to convert from moles to grams:

molar mass of LiOH = atomic mass of Li + 1 x atomic mass of O + 1 x atomic mass of H

= 6.94 g/mol + 15.99 g/mol + 1.01 g/mol

= 23.94 g/mol

mass of LiOH produced = moles of LiOH x molar mass of LiOH

= 0.03279 mol x 23.94 g/mol

= 0.785 g

Therefore, approximately 0.785 g of lithium hydroxide (LiOH) are produced when 0.38 g of lithium nitride (Li₃N) reacts with an excess of water.

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The word equation for the given skeleton equation is: Solid aluminum + gaseous oxygen → Solid aluminum oxide.

Chemical equations are a way of representing chemical reactions using chemical formulas and symbols. They provide a concise and standardized way of describing chemical reactions and are an important tool for chemists to communicate and understand chemical changes.

In any chemical equation, reactants are written on left side of an arrow, whereas the products are written on right side. The arrow indicates the direction of reaction, from the reactants to products.

Chemical equations use chemical formulas to represent the elements and compounds involved in a reaction.

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Sugar cane plantation owners needed cheap labor after the enslaved were freed.

We know that in the nineteenth century the era of slave trade was actually coming to a close and the concern of many of the slave owners both in Europe and America was the future of their agricultural businesses.

The slave owners mostly used the slaves to do the work on the farm and in the absence of the slaves, the work could not be done. This is why the owners of the slaves sought for cheap labor after the slaves were freed.

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Sugar was important in the late nineteenth century for both dietary and economic reasons. It was consumed by all social classes and significantly influenced the labor practices in agriculture.

Based on the information provided, one can infer that sugar was of significant importance in the late nineteenth century. It served as a staple in people's daily diets and was enjoyed by all social classes, from shopgirls to kings. In the economic sphere, sugar cane farming was a notable industry. Sugar canes were labor-intensive to grow and harvest, which required plantation owners to seek cheap labor. This implies that the sugar industry had a substantial influence on labor practices during this era. It is therefore clear that sugar was not only a dietary necessity but also a crucial factor in the broader economic and social structure of the community.

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Answer:  the mass percentage of C in codeine is 72.21%.

Explanation: Molar mass of C = 12.011 g/mol

Molar mass of CisHziNO = 299.368 g/mol (whole of molar masses of all molecules within the compound)

Number of C molecules in codeine = 18

Mass of C in codeine = 12.011 g/mol x 18 = 216.198 g/mol

Mass rate of C in codeine = (216.198 g/mol ÷ 299.368 g/mol) x 100% = 72.21%

2A. The mole of HCl used is 3.2 moles

2B. The moles of each product produced as:

The mole of HCl used can be obtained as follow:

NaHCO₃ + HCl -> NaCl + CO₂ + H₂O

From the balanced equation above,

1 mole of NaHCO₃ reacted with 1 mole of HCl

Therefore,

3.2 moles of NaHCO₃ will also react to 3.2 moles of HCl

Thus, the number of mole of HCl used is 3.2 moles

The mole of each product produced can be obtain as shown:

For NaCl

NaHCO₃ + HCl -> NaCl + CO₂ + H₂O

From the balanced equation above,

1 mole of NaHCO₃ reacted to produce 1 mole of NaCl

Therefore,

3.2 moles of NaHCO₃ will also react to produce 3.2 moles of NaCl

Thus, the mole of NaCl produced is 3.2 moles

For CO₂

NaHCO₃ + HCl -> NaCl + CO₂ + H₂O

From the balanced equation above,

1 mole of NaHCO₃ reacted to produce 1 mole of CO₂

Therefore,

3.2 moles of NaHCO₃ will also react to produce 3.2 moles of CO₂

Thus, the mole of CO₂ produced is 3.2 moles

For H₂O

NaHCO₃ + HCl -> NaCl + CO₂ + H₂O

From the balanced equation above,

1 mole of NaHCO₃ reacted to produce 1 mole of H₂O

Therefore,

3.2 moles of NaHCO₃ will also react to produce 3.2 moles of H₂O

Thus, the mole of H₂O produced is 3.2 moles

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Answer:

D. something's that's produced during a chemical reaction

Answer:

Explanation:

The answer is Option D.

A substance that is produced during a chemical reaction.

for example in a chemical reaction A+B -----> C

here, reactants A and B react together to produce product C.

a) Moles of HCI that were needed is 14.88 mol  ; b) Grams of HCI that were needed is 670.14 g.

Equation where number of atoms of each element in the reactant is equal to the number of atoms of the same element in the products is called as the balanced equation.

a.) 30 g / 2.016 g/mol = 14.88 mol of H₂

From the balanced equation, we can see that one mole of H₂ is produced for every mole of HCl consumed. Therefore, the number of moles of HCl needed is also 14.88 mol.

b.) 2 moles of HCl are consumed for every mole of PbCl₂ produced

So, the number of moles of HCl needed to produce 9.2 moles of PbCl₂ is:

9.2 mol PbCl₂ × 2 mol HCl/mol PbCl₂ = 18.4 mol HCl

18.4 mol × 36.46 g/mol = 670.14 g of HCl

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The half‑life of Ga67 is 3.67 days.

The half-life of a chemical reaction can be defined as the time taken for the concentration of a given reactant to reach 50% of its initial concentration (i.e. the time taken for the reactant concentration to reach half of its initial value). It is usually expressed in seconds.

Half-lives are characteristic properties of the various unstable atomic nuclei and the particular way in which they decay.

Given,

Let the initial amount be 100, so final amount will be 100 - 64 = 36

time = 2.33 days

= 2.33 × 24 × 60 × 60

= 201312 seconds

k = ( 1÷ t) log ( initial ÷ final)

k = 0.189 s⁻¹

half life = 0.693 ÷ k

= 315286.62 seconds

= 3.67 days

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Elements on the transition metal has the highest tendency to gain are chromium and manganese.

Transition metals typically have the ability to lose electrons to form positive ions, rather than to gain electrons to form negative ions. However, some transition metals can gain electrons under certain conditions.

The tendency to gain electrons (electron affinity) generally increases across a period from left to right on the periodic table, and decreases down a group from top to bottom.

Among the transition metals, the element with the highest electron affinity is usually considered to be either chromium [Cr] or manganese [Mn]. Both of these elements have a relatively high electron affinity due to their electronic configurations and the proximity of their valence electrons to the nucleus.

However, it is important to note that electron affinity values can vary depending on the specific experimental conditions used to measure them.

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The elements on the transition metal that have the highest tendency to gain are chromium and manganese.

Usually, transition metals form positive ions by losing electrons rather than gaining electrons to form negative ions. However, under certain conditions, some transition metals can gain electrons. The tendency to gain electrons is known as electron affinity.

This tendency generally increases as we go across a period from left to right in our periodic table and decreases while going down the group from top to bottom.

The element with the highest electron affinity, among the transition metals, is usually considered to be either manganese [Mn] or chromium [Cr]. Both of them have a relatively high electron affinity because of their electronic configurations and the proximity of their valence electrons to the nucleus.

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The number of atoms of each element is now the same on both sides, so the equation is balanced.

Nothing can be created or destroyed, according to the conservation of matter principle. The quantity of each element's atom in the reactants and the quantity of that same element in the products must match in chemical equations.

As soon as atoms distribute their electrons uniformly, a nonpolar covalent bond is formed. Normally, this happens when two atoms have comparable or identical electron affinities. The greater the attraction, the closer the values of their electron affinities are. This happens in gas molecules, also referred to as diatomic substances.

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Answer: Calcium: Alkaline earth metal

Vanadium: Transitional metal

Xenon: Noble gas

Iodine: Halogen

Potassium: Alkali metal

Strontium: Alkaline earth metal

Explanation:

Therefore, if 3.0 moles of Potassium chlorate are completely destroyed, 4.5 moles of oxygen will form.

Therefore, there will be 36=18 moles of water molecules in 3 moles of the material. Keep in mind that there are 6.02 x 1023 molecules in a mole of molecules. Thus, there will be 6.021023181.081025 water molecules in total here.

If 2 moles of Potassium chlorate decompose to form 3 moles of oxygen, then 1 mole of Potassium chlorate will decompose to form 3/2 moles of oxygen.

Therefore, to find out how many moles of oxygen will form from 3.0 moles of Potassium chlorate, we can use the following calculation:

moles of oxygen = moles of Potassium chlorate x (3/2)

[tex]moles of oxygen = 3.0 mol x (3/2)[/tex]

moles of oxygen = 4.5 mol.

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Answer:

Please mark brainlist

Explanation:

The equilibrium expression for this reaction is Kw = [H₃O⁺][OH⁻], where Kw is the autoionization constant for water. At 25°C, the value of Kw is 1.0 x 10⁻¹⁴.

The balanced equation for the given redox reaction in basic solution is:

NO2^- + Fe3+ + H2O → NO3^- + Fe2+ + OH^-

Here, NO2^- is oxidized to NO3^-, which means it loses electrons and undergoes an oxidation reaction. Fe3+ is reduced to Fe2+, which means it gains electrons and undergoes a reduction reaction.

To balance the equation in basic solution, we first balance the atoms that are not hydrogen or oxygen. We start with the Fe3+ ion, which is reduced to Fe2+. To balance the iron atoms, we add one electron to the left side of the equation:

Fe3+ + e^- → Fe2+

Next, we balance the nitrogen and oxygen atoms in NO2^- and NO3^- by adding H2O and OH^- to the appropriate sides of the equation:

NO2^- + H2O → NO3^- + OH^-

Finally, we balance the electrons by multiplying the Fe3+ reduction half-reaction by 2, and adding it to the oxidation half-reaction for NO2^-:

2Fe3+ + 2e^- → 2Fe2+

NO2^- + 2Fe3+ + 2H2O → NO3^- + 2Fe2+ + 2OH^-

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Answer: air

Explanation:would be the most likely candidate for the cans

When heated, ammonium nitrate decomposes explosively according to the balancing equation: 2 NH₄NO₃(s) + 2 N₂(g) + 4 H₂O(g) Calculate the total volume of gas generated by the full breakdown of 1.55 kg of ammonium nitrate (at 125 °C and 748 mmHg).

When ammonium nitrate is heated, it produces nitrous oxide and water molecules. When ammonium nitrate (NH₄NO₃) is heated, it produces nitrous oxide (N₂O) and water (H₂O).

Endothermic dissolution of ammonium nitrate in water occurs because more energy is consumed to separate the ions in the solid than is created when the ions establish new connections with water molecules.

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Answer:

The mass of the white solid calcium carbonate that forms is 0.300 g.

Explanation:

To calculate the mass of the white solid calcium carbonate that forms, we first need to determine the limiting reagent in the reaction between calcium nitrate and sodium carbonate. The balanced chemical equation for the reaction is:

Ca(NO3)2(aq) + Na2CO3(aq) → CaCO3(s) + 2NaNO3(aq)

From the equation, we can see that one mole of calcium nitrate reacts with one mole of sodium carbonate to produce one mole of calcium carbonate. Therefore, the limiting reagent is the one that produces the least amount of calcium carbonate.

To determine the limiting reagent, we need to calculate the moles of calcium nitrate and sodium carbonate used in the reaction:

Moles of calcium nitrate = volume of solution (L) x concentration (mol/L) = 25.0 L x 0.100 mol/L = 2.50 mol

Moles of sodium carbonate = volume of solution (L) x concentration (mol/L) = 0.0200 L x 0.150 mol/L = 0.00300 mol

Since the moles of sodium carbonate are much smaller than the moles of calcium nitrate, sodium carbonate is the limiting reagent.

The balanced chemical equation tells us that one mole of calcium carbonate is produced for every mole of sodium carbonate used. Therefore, the moles of calcium carbonate produced are also equal to 0.00300 mol.

Finally, we can calculate the mass of calcium carbonate produced using the molar mass of calcium carbonate:

Mass = moles x molar mass = 0.00300 mol x 100.1 g/mol = 0.300 g

Therefore, the mass of the white solid calcium carbonate that forms is 0.300 g.

0.0318 moles of acetic anhydride was used in the experiment.

To determine the number of moles of acetic anhydride (C4H6O3) used, we need to use the formula:

moles = mass / molar mass

First, we need to find the mass of 3.00 mL of acetic anhydride using its density:

mass = volume x density = 3.00 mL x 1.08 g/mL = 3.24 g

Next, we need to find the molar mass of acetic anhydride:

molar mass of C4H6O3 = 4(12.01 g/mol) + 6(1.01 g/mol) + 3(16.00 g/mol) = 102.08 g/mol

Now, we can use the formula to find the number of moles:

moles = mass / molar mass = 3.24 g / 102.08 g/mol = 0.0318 mol

Therefore, 0.0318 moles of acetic anhydride was used in the experiment.

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