Calculate the pH of a solution prepared by mixing 50 mL of a 0.10 M solution of HF with 25 mL of a 0.20 M solution of NaF. The pKa of HF is 3.14.
A) 3.14 B) 10.80 C) 5.83 D) 7.35 E) 12.00

The rate law for the overall reaction is: Rate = k[A][B]².  Option c is correct.

The rate-determining step in this mechanism is the slow step, which involves the collision of A and B to form AB. Therefore, the rate law for this step is Rate = k[A][B]. However, we still need to express the rate of the overall reaction in terms of the concentrations of the reactants. The first step, A₂ → 2A, is fast and does not affect the overall rate law. Thus, we can use the steady-state approximation to express the concentration of A in terms of [A₂] and [AB].

Since A₂ is consumed twice as fast as B in the overall reaction, we can assume that [A₂] = 2[AB]. Substituting this expression into the rate law for the slow step gives Rate = k[2AB][B] = 2k[AB][B] = k[A][B]², which is the rate law for the overall reaction.

C is the correct option.

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The complete question is

Given the following proposed mechanism, predict the rate law for the overall reaction.

A2 + 2B ? 2AB (overall reaction)

Mechanism A2 →2A fast A + B ? AB slow

Possible Answers: A. Rate = k[A2][B]

B. Rate = k[A2][B]1/2

C. Rate = k[A][B]

D. Rate = k [A2]1/2[B]

E. Rate = k[A2]

To prepare a 0.15 M solution using a 50 mL volumetric flask, the student needs to dissolve 0.15 moles of the salt in the flask. To find the mass of the salt needed, we can use the formula:
mass = moles x molar mass

So, mass = 0.15 moles x 20.163 g/mol = 3.02445 g
Therefore, the student should dissolve 3.02445 grams of the salt to prepare a 0.15 M solution in a 50 mL volumetric flask.To prepare a 0.15 M solution of the salt (molar mass = 20.163 g/mol) in a 50 mL volumetric flask, the student should dissolve:

grams of salt = (0.15 mol/L) x (20.163 g/mol) x (0.050 L) = 0.15195 g
The student should dissolve approximately 0.15195 grams of the salt.

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To determine which photo represents the ovum, we need more context or visual cues, such as descriptions or specific labeling, that are not provided. Without further information or visual guidance..

Similarly, without additional context or specific labeling, we cannot determine which photo represents the blastocyst.

Without the accompanying photos or more detailed information about the visual characteristics of each photo, it is not possible to identify which photo was taken on a specific day after fertilization (Day 1, Day 2, Day 3, Day 4, or Day 5).

To calculate the magnification of the structure in Photo B, we need to know the size of the structure in the photo and its actual size. The given information states that the structure in Photo B is 0.2 mm in actual life, but it does not provide the size of the structure in the photo. Without the size of the structure in the photo, we cannot calculate the magnification.

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The equations provided show the dissociation of various potassium salts in water, along with the ionization reactions of the weak acid anions they contain.

Potassium dihydrogen phosphate dissolving in water:

Potassium hydrogen sulfate dissolving in water:

Potassium hydrogen phthalate dissolving in water:

Potassium hydrogen tartrate dissolving in water:

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When determining the concentration of an unknown sulfuric acid solution, a titration can be performed with a known concentration of sodium hydroxide.

Here are some additional details that may be helpful in understanding the process of titration:

The titration involves adding small amounts of the sodium hydroxide solution to the sulfuric acid solution until the reaction between the two is complete, which is indicated by a change in color of the indicator used. The volume of the sodium hydroxide solution used in the reaction can then be used to calculate the concentration of the sulfuric acid solution.

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The solution contains 0.85 M Na+ ions, 0.25 M PO43- ions, and 0.10 M Cl- ions.

The concentration of each ion in a solution containing 0.25 M Na3PO4 and 0.10 M NaCl can be determined by breaking down the compounds into their individual ions. Na3PO4 dissociates into three Na+ ions and one PO43- ion, while NaCl dissociates into one Na+ ion and one Cl- ion.

Therefore, the concentration of Na+ ions in the solution is:

(3 x 0.25 M Na3PO4) + (1 x 0.10 M NaCl) = 0.85 M

The concentration of PO43- ions in the solution is:

1 x 0.25 M Na3PO4 = 0.25 M

The concentration of Cl- ions in the solution is:

1 x 0.10 M NaCl = 0.10 M

In summary, the solution contains 0.85 M Na+ ions, 0.25 M PO43- ions, and 0.10 M Cl- ions.

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The equilibrium concentration of Ag⁺ and PO₃⁻⁴ are 2.35 x 10⁻⁶ M and 7.05 x 10⁻⁶ M, respectively.

First, we need to write the balanced chemical equation for the precipitation of Ag₃PO₄;

3AgNO₃ + Na₃PO₄ → Ag₃PO₄ + 3NaNO₃

According to the stoichiometry of the equation, 3 moles of AgNO₃ are required to react with 1 mole of Na₃PO₄ to form 1 mole of Ag₃PO₄. So, we need to find out which reactant is limiting.

The number of moles of AgNO₃ present in 0.100 L of 0.270 M solution is:

0.100 L x 0.270 mol/L = 0.027 mol AgNO₃

The number of moles of Na₃PO₄ present in 0.100 L of 1.00 M solution is:

0.100 L x 1.00 mol/L = 0.100 mol Na₃PO₄

According to the stoichiometry of the equation, 0.100 mol Na₃PO₄ would require 0.300 mol AgNO₃ (3 times as many moles). However, we only have 0.027 mol AgNO₃, which is the limiting reactant.

Therefore, all 0.027 mol of AgNO will react to form Ag₃PO₄. The amount of Ag₃PO₄ that will precipitate can be calculated using its solubility product constant (Ksp);

Ksp = [Ag⁺]³ [PO₃⁻⁴]

Ksp = (x)(3x)³ = 8.89 x 10⁻¹⁷

Solving for x gives;

x = [Ag⁺] = 2.35 x 10⁻⁶ M

[PO₃⁻⁴] = 3x = 7.05 x 10⁻⁶ M

Therefore, the concentrations of Ag⁺ is 2.35 x 10⁻⁶ M and the concentration of PO3-4 is 7.05 x 10⁻⁶ M, respectively.

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To compute the applied load, we need to use the equation: Load = constant x (Diagonal Length)^2. The constant for a material with a measured hardness of 164 HK is typically 0.2.

To compute the applied load for a material with a measured hardness (HK) of 164 and an indentation diagonal length of 0.24 mm, please follow these steps:

Step 1: Recall the formula for Knoop hardness (HK):
HK = P / A, where P is the applied load in kgf, and A is the projected area of the indentation in mm².

Step 2: Calculate the projected area of the indentation (A) using the formula:
A = 0.0703 * L², where L is the indentation diagonal length in mm (0.24 mm in this case).
A = 0.0703 * (0.24)²
A ≈ 0.00403 mm²

Step 3: Rearrange the HK formula to solve for the applied load (P):
P = HK * A
P = 164 * 0.00403
P ≈ 0.66092 kgf

Therefore, the applied load for the material with a measured hardness of 164 and an indentation diagonal length of 0.24 mm is approximately 0.66092 kgf.

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The correct answer to the question is: the process will be spontaneous at any temperature.

ΔG is the amount of energy available to do useful work in a system. It is related to ΔH and ΔS through the equation ΔG = ΔH - TΔS, where T is the temperature in Kelvin.

If ΔG is negative, the process is spontaneous (meaning it will happen on its own without any external energy input), and if ΔG is positive, the process is nonspontaneous (meaning it will not happen on its own without external energy input).

Using the given values of ΔH = -214 kJ/mol and ΔS = 450 J/mol.k, we can calculate ΔG at different temperatures. However, we first need to convert ΔH from kJ/mol to J/mol by multiplying by 1000:

ΔH = -214,000 J/mol

Now we can calculate ΔG at different temperatures using the equation above:

At 298 K (room temperature):

ΔG = -214,000 J/mol - (298 K)(450 J/mol.K) = -349,100 J/mol

Since ΔG is negative, the process is spontaneous at room temperature.

At a high temperature (e.g. 1000 K):

ΔG = -214,000 J/mol - (1000 K)(450 J/mol.K) = 36,000 J/mol

Since ΔG is positive, the process is nonspontaneous at high temperatures.

At a low temperature (e.g. 100 K):

ΔG = -214,000 J/mol - (100 K)(450 J/mol.K) = -229,500 J/mol

Since ΔG is negative, the process is spontaneous at low temperatures.

Therefore, the correct answer to the question is: the process will be spontaneous at any temperature.

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The boiling point of the compound is approximately 457.9 K or 184.7°C. To determine the boiling point of the compound, we need to use the Clausius-Clapeyron equation: ln(P2/P1) = -δHvap/R * (1/T2 - 1/T1)

Here, P1 is the vapor pressure at temperature T1 (given as 20°C or 293.15 K), P2 is the vapor pressure at the boiling point, ΔHvap is the enthalpy of vaporization, and R is the gas constant (8.314 J/mol·K). We know that the vapor pressure of the compound at 20.°C (293.15 K) is 97.66 torr. We also know that δHvap = 37.8 kJ/mol. We can assume that the boiling point of the compound is much higher than 20.°C, so we can use 1 atm (760 torr) as P2.  ln(760/97.66) = -37.8*10^3 J/mol / (8.31 J/mol*K) * (1/T2 - 1/293.15 K)
Simplifying this equation gives: ln(7.78) = -4550.6 * (1/T2 - 1/293.15 K)
Solving for T2 gives: T2 = 457.9 K or 184.7°C
Therefore, the boiling point of the compound is approximately 184.7°C.

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To solve for the change in entropy, we can use the equation:

ΔS = nS°(products) - mS°(reactants)

where:

- ΔS is the change in entropy

- n and m are the stoichiometric coefficients of the products and reactants, respectively

- S° is the standard molar entropy of the substance

First, we need to write the balanced chemical equation for the combustion of silicon:

Si + O2 -> SiO2

From the equation, we can see that the stoichiometric coefficient of silicon is 1. Therefore, n = 1.

Next, we need to determine the standard molar entropy of silicon and silicon dioxide. According to standard tables, the values are:

S°(Si) = 18.8 J/(mol K)

S°(SiO2) = 41.8 J/(mol K)

Now we can substitute the values into the equation:

ΔS = nS°(SiO2) - mS°(Si)

Since all the silicon is consumed, m = 0.802 g / (28.09 g/mol) = 0.0286 mol.

ΔS = 1(41.8 J/(mol K)) - 0.0286 mol(18.8 J/(mol K))

ΔS = 0.919 J/K

Therefore, the change in entropy when 0.802 g of silicon is burned in excess oxygen to yield silicon dioxide at 298 K is 0.919 J/K.

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The solution for this question is A. Destructive; constructive

The weathering of a tall mountain down into a low-lying hill involves the breakdown and erosion of the mountain over time, which is a destructive process. This process typically occurs due to various factors such as wind, water, and ice erosion, which gradually wear away the mountain's structure.

On the other hand, the buildup of sand dunes through the deposition of sediment is a constructive process. This occurs when wind or water carries and deposits sand or sediment in a specific location, gradually forming dunes over time.

Therefore, the weathering of a tall mountain represents a landform being changed through a destructive process, while the creation of sand dunes through the deposition of sediment represents a landform being created through a constructive process.

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When the temperature in a room increases from 25°C to 33°C, a substance changes from a solid to a liquid. In a lab, methane and nitrogen are cooled from -170°C to -200°C, with methane freezing and nitrogen remaining as a gas. When gold is heated to 2,856°C, it changes from a liquid to a gas.

The temperature at which a substance changes its state depends on its melting point and boiling point. When the temperature in a room increases from 25°C to 33°C, a substance that was in the solid state may reach its melting point and change to the liquid state.

In the lab scenario, when methane and nitrogen are cooled from -170°C to -200°C, the temperature drops below the melting point of methane (-182.5°C), causing it to freeze and change from a gas to a solid. However, nitrogen remains in the gas state because its boiling point is much lower (-195.8°C).

When gold is heated to 2,856°C, it reaches its boiling point (2,856°C) and changes from a liquid to a gas. This high temperature causes the gold atoms to have enough energy to overcome the intermolecular forces and escape from the liquid phase, resulting in the conversion to a gas.

The state changes of substances are influenced by the balance between intermolecular forces and the thermal energy provided by the temperature. The specific temperature at which these changes occur depends on the unique properties of each substance.

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The solute is the substance being dissolved in a solution, the solvent is the medium in the solute is dissolved. The "universal" solvent refers to water, as it has the ability to dissolve a wide range of substances.

Not all solutions are solids dissolved in liquids. Other examples of solutions include gas dissolved in a liquid (e.g., carbonated drinks) and liquid dissolved in a liquid (e.g., ethanol dissolved in water). It is easiest to separate the component substances in a heterogeneous mixture because the different components can be physically separated based on their different properties, such as size, density, or solubility.

The three different types of mixtures are solutions, colloids, and suspensions. A solution is a homogeneous mixture where solute particles are dispersed and evenly distributed in a solvent. A colloid is a heterogeneous mixture where particles are dispersed but not dissolved in the medium. A suspension is a heterogeneous mixture where particles are larger and settle out upon standing.

Classification:

a. Salad - Heterogeneous mixture

b. Tap water - Homogeneous mixture

c. Muddy water - Heterogeneous mixture

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The spectator ions in a neutralization reaction involving this weak acid (HBrO2) and sodium hydroxide (NaOH) is Na+(aq) and BrO2-(aq).

The neutralization reaction between HBrO2 and NaOH can be represented as follows:

HBrO2  +  NaOH   →   NaBrO2   +  H2O

The complete ionic reaction is

H+BrO2-   +   Na+OH-   →   Na+BrO2-   +  H2O

The net ionic reaction is

H+    +   OH-    →       H2O

In this reaction, Na+ and OH- are the ions that combine to form NaOH and they are called the spectator ions because they do not participate in the formation of the products.

Therefore, the answer is option B. Na+(aq) and BrO2-(aq).

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Concentrated sodium hydroxide (NaOH) must be treated with caution because it is a highly corrosive and caustic substance. Proper protective equipment includes chemical-resistant gloves and safety goggles.

Handling concentrated sodium hydroxide requires strict safety measures due to its potential to cause severe burns and damage to the skin, eyes, and respiratory system. In addition to chemical-resistant gloves and safety goggles, other protective equipment such as a lab coat, closed-toe shoes, and even a face shield can be used to minimize the risk of exposure. In case of accidental contact, it is crucial to have an eyewash station and safety shower nearby to quickly rinse off any NaOH that comes into contact with the skin or eyes.

Furthermore, it is essential to work in a well-ventilated area to prevent the inhalation of harmful fumes, and proper storage guidelines must be followed. Sodium hydroxide should be stored in a tightly sealed, labeled container, away from any acidic or flammable materials. Lastly, it is important to be knowledgeable about emergency procedures and first-aid measures to handle any potential accidents or incidents involving concentrated NaOH.

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The predominant intermolecular force in [tex]NH_{3}[/tex] (ammonia) is hydrogen bonding.

This is because [tex]NH_{3}[/tex] contains a hydrogen atom bonded to a highly electronegative nitrogen atom, resulting in a highly polar molecule.

Hydrogen bonding occurs between a hydrogen atom in a polar molecule and a highly electronegative atom (in this case, the nitrogen atom in another [tex]NH_{3}[/tex] molecule).

This type of intermolecular force is stronger than the other two main types of intermolecular forces, which are London dispersion forces and dipole-dipole interactions.

Bromine ([tex]Br_{2}[/tex]) and iodine ([tex]I_{2}[/tex]) are both nonpolar molecules and only have London dispersion forces between them.

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In the type of hybridization known as sp³ hybridization, the angle between the hybrid orbitals is 109.5 degrees. In this hybridization, one s orbital and three p orbitals combine to form four equivalent sp³ hybrid orbitals, which are arranged in a tetrahedral geometry around the central atom, resulting in bond angles of approximately 109.5 degrees.

In sp³ hybridization, one s orbital and three p orbitals of the central atom combine to form four hybrid orbitals that are arranged in a tetrahedral shape. In order for an atom to be sp³ hybridized, it must have an s orbital and three p orbital. These hybrid orbitals are used to form bonds with other atoms or groups of atoms. Examples of molecules that exhibit sp³ hybridization include methane (CH₄), ethane (C₂H₆), and ammonia (NH₃).

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Increasing the amount of H3O+ does not directly affect C6H5COO- (the acetate ion).

[tex]H3O+[/tex] is a strong acid and acts as a proton donor in reactions. Acetate ions, on the other hand, are weak bases and can accept protons. However, in a typical scenario, increasing the amount of H3O+ does not directly influence the behavior of C6H5COO-. The reactivity of C6H5COO- is primarily determined by its specific reaction partners and the reaction conditions involved.

It's important to note that changes in the concentration of H3O+ may indirectly affect the overall reaction equilibrium or pH, which can influence the behavior of other species, including C6H5COO-. However, the direct impact of H3O+ on C6H5COO- is limited unless they are involved in a specific reaction where the acetate ion acts as a base.

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The solubility product (Ksp) for Ce(IO₃)₃ at the given temperature is 3.16×10⁻¹⁰

First, we shall determine the concentration of Ce³⁺. Details below:

Ce(IO₃)₃(aq) <==> Ce³⁺(aq) + 3IO₃⁻(aq)

From the above,

3 moles of IO₃⁻ is present in 1 mole of Ce(IO₃)₃

Therefore,

5.55×10⁻³ M IO₃⁻ will be present = 5.55×10⁻³ / 3 = 1.85×10⁻³ M Ce(IO₃)₃

Now, we can see from the above equation that Ce(IO₃)₃ and Ce³⁺ are in a ratio of 1:1.

Since the concentration of Ce(IO₃)₃ is 1.85×10⁻³ M. Thus, the concentration of Ce³⁺ is also 1.85×10⁻³ M

Finally, we can determine the solubility product (Ksp). This is illustarted below:

Ce(IO₃)₃(aq) <==> Ce³⁺(aq) + 3IO₃⁻(aq)

Ksp = [Ce³⁺ ] × [Ce³⁺]³

Ksp = 1.85×10⁻³ × (5.55×10⁻³)³

Ksp = 3.16×10⁻¹⁰

Thus, we can conclude that the solubility product (Ksp) is 3.16×10⁻¹⁰

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The most likely event to occur as a result of a prominence on the Sun is option 2: The auroras would become visible.

A prominence is a large, bright, and relatively cool plasma structure that extends outward from the Sun's surface into the corona. It is associated with magnetic fields and is often observed as a loop or curtain-like structure. When a prominence erupts or releases material, it can lead to the formation of a coronal mass ejection (CME). Coronal mass ejections are large bursts of plasma and magnetic fields from the Sun that can travel through space. When a CME interacts with Earth's magnetosphere, it can cause geomagnetic storms. These storms can trigger the phenomenon known as the auroras, which are displays of colorful lights in the Earth's polar regions. As the CME and its associated magnetic fields interact with Earth's magnetosphere, they can cause the charged particles in the atmosphere to emit light, leading to the formation of auroras. The auroras are typically seen in high-latitude regions such as the Arctic (Northern Lights) and Antarctic (Southern Lights). Therefore, when a prominence leads to a CME and subsequent interaction with Earth's magnetosphere, it is most likely that the auroras would become visible as a result of this solar event.

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Answer:

1.46 M

Explanation:

M = mol ÷ Liters

26.489 / 46 = .576 mol of ethanol

density of water is 1g/ml, so the amount of liters of water (L) is 395 ÷ 1000 = .395 Liters

.576 ÷ .395 = 1.46 M

The correct answer is:

B. The light is getting bigger when it's waxing and smaller when it's waning.

Waxing and waning are terms used to describe the changing appearance of the Moon's illuminated portion as seen from Earth.

Waxing refers to the phase of the Moon when the illuminated area is increasing, starting from a New Moon and progressing towards a Full Moon. During the waxing phase, the Moon appears to grow larger and brighter.

Waning, on the other hand, refers to the phase of the Moon when the illuminated area is decreasing, starting from a Full Moon and progressing towards a New Moon. During the waning phase, the Moon appears to shrink in size and become less bright.

Therefore, the key difference between waxing and waning lies in the direction of change in the illuminated portion of the Moon. Waxing means the illuminated area is getting larger, while waning means the illuminated area is getting smaller.

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The procedure instructs to not clean the Mg electrode with nitric acid because nitric acid can react with and dissolve the Mg metal. This is because Mg is a more active metal than hydrogen, and reacts with the acid to produce hydrogen gas and Mg2+ ions.

according to the following reaction :-

Mg(s) + 2HNO3(aq) → Mg(NO3)2(aq) + H2(g)

The reaction produces hydrogen gas which can interfere with the electrochemical measurements by creating additional voltage and current signals.

Therefore, instead of nitric acid, Mg electrode is typically cleaned using a mixture of water and methanol, followed by rinsing with distilled water, to remove any contaminants or impurities from its surface before use in electrochemical measurements.

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The change in energy required to make 50.0 mL of water from its elements under standard conditions is approximately -793.5 kJ.

To calculate the change in energy required to make 50.0 mL of water from its elements under standard conditions, we need to first determine the number of moles of water being formed.

Water has a density of 1 g/mL, so 50.0 mL of water weighs 50.0 g. The molar mass of water (H₂O) is 18.02 g/mol. To find the number of moles, divide the mass by the molar mass:

moles of water = 50.0 g / 18.02 g/mol ≈ 2.775 moles

The standard molar heat of formation of water is -285.8 kJ/mol. Multiply this value by the number of moles to find the total change in energy:

Change in energy = 2.775 moles × (-285.8 kJ/mol) ≈ -793.5 kJ

So, the change in energy required to make 50.0 mL of water from its elements under standard conditions is approximately -793.5 kJ.

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The hormone testosterone is known to act on receptors inside the target cell that directly activate specific genes. Testosterone is a steroid hormone that is produced in the testes in males and in smaller amounts in females in the ovaries and adrenal glands.

Once testosterone is produced, it can bind to specific receptors located in the cytoplasm of the target cell. This binding activates a process where the hormone-receptor complex moves into the nucleus and binds to specific DNA sequences, thereby regulating the expression of specific genes. This process is known as gene transcription and is essential for the proper development and function of various tissues and organs in the body. Therefore, testosterone can have significant effects on various physiological processes, such as growth, development, metabolism, and sexual function. In summary, testosterone is a hormone that can act on receptors inside the target cell to directly activate specific genes, resulting in a range of physiological effects.

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Essentially, "antiboil" is another term for the antifreeze's function of preventing the engine from overheating.

The antifreeze used in a car is a chemical mixture that is added to the engine's cooling system to prevent the engine from freezing in cold temperatures and overheating in hot temperatures, by raising the boiling point of the coolant.

This ensures that the car's cooling system maintains a stable and efficient temperature range, protecting the engine from overheating or freezing.

The term "antiboil" refers to the antifreeze's ability to prevent the engine's coolant from boiling and evaporating in high temperatures, which could cause the engine to overheat and potentially cause damage.

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Predicting the major and minor products of a chemical reaction requires a deep understanding of the reaction mechanism, the stereochemistry, and the reactivity of the reactants and the reaction conditions.

In chemical reactions, the major product is the most abundant product formed, while the minor product is the less abundant one. Predicting the major and minor products requires an understanding of the reaction mechanism, the stereochemistry, and the reactivity of the reactants and the reaction conditions.

One of the most important factors that determine the major and minor products is the regioselectivity of the reaction, which refers to the preference of the reaction to occur at a particular site of the molecule. In addition, the stereochemistry of the reactants and the reaction intermediates can also influence the outcome of the reaction.

Furthermore, the reaction conditions, such as the temperature, the solvent, and the presence of catalysts or other reagents, can also affect the selectivity of the reaction.

To predict the major and minor products of a chemical reaction, it is necessary to analyze the structure of the reactants and the expected intermediates, as well as to consider the factors that influence the selectivity of the reaction.

Computer simulations and experimental testing can also be used to verify the predictions and optimize the reaction conditions to achieve the desired products.

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The amount of Hydrogen Fluoride that can be form from the given reaction is 22.08 g.

The balanced chemical reaction is given as,

5F₂ (g)  +  2NH₃ (g)  -->  N₂F₄ (g)  +  6HF (g)

According to the stoichiometry of the reaction

5 moles of F₂ reacts with 2 moles of NH₃

Given,

Mass of NH₃ = 42 g

=> Moles of NH₃ = 42 / 17 = 2.75 moles

Mass of F₂ = 35 g

=> Moles of F₂ = 35 / 38 = 0.92 moles

5 moles of F₂ reacts with 2 moles of NH₃

=> 1 mole of F₂ reacts with 2/5 = 0.4 moles of NH₃

=> 0.92 moles of F₂ reacts with 0.4 x 0.92 = 0.368 moles of NH₃

We see form the above calculations that NH₃ is present in excess of 2.75 - 0.368 = 2.38 moles

Hence F₂ is the limiting reagent of the reaction

From the stoichiometry 5 moles of F₂ reacts to produce 6 moles of HF

Hence,

0.92 moles of F₂ reacts to produce 0.92 x 6 / 5 = 1.104 moles of HF

=> Moles of HF produced = 1.104

=> Mass of HF = 1.104 x 20 = 22.08 g

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