Which term is used to describe why we can smell air freshener across the room shortly after it has been sprayed? A. elastic collisions B. osmosis C. diffusion D. temperature​

Answer: 499.2g of O2

Explanation:CO2 + O2 ---> CO2 +H20

From the equation, we know 1 mole of oxygen requires 1 mole of CO2 to react completely.

Therefore it is sure that 15.6 mole of CO2 requires 15.6 moles of O2

To convert moles of O2 to grams, we need to use the molar mass of O2

i.e. 32.00g/mol

the mass of O2 required is

15.6 moles *32.00 g/mol = 499.2g

The balanced molecular equation when nitrosyl chloride gas, NOCI, is heated to produce gaseous chlorine and nitrogen monoxide is:

2NOCl(g) ⇔ Cl₂(g) + 2NO(g)

Chemical equation is simply a ritrogen monoxide epresentation of chemical reactions using the symbols and formula of elements. We must understand, that every reaction has two sides i.e reactants and products.

The balancing of chemical equation must obey the law of conservation of matter other wise, atoms will either be created or destroyed during the reaction.

Now we shall write the balanced molecular equation for the reaction invoving the heating of nitrosyl chloride gas, NOCI. Details below:

Nitrosyl chloride gas ⇔ Gaseous chlorine + Nitrogen monoxide

NOCl(g) ⇔ Cl₂(g) + NO(g)

There are 2 atoms of Cl on the right and 1 atom on the left. It can be balanced by writing 2 before NOCl as shown below:

2NOCl(g) ⇔ Cl₂(g) + NO(g)

There are 2 atoms of N on the left and 1 atom on the right. It can be balanced by writing 2 before NO as shown below:

2NOCl(g) ⇔ Cl₂(g) + 2NO(g)

Thus, the equation is balanced.

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the minimum mass of Mg(NO₃)₂ required to begin precipitation of MgF₂ is 6.37 × 10⁻³ g/L.

The balanced equation for the precipitation of MgF₂ from Mg(NO₃)₂ and HF is:

Mg(NO₃)₂ + 2HF → MgF₂(s) + 2HNO₃

The concentration of fluoride ions can be found using the equilibrium expression for the dissociation of HF:

HF + H₂O ⇌ H₃O⁺ + F⁻

Ka = [H₃O⁺][F⁻]/[HF]

Since the initial concentration of HF is 0.560 M, the concentration of [H₃O⁺] is negligible compared to [HF] because HF is a weak acid. Therefore, we can simplify the expression to:

Ka = [F⁻][HF]/[HF] = [F⁻]

[F⁻] = Ka = 7.2 × 10⁻⁴ M

Now we need to determine the minimum amount of Mg(NO₃)₂ required to provide enough fluoride ions to reach the saturation point of MgF₂.

Let's assume x moles of Mg(NO₃)₂ is added to 1.00 L of 0.560 M HF solution. This will result in the addition of 2x moles of F⁻ ions to the solution.

The molar solubility of MgF₂ is equal to the square root of Ksp because the stoichiometric coefficients are 1 for both MgF₂ and Mg(NO₃)₂.

Molar solubility of MgF₂ = sqrt(Ksp) = sqrt(7.4 × 10⁻⁹) = 8.6 × 10⁻⁵ M

To reach saturation, the concentration of F⁻ ions must be equal to the molar solubility of MgF₂. Therefore:

[F⁻] = 8.6 × 10⁻⁵ M = 2x moles/L

x = 4.3 × 10⁻⁵ moles/L

The mass of Mg(NO₃)₂ required is then:

mass = moles × molar mass = (4.3 × 10⁻⁵ mol/L) × (148.3 g/mol) = 6.37 × 10⁻³ g/L.

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Solution.

Henry's law states that the solubility of a gas in liquids at a constant temperature is proportional to its pressure. We write the formula:

�

=

�

×

�

S=K×P

The Henry's constant for each substance is individual, so we express it and find it through the solubility of the gas and its pressure.

�

=

�

�

K=

P

S

​

�

=

5.58

∗

1

0

−

6

K=5.58∗10

−6

And now we find the solubility of the gas at a different pressure.

S = 1.39 g/L

Answer:

S = 1.39 g/L

2.

Solution.

To begin with, we will translate the pressure from mm Hg of the column and from atmospheres to Pascals.

760 mmHg = 101324.72 Pa

2.5 atm = 253312.5 Pa

�

=

�

×

�

S=K×P

�

=

�

�

K=

P

S

​

�

=

1.58

∗

1

0

−

5

K=1.58∗10

−5

Now we find the solubility of the gas at a different pressure.

S = 4.00 g/L

Answer:

S = 4.00 g/L

3.

Solution.

To begin with, we will translate the pressure from atmospheres to Pascals.

0.750 atm = 75993.75 Pa

�

=

�

×

�

S=K×P

�

=

�

�

K=

P

S

​

�

=

3.22

∗

1

0

−

5

K=3.22∗10

−5

Now we find the gas pressure to get a solution with a concentration of 6.25 g / L.

�

=

�

�

P=

K

S

​

P = 194.1 kPa

Answer:

P = 194.1 kPa

The pulmonary diffusion capacity of carbon monoxide (DLCO) is used to determine the transfer of gases through the lungs. The amount of carbon monoxide absorbed by hemoglobin in the lungs is measured in this method.

Carbon monoxide has a higher diffusion coefficient than oxygen, and it is 20-30 times more soluble than oxygen. These differences affect the interpretation of DLCO results. Hemoglobin is a molecule that is critical in the transportation of oxygen and carbon monoxide. Hemoglobin is known to bind with carbon monoxide 240 times more readily than oxygen. In this method, hemoglobin plays a critical role. It is assumed that 100 percent of inhaled carbon monoxide is absorbed into the bloodstream and that the hemoglobin-CO complex forms reversible binding without causing an alteration in the blood's oxygenation level.

The DLCO measurement is based on the following assumptions: At sea level, the concentration of carbon monoxide is zero in the blood being analyzed. The solubility of carbon monoxide in blood is 20-30 times greater than that of oxygen, indicating that it is dissolved in the blood more easily. The alveolar-capillary barrier's thickness is consistent throughout the lung's entire surface area. The diffusion capacity is determined solely by the exchange of gases through the alveolar-capillary membrane and the available hemoglobin in the blood.

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The steep-rise interval in the weak acid-strong base curve is more pronounced than in the strong acid-strong base curve because the weak acid requires more titrant to be neutralized than the strong acid, so the pH changes more slowly initially.

A. The initial pH for the weak acid-strong base curve is higher than the initial pH for the strong acid-strong base curve.

This statement is inaccurate. In fact, the initial pH for the weak acid-strong base curve is lower than the initial pH for the strong acid-strong base curve. This is because a weak acid has a higher initial pH than a strong acid due to the lower concentration of H+ ions in solution.

B. At the equivalence points, the pH of the weak acid-strong base is greater than the pH of the strong acid-strong base.

This statement is accurate. At the equivalence point of a weak acid-strong base titration, the pH is greater than 7 due to the presence of the conjugate base of the weak acid in solution. At the equivalence point of a strong acid-strong base titration, the pH is 7.

C. At the half-equivalence points, the pH of the weak acid-strong base is greater than the pH of the strong acid-strong base.

This statement is accurate. At the half-equivalence point of a weak acid-strong base titration, the pH is greater than 7 because the solution contains both the weak acid and its conjugate base. At the half-equivalence point of a strong acid-strong base titration, the pH is 7.

D. The steep-rise interval in the weak acid-strong base curve is more pronounced than in the strong acid-strong base curve.

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The final state of the hydrogen atom is the n=2 energy level.

The initial state of the hydrogen atom is the ground state, where the electron is in the n=1 energy level. When it absorbs a photon of wavelength 93.73 nm, it jumps to a higher energy level. We can calculate the energy of the absorbed photon using the equation E = hc/λ, where h is Planck's constant, c is the speed of light, and λ is the wavelength of the photon.

E = (6.626 x 10^-34 J s)(2.998 x 10^8 m/s) / (93.73 x 10^-9 m) = 1.653 x 10^-18 J

This energy corresponds to the difference in energy between the ground state and some higher energy level, which we can calculate using the Rydberg equation:

1/λ = R_H(1/n_i^2 - 1/n_f^2)

where λ is the wavelength of the emitted photon, R_H is the Rydberg constant for hydrogen, and n_i and n_f are the initial and final energy levels, respectively. We know λ and R_H, and we can assume n_i = 1, so we can solve for n_f:

1/λ = R_H(1 - 1/n_f^2)

n_f^2 = 1 / (1 - λ/R_H) = 4

n_f = 2

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According to the Brønsted-Lowry model, an acid and its conjugate base differ by a single proton, while a base and its conjugate acid differ by the addition of a single proton. If a reaction produces conjugate acid-base pairs, it supports this model.

To determine if the reaction supports the Brønsted-Lowry model of acids and bases, there are several questions that could be asked. Two possible questions are:

1. Does the reaction involve the transfer of protons (H+ ions) from one molecule to another.This is a key concept in the Brønsted-Lowry model, which defines acids as proton donors and bases as proton acceptors. If a reaction involves the transfer of protons, it supports this model.

2. Do the reactants and products contain conjugate acid-base pairs.

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Answer:

1. Did an acid donate a hydrogen ion to become a conjugate base?

2. Did a base accept a hydrogen ion to become a conjugate acid?

When Muhammad died, his followers carried on the new religion. Islam prospered and grew under the Caliph Abu-Bakr and those who followed him. What cities became hubs of Muslim thought and power? How did word of this new religion spread? Did it appeal to people outside the Arabian Peninsula?

Explanation:

Both Ba and Sr belong to the same group of elements in the periodic table Oxygen and selenium both belong to group 16 of the periodic table Nitrogen and arsenic belong to the same periodic table group of elements.

Ba and Sr both have two valence electrons in their outermost shell and belong to the same group of elements in the periodic table, the alkaline earth metal group.

This similarity in electron arrangement leads to similar chemical properties.

Nitrogen and arsenic belong to group 15 of the periodic table, also known as the nitrogen group. Both elements have five valence electrons in their outermost shell, and their shared electron arrangement contributes to similar chemical properties such as their ability to form compounds with three other atoms.

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The solubility of AgCl in the given solution is 1.26 x 10⁻⁸ M.

The reaction is,

AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)

Ksp = [Ag⁺][Cl⁻] = 1.61 x 10⁻¹⁰

SrCl₂ dissociates as follows:

SrCl₂(s) ⇌ Sr²⁺(aq) + 2Cl⁻(aq)

Since the reaction between AgCl and SrCl₂ does not involve the Sr²⁺ ion, its concentration can be considered constant and omitted from the equilibrium expression.

The solubility of AgCl in a solution of 3.42 x 10⁻⁴ M SrCl₂ is given by the following equation:

Ksp = [Ag⁺][Cl⁻] = x(2x + 3.42 x 10⁻⁴)

where x is the molar solubility of AgCl.

Solving for x,

x = 1.26 x 10⁻⁸ M

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90.0 g of ethene is provided, which is more than the 14.025 g needed to react with 1.5 moles of oxygen in this reaction. There will therefore be some ethene left over once the reaction is finished.

In the process of respiration, six oxygen molecules combine with one glucose molecule to generate six carbon dioxide and water molecules. Equation: C6H12O6 + 6O2 6CO2 + 6H2O + Energy can be used to represent the respiration reaction.

To calculate the amount of ethene required to react with 1.5 moles of O2, we can now utilise stoichiometry:

3 moles of oxygen and 1 mole of ethene react.

1.5 moles of oxygen will react with (1/3) x 1.5 moles of ethene = 0.5 moles of ethene

Then, we can figure out how much 0.5 moles of ethene weigh:

m(ethene) = n(ethene) x Molar mass(ethene)

m(ethene) = 0.5 mol x 28.05 g/mol

m(ethene) = 14.025 g

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We get the dark blue color precipitate which contains  [tex]Cu(NH3)4]^{2+}[/tex](aq.) complex ions of the solution when 6 M Na OH is added.

When the 6M Na OH is used instead of the NH3 in a solution then [tex]Cu^{2+}[/tex]ions will be precipitate out as the [tex]Cu(OH)_{2}[/tex]  in the solution. When the NH3 is used the [tex]Cu^{2+}[/tex] ions will form a complex with the NH3 in the solution which can be expressed as,

[tex]Cu^{2+}[/tex] + [tex]4NH_{3}[/tex] ----> [[tex]Cu(NH3)4]^{2+}[/tex] (aq.) + 2[tex]OH^{-}[/tex](aq.)

The complex of the [tex]Cu^{2+}[/tex]  is soluble which gives dark blue color to the solution.

Therefore by using the 6 M Na OH we can get the blue precipitate of the  [tex]Cu(OH)_{2}[/tex] . Precipitation is defined as the process of transforming a dissolved substance of the solution into an insoluble solid. The solid formed is called the precipitate of the solution.

and by using the NH3 we can get the dark blue color of the solution that contains  [tex]Cu(NH3)4]^{2+}[/tex](aq.) complex ions of the solution.

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The correct question is,

Instead of conc. NH3 being added to the solution, 6 M Na OH is added (both are bases). How will this affect the test for the identification of copper(II) ion in the solution. Explain

The transition state for the following SN2 reaction CI + NaSH ---> SH + NaCI is option III.

Transition state in a chemical reaction can be referred to as a molecular configuration that describes the point of maximum energy in the reaction path during which the bonds that are involved in the chemical reaction are in a state of change.

Transition state theory provides detailed explanations of the reaction rates and reaction mechanisms occurring in the gas phase.

The reaction mechanism in question is a substitution reaction in which the hydroxyl ion attacks the alkyl halide, resulting in the creation of an intermediate complex.

The complex is created in a transition state involving the transfer of a negatively charged ion, which will lead to a new bond forming with the incoming nucleophile.

In conclusion, the transition state for the following SN2 reaction CI + NaSH ---> SH + NaCI is option III.

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The pH of the buffer will increase from 4.74 to 4.76 after adding 0.0020 mol of HCl. This increase in pH indicates that the buffer is able to resist the change in pH caused by the addition of an acid.

To determine the new pH of the buffer after adding 0.0020 mol of HCl, we need to use the Henderson-Hasselbalch equation, which relates the pH of a buffer to its acid dissociation constant (Ka) and the ratio of its conjugate base (A-) and acid (HA) concentrations. The equation is given as:

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

First, we need to determine the initial concentrations of CH3COOH and CH3COONa in the buffer solution. Since the buffer contains 0.50 M CH3COOH and 0.50 M CH3COONa, we can assume that the initial concentrations of CH3COOH and CH3COO- are both 0.50 M. We can then calculate the initial ratio of [CH3COO-]/[CH3COOH] as:

[CH3COO-]/[CH3COOH] = 0.50 M / 0.50 M = 1

Next, we need to determine the new concentrations of CH3COOH and CH3COO- after adding 0.0020 mol of HCl. Since the volume of the buffer is 100.0 mL or 0.100 L, the new concentration of CH3COOH ([HA]) can be calculated as:

[HA] = (0.50 M x 0.100 L) - 0.0020 mol = 0.0480 mol

Similarly, the new concentration of CH3COO- ([A-]) can be calculated as:

[A-] = (0.50 M x 0.100 L) + 0.0020 mol = 0.0520 mol

Now, we can calculate the new ratio of [CH3COO-]/[CH3COOH] as:

[CH3COO-]/[CH3COOH] = 0.0520 mol / 0.0480 mol = 1.083

Finally, we can use the Henderson-Hasselbalch equation to calculate the new pH of the buffer as:

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

pH = 4.74 + log(1.083)

pH = 4.76

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The molar mass corresponds to the chlorofluorocarbon CF3Cl (Freon-11), which has a molar mass of 137.37 g/mol. Therefore, the chlorofluorocarbon in the experiment is CF3Cl.

The rate of effusion of a gas through a porous barrier is inversely proportional to the square root of its molar mass. Therefore, we can use the rate of effusion to determine the relative molar mass of the two gases and identify which one is a chlorofluorocarbon.

The rate of effusion can be calculated using Graham's law:

Rate of effusion = Volume of gas / Time taken to effuse

For the chlorofluorocarbon, the rate of effusion is:

Rate of effusion (CFC) = 50 mL / 157 s = 0.3185 mL/s

For argon, the rate of effusion is:

Rate of effusion (Ar) = 50 mL / 76 s = 0.6579 mL/s

Using Graham's law, we can set up the following equation:

Rate of effusion (CFC) / Rate of effusion (Ar) = sqrt(Molar mass (Ar) / Molar mass (CFC))

Solving for the ratio of molar masses:

Molar mass (Ar) / Molar mass (CFC) = (Rate of effusion (Ar) / Rate of effusion (CFC))^2

Molar mass (Ar) / Molar mass (CFC) = (0.6579 mL/s / 0.3185 mL/s)^2

Molar mass (Ar) / Molar mass (CFC) = 4.294

Molar mass (CFC) = Molar mass (Ar) / 4.294

The molar mass of argon is 39.95 g/mol. Therefore, the molar mass of chlorofluorocarbon is:

Molar mass (CFC) = 39.95 g/mol / 4.294 = 9.30 g/mol

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In the following chemical reaction, the solution was supplemented with 8.01g of magnesium sulfate.

We must utilize the following chemical formula to determine how magnesium sulfate and sodium hydroxide react to solve this issue:

MgSO4 + 2 NaOH → Mg(OH)2 + Na2SO4

One mole of magnesium sulfate (MgSO4) interacts with two moles of sodium hydroxide (NaOH), resulting in one mole of magnesium hydroxide (Mg(OH)2) and one mole of sodium sulfate, according to the equation (Na2SO4).

We must utilize the solution's concentration and volume to determine how many moles of NaOH are present in the solution:

C = n/V

where n is the number of moles, V is the volume in liters, and C is the concentration in moles per liter.

n(NaOH) = C × V

               = 1 mol/L × 0.133 L  

               = 0.133 mol.

We require half as many moles of MgSO4 as we do of NaOH since their mole ratio is 1:2:

n(MgSO4) = 0.5 × n(NaOH)

                 = 0.5 × 0.133 mol

                 = 0.0665 mol

Using its molar mass, we can finally translate the quantity of MgSO4 from moles to grams:

m(MgSO4) = n(MgSO4) × M(MgSO4)

                  = 0.0665 mol × 120.4 g/mol = 8.01 g

Therefore the solution was supplemented with 8.01g of magnesium sulfate.

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The rate at which the water expands continues to increase until it reaches the freezing temperature of 0°C or 32°F.

When the temperature of ice-cold water is increased slightly, it undergoes a net contraction. In the case of water, the density of water is at a maximum at 4°C or 39°F. This implies that, as the temperature of water decreases, its density will increase until it reaches 4°C. As a result, the water will begin to expand again.

As a result, when the temperature of ice-cold water is increased slightly, it will undergo a net contraction.According to the Density Anomalies of Water experiment, as the temperature of water decreases, its density increases.

Water achieves its maximum density at 4°C or 39°F. When the temperature of water falls below 4°C, the water starts to expand. At this point, the water begins to freeze, causing a further expansion in volume.

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Answer: The heat capacity of a bomb calorimeter is 500 J°C.

The statements about diels alder reaction which are true are given in the explanation part.

The following statements about the Diels-Alder reaction are true:

- True: It is a reaction involving pi-electrons.
- True: It is a concerted reaction, meaning that it occurs in a single step without the formation of reaction intermediates.
- False: It is not necessarily favored by entropy considerations, as the reaction can be endothermic and enthalpy-driven.
- True: It is a [4+2] cycloaddition, meaning that it involves the formation of a 6-membered ring from a 4-carbon diene and a 2-carbon dienophile.
- False: While some Diels-Alder reactions can be thermal, others require specific conditions such as high pressure or the use of a catalyst to occur.
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Total 12 electrons are transferred in the reaction.

In the above reaction, the oxidation state of chlorine changes from +7 to +3. To determine the number of electrons transferred, we need to calculate the change in oxidation state of chlorine.

Each chlorate ion (ClO₄⁻) has a charge of -1, and each chlorite ion (ClO₃⁻) has a charge of -1 as well.

The oxidation state of oxygen is -2, so the oxidation state of the four oxygen atoms in the chlorate ion is -8, and the oxidation state of the three oxygen atoms in the chlorite ion is -6.

Let x be the oxidation state of chlorine in the chlorate ion, then we have;

x + 4(-2) = -1

x = +7

Let y be the oxidation state of chlorine in the chlorite ion, then we have;

y + 3(-2) = -1

y = +5

Therefore, each chlorine atom undergoes a reduction of 2 units, which corresponds to the gain of 2 electrons.

Since there are 6 chlorine atoms in total (3 in chlorate and 3 in chlorite), the total number of electrons transferred in the reaction is;

6 x 2 = 12 electrons.

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If I contain 3 moles of gas in a container with a volume of 60 liters and at a temperature of 400 K,The pressure inside the container is approximately 16.42 atm.

Using the Ideal Gas Law, we can calculate the pressure inside the container. The Ideal Gas Law is expressed as 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 in Kelvin.

In this case, you have 3 moles of gas (n = 3), a volume of 60 liters (V = 60 L), and a temperature of 400 K (T = 400 K). The gas constant (R) for this equation is 8.314 J/(mol·K), but since we are using liters and atm, we need to use the value 0.0821 L·atm/(mol· K).

Now, we can plug the values into the equation:

P × 60 L = 3 mol × 0.0821 L·atm/(mol·K) × 400 K

To find the pressure (P), we will rearrange the equation and solve for P:

P = (3 mol × 0.0821 L·atm/(mol·K) × 400 K) / 60 L

P ≈ 16.42 atm

So, the pressure inside the container is approximately 16.42 atm.

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The term absorbance is directly proportional to the concentration (c) of the solution which is used in the experiment of the spectra.

The absorbance is defined as the term which is said to be directly proportional to the length of the light path that is equal to the width of the cuvette. The relationship between concentration and absorbance can be explained by the Beer-Lambert law. This law relates the attenuation of light to the properties of a material. The Beer-Lambert law is defined as the law which states that the concentration of a chemical solution is directly proportional to its absorption of light.

The acid dissociation constants were determined in the spectra will be 2.97 pKa* and 9.47 pKa. The pKa and pKa* values are corresponded well with literature values of 9.45 and 3.0 with a 0.2 and 1.0% difference, respectively. This was determined that 2-naphthol is a weaker acid in the ground state than the excited state due to the value of  pKa > pKa.

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The new volume of the gas at 26.8°C and 742 mmHg is approximately 4.28 L.

To find the new volume of the gas at the given temperature and pressure, we can use the combined gas law formula:

P1 × V1 / T1 = P2 × V2 / T2

where P1 is the initial pressure, V1 is the initial volume, T1 is the initial temperature in Kelvin, P2 is the final pressure, T2 is the final temperature in Kelvin, and V2 is the final volume.

Convert the temperatures to Kelvin.
T1 = 25.6°C + 273.15 = 298.75 K
T2 = 26.8°C + 273.15 = 299.95 K

Convert the pressures to atm.
P1 = 748 mmHg × (1 atm / 760 mmHg) = 0.98421 atm
P2 = 742 mmHg × (1 atm / 760 mmHg) = 0.97632 atm

Rearrange the combined gas law formula to solve for V2.
V2 = P1 × V1 × T2 / (T1 × P2)

Substitute the given values into the formula.
V2 = (0.98421 atm) × (4.25 L) × (299.95 K) / (298.75 K × 0.97632 atm)

Calculate the final volume, V2.
V2 ≈ 4.28 L

The volume of the gas at 26.8°C and 742 mmHg is approximately 4.28 L.

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A solutions that is highly reactive & consists ions— molecules or atoms that also have received or exchanged electrons—is an electrolyte solution.

Since the final cost on I equal the net cost on (ii), the electrolytic mixture is always neutral (iii). Unlike with the superconductor, the sodium chloride conducts the electrical charge by virtue of progression of its (iv). The characteristic that causes a metal to prefer to dissolve in positive charge is defined as (v).

Substances known as electrolytes show a natural positive or negative static electricity when submerged in water. Several body activities are supported by them, including controlling nuclear reactions or preserving overall proportion of liquids both inside and outside your cells.

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When 2.24 mol NaOH and 1.20 mol [tex]CO2[/tex] are allowed to react, NaOH is the limiting reactant.

Let's try to figure out the limiting reactant of the reaction between NaOH and CO2 by using stoichiometry.

Stoichiometry is a branch of chemistry that deals with the calculation of quantities of reactants and products involved in a chemical reaction.

By using stoichiometry, we can calculate the limiting reactant of a reaction, which is the reactant that gets consumed entirely and determines the amount of product formed.

Limiting reactant calculation

[tex]NaOH + CO2 → Na2CO3 + H2O[/tex]

From the balanced chemical equation above, we can see that one mole of NaOH reacts with one mole of CO2 to produce one mole of Na2CO3 and one mole of H2O.

Using the mole ratio, we can calculate the number of moles of Na2CO3 that will be formed from 2.24 mol of NaOH.

[tex]2.24 mol NaOH × 1 mol Na2CO3/1 mol NaOH = 2.24 mol Na2CO3[/tex]

Similarly, we can calculate the number of moles of Na2CO3 that will be formed from 1.20 mol of CO2.1.20 mol CO2 × 1 mol Na2CO3/1 mol CO2 = 1.20 mol Na2CO3

Therefore, NaOH is the limiting reactant because it produces a smaller number of moles of Na2CO3 than CO2. NaOH is the limiting reactant, and we can say that 2.24 mol of NaOH reacts with 1.20 mol of CO2 to form 1.20 mol of Na2CO3.

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Indoor cooking with biomass fuel is the major source of indoor air pollution in developing countries, causing health risks.

The significant wellspring of indoor air contamination in emerging nations is indoor cooking with biomass as a fuel. Biomass fills, like wood, charcoal, and creature waste, are regularly utilized for cooking and warming in many non-industrial nations. Be that as it may, consuming these powers delivers a scope of poisons very high, including particulate matter, carbon monoxide, and nitrogen oxides, which can have serious wellbeing impacts.

Toxic paints on furnishings and walls, tobacco smoke, manufactured materials in development of structures, and arrival of radon-222 gas are additionally wellsprings of indoor air contamination, yet they are not as predominant in that frame of mind as indoor cooking with biomass fuel. Resolving this issue through clean cooking arrangements and advancing the utilization of cleaner powers can essentially further develop indoor air quality and decrease wellbeing gambles in these districts.

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