Which of the following ionic compounds would be expected to be insoluble in water?
NaOH
CaS
Ca(NO 3) 2
Na 2SO 4
KI

The ideal gas law may be used to determine the volume of oxygen created in the electrolysis that produces 548 litres of hydrogen at STP (Standard Temperature and Pressure). 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, according to the ideal gas equation.

The pressure is 1 atm, the temperature is 273 K, and the number of moles of hydrogen is 548/22.4 = 24.5 in this example. We may compute the volume of oxygen created by rearranging the ideal gas law: V = nRT/P = 24.5*0.082*273/1 = 483.3 litres.

As a result, the volume of oxygen created in the electrolysis at STP that produces 548 litres of hydrogen is 483.3 litres.

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Answer:

Because the food can fall and make a mess not only that but it can go into a experiment or something the scienctist are working on and ruin it, cause a reaction.

Explanation:

The solubility of carbon dioxide in water at the given pressure would be 8.45 g CO2 in 100 mL of water.  

According to Henry's law:

C = kH x P

where:

To solve this problem, we need to use the Henry's law constant for carbon dioxide in water at 20 [tex]^oC[/tex], which is 0.0349 mol/L/atm. Let's convert the given solubility from mass/volume units to molar concentration units as follows:

Now we can use Henry's law to calculate the solubility of carbon dioxide in water at 5.50 atm:

Concentration from mol/L to mass/volume units:

0.192 mol/L x 44.01 g/mol = 8.45 g CO2 / 100 mL

Therefore, the solubility of carbon dioxide in water at 5.50 atm pressure is approximately 8.45 g CO2 in 100 mL of water at 20 [tex]^oC[/tex].

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Answer:

The balanced chemical equation for the decomposition of solid lithium chloride into chlorine gas and solid lithium can be written as:

2LiCl(s) → 2Li(s) + Cl2(g)

This equation shows that for every 2 moles of solid lithium chloride, 2 moles of solid lithium and 1 mole of chlorine gas are produced. The equation is balanced because the number of atoms of each element is the same on both sides of the equation. There are 2 lithium atoms, 2 chlorine atoms, and 0 oxygen atoms on both the reactant and product side of the equation.

The correct option is C. Friction is the force that keeps your tires from sliding around on the road.

Friction is the force exerted by a surface when an object moves across it. It always acts in the opposite direction of the movement of the object. That is it resists the movement of the object and if there's no force acting on the moving object, the frictional force will eventually stop the motion of the object.

In absence of friction the tires of the vehicles and even humans would not have been able to cover a distance without slipping and falling.

Therefore, the correct option is C. Friction is the force that keeps your tires from sliding around on the road.

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There would be 10 moles of cobalt in 6.022 x [tex]10^{24[/tex] atoms of cobalt.

In order to calculate the number of moles in a given number of particles, we need to divide the number of particles by Avogadro's number.

6.022 x [tex]10^{24[/tex] atoms of cobalt:

Therefore, there are 10 moles of cobalt in 6.022 x [tex]10^{24[/tex] atoms of cobalt.

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Answer: 5,266

Explanation:

5,266

Answer: Yes

Explanation:

Convert [tex]30km[/tex] to [tex]min.[/tex] by using the conversion factor that is given which is [tex]\frac{50km}{1hr}[/tex].

[tex]30km/\frac{1hr}{50km}/\frac{60min}{1hr}=36min[/tex]

This means it would take 36 minutes for the driver to reach 30 km

The instantaneous rate of reaction at 17 minutes is approximately -0.178 mol dm⁻³

To find the instantaneous rate of reaction at 17 minutes, we can use the concept of differential calculus and estimate the slope of the tangent line at t=17 on the graph of rate versus time.

To do this, we can use the formula for the slope of a line

slope = (change in y) / (change in x)

In this case, the "y" values are the rates of reaction and the "x" values are the times. We want to find the slope at t=17, so we can choose two points that are very close to t=17, such as t=15 and t=20. Then, we can use these values to estimate the slope at t=17

slope = (rate at 20 min - rate at 15 min) / (20 min - 15 min)

slope = (0.135 - 0.223) / (20 - 15)

slope = -0.178

This slope represents the instantaneous rate of reaction at t=17. However, since it has a negative value, it means that the rate of reaction is decreasing at t=17.

Therefore, the instantaneous rate of reaction at 17 minutes is approximately -0.178 mol dm⁻³

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The exit gas stream has an average molecular weight of 29.0 g/mol and the amount of leftover Phenanthrene in the reactor is 5377 g.

Start by calculating the stoichiometric amount of air required to burn 10 kg of Phenanthrene. The balanced chemical equation for the combustion of Phenanthrene is:

C₁₄H₁₀ + 19O₂ → 14CO2 + 5H₂O

Therefore, to burn 10 kg (10000 g) of Phenanthrene:

nO₂ = 19 x (10000 g / 178.24 g/mol) = 1065.5 mol

So the actual amount of oxygen supplied will be:

nO₂, supplied = 0.7 x nO₂ = 745.9 mol

The amount of air required to supply this much oxygen can be calculated using the ideal gas law:

PV = nRT

where P = pressure, V = volume, n = number of moles, R = gas constant, and T = temperature.

Assuming standard temperature and pressure (STP):

P = 1 atm = 101.3 kPa

T = 273 K

R = 8.314 J/mol.K

The volume of air required is then:

Vair = nair × RT/P = (nO₂,supplied + nN₂,supplied) × RT/P

where nN₂,supplied = number of moles of nitrogen in the supplied air.

Since air is about 79% nitrogen by volume, assume that the number of moles of nitrogen is proportional to the number of moles of oxygen:

nN₂,supplied = (0.79/0.21) x nO₂,supplied = 2807.2 mol

Therefore,

Vair = (nO₂,supplied + nN₂,supplied) × RT/P

= (745.9 + 2807.2) × 8.314 × 273 / 101.3

= 63106 L

Calculate the average molecular weight of the exit gas stream using the ideal gas law again:

n = PV/RT

where n = number of moles of gas, P = pressure, V = volume, R = gas constant, and T = temperature.

Assuming that the combustion products are at the same temperature and pressure as the supplied air (STP):

nCO₂ = nH₂O = nO₂,supplied = 745.9 mol

nN₂ = nN₂,supplied = 2807.2 mol

The total number of moles of gas in the exit stream is then:

ntotal = nCO₂ + nH₂O + nN₂ = 745.9 + 745.9 + 2807.2 = 4298.0 mol

The volume of the exit stream can be calculated using the ideal gas law:

Vexit = ntotal × RT/P = 4298.0 × 8.314 × 273 / 101.3 = 36534 L

The average molecular weight of the exit gas stream is then:

M = mtotal/ntotal

where mtotal = total mass of gas in the exit stream.

Calculate mtotal by adding up the mass of each component in the exit stream:

mtotal = mCO₂ + mH₂O + mN₂

where mCO₂, mH₂O, and mN₂ = masses of carbon dioxide, water vapor, and nitrogen, respectively.

Calculate these masses using the molecular weights of the compounds and the number of moles:

mCO₂ = nCO₂ × MCO₂ = 745.9 × 44.01 g/mol = 32804 g

mH₂O = nH₂O × MH₂O = 745.9 × 18.02 g/mol = 13419 g

mN₂ = nN₂ × MN₂ = 2807.2 × 28.01 g/mol = 78617 g

Therefore,

mtotal = mCO₂ + mH₂O + mN₂ = 32804 + 13419 + 78617 = 124840 g

Substituting into the equation:

M = mtotal/ntotal = 124840 g/4298.0 mol = 29.0 g/mol

So the exit gas stream has an average molecular weight of 29.0 g/mol.

The leftover Phenanthrene amount can be calculated as follows:

mPhenanthrene,leftover = mPhenanthrene,initial - mCO₂ - mH₂O

where mPhenanthrene,initial = initial mass of Phenanthrene, which is 10 kg (10000 g).

Substitute these values into the equation:

mPhenanthrene,leftover = 10000 - 32804 - 13419 = 5377 g

Therefore, the amount of leftover Phenanthrene in the reactor is 5377 g.

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If the object takes 3.62s to fall from a height of 65.0m then the speed of the object is 17.95m/s

According to the scenario, the object is falling from a height, so it is likely to follow a straight path.

When the motion of a body is in a straight line, it's speed will be numerically equal to it's velocity.

Mathematically, velocity could be calculated as follows

 [tex]\rm Velocity\ =\ \frac{displacement}{time}[/tex]

Here, the displacement of the object is the total height from where it is dropped.

Therefore, [tex]\rm Velocity\ =\ \frac{65.0}{3.62}[/tex]

                                 [tex]=\ 17.95\ m/s[/tex]

So if the object takes 3.62s to fall from a height of 65.0m then the speed of the object is 17.95m/s

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The mass of the phosphoric(V) acid, H₃PO₄ , is formed in the reaction is 7 g.

The chemical equation is as :

PCl₅ + H₂O  --->   H₃PO₄ + HCl

The mass of the PCl₅ = 15 g

The moles of the PCl₅ = mass / molar mass

The moles of the PCl₅ = 15 / 208.24

The moles of the PCl₅ = 0.0720 mol

The 1 mol of the PCl₅ forms the 1 mole of the H₃PO₄

The moles of the H₃PO₄ = 0.0720 mol

The mass of the H₃PO₄ = moles ×  molar mass

The mass of the H₃PO₄ = 0.0720 × 97.99

The mass of the H₃PO₄ = 7 g.

The mass of the Phosphoric acid, H₃PO₄ is 7 g.

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This question is incomplete, the complete question is :

Calculate the mass of phosphoric(V) acid, H3PO4 , formed in the reaction. The mass of the PCl₅ is 15 g.

PCl₅ + H₂O  --->   H₃PO₄ + HCl

The pH of the original buffer solution is 43.85 mM.

To calculate the pH of the original buffer solution, use the Henderson-Hasselbalch equation:

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

Where

pKa = acid dissociation constant of the weak acid (acetic acid in this case),

[A^-] = concentration of the conjugate base (acetate ion), and

[HA] = concentration of the weak acid (acetic acid).

Calculate the concentrations of acetate ion and acetic acid in the original buffer solution. Therefore:

(3509.9 mg) / (136 g/mol) = 25.81 mmol NaC₂H₂O₂·3H₂O

Since NaC₂H₂O₂·3H₂O dissociates completely into Na⁺ and C₂H₃O₂⁻ in solution:

[NaC₂H₂O₂·3H₂O] = [C₂H₃O₂⁻] = 25.81 mM

The pH of the original buffer solution is 4.80 means that:

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

4.80 = pKa + log([C₂H₃O₂⁻]/[HA])

Rearrange this equation to solve for pKa:

pKa = pH - log([C₂H₃O₂⁻]/[HA])

Substituting the given values:

pKa = 4.80 - log(25.81 mM / [HA])

Calculate [HA]:

Mass of NaC₂H₂O₂·3H₂O = Mass of C₂H₄O₂ (acetic acid) + Mass of C₂H₃O₂⁻ (acetate ion)

3509.9 mg = Mass of C₂H₄O₂ + (25.81 mM) × (136 g/mol) × (1 L / 1000 mM) × (1000 g / 1 kg)

Solving for the mass of C₂H₄O₂:

Mass of C₂H₄O₂ = 3509.9 mg - (25.81 mM) × (136 g/mol) × (1 L / 1000 mM) × (1000 g / 1 kg)

Mass of C₂H₄O₂ = 2634.4 mg

Now calculate [HA]:

[HA] = (2634.4 mg / 60.05 g/mol) × (1 L / 1000 g) × (1000 mM / 1 L)

[HA] = 43.85 mM

Substituting this into the equation for pKa:

pKa = 4.80 - log(25.81 mM / 43.85 mM)

pKa = 4.80 + 0.222

pKa = 5.02

Use the Henderson-Hasselbalch equation to calculate the pH of the original buffer solution with [A^-] = [C₂H₃O₂⁻] = 25.81 mM and [HA] = 43.85 mM

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At a pressure of 939 mbar, the corresponding height of the mercury in the column of a mercury barometer is approximately 141.9 cm.

The height of the mercury column in a barometer is proportional to the atmospheric pressure. The proportionality constant depends on the density of mercury and the acceleration due to gravity. At standard conditions, the height of the mercury column in a barometer is about 760 mm.

To find the height of the mercury column at a pressure of 939 mbar, we can use the following proportion;

pressure / height = constant

The constant is the product of the density of mercury and the acceleration due to gravity, which we can take as 13,595 kg/m^3 * 9.81 m/s^2 = 133,322 Pa/m.

Substituting the given pressure and the constant, we get;

939 mbar / height = 133,322 Pa/m

Solving for height, we get;

height = 133,322 Pa/m / 939 mbar

= 141.9 cm

Therefore, the height of the mercury in the column of a mercury barometer is 141.9 cm.

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The ΔG for the thermite reaction at 447 K is -816.894 kJ/mol.

The Gibbs free energy change (ΔG) of a reaction can be calculated using the equation:

ΔG = ΔH - TΔS

where ΔH is the enthalpy change, ΔS is the entropy change, and T is the temperature in Kelvin.

First, calculate ΔG at 298.15 K:

ΔG(298.15 K) = -807.1 kJ/mol - 298.15 K x (-0.0373 kJ/mol-K)

ΔG(298.15 K) = -807.1 kJ/mol + 11.146 kJ/mol

ΔG(298.15 K) = -795.954 kJ/mol

Using the following equation to calculate ΔG at 447 K:

ΔG(447 K) = ΔG(298.15 K) + ΔH(T2-T1)/T2 + ΔS ln(T2/T1)

where T1 = 298.15 K and T2 = 447 K

ΔG(447 K) = -795.954 kJ/mol + (-807.1 kJ/mol)(447 K - 298.15 K)/447 K + (-0.0373 kJ/mol-K)ln(447 K/298.15 K)

ΔG(447 K) = -795.954 kJ/mol - 24.902 kJ/mol + 3.962 kJ/mol

ΔG(447 K) = -816.894 kJ/mol

Therefore, the ΔG for the thermite reaction at 447 K is -816.894 kJ/mol.

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To neutralize 11.3 mL of 0.16 M HCl in stomach acid, 0.001632 grams of KOH required are 0.10145

Potassium hydroxide, also known as caustic potash, is an inorganic compound with the formula KOH. It's a strong base, dissolving easily in water, creating a strongly alkaline solution. The reaction between an acid and a base that produces water and a salt is called neutralization. A salt is a compound that is formed when the hydrogen ion in an acid is replaced by a metal ion or another positive ion.

The balanced chemical equation for the reaction is:

HCl(aq) + KOH(aq) → H₂O(l) + KCl(aq)

The equation tells us that 1 mole of HCl is neutralized by 1 mole of KOH, thus the amount of KOH required is equal to the amount of HCl. For this specific problem, we are given 11.3 mL of 0.16 M HCl, which we will need to convert to moles of HCl.

11.3 mL of 0.16 M HCl = 11.3 × 0.16/1000 = 0.001808 mol of HCl

To neutralize this much HCl, we need 0.001808 mol of KOH.

The molar mass of KOH is 56.11 g/mol.

Thus, the mass of KOH needed is:

0.001808 mol × 56.11 g/mol

= 0.10144688 g or 0.10145 g of KOH (rounded to four significant figures).

Therefore, to neutralize 11.3 mL of 0.16 M HCl in stomach acid, 0.10145 grams of KOH are required.

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There are different reasons for and against the inclusion of hydrogen in the main groups of the periodic table.

Hydrogen has properties that justify its inclusion in the main groups of the periodic table, including its chemical behavior, abundance in the universe, and the structure of its atoms.

However, its ambiguity in properties, existence in three isotopes, and lower abundance on Earth can be arguments against its classification. The unique chemistry and abundance of hydrogen in the universe support its inclusion in the periodic table.

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6 moles of oxygen gas will produce 134.4 liters of water vapor.

According to the balanced equation for cellular respiration:

[tex]C_6H_{12}O_6 + 6O_2[/tex] → [tex]6H_2O + 6CO_2[/tex]

The molar volume of any gas at standard temperature and pressure (STP) is 22.4 L.

Therefore, to calculate the volume of water vapor produced from 6 moles of oxygen gas, we need to determine the number of moles of water vapor produced and then convert that to volume:

6 moles of oxygen gas produce 6 moles of water vapor

1 mole of water vapor at STP = 22.4 L

Therefore, 6 moles of water vapor at STP = (6 moles) x (22.4 L/mole) = 134.4 L

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The given statement "A parasitic way of life can be best demonstrated by the feeding adaptations of the spider." is False. Because, While some spiders are parasites, not all spiders are parasitic.

Additionally, many spiders are not even true parasites, as they typically do not harm their host organism. Spiders are typically classified as predators, as they feed on other insects and arthropods. While some spiders may occasionally feed on the blood of larger animals, such as birds or mammals, this behavior is not typically considered parasitic, as it does not involve a long-term relationship between the spider and the host organism.

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Part 1. The combustion of methane ([tex]CH_4[/tex]) to form carbon dioxide ([tex]CO_2[/tex]) and water ([tex]H_2O[/tex]) is an exothermic reaction. Part 2. It means that it releases energy.

1. The potential energy change for a reaction pathway in this reaction can be calculated by comparing the potential energy of the reactants ([tex]CH_4[/tex] and [tex]O_2[/tex]) to the potential energy of the products ([tex]CO_2[/tex] and [tex]H_2O[/tex]). In this case, the potential energy of the reactants is higher than the potential energy of the products. 2. It means that energy is released during the reaction. The potential energy change for a reaction pathway in the combustion of methane is approximately -890 kJ/mol. This energy release is what makes combustion reactions useful for heating and energy production.

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--The complete Question is, What is the potential energy change for a reaction pathway in the combustion of methane (CH4) to form carbon dioxide (CO2) and water (H2O)--

A major purchase refers to a high-cost item or service that is considered a significant investment for an individual or household, such as a car, home, or college education. Major purchases usually involve a large sum of money and require careful planning and consideration before a final decision is made.

Any item that a person or a household buys for their own use or consumption is called a consumer good. Durable and non-durable goods are two types of consumer goods that can be classified. Consumer goods are tangible items that people or households buy for their own use or consumption.

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Your question is incomplete, most probably the complete question is:

The difference between Major purchase and Consumer good​?

When two oppositely charged particles are closer together, they have a stronger attraction toward one another than if they are farther apart.

The electrostatic force, which is the force of attraction or repulsion between two charged particles, is the cause of this attraction. The magnitude of the electrostatic force between two charged particles is inversely proportional to the square of the distance between them and directly proportional to the product of their charges.

The inverse square law states that as the distance between two charged particles moves closer together, the electrostatic force between them grows as a result of a smaller denominator. Therefore, the stronger the attraction between two oppositely charged particles will be the closer they are.

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For a chemical reaction C₂H₆ + 137 kJ → C₂H₄ + H₂, the chemical energy of the reactant and the chemical energy of the products are equal. Option D is correct.

In an exothermic reaction, energy is released as the reactants are converted to products. The negative sign on the enthalpy change (ΔH) indicates that the reaction is exothermic, meaning that energy is released by the system to the surroundings. This energy comes from the chemical energy of the reactants, and in this case, 137 kJ of energy are released.

The chemical energy of a substance is the energy stored in the bonds between its atoms. In this reaction, C₂H₆ (ethane) is converted to C₂H₄ (ethylene) and H₂ (hydrogen gas), which means that some of the bonds in the reactant molecule are broken and new bonds are formed in the product molecules.

The total amount of energy stored in the bonds of the reactants is equal to the total amount of energy stored in the bonds of the products, since energy cannot be created or destroyed, only converted from one form to another. Therefore, the chemical energy of the reactant and the chemical energy of the products are equal.

Hence, D. is the correct option.

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Considering the definition of empirical and molecular formula, if the molecular formula mass of the compound is 60.0 g/mol, the molecular formula is N₂O₂.

The empirical formula is the simplest expression to represent a chemical compound, which indicates the elements that are present and the minimum proportion in whole numbers that exist between its atoms.

The molecular formula is the chemical formula that indicates the number and type of distinct atoms present in the molecule. The molecular formula is the actual number of elements that make up a molecule.

The empirical formula and the molecular formula are generally related in the following way:

Molecular Formula = n× Empirical Formula

where n is n=molecular mass÷ empirical mass

In this case, in first place you need to convert the number of grams given into percentage as follow:

Assuming a 100 grams sample, the percentages match the grams in the sample. So you have:

Then it is possible to calculate the number of moles of each atom in the molecule, taking into account the corresponding molar mass:

The empirical formula must be expressed using whole number relationships, for this the numbers of moles are divided by the smallest result of those obtained. In this case:

Therefore the N: O mole ratio is 1: 1

Finally, the empirical formula is N₁O₁= NO

The molecular mass is 60.0 g/mol and the empirical mass is 30 g/mole. Then, the value of n is:

n=60 g/mole÷ 30 g/mole

Solving:

n=2

Being the Molecular Formula = n× Empirical Formula, the you can calculate:

Molecular Formula = 2× (NO)

Molecular formula= N₂O₂

Finally, the molecular formula is N₂O₂.

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We can not use a salt bridge saturated with potassium chloride solution in an electro-chemical cell made from a Ag/Ag+ cathode and a Cu/Cu²+ anode because it will precipitate.

An electrochemical cell that uses a weak electrolyte and a salt bridge to connect oxidation as well as reduction half cells. A junction that joins the anodic with cathodic compartments of a cell and electrolytic solution is referred to as a salt bridge.

Because both chloride and potassium ions have very similar diffusion coefficients and minimise junction potential, the inactive minerals potassium chloride (KCl) frequently used. We can not use a salt bridge saturated with potassium chloride solution in an electro-chemical cell made from a Ag/Ag+ cathode and a Cu/Cu²+ anode because it will precipitate.

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There are approximately 189.3 liters of hydrogen gas produced at STP.

To solve this problem, we need to use the ideal gas law:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

At STP, the pressure is 1 atm and the temperature is 273 K. We can use Avogadro's number to convert the number of particles to moles:

[tex]n = 5.02*10^{30}\ particles / 6.022*10^{23}\ particles/mol =[/tex] 8.34 mol of [tex]H_2[/tex]

From the balanced chemical equation, we know that 1 mole of aluminum reacts to produce 1 mole of [tex]H_2[/tex]. Therefore, 8.34 moles  [tex]H_2[/tex] were produced.

Using the ideal gas law, we can solve for the volume of [tex]H_2[/tex]:

V = nRT/P =[tex](8.34 mol)(0.08206 L.atm/(mol.K))(273 K)/(1 atm)[/tex] = 189.3 L

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The mass of HF produced from the given amounts of BF₃ and H₂ is 99.5 grams.

To solve this problem, we need to use stoichiometry to determine the amount of HF produced from the given amounts of BF₃ and H₂.

From the balanced equation;

2 BF₃ + 3 H₂ → 2 B + 6 HF

We can see that 2 moles of BF₃ react with 3 moles of H₂ to produce 6 moles of HF. This means that the ratio of BF₃ to HF is 2:6, or 1:3.

First, we need to calculate the number of moles of BF₃ and H₂ from their given masses and molar masses;

n(BF₃) = m(BF₃) / M(BF₃) = 40 g / 67.81 g/mol = 0.59 mol

n(H₂) = m(H₂) / M(H₂) = 5 g / 2.02 g/mol = 2.48 mol

Next, we can determine which reactant is limiting by comparing their mole ratios. The ratio of BF₃ to H₂ is 1:1.67, which means there is not enough H₂ to react with all of the BF₃. This makes H₂ the limiting reactant.

To find the number of moles of HF produced, we use the mole ratio from the balanced equation;

n(HF) = n(H₂) × (6 mol HF / 3 mol H₂) = 2.48 mol × 2

= 4.96 mol

Finally, we calculate the mass of HF produced from its number of moles and molar mass;

m(HF) = n(HF) × M(HF) = 4.96 mol × 20.01 g/mol

= 99.5 g

Therefore, the mass of HF is 99.5 grams.

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Total, 0.0137 moles of carbon dioxide gas is produced when 2.3 g of sodium bicarbonate decomposes to produce carbon dioxide and sodium hydroxide.

To determine the number of moles of carbon dioxide gas produced, we first need to find the balanced chemical equation for the reaction;

2NaHCO₃ → Na₂CO₃ + CO₂ + H₂O

From this equation, we can see that for every 2 moles of sodium bicarbonate, 1 mole of carbon dioxide is produced.

Next, we need to calculate the number of moles of sodium bicarbonate in 2.3 g using its molar mass;

molar mass of NaHCO₃ = 84.0066 g/mol

moles of NaHCO₃ = mass/molar mass

= 2.3 g / 84.0066 g/mol = 0.0274 mol

Finally, we can use the mole ratio from the balanced equation to find the number of moles of carbon dioxide produced;

moles of CO₂ = (0.0274 mol NaHCO₃) x (1 mol CO₂ / 2 mol NaHCO₃)

= 0.0137 mol CO₂

Therefore, 0.0137 moles of carbon dioxide gas is produced.

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