25g of NH3 is mixed with 4 moles of O2 in the given reaction
a) Which is limiting reactant?
b) What mass of NO is formed?
c) What mass of H2O is formed?

The new pressure of the gas with a solubility of 24 g/L will be222 kPa.

The solubility of the gas in the liquid is directly proportional to the partial pressure of a gas above the liquid. This relationship is described by Henry's Law, which states that the solubility of a gas is proportional to its partial pressure:

C = kP

where C is the concentration of the gas in the liquid (in units of g/L), P is the partial pressure of the gas above the liquid (in units of kPa), and k is the proportionality constant.

If we know the solubility of the gas at one pressure, we can use Henry's Law to calculate the solubility at a different pressure. For example, if the solubility of the gas is 48 g/L at 444 kPa, and we want to know the solubility at a new pressure, we can use the following formula:

P₁ / P₂ = C₁ / C₂

where P₁ is the initial pressure, P₂ is the new pressure, C₁ is the initial concentration (solubility) at P₁, and C₂ is the new concentration (solubility) at P₂.

Rearranging this formula, we get:

C₂ = C₁ x (P₂ / P₁)

Substituting the given values, we have:

C₁ = 48 g/L (solubility at 444 kPa)

P₁ = 444 kPa

C₂ = 24 g/L (desired solubility at new pressure)

P₂ = ?

Solving for P₂, we get:

P₂ = (C₂ x P₁) / C₁

P₂ = (24 g/L x 444 kPa) / 48 g/L

P₂ = 222 kPa

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Therefore, 0.00723 moles of Na3C6H5O7 can be produced if one tablet containing 0.0217 mol of NaHCO3 is dissolved.

The balanced chemical equation shows that 3 moles of NaHCO3 react with 1 mole of H2C6H6O7 to produce 1 mole of Na3C6H5O7. Therefore, the number of moles of Na3C6H5O7 produced is directly proportional to the number of moles of NaHCO3.

If one tablet containing 0.0217 mol of NaHCO3 is dissolved, then according to the balanced chemical equation, the number of moles of Na3C6H5O7 produced is:

(0.0217 mol NaHCO3) x (1 mol Na3C6H5O7 / 3 mol NaHCO3) = 0.00723 mol Na3C6H5O7

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The percentage by mass of water in the hydrate [tex]CuSO^{4}.5H^{2}O[/tex] is 47.6%.

The percentage composition of each of the following compounds are

1.a. NaClO:

The molar mass of NaClO =

22.99 g/mol + 35.45 g/mol + 15.99 g/mol = 74.43 g/mol

Percentage composition of Na =

(22.99 g/mol / 74.43 g/mol) *  100 \ percent \ = 30.9percent

Percentage composition of Cl =

(35.45 g/mol / 74.43 g/mol) *  100percent \ =  47.6percent

Percentage composition of O =

(15.99 g/mol / 74.43 g/mol) * 100percent \ =  21.5percent

Therefore, the percentage composition of NaClO is approximately:

30.9% Na, 47.6% Cl, and 21.5% O.

1.b. [tex]Al^{2}(SO^{3})^{3}[/tex]:

The molar mass of [tex]Al^{2}(SO^{3})^{3}[/tex] = [tex]2 * (26.98 g/mol) + 3 * (32.06 g/mol + 3 * 16.00 g/mol) = 342.14 g/mol[/tex]

Percentage composition of Al =

[tex](2 * 26.98 g/mol / 342.14 g/mol) * 100 \ = 3.99[/tex]%

Percentage composition of S =

[tex](3 * 32.06 g/mol / 342.14 g/mol) * 100 \ = 8.99[/tex]%

Percentage composition of O =

[tex](9 * 16.00 g/mol / 342.14 g/mol) * 100 \ = 10.47[/tex]%

Therefore, the percentage composition of [tex]Al^{2}(SO^{3})^{3}[/tex] is approximately:

3.99% Al, 8.99% S, and 10.47% O.

1.c. [tex]C^{2}H^{5}COOH[/tex]:

The molar mass of [tex]C^{2}H^{5}COOH[/tex] = [tex]2 * (12.01 g/mol) + 6 * (1.01 g/mol) + 12.01 g/mol + 16.00 g/mol = 60.05 g/mol[/tex]

Percentage composition of C =

[tex](2 * 12.01 g/mol / 60.05 g/mol) * 100 \ = 40.0[/tex]%

Percentage composition of H =

[tex](6 * 1.01 g/mol / 60.05 g/mol) * 100\ \ = 10.1[/tex]%

Percentage composition of O = [tex](2 * 16.00 g/mol + 12.01 g/mol / 60.05 g/mol) * 100 \ = 49.9[/tex]%

Therefore, the percentage composition of [tex]C^{2}H^{5}COOH[/tex] is approximately:

40.0% C, 10.1% H, and 49.9% O.

1.d. [tex]BeCl^2[/tex]:
The molar mass of [tex]BeCl^2[/tex] = [tex]9.01 g/mol + 2 * 35.45 g/mol = 79.91 g/mol[/tex]

Percentage composition of Be = [tex](9.01 g/mol / 79.91 g/mol) * 100 \ = 11.3[/tex]%

Percentage composition of Cl =

[tex](2 * 35.45 g/mol / 79.91 g/mol)* 100 \ = 88.7[/tex]%

Therefore, the percentage composition of [tex]BeCl^2[/tex] is approximately:

11.3% Be and 88.7% Cl.

2. [tex]CuSO^{4}.5H^{2}O[/tex]:

Mass of CuSO4.5H2O = 8.40 g

Mass of anhydrous [tex]CuSO^{4}[/tex] = mass of [tex]CuSO^{4}.5H^{2}O[/tex] - mass of water

Mass of water = mass of [tex]CuSO^{4}.5H^{2}O[/tex] - mass of anhydrous CuSO4

[tex]= 8.40 g - 4.40 g= 4.00 g[/tex]

So, the mass of water in the hydrate is 4.00 g.

To determine the percentage by mass of water in the hydrate, we need to divide the mass of water by the mass of the entire compound ([tex]CuSO^{4}.5H^{2}O[/tex]) and multiply by 100%:

Percentage by mass of water = (mass of water/mass of [tex]CuSO^{4}.5H^{2}O[/tex]) x 100%

[tex]= (4.00 g/8.40 g) * 100= 47.6[/tex]%

Therefore, the percentage by mass of water in the hydrate [tex]CuSO^{4}.5H^{2}O[/tex] is 47.6%.

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Enzymes speed up chemical reactions by lowering the activation energy needed for the reaction. The correct option is d.

Enzymes are biological molecules, typically proteins, that act as catalysts for chemical reactions. Catalysts are substances that can speed up a chemical reaction without being consumed in the process.

Enzymes work by lowering the activation energy needed for a chemical reaction to occur. Activation energy is the energy required to break the bonds of the reactants and initiate the chemical reaction. By lowering the activation energy, enzymes make it easier for reactants to come together and form products.

Enzymes achieve this by binding to the reactants, or substrates, and bringing them into close proximity to each other. This allows the substrates to interact more easily and lowers the energy barrier for the reaction to occur. Enzymes also stabilize the transition state of the reaction, further reducing the activation energy needed for the reaction to proceed.

Therefore, the correct answer is d) they lower the activation energy needed for the reaction.

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All the statements are about the acidity of compounds, of which statement 1, 3 & 4 are true and statement 2 is false.

Statement 1: "the strength of an acid depends primarily on the conjugate base's ability to support a negative charge" is true because, a strong acid has a weak conjugate base, which means it is less able to support a negative charge.

Statement 2: "induction is long range electronegativity, and the presence of electronegative atoms improves acidity of distant H" is false.

Because, induction is the ability of an electronegative atom or group to withdraw electron density from a neighbouring atom or group, leading to an increase in acidity.

However, this effect is not long range, and typically only affects neighbouring atoms.

Statement 3: "H bonded to larger atoms are more acidic due to poor orbital overlap, and polarizability of large orbitals" is true.

Because larger atoms have larger orbitals, which leads to poor orbital overlap and weaker bonds with hydrogen.

This makes it easier for the hydrogen to dissociate and increases the acidity of the compound.

Statement 4: "increasing the % of s-orbital character in a bond increases acidity" is true.

S-orbitals are closer to the nucleus and have a lower energy than p-orbitals, leading to stronger bonds and increased acidity.

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Factors that make the higher yield of crop possible are as follows: enough water for irrigation. access of organic manure of high quality . Fertilizers are a factor in the higher yields of high-cost farming.

Fertilizer is used to ensure good vegetative growth (leaves, branches and flowers), giving rise to healthy plants. They are artificially made but have a rapid response. Overuse of fertilisers can cause negative effects like loose soil fertility. There's a risk with any bond that the issuing company might not be able to meet its obligations.

However, the risks of default are typically higher for companies that issue high-yield bonds. Interest rate . If the octet is expanded with preferred number of bonds, formal charge = 0. The S-atom will gain two electrons in order to complete its octet.

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Answer:

0.9299 x [tex]10^-7[/tex] moles of [tex]OH^-[/tex]

Explanation:

[tex]OH^-[/tex]: 5.6 x 10^(15)  (brainly doesn't allow exponents greater than 9)

This number of particles can also be represented as 5.6e15.

'e' in this case means 'times 10 to the n' where n is the number that follows e.

Use stoichiometry to convert between different units.

[tex]\frac{5.6e15 particles}{1 sample} *\frac{1 mole}{6.022 e22 particles}= 0.9299 e -7 moles[/tex]

So in that sample of particles, there are 0.9299 x [tex]10^-7[/tex] moles of [tex]OH^-[/tex].

Answer: Its the bottom one

There are approximately 6.031 x 10^23 molecules of O2 gas contained in 1357 mL of O2 gas at −14°C and 1437 torr.

we can use the ideal gas law:

PV = nRT

where, P = pressure of the gas in torr,

V = volume of the gas in liters,

n = number of moles of gas,

R = ideal gas constant (0.08206 L·atm/mol·K),

T = temperature of the gas in Kelvin.

the volume of the gas from milliliters (mL) to liters (L):

V = 1357 mL / 1000 mL/L

V = 1.357 L

Now to convert the temperature of the gas from Celsius (°C) to Kelvin (K):

T = -14°C + 273.15

T = 259.15 K

Substituting the given values:

(1437 torr) (1.357 L) = n (0.08206 L·atm/mol·K) (259.15 K)

n = (1437 torr x 1.357 L) / (0.08206 L·atm/mol·K x 259.15 K)

n = 1.0008 mol

the number of moles of O2 gas contained in 1357 mL of O2 gas at −14°C and 1437 torr is 1.0008 mol. To find the number of molecules,

we can use Avogadro's number:

1 mol of O2 gas contains 6.022 x 10^23 molecules

So, the number of molecules in 1.0008 mol of O2 gas is:

(1.0008 mol) x (6.022 x 10^23 molecules/mol) = 6.031 x 10^23 molecules

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The grams of water needed to create a 4 molal solution with 9 moles of NaOH is 2250 grams.

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Answer:

burn

Explanation:

we burn fossil fuels, which releases carbon into the atmosphere

Answer:

polution

Explanation:

if the global of earth all polution in earth so,use electric bike

It stays much colder than traditional ice packs with the added bonus of no condesation building up on the inside of the travel box, like no water to make everything soggy &wet? However never take a huge breath of the contents as you open the box or container because inhaling trapped dry ice gas can be very harmful to your lungs! Simply let it mix with the air then carefully use gloves to move any dry ice away from your product and it should still be very frozen. We do this for long distance wedding cake delivery and I've never had one melt, 2 hours or more of driving to the events! Hope this helps!

The pH of the solution is 1.8. Option C is the right answer

First, we need to calculate the moles of LiOH and HNO3:

Moles of LiOH = 0.1 mol/L × 0.025 L = 0.0025 mol

Moles of HNO3 = 0.1 mol/L × 0.035 L = 0.0035 mol

Next, we need to determine which of the reactants will be limiting. LiOH and HNO3 react in a 1:1 stoichiometric ratio, so whichever reactant is present in the smaller amount will be limiting. In this case, LiOH is limiting because it has fewer moles than HNO3.

The balanced chemical equation for the reaction between LiOH and HNO3 is:

LiOH + HNO3 → LiNO3 + H2O

Since LiOH is limiting, all of it will react with HNO3. This means that all of the moles of HNO3 will be consumed, and there will be some excess LiOH remaining.

Moles of LiOH remaining = Moles of LiOH - Moles of HNO3 = 0.0025 mol - 0.0035 mol = -0.0010 mol

Note that the result is negative, which means that all of the HNO3 has been consumed, and there is still some LiOH remaining.

The concentration of LiOH in the final solution is:

[LiOH] = Moles of LiOH remaining / Total volume of solution = -0.0010 mol / (0.025 L + 0.035 L) = -0.010 mol/L

Again, the result is negative, which is not physically meaningful. This indicates that our assumption that LiOH is limiting is incorrect.

To correct this, we can assume that HNO3 is limiting and repeat the calculations. In this case, all of the HNO3 will react with LiOH, and there will be some excess HNO3 remaining.

Moles of HNO3 remaining = Moles of HNO3 - Moles of LiOH = 0.0035 mol - 0.0025 mol = 0.0010 mol

The concentration of HNO3 in the final solution is:

[HNO3] = Moles of HNO3 remaining / Total volume of solution = 0.0010 mol / (0.025 L + 0.035 L) = 0.01 mol/L

The balanced chemical equation for the reaction between HNO3 and LiOH is:

HNO3 + LiOH → LiNO3 + H2O

The stoichiometry of this reaction tells us that one mole of HNO3 reacts with one mole of LiOH to produce one mole of water. Therefore, the moles of water produced in the reaction are equal to the moles of HNO3 remaining:

Moles of water = 0.0010 mol

The volume of the final solution is:

Volume = 0.025 L + 0.035 L = 0.06 L

The molarity of the water in the final solution is:

Molarity of water = Moles of water / Volume of solution = 0.0010 mol / 0.06 L = 0.0167 mol/L

The pH of water is 7.0, and the pH of a solution is related to the concentration of H+ ions in the solution. Since water is neutral and does not contribute any H+ ions, the pH of the final solution is:

pH = 7.0 - log[H+] = 7.0 - log(0.0167) ≈ 1.8

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Determine the pH of the solution when 25ml of 0.1mLiOH is added to 35ml of 0.1m nitric acid

(A) 0.23

(B) 0.47

(C) 1.8

(D) 7.0

T = 23 C + 273 = 296 K V(H2) = (1.7 mol*62360 mmHg/mol*K*296 /850 mmHg) = 36917.1 ml = 36.9 L if volume, per liters, of H2 gas is created at 26 C and 800 mmHg.

For instance, hydrogen is not harmful. Also, as hydrogen is that much lighter than the atmosphere, it evaporates quickly when it is discharged, enabling the fuel to disperse relatively quickly in the event of a leak. To ensure its safe use, some of hydrogen's features necessitate additional technical restrictions.

When two atoms of hydrogen join forces to form a hydrogen molecule, the result is a gas known as H2. Molecular hydrogen is another name for H2. There are two protons & two electrons in it. Hence, it has a neutral charge and is stable, making it the most prevalent type of hydrogen.

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The red-colored blocks indicate non metals in the periodic table. Nonmetals are located on the right side of the periodic table, in the p-block.

Non-metals are composed of a variety of elements, such as carbon, nitrogen, oxygen, sulfur, phosphorus, and selenium. These elements are generally non-reactive, but some of them can form compounds with other elements. They are also used in various industries, such as the production of plastics, fertilizers, and explosives. Non-metals are divided into two categories: metalloids and non-metalloids. Nonmetals are characterized by their lack of luster and their low thermal conductivity.

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The pH of the 0.30 M solution of sodium formate is 2.13.
The correct answer is E. 2.13.

Formic acid dissociates in water to form hydrogen formate (HCOO-) and a proton (H+).

Since the Ka of HCOOH is [tex]1.8 * 10^{-4}[/tex], the Kb of HCOO- is equal to Ka x Kb = ([tex]1.8 * 10^{-4}[/tex])([tex]1.0 * 10^{-14}[/tex]) = [tex]1.8 * 10^{-18}[/tex].

The ionization of sodium formate (NaHCOO) can be represented as:

[tex]NaHCOO + H^2O < = > HCOO^- + Na^+ +[/tex] [tex]H^3O^+[/tex]

We can use the Kb of HCOO- to calculate the equilibrium concentrations of the species in the solution.

[HCOO-] = [NaHCOO] = 0.30 M

[Na+] = [[tex]H^3O^+[/tex]] = 0

Kb = [HCOO-][[tex]H^3O^+[/tex]]/[NaHCOO]

1.8 x 10-18 = (0.30)(x)/(0.30)

x = 1.8 x 10-18

The pH of the solution can then be calculated using the equation:

pH = -log[[tex]H^3O^+[/tex]]

pH = -log(1.8 x 10-18)

pH = 2.13

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Question should be like

Calculate the pH of a 0. 30 M solution of sodium formate (NaHCOO) given that the Ka of formic acid (HCOOH) is 1. 8 x 10^-4.

A. 8. 61

B. 10. 26

C. 11. 87

D. 5. 39

E. 2. 13

The synthesis of carboxylic acid can be done by hydrolysing the nitrile which is substituted in place of halide by using cyanide ion. The structure that contains a nitrile is OCH2C≡N.

A nitrile is an organic compound that contains a carbon triple bonded to a nitrogen atom.

The general formula for a nitrile is R–C≡N, where R is an alkyl or aryl group. In the structure OCH2C≡N, the OCH2 group is the alkyl group and the C≡N is the nitrile functional group. Therefore, this structure contains a nitrile.

To synthesize a carboxylic acid from an alkyl halide, you can first convert the alkyl halide to a nitrile through a substitution reaction with a cyanide ion. The nitrile can then be hydrolyzed to form a carboxylic acid. The overall reaction is:

R–X + CN– → R–C≡N → R–COOH

Where R is an alkyl or aryl group, X is a halogen, and CN– is a cyanide ion.

In conclusion, the structure that contains a nitrile is OCH2C≡N and it can be used to synthesize a carboxylic acid from an alkyl halide through a substitution reaction followed by hydrolysis.

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The balanced equation for the reaction would be 2AlPO4 + 3CuCrO4 → Al2(CrO4)3 + 3Cu3(PO4)2.

The formulas of the reactants are:

The formulas of the products are:

The balanced equation indicates that 2 moles of Aluminum Phosphite react with 3 moles of Copper (II) Chromate to produce 1 mole of Aluminum Chromate and 3 moles of Copper (II) Phosphate.

Thus: 2AlPO4 + 3CuCrO4 → Al2(CrO4)3 + 3Cu3(PO4)2

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the area of one of the cat's paws is 4 square centimeters.

The weight of the cat is 4.0 kilograms, which is the force exerted on the scale by the cat's body. The weight per unit area on the scale due to the cat's paws is 0.25 kilograms per square centimeter.

Let A be the area of one of the cat's paws in square centimeters. The weight of the cat is supported by all four paws, so the total force exerted by the paws on the scale is:

force = weight = 4.0 kg

The force exerted by one paw is one-fourth of the total force:

force per paw = force / 4 = 4.0 kg / 4 = 1.0 kg

This force is distributed over the area of one paw, so we can write:

pressure = force per paw / A

where pressure is the weight per unit area, which is given as 0.25 kg/cm

Substituting the values and solving for A, we get:

0.25 kg/cm = 1.0 kg / A

A = 1.0 kg / 0.25 kg/cm = 4 cm

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The process that could result in the net movement of a substance into a cell, if the substance is more concentrated in the cell than in the surroundings is called osmosis.

Osmosis is a passive transport process where water molecules move from an area of high concentration to an area of low concentration through a selectively permeable membrane. In this process, a solute moves from a region of high concentration to a region of low concentration. A selectively permeable membrane is a type of biological membrane that allows certain molecules or ions to pass through it by diffusion and occasionally specialized "facilitated diffusion" and passive or active transport processes, while preventing other molecules or ions from passing through it. Therefore, the correct answer is the option C - Osmosis.

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Answer:

We can begin by using the balanced chemical equation for the reaction between carbon monoxide and oxygen:

2CO + O2 -> 2CO2

This tells us that 2 volumes of CO react with 1 volume of O2 to produce 2 volumes of CO2. Since we have equal volumes of CO and O2, the limiting reactant will be the one that requires more volume, which is the CO:

2 volumes of CO + 1 volume of O2 -> 2 volumes of CO2

Using the ideal gas law, we can convert the volumes of CO and O2 at STP (standard temperature and pressure, which is 0°C and 1 atm) to moles:

n(CO) = V(CO) / Vm = 60 cm3 / 22.4 L/mol = 0.00268 mol

n(O2) = V(O2) / Vm = 30 cm3 / 22.4 L/mol = 0.00134 mol

Since 2 volumes of CO react with 1 volume of O2, we can say that the reaction will use up 2 x 0.00134 = 0.00268 mol of O2. Since we have 0.00134 mol of O2 initially, this means that all of the O2 will be used up in the reaction, leaving none in the residual gases. The reaction will also produce 2 x 0.00268 = 0.00536 mol of CO2.

Using the ideal gas law again, we can convert the volume of CO2 produced to a volume at STP:

V(CO2) = n(CO2) x Vm = 0.00536 mol x 22.4 L/mol = 0.120 cm3

Therefore, the volume of residual gases after the reaction is:

V(residual) = V(total) - V(CO) - V(O2) - V(CO2) = 90 cm3 - 60 cm3 - 30 cm3 - 0.120 cm3 = 0.880 cm3

The various types of chemical reaction are Combination Reaction,Decomposition Reaction,Displacement Reaction,Double Displacement Reaction, and Precipitation Reaction.

A chemical reaction is defined as the reaction that involves the combination of two or more substances leading to the formation of a new substance called the product of the reaction.

The various types of chemical reaction include the following:

There are three parts of a chemical reaction which is the reactant part, the arrow and the product part.

That is;

A. + B ---------> D + E.

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Answer:

10. methanol would be more.

11. methanol would be more.

12. methanol would remain the same.

13. CO would be less.

Explanation:

10. Methanol increases because the forward reaction is favoured with increase in temperature and thus it is endothermic.

11. Methanol increases because the forward reaction is favoured with decrease in volume thus increasing pressure decreases the volume of the vessel.

12.Methanol remains the same because the hydrogen added is its raw material but the CO is acting as a limiting reactant for the formation of methanol.

13. Carbon monoxide becomes less because it will rapidly combine with hydrogen added to form methanol and then later be a limiting reactant when hydrogen is still present

Answer:

33.67g

Explanation:

To calculate the mass of carbon monoxide, we can use the ideal gas law, which relates the pressure, volume, number of moles, and temperature of a gas:

PV = nRT

Where:

P = pressure

V = volume

n = number of moles

R = ideal gas constant

T = temperature

At STP (standard temperature and pressure), the pressure is 1 atm and the temperature is 273.15 K. The ideal gas constant is R = 0.08206 L atm/(mol K).

To find the number of moles of carbon monoxide in 29.3 L at STP, we can rearrange the ideal gas law to solve for n:

n = PV/(RT)

n = (1 atm)(29.3 L)/(0.08206 L atm/(mol K) * 273.15 K)

n = 1.203 mol

The molar mass of carbon monoxide (CO) is 28.01 g/mol. Therefore, the mass of 1.203 mol of CO is:

mass = n * molar mass

mass = 1.203 mol * 28.01 g/mol

mass = 33.67 g

Therefore, the mass of 29.3 L of carbon monoxide at STP is approximately 33.67 grams.

Answer:

6 moles of H2O

Explanation:

We need the balanced equation for this reaction.

   C2H2 + O2 = CO2 + H2O

   2C2H2 + 5O2 = 2CO2 + 4H2O   [Balanced reaction]

The balanced reaction tells us that we should expect to obtain 4 moles of water, H2), for every 2 moles of C2H2 (ethylene).

That is a ratio:  (4 moles H2O)/(2 moles C2H2) or (2/1)(moles H2O/moles C2H4)

If we start with 3 moles of C2H4:

 (3 moles C2H4)*((2 moles H2O)/(1 moles C2H2)) = (2*3/1) moles H2O [the moles C2H2 cancel]

(2*3/1) moles H2O = 6 moles of H2O

If 9.7 mol of CO₂ are produced, 14.55 moles of O₂ were reacted.

In chemistry, a mole is a unit of measurement that represents an amount of substance. This number of entities is known as Avogadro's number, which is approximately equal to 6.02 x 10^23.

The mole is a convenient unit for measuring the amount of substance in chemical reactions, because it allows chemists to predict and calculate the quantities of reactants and products involved in a chemical reaction. For example, if we know the balanced chemical equation for a reaction, we can use the stoichiometry of the reaction to determine the amount of reactants needed to produce a given amount of product, or the amount of product that can be produced from a given amount of reactants.

The balanced chemical equation for the combustion of ethylene (C2H4) in oxygen is:

C₂H₄ + 3O₂ → 2CO₂ + 2H₂O

From the balanced equation, we can see that 3 moles of O₂ react with 1 mole of C₂H₄ to produce 2 moles of CO₂. Therefore, the number of moles of O₂ that reacted can be calculated as:

moles of O₂ = (moles of CO₂ produced) × (3/2)

moles of O₂ = 9.7 mol × (3/2)

moles of O₂ = 14.55 mol

Therefore, 14.55 moles of O₂ were reacted.

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This means that for every liter of water, there is 6.15 moles of the substance dissolved in it.

Substance is a concept that refers to a physical material or thing that has mass and occupies space. It is a fundamental concept of physics that applies to all physical and visible things in the universe. In philosophy, substance is a primary category of ontology that refers to the physical or material existence of things.

The molarity of the saturated solution can be calculated by dividing the solubility (360 g/L) by the molar mass of the substance (58.5 g/mol).
The molarity of the saturated solution is thus 6.15 mol/L.
This means that for every liter of water, there is 6.15 moles of the substance dissolved in it.

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