If 8.25
mol of C5H12
reacts with excess O2,
how many moles of CO2
will be produced by the following combustion reaction?

C5H12+8O2⟶6H2O+5CO2

Answer:

Explanation:

a) The balanced equation for the double replacement reaction between lead (II) nitrate and potassium bromide is:

Pb(NO₃)₂(aq) + 2KBr(aq) → PbBr₂(s) + 2KNO₃(aq)

In this reaction, lead (II) bromide (PbBr₂) will precipitate, while potassium nitrate (KNO₃) will remain in solution.

b) To determine the amount of precipitate produced, we need to first determine the limiting reactant. We can do this by calculating the number of moles of each reactant and comparing it to the stoichiometry of the balanced equation.

The molar mass of lead (II) nitrate is 331.21 g/mol and the molar mass of potassium bromide is 119.00 g/mol.

The number of moles of lead (II) nitrate is 32.5 g / 331.21 g/mol = 0.0981 mol The number of moles of potassium bromide is 38.75 g / 119.00 g/mol = 0.3256 mol

According to the balanced equation, one mole of lead (II) nitrate reacts with two moles of potassium bromide to produce one mole of lead (II) bromide. This means that if all the lead (II) nitrate were to react, it would require 0.0981 mol * 2 = 0.1962 mol of potassium bromide.

Since we have more than enough potassium bromide (0.3256 mol > 0.1962 mol), lead (II) nitrate is the limiting reactant.

The number of moles of lead (II) bromide produced will be equal to the number of moles of lead (II) nitrate consumed, which is 0.0981 mol.

The molar mass of lead (II) bromide is 367.01 g/mol, so the mass of lead (II) bromide produced will be 0.0981 mol * 367.01 g/mol = 36.0 g.

c) To determine the amount of excess reactant remaining, we need to subtract the amount consumed from the initial amount.

The number of moles of potassium bromide consumed is half the number of moles of lead (II) nitrate consumed, which is 0.0981 mol / 2 = 0.04905 mol.

The mass of potassium bromide consumed is 0.04905 mol * 119.00 g/mol = 5.84 g.

The mass of potassium bromide remaining is 38.75 g - 5.84 g = 32.91 g.

The volume of the balloon once they have cooled to the outside temperature of -34.0 c is 1.53 L.

Charles' law predicts the relationship between the volumes and the temperatures of a sample of an ideal gas at different conditions. For this equation to hold true, the number of molecules and the pressure must remain constant despite changes in the environment.

Determine the volume of the balloon outside, V2. We do this by applying Charles' law, such that we relate the volume, V, and the temperature, T, of a sample of gas as

V₁ /T₁ = V₂/ T₂

at two conditions. We are given the following values for the variables:

• V₁ = 1.90 L

T₁ = 22.0+ 273.15 = 295.15 K

T₂= 34.0+273.15= 239.15 K

We proceed with the solution.

V₁/T₁ = V₂/T₂

V₁ /T₁ × T₂ = V₂

1.90 L/295.15 K x 239.15 K = V₂

1.53 L =V₂

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The complete question is

Balloons for a New Year's Eve party in Fargo, ND, are filled to a volume of 1.90 L at a temperature of 22.0 degrees Celsius and then hung outside where the temperature is -34.0 degrees Celsius. What is the volume of the balloons after they have cooled to the outside temperature? Assume that atmospheric pressure inside and outside the house is the same.

The ΔH°vap of water at the critical point is approximately 0.04614 kJ/mol.

To calculate the ΔH°vap (enthalpy of vaporization) of water at the critical point, we can use the Clausius-Clapeyron equation;

ln(P₂/P₁) = ΔH°vap/R [1/T₁ - 1/T₂]

where P₁ and T₁ are the pressure and temperature at which the enthalpy of vaporization is known (usually at standard conditions of 1 atm and 100 °C), P₂ and T₂ are the pressure and temperature at the critical point, ΔH°vap is the enthalpy of vaporization, and R is the gas constant (8.314 J/mol∙K).

Using the given values, we can plug them into the equation and solve for ΔH°vap;

ln(3.2 atm / 1 atm) = ΔH°vap / R [1/373 K - 1/275 K]

Simplifying;

ln(3.2) = ΔH°vap / R [0.0026819]

ΔH°vap / R = ln(3.2) / 0.0026819

ΔH°vap / R = 5.552

Multiplying both sides by R:

ΔH°vap = 5.552 x R

ΔH°vap = 5.552 x 8.314 J/mol∙K

ΔH°vap = 46.14 J/mol

Converting to kJ/mol;

ΔH°vap = 0.04614 kJ/mol

Therefore, the ΔH°vap of water is 0.04614 kJ/mol.

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The centripetal acceleration experienced by the object can be calculated using the formula a = v^2/r, where v is the speed of the object and r is the radius of the circle. Substituting the given values, we get:

a = (50 cm/s)^2 / (250 cm)
a = 10 cm/s^2

Therefore, the centripetal acceleration experienced by the object is 10 cm/s^2.
To calculate the centripetal acceleration experienced by the object, you can use the formula:
Centripetal acceleration (a_c) = (velocity^2) / radius
Here, the velocity (v) is 50 cm/s and the radius (r) is 250 cm. Plugging in these values, we get:
a_c = (50^2) / 250 = 2500 / 250 = 10 cm/s²
So, the centripetal acceleration experienced by the object is 10 cm/s².

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20 g of calcium would release the same number of valence electrons as 23 g of sodium.

The atomic number of calcium (Ca) is 20, which means it has 20 electrons in its neutral state. When calcium loses two electrons, it becomes a Ca2+ ion with 18 electrons.

On the other hand, the atomic number of sodium (Na) is 11, which means it has 11 electrons in its neutral state. When sodium loses one electron, it becomes a Na+ ion with 10 electrons.

To release the same number of valence electrons as 23 g of Na, we need to calculate how many moles of Na there are in 23 g:

Molar mass of Na = 23 g/mol

Number of moles of Na = 23 g / 23 g/mol = 1 mol

Since each Na+ ion has lost one electron, 1 mol of Na+ ions has lost 1 mol of valence electrons.

To release the same number of valence electrons as 1 mol of Na+ ions, we need to calculate how many moles of Ca2+ ions are required:

1 mol of Na+ ions = 1 mol of valence electrons

1 mol of Ca2+ ions = 2 mol of valence electrons

Therefore, we need 0.5 mol of Ca2+ ions to release the same number of valence electrons as 1 mol of Na+ ions.

Finally, we can calculate the mass of calcium that would release the same number of valence electrons as 23 g of Na:

Molar mass of Ca = 40 g/mol

Mass of Ca required = 0.5 mol x 40 g/mol = 20 g

Therefore, 20 g of calcium would release the same number of valence electrons as 23 g of sodium.

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The molecular formula of the strychnine, given that it is composed of 75.45% carbon, 6.59% hydrogen, 8.38% nitrogen, and the balance oxygen is C₂₁H₂₂N₂O₂

First, we shall obtain the empirical formula of compound. Details below:

Divide by their molar mass

C = 75.45 / 12 = 6.2875

H = 6.59 / 1 = 6.59

N = 8.38 / 14 = 0.5986

O = 9.58 / 16 = 0.59875

Divide by the smallest

C = 6.2875 / 0.5986 = 10.5

H = 6.59 / 0.5986 = 11

N = 0.5986 / 0.5986 = 1

O = 0.59875 / 0.5986 = 1

Multiply through by 2 to express in whole number

C = 10.5 × 2 = 21

H = 11 × 2 = 22

N = 1 × 2 = 2

O = 1 × 2 = 2

Thus, we can conclude that the empirical formula is C₂₁H₂₂N₂O₂

Now, we shall determine the molecular formula of strychnine. Details below

Molecular formula = empirical × n = mass number

[C₂₁H₂₂N₂O₂]n = 140.22

[(12×21) + (1×22) + (14×2) + (16×2)]n = 334

334n = 334

Divide both sides by 334

n = 334 / 334

n = 1

Molecular formula = [C₂₁H₂₂N₂O₂]n

Molecular formula = [C₂₁H₂₂N₂O₂]1

Molecular formula = C₂₁H₂₂N₂O₂

Thus, we can conclude that the molecular formula of strychnine is C₂₁H₂₂N₂O₂

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The mass of hydrogen produced in the reaction is 2g.

The mole is an amount unit similar to familiar units like pair, dozen, gross, etc. It provides a specific measure of the number of atoms or molecules in a bulk sample of matter.

A mole is defined as the amount of substance containing the same number of atoms, molecules, ions, etc. as the number of atoms in a sample of pure 12C weighing exactly 12 g.

Given,

Mass of Zn = 65g

Mass of HCl = 72g

Moles of Zn = mass / molar mass

= 65 / 65 = 1 mole

Moles of HCl = 72 / 36.5

= 1.97 moles

Since moles of Zn is lesser, therefore it is the limiting reagent.

From the reaction, 1 mole of Zn gives 1 mole of hydrogen

Moles of hydrogen = 1 mole

mass of hydrogen = moles × molar mass

= 1 × 2 = 2g

Therefore, the mass of hydrogen produced in the reaction is 2g.

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When 28.2 grams of Mg is burned, 46.7 grams of MgO will be formed.

The balanced chemical equation for the combustion of magnesium is:

2 Mg + O2 --> 2 MgO

This equation shows that 2 moles of Mg react with 1 mole of O2 to produce 2 moles of MgO.

To determine the amount of MgO formed when 28.2 grams of Mg is burned, we first need to convert the given mass of Mg to moles:

molar mass of Mg = 24.31 g/mol

moles of Mg = mass of Mg / molar mass of Mg

moles of Mg = 28.2 g / 24.31 g/mol

moles of Mg = 1.16 mol

According to the balanced chemical equation, 2 moles of Mg produce 2 moles of MgO. Therefore, we can use the mole ratio to calculate the moles of MgO formed:

moles of MgO = moles of Mg x (2 moles of MgO / 2 moles of Mg)

moles of MgO = 1.16 mol x 1

moles of MgO = 1.16 mol

Finally, we can convert the moles of MgO to grams using its molar mass:

molar mass of MgO = 40.31 g/mol

mass of MgO = moles of MgO x molar mass of MgO

mass of MgO = 1.16 mol x 40.31 g/mol

mass of MgO = 46.7 g

Therefore, when 28.2 grams of Mg is burned, 46.7 grams of MgO will be formed.

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Answer:

Unsaturated

Explanation:

A solution is unsaturated when it contains less than the maximum amount of solute that is capable of being dissolved.

Answer: Mixtures can be physically separated by using methods that use differences in physical properties to separate the components of the mixture, such as

Explanation:

The concentration of the solution is  0.000435 mol/dm³.

The concentration of a solution is calculated as follows;

Concentration = (Absorbance) / (Molar absorptivity x path length)

the path length =  3cm

the molar absorptivity (E) = 400 dm/mol/cm.

if the solution transmits 30% of the light, it absorbs 70% of the incident light.

Absorbance = log (1/Transmittance)

Absorbance  = log (1/0.3)

Absorbance  = 0.523

Concentration = (0.523) / (400 dm/mol/cm x 3 cm)

= 0.000435 mol/dm³

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1. The volume of 0.350 M NH₄I solution is required is 286 mL

2. The mole of PbI₂ formed is 0.0498

First, we shall determine the mole in 415 mL of 0.120 M Pb(NO₃)₂. Details below:

Mole = molarity × volume

Mole of Pb(NO₃)₂ = 0.120 × 0.415

Mole of Pb(NO₃)₂ = 0.0498 mole

Next, we shall determine the mole of NH₄I that reacted. Details below:

Pb(NO₃)₂(aq) + 2NH₄I(aq) ⟶ PbI₂(s) + 2NH₄NO₃(aq)

From the balanced equation above,

1 moles of Pb(NO₃)₂ reacted with 2 moles of NH₄I

Therefore,

0.0498 mole of Pb(NO₃)₂ will react with = 0.0498 × 2 = 0.1 mole of NH₄I

Finally, we shall determine the volume of NH₄I required for the reaction. Details below:

Volume = mole / molarity

Volume of NH₄I = 0.1 / 0.350

Volume of NH₄I = 0.286 L

Multiply by 1000 to express in mL

Volume of NH₄I = 0.286  × 1000

Volume of NH₄I = 286 mL

The mole of PbI₂ formed can be obtain as follow:

Pb(NO₃)₂(aq) + 2NH₄I(aq) ⟶ PbI₂(s) + 2NH₄NO₃(aq)

From the balanced equation above,

1 mole of Pb(NO₃)₂ reacted to produced 1 mole of PbI₂

Therefore,

0.0498 mole of Pb(NO₃)₂ will also react to produce 0.0498 mole of PbI₂

Thus, the number of mole of PbI₂ formed is 0.0498 mole

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1. To solve for the grams of H2 needed, we need to use the given ΔH value to calculate the amount of moles of N2 that reacted. From the balanced chemical equation, we know that for every 3 moles of H2 that reacts, 1 mole of N2 reacts. Therefore, we can use the mole ratio to convert the moles of N2 to moles of H2 and then use the molar mass of H2 to convert to grams.

First, we need to calculate the moles of N2 that reacted to produce 150.9kJ of heat:
ΔH = -92.40 kJ/mol N2
150.9 kJ = (1 mol N2 / -92.40 kJ) x (-150.9 kJ)
mol N2 = 1.63 mol

Using the mole ratio from the balanced chemical equation:

1 mol N2 : 3 mol H2

We can calculate the moles of H2 needed:3 mol H2 = 1 mol N2

3 mol H2 = 1.63 mol N2
mol H2 = 0.543 mol

Finally, we can convert moles of H2 to grams:

mol H2 = 0.543 mol
molar mass of H2 = 2.02 g/mol
grams of H2 = (0.543 mol) x (2.02 g/mol)
grams of H2 = 1.10 g

Therefore, 1.10 grams of H2 are needed to involve 150.9kJ of heat.

2. To solve for the moles of NH3 produced, we can use the same mole ratio from the balanced chemical equation:

1 mol N2 : 2 mol NH3

From the moles of N2 that reacted calculated in part 1, we can calculate the moles of NH3 produced:
1 mol N2 = 2 mol NH3
1 mol N2 = 1.63 mol N2
mol NH3 = (2 mol NH3 / 1 mol N2) x (1.63 mol N2)
mol NH3 = 3.26 mol

Therefore, 3.26 moles of NH3 were produced in the process.

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1. To find the volume and units of a solution that contains 0.50 moles of NaOH dissolved in enough distilled water to make 3.0 M NaOH solution:

We first need to use the formula:

moles = concentration (in moles/L) x volume (in L)

Rearranging the formula to solve for volume, we get:

volume = moles / concentration

Substituting the given values, we get:

volume = 0.50 moles / 3.0 M = 0.17 L

Since the volume is given in liters, the units of the solution are L. Therefore, the solution contains 0.50 moles of NaOH dissolved in 0.17 L of distilled water, which makes a 3.0 M NaOH solution.

2. To find the molarity of a solution that contains 60.0 g of Ca(OH)2 dissolved in 150 mL of solution:

We first need to convert the mass of Ca(OH)2 to moles using the molar mass:

molar mass of Ca(OH)2 = 40.08 g/mol + 2 x 16.00 g/mol + 2 x 1.01 g/mol = 74.10 g/mol

moles of Ca(OH)2 = 60.0 g / 74.10 g/mol = 0.810 moles

Next, we need to convert the volume of the solution from milliliters to liters:

volume of solution = 150 mL / 1000 mL/L = 0.150 L

Finally, we can use the formula:

molarity = moles / volume

Substituting the given values, we get:

molarity = 0.810 moles / 0.150 L = 5.4 M

Therefore, the molarity of the solution is 5.4 M.

A pair of molecules which exist in two forms that are mirror images of each other but cannot be superimposed one upon the other are called the enantiomers. They are present in pairs and have similar molecular shape.

The compounds with the same molecular formula but are non-superimposable non-mirror images are called diastereomers. They have distinct physical properties and molecular shape.

The constitutional isomers have the same molecular formula but have different bonding atomic organization and bonding patterns.

So here:

1st structure is constitutional isomers (c), 2nd structures are enantiomers (b) and the 3rd are completely different not isomers (d).

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The concentration of the ammonia from the calculation is  [tex]1.1 * 10^-15[/tex]M.

The ratio of the concentrations of products to reactants in chemical equilibrium, for a given chemical reaction at a given temperature, is described by the equilibrium constant, abbreviated as Kc or Keq.

We can see that;

[tex]Keq = [NH_{3} ]^2/[N_{2} ] [ H_{2}]^3\\Keq[N_{2} ] [ H_{2}]^3 = [NH_{3}]^2\\\\N_{2} ] = [NH_{3}]^2/Keq[ H_{2}]^3[/tex]

=[tex](0.000123 )^2/ 0.00659 * (0.0275)^3= 1.1 * 10^-15 M[/tex]

Thus we would have the nitrogen concentration as [tex]1.1 * 10^-15[/tex] M

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The final temperature of the copper and water after they come to thermal equilibrium is 109.8°C.

Temperature is a measure of the amount of thermal energy present in a substance or object. It is measured in degrees on a scale such as Fahrenheit, Celsius, or Kelvin. Temperature is important in determining the physical and chemical properties of a substance, such as its melting point, boiling point, and specific gravity. Temperature also affects the rate of a chemical reaction and the speed of diffusion.

The change in temperature of the copper can be calculated using the equation
ΔT = (Q/mc), where Q is the heat transferred, m is the mass of the copper, and c is the specific heat of copper.
Q = mcΔT = (192 g)(0.385 J/g °C)(100°C) = 74080 J
The heat transferred from the copper must equal the heat transferred to the water. Therefore,
(74080 J) = (0.85 L)(4.184 J/g°C)(ΔT)
ΔT = (74080 J)/[(0.85 L)(4.184 J/g°C)] = 109.8°C
The final temperature of the copper and water after they come to thermal equilibrium is 109.8°C.

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The molarity of the product is  0.00368 M. Option B

When the rates of the forward and reverse reactions in a chemical reaction are equal and the concentrations of reactants and products are stable over time, this condition is referred to as reaction equilibrium.

The equilibrium constant is a measure of the relative concentrations of reactants and products at equilibrium.

[tex]Keq = [H_{2} O] [CO]/[H_{2}] [CO_{2} ]\\0.106 = x^2/(0.0113)^2\\x = \sqrt{} 0.106 (0.0113)^2\\x = 0.00368 M[/tex]

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Using the Ideal Gas Law, the pressure expected for 30 mol of  [tex]C_2H_4[/tex] gas in a 6.00-L container at 20 °C is 1210.07 atm, and using the van der Waals equation, the pressure is 1179.71 atm.

To calculate the pressure expected for 30 mol of [tex]C_2H_4[/tex] gas in a 6.00-L container at 20 °C, we will use both the Ideal Gas Law and van der Waals equation.

Ideal Gas Law: PV = nRT
P = pressure
V = volume (6.00 L)
n = moles (30 mol)
R = ideal gas constant (0.0821 L•atm/mol•K)
T = temperature (20 °C + 273.15 = 293.15 K)

Solve for P (pressure):

P = nRT / V

P = (30 mol)(0.0821 L•atm/mol•K)(293.15 K) / 6.00 L

P = 1210.07 atm

Van der Waals equation:

(P + a(n/V)²)(V - nb) = nRT
a = 4.612 atm•L²/mol²
b = 0.0582 L/mol

Solve for P (pressure):
(P + (4.612)(30/6)²) (6 - 0.0582 * 30) = (30)(0.0821)(293.15)

P = 1179.71 atm

Using the Ideal Gas Law, the pressure is 1210.07 atm, and using the van der Waals equation, the pressure is 1179.71 atm.

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The partial pressure of the oxygen is 0.236 atm.

The pressure that one gas component in a mixture of gases exerts is known as partial pressure. It is the pressure that the gas would experience if it took up the same amount of space in the mixture at the same temperature on its own.

We know that;

P[tex]CO_{2}[/tex] = 0.285 torr or 0.000375 atm

P[tex]N_{2}[/tex] = 580.502 torr or 0.764 atm

P[tex]O_{2}[/tex] = ?

Total pressure = 1 atm

Then we have that;

PT =P[tex]CO_{2}[/tex]  +P[tex]N_{2}[/tex]+ P[tex]O_{2}[/tex]

P[tex]O_{2}[/tex] = PT - (P[tex]CO_{2}[/tex]  + P[tex]N_{2}[/tex])

P[tex]O_{2}[/tex] = 1 - (0.000375  + 0.764)

P[tex]O_{2}[/tex]= 0.236 atm

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Step-by-step Explanation:

8.51 moles is to 5.12 L  as  'x'  moles is to (13.3-5.12) L

8.51 moles / 5.12 L    =   x / ( 13.3-5.12)

x = 13.6 moles

To determine the change in enthalpy associated with the generation of 155.3 g of [tex]C_5H_1_0O[/tex] we must first calculate the moles of [tex]C_5H_1_0O[/tex] produced using its molar mass.

The molar mass of C₅H₁₀O is:

5(12.01 g/mol) + 10(1.01 g/mol) + 16.00 g/mol = 88.15 g/mol

Moles of C₅H₁₀O produced:

155.3 g / 88.15 g/mol = 1.763 mol C₅H₁₀O

The balanced chemical equation states that the formation of 1 mol of C₅H₁₀O results in an enthalpy change of -191.3 kJ.

As a result, the enthalpy change during the formation of 1.763 mol of C₅H₁₀O is: -191.3 kJ/mol x 1.763 mol = -337.8 kJ

The enthalpy change for the production of 155.3 g of C5H10O is -337.8 kJ.

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To determine the enthalpy of reaction for the given reaction, use Hess's Law, which states that the total enthalpy change of a reaction is independent of the pathway between the initial and final states.

Break the given reaction into a series of steps for which the enthalpy changes are known or can be measured experimentally.

Add the enthalpy changes of each step to determine the overall enthalpy change for the reaction.

For example, determine the enthalpy of formation for A₂X₄(l), X₂(g), and AX₃(g) and use them to calculate the enthalpy of reaction for the given equation.

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Answer:

All is Correct

Explanation:

Fossil fuels have the following properties:

Therefore, the answer is to select all

Answer:

It's A, B, and D

Explanation:

Maybe not D, because that is burning wood like charcoal. Not sure about that. Hope this helps!

Since the ΔH for the given reaction has negative value, the reaction is exothermic reaction.

A reaction that is exothermic is one in which power is given off as heat or light. In contrast to an endothermic process, which draws energy from its surroundings, an exothermic reaction transfers energy into the environment. The alteration in enthalpy (H) during an exothermic reaction will be negative. Since the ΔH for the given reaction has negative value, the reaction is exothermic reaction.

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The rate constant (k) for a chemical reaction can be determined experimentally by measuring the reaction rate (v) at different concentrations of reactants and plotting the data using a suitable rate law equation.

There are different methods to determine the rate constant depending on the type of reaction, but a general approach is as follows:

Conduct the reaction under different initial concentrations of reactants while keeping other variables constant such as temperature, pressure, and pH.

Measure the reaction rate at each concentration by monitoring the change in concentration of reactants or products over time.

An overview was given based on incomplete information.

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The ratio of the concentrations at equilibrium is as follows:

Chemical equilibrium is the point in a chemical reaction where both the forward and backward processes are occurring at the same rate.

The concentrations of the reactants and products are constant at equilibrium because the forward and reverse speeds are equal.

Considering the given statements based on the reaction equilibrium concentrations, the correct options are:

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The pressure of the gas when the volume is 1.09 L and the temperature is 308 K is 2.36 atm.

The final pressure of the gas is calculated by applying ideal gas law as follows;

(P₁V₁)/T₁ = (P₂V₂)/T₂

where

P₂ = (P₁V₁ x T₂)/(V₂ x T₁)

P₂ = (1.01 atm x 2.31 L x 308 K) / (1.09 L x 279 K)

P₂ = 2.36 atm

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