How does a Sulfur 2 ion differ from a neutral Sulfur atom?
O
Mass number
Atomic number
Number of electrons
Proton number

Since vanadium is a transition metal and sulfate is an anion, we can insist that V(SO4)2

is an ionic compound.

Answer:

V(SO4)2 is ionic

Explanation:

In this compound, Vanadium (V) is a transition metal with an oxidation state of +5, and sulfate (SO4) is a polyatomic ion with a charge of -2. The compound is formed by the transfer of two electrons from each sulfur atom to the vanadium atom. This results in the formation of two V3+ cations and one SO42- anion, which combine to form V(SO4)2.

Ionic compounds are formed by the transfer of electrons between atoms or ions, resulting in the formation of positively charged cations and negatively charged anions. These oppositely charged ions are held together by strong electrostatic forces, forming a crystalline lattice structure.

In conclusion, V(SO4)2 is an ionic compound formed by the transfer of electrons from the sulfate ion to the vanadium ion.

At 25 oC, the reaction's equilibrium constant is Cu(s) + 2Ag+ (aq) Cu+2 (aq) + 2Ag. (s). Eocell is equal to 0.47 V, R is 8.134 JK-1, and F is 96500 C at 25oC.

Antilogarithm is the value that a certain logarithm represents. For instance, x corresponds to the antilogarithm for y if log x = y. An antilogarithm of the stated number is an integer whose logarithm equals the specified number. Since 10,000 (104) has a logarithm of 4, 10,000 has an antilogarithm of 4.

Antilog of 7.06 with K equal to 1164

You must calculate K using the relationship Go = -RTlnK.  In this example, 8.315 j K-1mol-1 should be used as R, therefore lnK = Go/ -RT = -17.5 x 103 jmole-1 / (-8.315 j K-1mol-1 x 298 °K) = 7.06

An antilog on a calculator is what?

By elevating a logarithm above its base, one can find an antilog, which is the opposite of a logarithm. For instance, 10y = 5 is the antilog with y = log10(5). If y = ln(x), where y is time, and x is the value that needs to grow, then the natural logarithm can be used to determine how long it will take to reach a particular level of growth.

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Answer:

The answer is B: 0.0025 M

According to molar concentration and dilution concept, the  [H₃O+] in the final solution is closest to 0.05 M.

Molar concentration is defined as a measure by which concentration of chemical substances present in a solution are determined. It is defined in particular reference to solute concentration in a solution . Most commonly used unit for molar concentration is moles/liter.

The molar concentration depends on change in volume of the solution which is mainly due to thermal expansion. Molar concentration is calculated by the formula, molar concentration=mass/ molar mass ×1/volume of solution in liters.

In terms of moles, it's formula is given as molar concentration= number of moles /volume of solution in liters.In case of 2 solutions concentrated and diluted it is calculated as, M₁V₁=M₂V₂ substitution gives M₂=0.25×100/500=0.05

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In the peroxodisulfate ion, Peroxydisulfate-, all the atoms have oxidation values that add up to +6.

Most often, the polymerization of different alkenes, such as styrene, acrylonitrile, and fluoroalkenes, is started using salts of peroxydisulfate. The homolysis of peroxydisulfate causes polymerization to begin. 2 [Peroxydisulfate] [Sulfate]

Ag: +1

Sulfur: 0

Nitrogen: +2 (in Nitric oxide)

Oxygen: -2 (in Water)

Hydrogen: +1 (in Water)

The peroxodisulfate ion, Peroxydisulfate-, consists of one disulfate (Peroxydisulfate-) group and two peroxy (Oxygen2-) groups. Normal oxygen oxidation number is 2, however in a peroxy group it is -1. The charge of the ion, which is -2, is equal to the total sum of the oxidation numbers of all the atoms in the ion. As a result, we may write:

2(-1) + 2x + 8(-2) = -2

-2 + 2x - 16 = -2

2x = 12

x = +6

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The pH of .3 M NH, where is K = 1.7 x 10^-5 is 11.87 calculated from the equation of dissociation constant.

Kb= [A] /[A + ][X− ]

​1.7×10 −5 = x ^2 /0.3

⇒x= 7.5 ×10 −3

∴[OH − ][H + ]=7.5 ×10 −3

[H + ] =10 ^−14 ⇒pH=11.87

When describing the acidity or basicity of an aqueous solution, chemists use the pH scale, which is also known as acidity and previously stood for "potential of hydrogen". Greater pH values are seen in basic or alkaline solutions than acidic solutions.

Potential hydrogen is the meaning of the acronym pH, which indicates how much hydrogen is present in liquids and how active the hydrogen ion is.

As a first step, we shall ascertain the pKa of the solution before calculating its Ka. When a solution reaches the equivalence point, its pH and pKa are equal. So, by using a titration curve and the Ka = - log pKa equation, we may rapidly ascertain the value of Ka.

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Answer silver chloride (AgCl) and lead chloride (PbCl2).

Explanation:

Two salts that have the same solubility at approximately 23°C are silver chloride (AgCl) and lead chloride (PbCl2).

Both AgCl and PbCl2 have very low solubilities in water at room temperature, and their solubilities are similar at around 23°C. They are both sparingly soluble salts, meaning they dissolve only to a limited extent in water to form a saturated solution.

It's important to note that solubility can vary depending on the specific conditions, such as temperature, pressure, and presence of other substances. The solubility of salts can also be affected by factors such as pH and the presence of other ions in solution. Therefore, it's always best to consult reliable sources, such as reference tables or experimental data, for accurate solubility information at a given temperature.

For chemical reaction (3), the standard enthalpy change is 279.0 kJ.

To find the standard enthalpy change for reaction (3), use Hess's Law which states that the overall enthalpy change of a reaction is independent of the pathway between the initial and final states.

Obtain the required reaction by subtracting the enthalpy change of reaction (2) from that of reaction (1) as follows:

(1) 2C(s) + H₂(g) → C₂H₂(g) ΔH° = 226.7 kJ

(2) 2C(s) + 2H₂(g) → C₂H₄(g) ΔH° = 52.3 kJ

(3) C₂H₂(g) + H₂(g) → C₂H₄(g) ΔH° = ?

To get the enthalpy change for reaction (3), flip the reaction (2) and multiply by 1/2 so that the reactants match those in reaction (1):

(2) C₂H₄(g) → 2C(s) + 2H₂(g) ΔH° = -52.3 kJ

Now write the reaction (3) as the difference between (1) and (2):

(1) 2C(s) + H₂(g) → C₂H₂(g) ΔH° = 226.7 kJ

(2) C₂H₄(g) → 2C(s) + 2H₂(g) ΔH° = -52.3 kJ

(3) C₂H₂(g) + H₂(g) → C₂H₄(g) ΔH° = 279.0 kJ

Therefore, the standard enthalpy change for reaction (3) is 279.0 kJ.

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To solve this problem, we can use the following equation that relates the pH of a solution to the acid dissociation constant (Ka) and the concentration of the acid:

pH = -log[H+]

where [H+] is the concentration of hydrogen ions in the solution.

(a) To find the Ka of the unknown acid, we need to first find the concentration of hydrogen ions in the solution. We can do this by taking the inverse of the pH and converting it to a concentration:

[H+] = 10^(-pH) = 10^(-3.56) = 2.17 × 10^(-4) M

The acid dissociation constant (Ka) can then be calculated using the equation:

Ka = [H+][A-]/[HA]

where [A-] is the concentration of the conjugate base of the acid and [HA] is the concentration of the undissociated acid. Since we don't know the values of these concentrations, we need to use the fact that the solution is 0.500 M to make an assumption about the degree of dissociation (α) of the acid:

α = [A-]/[HA]

Since the solution is not extremely dilute, we can assume that the degree of dissociation is small and that the concentration of the undissociated acid is approximately equal to the initial concentration of the acid. Therefore, we can write:

[A-] ≈ 0.500α

[HA] ≈ 0.500 - 0.500α

Substituting these expressions into the equation for Ka, we get:

Ka = [H+][A-]/[HA] ≈ ([H+][A-])/0.500α

≈ ([H+]/Ka)(0.500α)/(1-α)

Solving for Ka, we get:

Ka ≈ H+/0.500α

Substituting the values we have calculated, we get:

Ka ≈ (2.17 × 10^(-4))(1-α)/(0.500α) = 4.37 × 10^(-5)

Therefore, the acid dissociation constant of the unknown acid is approximately 4.37 × 10^(-5).

(b) To find the percentage of acid that is ionized in the solution, we can use the equation:

α = [A-]/[HA] = 10^(-pKa + pH)/(1 + 10^(-pKa + pH))

where pKa is the negative logarithm of the acid dissociation constant. Substituting the values we have calculated, we get:

α = 10^(-(-4.36) + 3.56)/(1 + 10^(-(-4.36) + 3.56)) ≈ 0.008

Therefore, the percentage of acid that is ionized in the solution is approximately 0.8%.

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To solve this problem, we can use the following equation that relates the pH of a solution to the acid dissociation constant (Ka) and the concentration of the acid:

pH = -log[H+]

where [H+] is the concentration of hydrogen ions in the solution.

(a) To find the Ka of the unknown acid, we need to first find the concentration of hydrogen ions in the solution. We can do this by taking the inverse of the pH and converting it to a concentration:

[H+] = 10^(-pH) = 10^(-3.56) = 2.17 × 10^(-4) M

The acid dissociation constant (Ka) can then be calculated using the equation:

Ka = [H+][A-]/[HA]

where [A-] is the concentration of the conjugate base of the acid and [HA] is the concentration of the undissociated acid. Since we don't know the values of these concentrations, we need to use the fact that the solution is 0.500 M to make an assumption about the degree of dissociation (α) of the acid:

α = [A-]/[HA]

Since the solution is not extremely dilute, we can assume that the degree of dissociation is small and that the concentration of the undissociated acid is approximately equal to the initial concentration of the acid. Therefore, we can write:

[A-] ≈ 0.500α

[HA] ≈ 0.500 - 0.500α

Substituting these expressions into the equation for Ka, we get:

Ka = [H+][A-]/[HA] ≈ ([H+][A-])/0.500α

≈ ([H+]/Ka)(0.500α)/(1-α)

Solving for Ka, we get:

Ka ≈ H+/0.500α

Substituting the values we have calculated, we get:

Ka ≈ (2.17 × 10^(-4))(1-α)/(0.500α) = 4.37 × 10^(-5)

Therefore, the acid dissociation constant of the unknown acid is approximately 4.37 × 10^(-5).

(b) To find the percentage of acid that is ionized in the solution, we can use the equation:

α = [A-]/[HA] = 10^(-pKa + pH)/(1 + 10^(-pKa + pH))

where pKa is the negative logarithm of the acid dissociation constant. Substituting the values we have calculated, we get:

α = 10^(-(-4.36) + 3.56)/(1 + 10^(-(-4.36) + 3.56)) ≈ 0.008

Therefore, the percentage of acid that is ionized in the solution is approximately 0.8%.

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To convert to Celsius, we subtract 273.15 from the Kelvin temperature, giving us a final answer of 42.1°C.

Temperature is a physical quantity that measures the average kinetic energy of the particles in a system. It is an important parameter for understanding the behavior of matter and the underlying physical processes at work. Temperature is measured in units such as degrees Celsius (°C), Fahrenheit (°F), Kelvin (K), or Rankine (°R). Temperature affects the rate at which chemical reactions occur and the movement of particles in solids, liquids, and gases.

The ideal gas law states that PV = nRT,
where n is the number of moles,
P is the pressure,
V is the volume, R is the ideal gas constant (8.314 J/molK), and
T is the temperature in Kelvin.
Rearranging the equation, we get T = (PV)/(nR).
Plugging in our values, we get T = (89.9 atm * 4191 mL)/(2.6 mol * 8.314 J/molK) = 115.2 K.
To convert to Celsius, we subtract 273.15 from the Kelvin temperature, giving us a final answer of 42.1°C.

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5,250. The number was rounded up from 5,249 because the last digit, 9, is greater than or equal to 5.

Rounding up is a mathematical operation that involves increasing a number to its nearest whole number. It is commonly used when dealing with money, measurements, or statistics. When rounding up, the number is increased to the next highest whole number. For example, if a number is 6.7, it would be rounded up to 7. Rounding up is often used when dealing with exact measurements or estimates to simplify the calculations. It can also be used to make the results of a calculation easier to understand. In the case of money, rounding up can be used to round a number to the nearest dollar. This prevents dealing with fractional amounts of money. Rounding up can also be utilized in statistical analysis, such as in the calculation of mean or median. This simplifies the data and prevents dealing with fractions or decimals.

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According to the question the standard change in Gibbs free energy is 2818.4 kJ/mol.

The capacity to perform work is energy. It is a characteristic of all matter and can assume many different shapes. It exists in a variety of shapes, including those of light, heat, chemical, electrical, mechanical, and nuclear. Energy is the ability to accomplish work and is measured in joules, which are equivalent to the amount of work completed when one newton of force is applied over a one metre distance.

The following equation can be used to get the reaction's standard change in Gibbs free energy (ΔG°) at 25°C:

ΔG° = [4 ΔG°f ([tex]Co_2[/tex]) + 6 ΔG°f ([tex]H_2o[/tex])] - [2 ΔG°f ([tex]C_2H_6[/tex]) + 7 ΔG°f ([tex]o_2[/tex])]

At 25°C, ΔG°f ([tex]Co_2[/tex]) = -393.5 kJ/mol, ΔG°f ([tex]H_2o[/tex]) = -237.2 kJ/mol, ΔG°f ([tex]C_2H_6[/tex]) = -85.2 kJ/mol, and ΔG°f ([tex]o_2[/tex]) = 0 kJ/mol.

As a result, the typical variation in Gibbs free energy is:

ΔG° = [-393.5 kJ/mol × 4] + [-237.2 kJ/mol × 6] - [-85.2 kJ/mol × 2] - [0 kJ/mol × 7]

ΔG° = -2818.4 kJ/mol.

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2818.4 kJ/mol is  the standard change in Gibbs free energy for the reaction at 25 °C.

What does the name Gibbs free energy mean?

Because it is readily accessible at all times, Gibb's free energy is known as free energy. If necessary, the reaction can obtain this energy without exerting any effort. Enthalpy and the system's product of temperature and entropy are added to determine the change in Gibb's free energy.

Enthalpy and entropy are combined into a single quantity known as Gibbs free energy, or G. The product of the system's temperature and entropy, added to the enthalpy, equals the change in free energy, or G.

2C2H6(g)+7O2(g)⟶4CO2(g)+6H2O(g)

ΔG° = [4 ΔG°f (CO2) + 6 ΔG°f (H2O)] - [2 ΔG°f (C2H6) + 7 ΔG°f (O2)]

At 25°C, ΔG°f (CO2) = -393.5 kJ/mol,

ΔG°f (H2O) = -237.2 kJ/mol,

ΔG°f (C2H6) = -85.2 kJ/mol,

and ΔG°f (O2) = 0 kJ/mol.

ΔG° = [-393.5 kJ/mol × 4] + [-237.2 kJ/mol × 6] - [-85.2 kJ/mol × 2] - [0 kJ/mol × 7] = -2818.4 kJ/mol.

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40 grams of KCl are dissolved in 100 mL of water at 45C. 5g of  additional grams of KCI are needed to make the solution saturated at 80 C as the solubility of KCl is 45g/ml

A uniform combination of a number of solutes within a solvent is referred to as a solution. One frequent illustration of a Solution is adding sugar cubes into your cup of tea and coffee. Solubility is the quality that makes sugar molecules more soluble.

In water, potassium chloride (KCl) dissolves. Its water solubility, like that of all other solutes, depends on temperature. The solubility of a salt increases as the solvent's temperature rises. This is fairly simple to experience with sugar. 40 grams of KCl are dissolved in 100 mL of water at 45C. 5g of  additional grams of KCI are needed to make the solution saturated at 80 C as the solubility of KCl is 45g/ml.

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Percent yield = 1566%, The percent yield is greater than 100%, which means that the student obtained more product than the theoretical yield.

The theoretical yield is the maximum amount of product that could be produced in a chemical reaction, based on stoichiometric calculations. The actual yield is the amount of product that is actually obtained from the reaction in a laboratory experiment.

To calculate the percent yield of the experiment, we need to compare the actual yield to the theoretical yield.

The balanced chemical equation for the formation of alum from aluminum and potassium hydroxide is:

2 Al + 2 KOH + 4 H₂O + 1/2 O₂ → KAl(SO₄)₂·12H₂O

From this equation, we can see that the mole ratio of aluminum to alum is 2:1. This means that for every 2 moles of aluminum used, we should theoretically obtain 1 mole of alum.

First, we need to convert the mass of aluminum foil used to moles of aluminum:

0.0432 g Al × 1 mol Al/26.98 g Al = 0.001600 mol Al

Next, we can use the mole ratio to calculate the theoretical yield of alum:

Theoretical yield = 0.001600 mol Al × 1 mol KAl(SO₄)₂·12H₂O/2 mol Al × 474.39 g KAl(SO₄)₂·12H₂O/1 mol KAl(SO₄)₂·12H₂O

= 0.0382 g KAl(SO₄)₂·12H₂O

Now we can calculate the percent yield:

Percent yield = actual yield/theoretical yield × 100%

Actual yield = 0.5987 g

Percent yield = 0.5987 g/0.0382 g × 100% = 1566%

The percent yield is greater than 100%, which means that the student obtained more product than the theoretical yield. This could be due to experimental error, such as incomplete drying of the crystals or loss of product during the experiment. It's also possible that the student accidentally used more aluminum foil than intended.

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Answer:

acid +metal ----->salt +hydrogen

C. The pH stays about the same.

Answer:

C.The pH stays about the same.

Explanation:

Buffer reactions maintain stable pH of solutions.

The statement "Plasmas can change shape" and "Plasmas contain ionized particles" best describe plasmas.

Plasma is a state of matter that is similar to gas but differs in that it contains ionized particles, which are atoms or molecules that have lost or gained one or more electrons. This results in a mixture of positively charged ions and negatively charged electrons, making plasma electrically conductive.

Plasma can be found in many natural phenomena such as lightning, stars, and the aurora borealis, and it is also used in various technological applications such as plasma TVs, fusion reactors, and fluorescent lights. Because of its unique properties, plasma has many interesting and useful applications in fields such as physics, chemistry, and engineering.

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The total mass of protons and neutrons makes up an element's atomic mass. Lithium is the element X; it has a mass of 6.941 u.

When an atom has the same number of protons in its atomic nucleus, it is said to be an element. The number of protons in the nucleus of each element's atoms, or atomic number, serves as the element's sole means of identification.

Seven hydrogen atoms will weigh 7.056 g as each hydrogen atom has an atomic mass of 1.008. The periodic chart shows that an atomic mass of 7.056 g is the one that is closest to that of a lithium atom (6.941 u).

Seven hydrogen atoms have a mass comparable to one lithium atom in the periodic table, based on their mass. With an atomic mass of 6.941 g/mol, the lithium atom bears the atomic number 3.

Lithium is  the name of element X.

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Enantiomers rotate the plane of polarized light in opposite directions, and this property is used to distinguish between them in a process called optical rotation.

Enantiomers are pairs of molecules that are non-superimposable mirror images of each other.

They are isomers, meaning they have the same molecular formula and connectivity but differ in their three-dimensional arrangement of atoms in space.

Enantiomers exhibit identical physical and chemical properties, except for their interaction with plane-polarized light (a type of light that oscillates in a single plane).

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The molecular formula of the compound, given that it contains 51.39% carbon, 8.64% hydrogen, and 39.97% nitrogen is C₆H₁₂N₄

To obtain the molecular formula, we must first determine the empirical formula. Details on how to obtain the empirical formula is given beloww:

Divide by their molar mass

C = 51.39 / 12 = 4.283

H = 8.64 / 1 = 8.64

N = 39.97 / 14 = 2.855

Divide by the smallest

C = 4.283 / 2.855 = 1.5

H = 8.64 / 2.855 = 3

N = 2.855 / 2.855 = 1

Multiply through by 2 to express in whole number

C = 1.5 × 2 = 3

H = 3 × 2 = 6

N = 1 × 2 = 2

Thus, we can conclude that the empirical formula is C₃H₆N₂

Finally, we shall determine the molecular formula. Details below

Molecular formula = empirical × n = mass number

[C₃H₆N₂]n = 140.22

[(12×3) + (1×6) + (14×2)]n = 140.22

70n = 140.22

Divide both sides by 70

n = 140.22 / 70

n = 2

Molecular formula = [C₃H₆N₂]n

Molecular formula = [C₃H₆N₂]₂

Molecular formula = C₆H₁₂N₄

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At 40°C is the temperature at which saturated solutions of potassium nitrate and sodium nitrate contain the same weight of solute per 100 mL of water contain the same weight of solute per 100 mL of water.

Temperature is a unit of warmth or coldness that can be defined in the context of any number of arbitrary scales. It indicates the direction that heat energy will naturally flow, i.e., through a hotter (body) to a colder (body) body.

Temperature is not the same as the energy in a thermodynamic system; for instance, an iceberg has a significantly larger total heat energy than a match, despite the fact that a match is burning at an extremely high temperature. At 40°C is the temperature at which saturated solutions of potassium nitrate and sodium nitrate contain the same weight of solute per 100 mL of water.

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The final volume of the ideal gas is 156.6 L.

A sample of an ideal gas is allowed to expand adiabatically while producing external work (W) at first. The volume is then maintained at its new value with the help of heat Q until the pressure returns to its initial level.

We can use the combined gas law to solve this problem:

(P1 × V1) / T1 = (P2 × V2) / T2

where P1, V1, and T1 are the initial pressure, volume, and temperature, respectively, and P2 and V2 are the final pressure and volume, respectively. Since the gas is kept at a constant temperature, we can simplify this to:

P1 × V1 = P2 × V2

Plugging in the given values, we get:

2.01 bar × 78.3 L = 1.00 bar × V2

Solving for V2, we get:

V2 = (2.01 bar × 78.3 L) / 1.00 bar = 157 L

Therefore, the final volume of gas is 157 L.

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Answer: 40. g

Explanation:

To find the grams in 0.89 moles of CO2, we just need to use the molar mass of CO2. The molar mass which tells us how many grams are in a mole for an element or compound.

The molar mass of CO2 is equal to the molar mass of carbon, 12.0107, and 2 oxygens, 2*15.9994 (you can find the molar mass of an element on any periodic table). Add these, and you get the molar mass of CO2 to be 44.01 g/mol, a helpful value to remember.

Now, just multiply the molar mass by the amount of moles to find grams.

[tex]\frac{44.01g}{mole} * 0.89mole[/tex], moles cancel out, [tex]\frac{44.01g}{mole} * 0.89mole=40.48g[/tex]

There are 2 significant figures in the question, so I will round this answer to 2 significant figures, 40. g

The maximum mass of carbon dioxide that could be produced from 6.9 g of octane and 42.2 g of oxygen is 21.3 g, rounded to 2 significant digits.

What is Octane?

Octane is a hydrocarbon with the chemical formula [tex]C_{8} H_{18}[/tex] It is an organic compound belonging to the alkane group, which means it consists of only carbon (C) and hydrogen (H) atoms bonded together by single covalent bonds. Octane is a colorless liquid with a molecular weight of approximately 114 g/mol and is commonly used as a component in gasoline or fuel for internal combustion engines.

From the balanced equation, we know that 1 mole of octane reacts with 12.5 moles of oxygen to produce 8 moles of carbon dioxide. Therefore, 0.0605 mol of octane would require 0.0605 mol x 12.5 = 0.75625 mol of oxygen to fully react.

Since we have only 1.32 mol of oxygen, which is in excess compared to the 0.75625 mol required by octane, oxygen is the excess reactant, and octane is the limiting reactant.

Now, we can use the stoichiometry of octane to carbon dioxide to calculate the maximum mass of carbon dioxide produced:

From the balanced equation, we know that 1 mole of octane produces 8 moles of carbon dioxide.

Molar mass of carbon dioxide (CO2) = 44.01 g/mol

Maximum moles of carbon dioxide produced from octane = 0.0605 mol x 8 = 0.484 mol

Maximum mass of carbon dioxide produced from octane = 0.484 mol x 44.01 g/mol = 21.3 g

Remember to round the final answer to 2 significant digits as requested.

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Therefore, the standard enthalpy change for reaction (3) is -152.6 kJ.

To determine the standard enthalpy change for reaction (3), we can use Hess's law, which states that the overall enthalpy change for a reaction is the sum of the enthalpy changes for each step of the reaction.

We can see that reaction (1) has the same products as reaction (3) but in reverse order. We can use reaction (1) to reverse the formation of FeO:

2Fe(s) + O2(g) → 2FeO(s) ΔH° = -544.0 kJ

2FeO(s) → 2Fe(s) + O2(g) ΔH° = +544.0 kJ (reversed)

We can also use reaction (2) to form ZnO, which is a product in reaction (3):

2Zn(s) + O2(g) → 2ZnO(s) ΔH° = -696.6 kJ

Now, we can add the two reactions to get the overall reaction (3):

2FeO(s) + 2Zn(s) → 2Fe(s) + 2ZnO(s)

We add the enthalpies of the two reactions to get the enthalpy change for reaction (3):

ΔH° = (+544.0 kJ) + (-696.6 kJ)

ΔH° = -152.6 kJ

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