How many grams of sulfuric acid (h2so4) are dissolved in a 2 liter solution that is 18 molar?

Answer:

0.25g

Explanation:

Dimensional analysis.

Assuming the reaction is taking place at standard temperature and pressure (STP, 1 atm at 298.15K or 25 C), 1 mol of gas occupies 22.4L.

We are given the volume of the gas, with this we are able to find its number of moles.

125mL = 0.125L

[tex]0.125 L * \frac{1 mol}{22.4 L}[/tex]

= 0.0056mol

With the number of moles we can simply multiply by the molecules molar mass.

CO2 = 12.011 g/mol+ 2*15.999 g/mol

CO2 = 44.009g / mol

[tex]44.009 \frac{g CO2}{mol} * 0.0056mol CO2\\\\=0.25 g CO2[/tex]

Paleontologists use a variety of methods to determine the placement of a fossil for display. One important factor is the diagnostic structure of the fossil, which refers to unique features that help to identify the species and its evolutionary relationships. For example, if a fossil has a particular shape or pattern on its shell, this could indicate a specific genus or species.

To accurately place a fossil for display, paleontologists will carefully examine its diagnostic structures and compare them to other specimens in their collection or in published research. They may also consult with experts in the field or use advanced imaging techniques to better understand the fossil's characteristics.

Once the paleontologists have identified the species and determined its placement, they can design a display that showcases the fossil in a way that is both educational and visually appealing. This may involve creating a custom mount or exhibit case, selecting appropriate lighting and text labels, and considering the context in which the fossil was found.

Overall, the accurate placement of a fossil for display is crucial for conveying its scientific significance to the public and helping people to better understand the history of life on Earth. By using diagnostic structure as a key tool in this process, paleontologists can ensure that the fossils are correctly identified and presented in a way that is both informative and engaging.

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the molar solubility of lead phosphate in a 0.202 M sodium phosphate solution is approximately 1.27 × 10^-7 M.

To calculate the molar solubility of lead phosphate in a sodium phosphate solution, we need to use the solubility product constant (Ksp) of lead phosphate and the common ion effect of sodium phosphate.

The balanced equation for the dissolution of lead phosphate (Pb3(PO4)2) is:

Pb3(PO4)2(s) ⇌ 3Pb2+(aq) + 2PO42-(aq)

The Ksp expression for lead phosphate is:

Ksp = [Pb2+]^3[PO42-]^2

The balanced equation for the dissociation of sodium phosphate (Na3PO4) is:

Na3PO4(s) ⇌ 3Na+(aq) + PO42-(aq)

In a 0.202 M sodium phosphate solution, the concentration of the PO42- ion is [PO42-] = 3 × 0.202 M = 0.606 M, due to the dissociation of sodium phosphate.

To calculate the molar solubility of lead phosphate, we can assume that x mol/L of Pb3(PO4)2 dissolves and forms 3x mol/L of Pb2+ and 2x mol/L of PO42-. Using the Ksp expression and the common ion effect, we can write:

Ksp = [Pb2+]^3[PO42-]^2
Ksp = (3x)^3(2x)^2 = 108x^5

Since the concentration of PO42- is 0.606 M, the concentration of Pb2+ is also 3x = 3(0.202 M - x). Substituting this into the Ksp expression gives:

Ksp = (3x)^3(2x)^2 = 108x^5
4.8 × 10^-27 = (3(0.202 - x))^3(2x)^2

Solving for x, we get:

x = 1.27 × 10^-7 M

Therefore, the molar solubility of lead phosphate in a 0.202 M sodium phosphate solution is approximately 1.27 × 10^-7 M.
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a) The temperature of the water in the beaker is likely to increase as heat flows from the metal to the water until they reach thermal equilibrium.

b) The metal will likely lose heat to the water until it reaches thermal equilibrium with the water.

a) The temperature of the water in the beaker is likely to increase due to the transfer of heat from the metal to the water. This process is known as conduction, and it occurs because heat always flows from hotter objects to cooler objects. The metal, being at a higher temperature than the water, will transfer heat to the water until both reach a state of thermal equilibrium.

b) The amount of heat lost by the metal will depend on its mass, specific heat capacity, and initial temperature. The metal may also undergo physical changes due to the change in temperature, such as contraction or expansion. The type of metal and its properties will influence how much it changes in response to the temperature change.

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The heat evolved when 27.5 g of ammonia gas condenses to a liquid at its boiling point is -37.8 kJ.

First, we need to calculate the amount of heat required for the ammonia gas to condense. The heat of vaporization of ammonia is 23.4 kJ/mol. The molar mass of ammonia is 17.03 g/mol, so we have:

23.4 kJ/mol x (27.5 g / 17.03 g/mol) = 37.8 kJ

This means that 37.8 kJ of heat is required for 27.5 g of ammonia gas to condense. However, since the question asks for the heat evolved, we need to reverse the sign of the answer.

Thus, the amount of heat released as 27.5 grams of gaseous ammonia undergoes condensation at its boiling point is equal to -37.8 kJ.

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Explanation:

You will need the specific heat of Mg which I found to be 1.02 J / (g C)

m * 45 C  * 1.02 J . (g C) = 843

m = 843 / (45* 1.02) = 18.4 g  of Magnesium

The standard free-energy change for the reaction H₂(g) + Br₂(l) → 2HBr(g) at 25°C can be calculated using the equation ΔG° = ΔH° - TΔS°, where ΔH° is the standard enthalpy change, T is the temperature, and ΔS° is the standard entropy change.

To calculate the standard free-energy change for the reaction H₂(g) + Br₂(l) → 2HBr(g) at 25°C, you need to use the equation: ΔG° = ΔH° - TΔS°. Follow these steps:

1. Determine the standard enthalpy change (ΔH°) for the reaction.
2. Determine the standard entropy change (ΔS°) for the reaction.
3. Calculate ΔG° using the equation and the given temperature (25°C = 298.15 K).

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When 142 g Cl₂ combines with lots of aluminium, 1.33 moles of AlCl₃ are formed.

To determine the number of moles of AlCl₃ formed when 142 g Cl₂ reacts with plenty of aluminum, we first need to write a balanced chemical equation for the reaction:

2 Al + 3 Cl₂ → 2 AlCl₃

From the balanced equation, we can see that 3 moles of Cl₂ react with 2 moles of Al to form 2 moles of AlCl₃.

Next, we need to calculate the number of moles of Cl₂ present in 142 g:

n(Cl₂) = m/M

n(Cl₂) = 142 g / 70.9 g/mol

n(Cl₂) = 2.00 moles

Since the reaction consumes 3 moles of Cl₂ for every 2 moles of AlCl₃ formed, we can determine the number of moles of AlCl₃ formed as:

n(AlCl₃) = (2/3) x n(Cl₂)

n(AlCl₃) = (2/3) x 2.00 moles

n(AlCl₃) = 1.33 moles

Therefore, 1.33 moles of AlCl₃ form when 142 g Cl₂ reacts with plenty of aluminum.

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When the force of tectonic plates pushing against each other become too strong the plates break and create faults.

Tectonic plates are large, rigid pieces of the Earth's lithosphere that move relative to one another. The lithosphere is the rocky outer layer of the Earth that includes the crust and part of the upper mantle. Tectonic plates can vary in size from a few hundred to thousands of kilometers in diameter. They interact at their boundaries and are responsible for the formation and destruction of the Earth's surface features, such as mountains and ocean trenches. Movement of the plates helps explain the occurrence of earthquakes, volcanic eruptions, mountain building, and oceanic trench formation. In plate tectonics, the lithosphere is divided into seven major plates and several minor ones. The major plates include the African, Antarctic, Eurasian, Indo-Australian, North American, Pacific, and South American plates. The plates move at a rate of several centimeters per year in response to convection currents in the mantle.

When the force of tectonic plates pushing against each other become too strong the plates break and create faults. This can cause folding of the crust to create mountains or reverse faults can create deep valleys. Volcanoes are also involved in the formation of landforms as they create mountains, ridges and canals. Earthquakes can also cause landforms to form in areas that were previously flat.

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The best answer to the question "What is volume?" is b. The amount of space occupied by matter. Volume is a physical property of matter that refers to the amount of space that an object or substance takes up. It is often measured in cubic units such as cubic meters, cubic feet, or cubic centimeters.

It is important to note that volume is not the same as mass or weight, as it refers to the amount of space that matter occupies rather than the amount of matter itself. In summary, volume is the amount of space occupied by matter and is an important concept in the study of physics and chemistry.

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Small peptides buffer stomach acid, so the pH does not fall excessively low.

The stomach produces hydrochloric acid, which helps in the digestion of food by breaking down complex molecules into simpler ones. However, excessive production of stomach acid can lead to various digestive disorders, such as acid reflux, ulcers, and gastritis.

Small peptides are short chains of amino acids that are produced during the digestion of proteins. They have a buffering effect on stomach acid by neutralizing the excess acid, which helps to maintain the pH of the stomach within a healthy range.

This buffering action is important for protecting the stomach lining from the harmful effects of excess acid, as well as for ensuring efficient digestion and absorption of nutrients from food.

Therefore, consuming protein-rich foods that can be broken down into small peptides may help to buffer stomach acid and prevent digestive problems associated with excess acid.

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In the field of chemistry, the term "conjugate" is used to describe pairs of molecules or ions that are connected through the transfer of a proton, which is represented as H⁺. Conjugate acids and bases, specifically, are pairs of molecules or ions that vary by the presence or absence of one proton.

These equilibrium equations represent the transfer of a proton between a weak acid or base and water, resulting in the formation of its conjugate acid or base.

Answer of the given questions are as follows :

1. The formula of the conjugate acid: HCO₂H

2. The formula of the conjugate base: C₆HNH₃⁺

3. The formula of the conjugate acid of the brønsted-lowry base: H₂CO₃

4. The formula of the conjugate acid of the brønsted-lowry base:

C₆H₅NH₃⁺

5. The acidic equilibrium equation for HC₂H₃O₂: HC₂H₃O₂ + H₂O ⇌ H₃O⁺ + C₂H₃O²⁻

6. The basic equilibrium equation for C₆H₅NH₂

C₆H₅NH₂ + H₂O ⇌ C₆H₅NH₃⁺ + OH⁻

7. The basic equilibrium equation for NH₃

NH₃ + H₂O ⇌ NH₄⁺ + OH⁻

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Diiodine monoselenide is an inorganic compound with the chemical formula I2Se. It is a dark red solid that is sparingly soluble in water. The structure of diiodine monoselenide consists of a linear Se-I-I unit, with the selenium atom in the middle and the two iodine atoms on either side. This arrangement gives the compound a linear, V-shaped structure.

Diiodine monoselenide is an important compound in the field of materials science, as it exhibits some interesting properties. For example, it can be used as a precursor for the synthesis of various selenium-containing compounds, including organoselenium compounds, which are used in catalysis and medicine.

Additionally, diiodine monoselenide has been studied as a potential material for use in electronic devices, due to its semiconducting properties. In conclusion, diiodine monoselenide is an important inorganic compound that exhibits some interesting structural and material properties.

Its linear, V-shaped structure is due to the arrangement of the selenium and iodine atoms in a linear Se-I-I unit. This compound is used in the synthesis of various selenium-containing compounds and has potential applications in the field of electronics.

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At a constant pressure, the gas occupies a volume of 500 ml when the temperature is increased to 250k.

At a constant pressure, the volume of a gas is directly proportional to its temperature. This relationship is known as Charles' Law. According to the problem, the sample of gas occupies 420 ml at a temperature of 210k. We need to find out the volume of the gas when the temperature is increased to 250k.

To solve this problem, we can use the formula V1/T1 = V2/T2, where V1 is the initial volume, T1 is the initial temperature, V2 is the final volume, and T2 is the final temperature. Plugging in the given values, we get:

420 ml/210k = V2/250k

Simplifying this equation, we get:

V2 = (420 ml/210k) x 250k
V2 = 500 ml

Therefore, at a constant pressure, the gas occupies a volume of 500 ml when the temperature is increased to 250k.

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To convert moles of sodium chloride to grams, we multiply by its molar mass of 58.44 g/mol. Therefore, the amount of sodium chloride produced is 0.363 mol x 58.44 g/mol = 21.2 grams.

To determine the limiting reagent in this reaction, we need to calculate the moles of both reactants. From the given information, we know that the mass of iron (II) chloride is 23 grams, and its molar mass is 126.75 g/mol.

Therefore, the number of moles of iron (II) chloride is 23 g/126.75 g/mol = 0.1815 mol.

Next, we calculate the number of moles of sodium phosphate. Since there are two molecules of sodium phosphate for every three molecules of iron (II) chloride, we need to multiply the moles of iron (II) chloride by the ratio of the coefficients. Therefore, the number of moles of sodium phosphate is (0.1815 mol x 2/3) = 0.121 mol.

Since there are fewer moles of sodium phosphate than iron (II) chloride, sodium phosphate is the limiting reagent. This means that all of the sodium phosphate will be used up in the reaction, and any remaining iron (II) chloride will be left over.

To calculate the amount of sodium chloride produced, we need to use the stoichiometric coefficients from the balanced equation.

For every 2 moles of sodium phosphate used, 6 moles of sodium chloride are produced. Therefore, since we have 0.121 mol of sodium phosphate, we can produce (0.121 mol x 6/2) = 0.363 mol of sodium chloride.

Finally, to convert moles of sodium chloride to grams, we multiply by its molar mass of 58.44 g/mol. Therefore, the amount of sodium chloride produced is 0.363 mol x 58.44 g/mol = 21.2 grams.

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0.72 L (or 720 mL) of the 4.5 M HCl solution is required to prepare a 65 L solution of 0.050 M HCl.

We need to dilute a concentrated 4.5 M HCl solution. We can use the dilution equation to calculate the volume of the concentrated solution required:

[tex]M_1V_1 = M_2V_2[/tex]

where M1 is the concentration of the concentrated solution, V1 is the volume of the concentrated solution required, M2 is the concentration of the diluted solution, and V2 is the final volume of the diluted solution.

In this case, we want to prepare a 65 L solution of 0.050 M HCl from a 4.5 M HCl solution. Therefore:

[tex]M_1[/tex] = 4.5 M

[tex]V_2[/tex] = 65 L

[tex]M_2[/tex] = 0.050 M

Solving for V1:

[tex]V_1 = (M_2 * V_2) / M_1 \\V_1 = (0.050 M * 65 L) / 4.5 M \\V_1 = 0.72 L[/tex]

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a. HNO3 + KOH --> KNO3 + H2O

b. HCl + LiOH --> LiCl + H2O

c. H2S + 2NaOH --> Na2S + 2H2O

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COCl2(g) will deviate most from the ideal gas law at low temperature. The other name for COCl2(g) is Phosgene. This is because COCl2(g) is a larger molecule with stronger intermolecular forces than Cl2(g). At low temperatures, these intermolecular forces become significant and cause the molecules to be closer together, resulting in a smaller molar volume than predicted by the ideal gas law.

Additionally, COCl2(g) is a polar molecule, which also contributes to the deviation from the ideal gas law as the polar interactions between molecules become stronger at low temperatures. Thus COCl2(g) will be the one deviating from the ideal gas law at low temperature.

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4.85 grams of Zn are needed to produce 0.15 grams of H2.

The balanced chemical equation for the reaction between zinc and phosphoric acid is:

[tex]3 Zn + 2 H_3PO_4[/tex] → [tex]3 H_2 + Zn_3(PO4)2[/tex]

Step 1: Calculate the number of moles of [tex]H_2[/tex] produced

We can use the molar mass of hydrogen gas ([tex]H_2[/tex]) to calculate the number of moles produced:

n([tex]H_2[/tex]) = mass of [tex]H_2[/tex] / molar mass of [tex]H_2[/tex]

n([tex]H_2[/tex]) = 0.15 g / 2.016 g/mol = 0.0743 mol

Step 2: Calculate mass  [tex]Z_2[/tex] needed

We can use the molar mass of zinc to convert moles of Zn to grams of Zn:

mass of Zn = n(Zn) x molar mass of Zn

mass of Zn = 0.0743 mol x 65.38 g/mol

mass of Zn = 4.85 g

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6. The mass (in grams) present is 9.72×10⁸ grams

7. The number of atoms is 8.50×10²⁴ atoms

8. The mass (in grams) present is 3.73×10¹⁰ grams

First, we shall determine the mole of LiNO₃. Details below:

6.022×10²³ atoms = 1 mole of LiNO₃

Therefore,

8.48×10³⁰ atoms = 8.48×10³⁰ / 6.022×10²³

8.48×10³⁰ atoms = 1.41×10⁷ moles of LiNO₃

Finally, we shall determine the mass of LiNO₃. Details below:

Mass = Mole × molar mass

Mass of LiNO₃ = 1.41×10⁷ × 68.95

Mass of LiNO₃ = 9.72×10⁸ grams

First, we shall determine the mole in 2105 g of (NH₄)₃PO₃. Details below:

Mole = mass / molar mass

Mole of  (NH₄)₃PO₃ = 2105 / 149.09

Mole of  (NH₄)₃PO₃ = 14.12 moles

Finally, we shall determine the number of atoms. Details below:

From Avogadro's hypothesis,

1 mole of (NH₄)₃PO₃ = 6.022×10²³ atoms

Therefore,

14.12 moles of (NH₄)₃PO₃ = 14.12 × 6.022×10²³

14.12 moles of (NH₄)₃PO₃ = 8.50×10²⁴ atoms

Thus, the number of atoms is 8.50×10²⁴ atoms

First, we shall determine the mole of (NH₄)₂SO₄. Details below:

6.022×10²³ atoms = 1 mole of (NH₄)₂SO₄

Therefore,

1.7×10³² atoms = 1.7×10³² / 6.022×10²³

1.7×10³² atoms = 2.82×10⁸ moles of (NH₄)₂SO₄

Finally, we shall determine the mass of (NH₄)₂SO₄. Details below:

Mass = Mole × molar mass

Mass of (NH₄)₂SO₄ = 2.82×10⁸ × 132.14

Mass of (NH₄)₂SO₄ = 3.73×10¹⁰ grams

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The statements that correctly describe the strength of an acid or base are:

An acid is a chemical that donates hydrogen ions, whose addition to an existing solution results in increased acidity.

According to the conventional definition of acids, they are compounds which discharge positively charged hydrogen ions when mixed with water. Acids have a sour flavor and possess pH levels below 7.

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3.86 grams of chlorine would exert a pressure of 610 torr in a 3.26-liter container at standard temperature.

To calculate the number of grams of chlorine required to exert a pressure of 610 torr in a 3.26-liter container at standard temperature, we need to use the ideal gas law equation: PV = nRT.

Where,

P = pressure = 610 torr

V = volume = 3.26 L

n = number of moles

R = gas constant = 0.0821 Latm/(molK) (standard value)

T = temperature = 273 K (standard temperature)

n = PV ÷ RT

Substituting the given values, we get:

n = (610 torr × 3.26 L) ÷ (0.0821 Latm/(molK) × 273 K)

n = 0.109 mol

Now, to convert moles to grams, we need to use the molar mass of chlorine, which is 35.45 g/mol.

Thus, number of grams of chlorine required is:

0.109 mol × 35.45 g/mol = 3.86 g

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The molarity of the solution is 5.06 M

To find the molarity of a solution, we use the formula:

Molarity = moles of solute / liters of solution

First, we need to find the moles of[tex]NaOH[/tex]in the solution:

moles of [tex]NaOH[/tex] = mass / molar mass

The molar mass of [tex]NaOH[/tex] is 40.00 g/mol (sodium = 22.99 g/mol, oxygen = 15.99 g/mol, hydrogen = 1.01 g/mol).

moles of[tex]NaOH[/tex] = 72.0 g / 40.00 g/mol = 1.80 mol

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

356 mL = 0.356 L

Now we can calculate the molarity of the solution:

Molarity = 1.80 mol / 0.356 L = 5.06 M

Therefore, the molarity of the solution is 5.06 M

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The term molarity is an important method which is used to calculate the concentration of a solution. By dissolving the same number of moles of each substance in the same volume of water we can make solutions of same molarity. The correct option is B.

Molarity is defined as the number of moles of the solute present per litre of the solution. It is represented as 'M' and its unit is mol / L. The term molarity is also called the molar concentration.

When same number of moles of substances are dissolved in the same volume of water, then the two solutions have same molarity.

Thus the correct option is B.

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The process described in the question is known as a precipitation reaction.

In a precipitation reaction, two aqueous solutions of ionic compounds are mixed together to form a solid compound called a precipitate. This occurs because the ions in the two solutions react with each other to form an insoluble product, which separates from the solution as a solid.

Precipitation reactions are commonly used in analytical chemistry to determine the presence or absence of certain ions in a solution. The reaction is usually identified by observing a change in the appearance of the solution, such as the formation of a cloudy or milky precipitate.

The chemical equation for a precipitation reaction can be written as:

[tex]AB(aq) + CD(aq) → AD(s) + CB(aq)[/tex]

where A, B, C, and D are ions, and (aq) and (s) denote aqueous and solid states, respectively.

Overall, precipitation reactions play an important role in chemical analysis and in the formation of minerals and other solids in natural processes.

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The amount of energy required to boil water depends on the initial temperature of the water, the mass of the water, and the heat of vaporization of water.

The heat of vaporization of water is 40.7 kJ/mol or 2.26 kJ/g.

To completely boil 500g of water that is at 15°C, we need to first heat the water to its boiling point (100°C), and then provide the energy required for the phase change from liquid to gas.

The amount of energy required to heat the water from 15°C to 100°C can be calculated using the specific heat capacity of water, which is 4.184 J/g°C:

Q1 = m * c * ΔT

Q1 = 500g * 4.184 J/g°C * (100°C - 15°C)

Q1 = 191,020 J

The amount of energy required for the phase change from liquid to gas can be calculated as follows:

Q2 = m * Hv

Q2 = 500g * 2.26 kJ/g

Q2 = 1,130 kJ

Therefore, the total amount of energy required to completely boil 500g of water that is at 15°C is:

Qtotal = Q1 + Q2

Qtotal = 191,020 J + 1,130 kJ

Qtotal = 1,321,020 J

So, it would require 1,321,020 joules of energy to completely boil 500g of water that is at 15°C.

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No, it is not true that the hospital accidentally switched babies. The trait is most likely due to the inheritance of two recessive alleles.

Toad's family being purebred dominant for red spots on their white cap means that they have two copies of the dominant allele for red spots on their cap.

However, Toadette may carry one copy of the dominant allele and one copy of the recessive allele for white spots on the cap. If Toad also carries one copy of the recessive allele, there is a chance that their offspring may inherit the recessive allele from both parents, resulting in a white cap with white spots.

Therefore, it is entirely possible for Little Toad to inherit the recessive allele for white spots from Toadette and Toad and display the trait. There is no need to suspect the hospital of switching babies as the genetics of the situation explains the observed outcome.

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175.16 grams of[tex]CaCO3[/tex]will be produced when 98.2 grams of [tex]CaO[/tex] are reacted with an excess of [tex]CO2[/tex].

The balanced chemical equation for the reaction between[tex]CaO and CO2[/tex]is:

[tex]CaO + CO2 → CaCO3[/tex]

According to the equation, one mole of[tex]CaO[/tex] reacts with one mole of [tex]CO2[/tex]to produce one mole of [tex]CaCO3[/tex].

The molar mass of [tex]CaO[/tex]is 56.08 g/mol, and the molar mass of [tex]CO2[/tex] is 44.01 g/mol. Therefore, the number of moles of [tex]CaO[/tex] present in 98.2 g can be calculated as:

moles of [tex]CaO[/tex] = mass / molar mass = 98.2 g / 56.08 g/mol = 1.75 mol

Since the reaction is with an excess of [tex]CO2[/tex], all the [tex]CaO[/tex]will react. Therefore, the number of moles of CaCO3 produced will be the same as the number of moles of [tex]CaO[/tex] used, which is 1.75 mol.

The molar mass of [tex]CaCO3[/tex]is 100.09 g/mol. Therefore, the mass of [tex]CaCO3[/tex] produced can be calculated as:

mass of [tex]CaCO3[/tex] = moles of [tex]CaCO3[/tex] × molar mass = 1.75 mol × 100.09 g/mol = 175.16 g

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Total, 0.77 g of I2 is the amount of the excess reagent that remains after the reaction.

To determine the excess reagent remaining, we first need to find the limiting reagent.

The balanced equation tells us that 2 moles of Al react with 3 moles of I₂ to form 2 moles of AlI₃. We can use this information to calculate the theoretical yield of AlI3 based on the amount of each reactant;

moles of Al = 1.80 g / 26.98 g/mol = 0.067 moles

moles of I₂  = 2.30 g / 253.81 g/mol = 0.009 moles

Since the stoichiometry of the reaction is 2:3 for Al and I₂ , respectively, we can see that I₂ is the limiting reagent. Thus, all of the Al will react, while some of the I₂ will be left over.

The amount of AlI₃ that can be formed from the limiting reagent (I2) is:

moles of AlI₃ = 0.009 moles I₂  × (2 moles AlI₃ / 3 moles I₂ )

= 0.006 moles AlI₃

The mass of AlI₃ that can be formed is;

mass of AlI₃ = 0.006 moles × 407.82 g/mol

= 2.47 g

Since we know that only 2.30 g of I₂  was present initially, we can calculate the amount of excess I₂ remaining after the reaction;

excess I₂  = 2.30 g - (0.009 moles I₂  × 253.81 g/mol)

= 0.77 g

Therefore, 0.77 g of reagent that remains after the reaction.

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The number of moles in 1.16 × 10³g of Fe₂O₃ is 7.26 moles.

The number of moles in a substance can be calculated by dividing the mass of the substance by its molar mass as follows:

no of moles = mass ÷ molar mass

According to this question, 1.16 × 10³ grams of iron (II) oxide is given. The molar mass of this compound is 159.69 g/mol.

no of moles in Fe₂O₃ = 1160g ÷ 159.69g/mol = 7.26 moles.

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