PLS HELP ITS DUE IN 10 MINS!! How many grams of argon (Ar) are present in 2.35 x 10^24 atoms of argon?

The per cent ionization of the solution is 0.017%. The steps involved are;

Write the reaction equation

Setup the ICE table

make the necessary calculation

The equation of the reaction is as follows;

CH₃CH₂CH₂COOH(aq)  ⇄ [tex]H^{aq}[/tex]  + CH₃CH₂CH₂COO[tex]^{-aq}[/tex]

I       0.0075                               0                0.085

C    -x                                        +x             + x

E     0.0075 - x                               x               0.085 + x

The Ka of the acid = 1.5 x 10⁻⁵

Hence;

1.5 x 10⁻⁵ = x(0.085 + x)/0.0075 - x

1.5 x 10⁻⁵ (0.0075 - x ) =  x(0.085 + x)

1.1 x 10⁻⁷ - 1.5 x 10⁻⁵ˣ = 0.085x + x²

x² + 0.085x - 1.1 x 10⁻⁷ = 0

x = 0.0000013 M

Percent ionization=  0.0000013 M/0.0075 × 100/1

= 0.017%

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The given statement "at low temperatures, intermolecular forces become important and pressure of the gas will be higher than predicted by ideal gas law" is false. Because, at low temperatures, intermolecular forces become more significant and cause gas molecules to come closer together, which reduces the volume of the gas.

As a result, the pressure of the gas will be lower than predicted by the ideal gas law, which assumes that gas molecules have no volume and do not interact with each other.

The ideal gas law is only accurate for gases that behave like ideal gases, meaning that they have negligible intermolecular forces and occupy no volume. Real gases, on the other hand, deviate from the ideal gas law at high pressures and low temperatures, where intermolecular forces become more significant.

At these conditions, the volume of the gas becomes significant compared to the volume of the container, and the gas behaves less ideally. In this case, we need to use more complex equations of state to accurately predict the behavior of the gas.

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When we add more carbon in the atmosphere it warms the planet and helps the plants on land grow more. Excess of carbon can makes the water more acidic.

When carbon dioxide is added to atmosphere it reacts with water to form carbonic acid from which hydrogen ions dissociate and results in increasing the acidity of the system. So we can say that in addition to any greenhouse effect the anthropogenic carbon dioxide emissions to the atmosphere can increase the acidity of the atmosphere and precipitation of the atmosphere. Most of the plant species show higher rates of photosynthesis which cause increased growth and the decreased water use and lowered tissue concentrations of nitrogen and protein of the water.

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The correct question is,

What happens when too much CO2 enters the atmosphere?

The higher temperature will witness more volume in gas when the amount of reactants is same.

The Charle's Law states that the volume is directly proportional to the temperature when other parameters are kept constant. Thus, the increase in temperature will accompany increase in volume.

The reason for such phenomenon is expansion of gases on heating due to excess absorption of energy. It is available in the form of kinetic energy which subsequently is evident through square of velocity.

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The replacement of the polar amino acid residue His 57 with a nonpolar residue in the chymotrypsin active site could impair the proper positioning of catalytic residues, potentially reducing or eliminating the enzyme's activity.

His 57 forms a hydrogen bond with Ser 195 in the chymotrypsin active site. His 57 is a polar amino acid residue, and its interaction with Ser 195 contributes to the proper positioning of the catalytic residues in the active site.

If His 57 were replaced by a nonpolar amino acid residue, such as alanine or valine, the hydrogen bond with Ser 195 would be lost. This could result in a change in the positioning of the catalytic residues in the active site, which could impair the ability of chymotrypsin to properly interact with its substrate and perform its enzymatic function. Therefore, the replacement of His 57 with a nonpolar amino acid residue could potentially reduce or eliminate the enzymatic activity of chymotrypsin.

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Answer:

Yes, the synthesis strategy can give a substitution reaction with the expected regiochemistry and stereochemistry

Explanation:

The expected product of the forward reaction would be a single enantiomer of the substituted product, depending upon the starting materials used. For example, if the starting material is an unsubstituted chiral compound, then the product would be a single enantiomer of the desired product. If the starting material is a racemic mixture, then the product would be a racemic mixture of the desired product. In detail, a substitution reaction involves the replacement of a functional group in a molecule with a different functional group. This process can be catalyzed by a transition metal catalyst, such as a palladium or nickel complex. The reaction is usually carried out in the presence of a base, such as sodium hydroxide or potassium hydroxide, and a nucleophile, such as an alcohol or an amine.

When it comes to synthesis, the goal is to create a compound or molecule from simpler components through chemical reactions. In terms of regiochemistry and stereochemistry, these refer to the specific spatial orientation of atoms or groups in a molecule and the effects this has on reactivity and other properties.

In regards to the question, it is difficult to provide a specific answer without more information about the specific reaction and compounds involved. However, in general, the expected product of a substitution reaction will depend on factors such as the type of substitution reaction (e.g. nucleophilic, electrophilic), the nature of the substituent groups involved, and the reaction conditions. For example, in a nucleophilic substitution reaction, the incoming nucleophile will typically replace a leaving group on the substrate molecule. The regiochemistry and stereochemistry of the product will depend on the location and orientation of the leaving group and the incoming nucleophile, as well as any other factors that may influence the reaction. In summary, the expected product of a forward reaction in a substitution reaction will depend on a variety of factors related to the reaction and the compounds involved.

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A polar solvent is prepared by mixing 27.5 mL of propanone with 222.5 mL of water the percentage by volume of propanone in the mixture is 11%.

The percentage by volume of propanone in the mixture can be calculated as follows:

Find the total volume of the mixture:

Total volume = volume of propanone + volume of water

Total volume = 27.5 mL + 222.5 mL

Total volume = 250 mL

Calculate the volume fraction of propanone:

Volume fraction of propanone = volume of propanone / total volume

Volume fraction of propanone = 27.5 mL / 250 mL

Volume fraction of propanone = 0.11

Convert the volume fraction to a percentage:

Percentage by volume of propanone = volume fraction of propanone x 100%

Percentage by volume of propanone = 0.11 x 100%

Percentage by volume of propanone = 11%

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The measured Kc is different from the calculated Kc, the system is not at equilibrium. There is either too much hydrogen iodide and/or too little hydrogen and/or iodine in the cylinder.

To determine whether the system is at equilibrium, we can use the equilibrium constant expression for the reaction;

H₂(g) + I₂(g) ⇌ 2HI(g)

The equilibrium constant expression for this reaction is;

Kc = [HI]² / [H₂] [I₂]

We are given the partial pressures of each component in the cylinder, so we can calculate the concentrations using the ideal gas law.

[P] = n/V = (m/M) / V

where P is the partial pressure, n is the number of moles, V is the volume, m is the mass, M is the molar mass, and the square brackets denote concentration.

For hydrogen (H₂)

[P] = n/V = (m/M) / V

[m/M] = [P] x V

[m/M] = 0.127 atm x V / R x T

For iodine (I₂)

[P] = n/V = (m/M) / V

[m/M] = [P] x V

[m/M] = 0.134 atm x V / R x T

For hydrogen iodide (HI)

[P] = n/V = (m/M) / V

[m/M] = [P] x V

[m/M] = 1.055 atm x V / R x T

Switching these expressions into the equilibrium constant expression, we get;

Kc = ([HI]/V)² / ([H₂]/V) x ([I2]/V)

Kc = ([HI]² / V²) / ([H₂] x [I2] / V²)

Put in the values we get:

Kc = [(1.055 atm / V)² / (0.127 atm / V) x (0.134 atm / V)]

Kc = 8.37 atm² / V²

If the system is at equilibrium, the measured concentrations should give the same value for Kc as calculated above. If the measured Kc is different, then the system is not at equilibrium.

Therefore, we can calculate the measured Kc as;

Kc = ([HI]² / V²) / ([H₂] x [I2] / V²)

Kc = [(1.055 atm / V)² / (0.127 atm / V) x (0.134 atm / V)]

Kc = 8.87 atm² / V²

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The heat released during the reaction is -212 kJ. Heat is a fundamental concept in physics, chemistry, and engineering and plays a critical role in many natural phenomena, such as thermodynamics, phase transitions, and thermal radiation.

What is Heat?

Heat is a form of energy that can be transferred from one object to another as a result of a difference in temperature. Heat flows from hotter objects to colder objects until they reach thermal equilibrium, meaning that their temperatures become equal. The amount of heat transferred is typically measured in joules (J) or calories (cal) and is related to the mass of the object, the specific heat capacity of the material, and the temperature change experienced.

The ΔH for the reaction is -847 kJ.

The stoichiometric coefficient of Al in the balanced equation is 2. This means that 2 moles of Al are required to produce 2 moles of Fe and 1 mole of [tex]Al_{2} O^{3}[/tex].

Since the reaction uses 4.0 mol of Al, it will produce 2.0 mol of Fe and 1.0 mol of [tex]Al_{2} O^{3}[/tex].

The amount of heat released during the reaction can be calculated using the equation:

ΔH = q/n

where ΔH is the enthalpy change, q is the heat released, and n is the number of moles of the limiting reactant (in this case, Al).

Substituting the values gives:

ΔH = (-847 kJ) / 4.0 mol = -212 kJ/mol

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Answer:

The main places you would find glaciers on in today's world are places like Alaska, Antarctica, and Greenland. Places that are cold either year-round or most of the year and are in an ocean area are typically where glaciers form.

Explanation:

These places typically are cold year-round and are the perfect area for glaciers to form, but not everywhere forms glaciers as they need certain weather conditions to form.

Answer: 0.5 atm

Explanation: To calculate the partial pressure of carbon dioxide, we need to use the mole fraction of carbon dioxide and the total pressure of the container.

The mole fraction of carbon dioxide is:

X(CO2) = n(CO2) / n(total)

X(CO2) = 0.5 moles / (0.5 moles + 0.32 moles + 0.2 moles)

X(CO2) = 0.4762

The partial pressure of carbon dioxide is:

P(CO2) = X(CO2) * P(total)

P(CO2) = 0.4762 * 1.05 atm

P(CO2) = 0.5 atm (rounded to one decimal place)

Therefore, the partial pressure of carbon dioxide in the container is 0.5 atm.

Answer:

0.58 atm

Explanation:

5 to 10 ml urine should be collected for clean-catch procedure. Thus Option A is the correct answer.

In the clean-catch procedure, 5 to 10 ml is adequate amount of urine because more than this quantity will cause a difficulty in sampling the solution. For this process to work there is a set of criteria that needs to be followed to ensure that the given sample is tested with readiness with accuracy.

The given criteria are the sample must be stored away in an compact space, the sample must be free from any impurities, the sample must be fresh for the process to run smoothly. Furthermore, this set of process is crucial for finding the cause of the disease that are caused by bacteria.

This type of process is adequate for all age groups ranging from infants to adults.

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When a gas expands from 271 ml to 903 ml at a constant temperature, the work done (in joules) by the gas if it expands in vacuum is O J

According to the given data:

Initial volume occupied by the gas, V₁=271 ml =0.271 L

Final volume occupied by the gas, V₂= 903 ml =0.903 L

To calculate the work done (in joules) by the gas if it expands against a vacuum, We can use the following expression.

w = -P × ΔV

Since the gas expands against a vacuum, pressure will be equal to zero

P = 0.

Thus, w = 0 J

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The fossil is approximately 17,100 years old.

The age of a fossil can be estimated by using the half-life of carbon-14. The half-life of carbon-14 is approximately 5,700 years, which means that after this time, half of the carbon-14 in a sample will have decayed into nitrogen-14.

If a fossil contains 1 part carbon-14 to 7 parts nitrogen-14, this means that the carbon-14 has decayed to one-eighth (1/8) of its original amount. Since we know the half-life of carbon-14, we can use this information to estimate the age of the fossil.

If we assume that the fossil originally contained only carbon-14, then the number of half-lives that have passed can be calculated as follows:

1/2n = 1/8

where n is the number of half-lives that have passed.

Simplifying this equation, we get:

2n = 8

n = 3

Therefore, the fossil has undergone 3 half-lives of carbon-14 decay. Since the half-life of carbon-14 is approximately 5,700 years, we can estimate the age of the fossil by multiplying the half-life by the number of half-lives:

age = 5,700 years/half-life * 3 half-lives

age = 17,100 years.

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The mixture is placed in an ice-bath during the addition of the concentrated sulfuric acid for multiple reasons. Firstly, the concentrated sulfuric acid is highly reactive and exothermic.

When added to the mixture, it can cause a sudden rise in temperature, leading to undesired side reactions. Placing the mixture in an ice-bath helps to control the reaction temperature and avoid any unwanted reactions.

Secondly, the ice-bath helps to reduce the sulfonation of the 1,4-dimethoxybenzene. Sulfonation can lead to the formation of undesired products and reduce the yield of the desired product.

Finally, placing the mixture in an ice-bath can also help to reduce the formation of butene, which is an undesired byproduct that can form at higher temperatures. The cooling effect of the ice-bath helps to prevent the formation of butene and favor the formation of the FC alkylation product.

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Full Question ;

Why is the mixture placed in an ice-bath during the addition of the concentrated sulfuric acid? (Check all that apply.) MULTIPLE ANSWER A. To avoid the evaporation of the concentrated sulfuric acid. B. To increase the rate of the reaction. C. To favor the formation of the FC alkylation product. D. To reduce the sulfonation of the 1,4-dimethoxybenzene E. To reduce the formation of butene

The limiting reagent is nitrogen. The number of molecules of ammonia formed are 8.97. The number of hydrogen molecules in excess is 3.

You might have learnt that the formation of ammonia can be represented by:

N₂ + 3H₂ → 2NH₃

From the image provided by you, you have 6 molecules of hydrogen and 3 molecules of nitrogen. To determine the limiting reagent, you'll need to compare the number of moles of each reactant with the stoichiometric ratio in the balanced equation.

For nitrogen, you have:

3 molecules N₂ × 1 mole N₂/6.022 × 10²³ molecules N₂ = 4.98 × 10⁻²⁴ moles N₂

For hydrogen:

6 molecules H₂ × 1 mole H₂/6.022 × 10²³ molecules H₂ = 9.96 × 10⁻²⁴ moles H₂

Using the stoichiometric ratio from the balanced equation, you'll see that 1 mole of N₂ reacts with 3 moles of H₂ to produce 2 moles of NH₃.

Based on the calculations above, you can deduce easily that the amount of nitrogen is limiting because you only have 4.98 × 10⁻²⁴ moles of N₂, which is less than the amount of hydrogen required to react with it.

To find out how many molecules of NH₃ are formed, you'll have to use the amount of limiting reagent in your calculation. In this case, you're with 3 moles of NH₃ produced for every 1 mole of N₂, so calculate:

4.98 × 10⁻²⁴ moles N₂ × 3 moles NH₃/1 mole N₂ = 1.49 × 10⁻²³ moles NH₃

To convert this to molecules of NH₃, can use Avogadro's number:

1.49 × 10⁻²³ moles NH₃ × 6.022 × 10²³ molecules/mole = 8.97  molecules NH₃

So, 8.97 molecules of NH₃ are formed.

To determine the excess reactant, you first calculate the amount of hydrogen remaining after the reaction:

Amount of H₂ consumed = 3 moles NH₃ × 3 moles H₂/1 mole NH₃ = 9 moles H₂

Amount of H₂ remaining = 6 moles H₂ - 9 moles H₂ = -3 moles H₂

The negative value for the amount of hydrogen remaining indicates that there is an excess of hydrogen by 3 moles.

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Based on the solubility rules, the compound D. BaSO4 is insoluble in water.

This is because barium sulfate is one of the exceptions to the general solubility rule for sulfates (SO4^2-). While most sulfates are soluble in water, those containing barium (Ba^2+), along with lead (Pb^2+), and calcium (Ca^2+) are not soluble.

In contrast, the other compounds listed are soluble:
A. Rb2SO4 - Rubidium sulfate is soluble because alkali metal salts (like rubidium) are generally soluble in water.
B. Na2SO4 - Sodium sulfate is soluble for the same reason as rubidium sulfate - sodium is an alkali metal.
C. CuSO4 - Copper sulfate is soluble because most sulfate salts are soluble in water.
E. MgSO4 - Magnesium sulfate is soluble, as it is also a sulfate salt.

In summary, BaSO4 is the only compound among the given choices that is insoluble in water due to the specific solubility rules for sulfate compounds. Therefore the correct option is D

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The correct options are AB+CD →AD+CB, Oxygen, HF, Ba(OH)₂, Na₂SO₄+H₂O, They are double displacement reactions.

When pieces of two ionic compounds are exchanged, two new compounds are created. These reactions are known as double displacement reactions. Chemical reactions known as double displacement reactions occur when the reactant ions move around to create new products. Precipitate production often happens as a result of a double displacement process. Covalent or ionic chemical bonds may be present between the reactants. Iron sulphate is produced as a result of the interaction between iron and copper sulphate. Because iron is more reactive than copper in this situation, copper is replaced. Zinc and iron sulphate react, producing zinc sulphate as a byproduct.

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It will take approximately 9340 seconds or 2.59 hours to deposit 66.5 g of Zn on the steel gate when 21.0 A of current is passed through the ZnSO₄ solution.

To calculate the time required to deposit 66.5 g of Zn on a steel gate, we can use Faraday's laws of electrolysis, which states that the amount of substance produced (in moles) at an electrode is directly proportional to the amount of electric charge passed through the electrode.

The formula for calculating the amount of substance produced is;

n = Q / (zF)

where; n will be the amount of substance produced (in moles)

Q is the electric charge passed through the electrode (in coulombs)

z is the number of electrons transferred per molecule of Zn (2 for Zn)

F is Faraday's constant (96485 C/mol)

We can first calculate the number of moles of Zn that are deposited on the gate using the given mass and molar mass of Zn

n(Zn) = m(Zn) / M(Zn)

n(Zn) = 66.5 g / 65.38 g/mol

n(Zn) = 1.018 mol

Next, we can calculate the electric charge required to deposit this amount of Zn using the formula;

Q = n × z × F

Q = 1.018 mol × 2 × 96485 C/mol

Q = 196293 C

Finally, we can use the formula for electric current (I = Q / t) to calculate the time required (t) to pass an electric charge of 196293 C through the solution at a current of 21.0 A;

t = Q / I

t = 196293 C / 21.0 A

t = 9340 s or 2.59 hours

Therefore, it will take 9340 seconds.

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If you use Copper (Cu) as the metal , Oxygen (O) and Carbon (C) as the c, then you can make 2 new compounds in five minutes.

The two possible compounds are:

Copper oxide (CuO) is a chemical compound composed of copper and oxygen. It is a black, solid material with a high melting point and is commonly referred to as cupric oxide or black oxide of copper.

CuO can be produced by heating copper in the presence of oxygen, or by reacting copper sulfate with sodium hydroxide. It is commonly used as a raw material in the production of copper salts, in the manufacturing of ceramics, as a pigment, and as a catalyst in various chemical reactions.

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Answer:

≈15

Explanation:

1.43-.96=.57

.57/1.43 (change / original) * 25 ≈ 9.96

25 - 9.96 ≈ 15

SCUBS

Explanation:

Scubs is the form of pop by nayeon and Liz doesn't get line kitch

The white solid could be silver chloride (AgCl) which is insoluble in water, insoluble in excess ammonia solution and soluble in dilute HCl.

AgCl is insoluble in water, and when it reacts with excess ammonia solution, it forms a complex with the formula [Ag(NH₃)²⁺]. Therefore, AgCl is insoluble in excess ammonia solution. When AgCl is treated with dilute hydrochloric acid (HCl), it dissolves to form a solution containing silver ions (Ag⁺) and chloride ions (Cl⁻). AgCl is a white crystalline solid with a high melting point (455°C) and boiling point (1,155°C). It is sparingly soluble in water, and its solubility decreases as the temperature decreases. AgCl is also insoluble in organic solvents.

AgCl is used in various applications, such as in the production of silver electrodes, in electroplating, and as a reagent in chemical analysis. AgCl is considered to be relatively non-toxic, but it can be harmful if ingested or inhaled in large quantities. It is also considered an environmental hazard and should be handled and disposed of with care.

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Answer:

A. Physical B.Chemical C. Physical

Explanation:

Shiny is a physical Mercury is liquid which is physical and sliver added into calcium is a chemical property.

The pH of the solution after 50.0 mL of 0.500 M HCl has been added is 4.74.

The balanced chemical equation for the reaction between NH₃ and HCl is:

NH₃ + HCl → NH₄Cl

This reaction is a neutralization reaction, in which an acid (HCl) reacts with a base (NH₃) to form a salt (NH₄Cl) and water.

Since NH₃ is a weak base and HCl is a strong acid, the reaction will go to completion, and all of the NH₃ will react with the HCl. Before any HCl is added, the solution contains only NH₃, so its pH is basic. As HCl is added, it reacts with the NH₃ to form NH₄Cl, which is a neutral salt. The pH of the solution will decrease as more HCl is added, until all of the NH₃ has reacted and the solution becomes acidic.

To find the pH of the solution after 50.0 mL of 0.500 M HCl has been added to 100.0 mL of 0.500 M NH₃, we need to calculate the moles of NH₃ and HCl that have reacted.

Moles of NH₃ = (100.0 mL)(0.500 M) = 0.0500 moles

Moles of HCl = (50.0 mL)(0.500 M) = 0.0250 moles

Since NH₃ and HCl react in a 1:1 ratio, all of the HCl has reacted with half of the NH₃, leaving 0.0250 moles of NH₃ unreacted.

To calculate the concentration of NH₄⁺ ions in the solution, we need to consider the equilibrium reaction between NH₃  and NH₄⁺ :

NH₃ + H₂O ⇌ NH₄⁺ + OH⁻

The equilibrium constant for this reaction is:

Kb = [NH₄⁺][OH⁻] / [NH₃]

At equilibrium, the concentration of NH₃ will be 0.0250 moles / 0.150 L = 0.167 M (since the total volume of the solution is 150.0 mL). The concentration of OH- ions can be calculated using the Kb value for NH₃, which is 1.8 × 10⁻⁵:

Kb = [NH₃⁺][OH⁻] / [NH₃]

1.8 × 10⁻⁵ = [NH₄⁺][OH⁻] / 0.167

[NH₄⁺][OH⁻] = 3.006 × 10⁻⁶

Since the solution is neutral at equilibrium, the concentration of H+ ions is equal to the concentration of OH- ions. Therefore, the pH of the solution is:

pH = -㏒[H⁺]

[H⁺] = [OH⁻] = √(Kw / [NH₄⁺])

                     = √(1.0 × 10⁻¹⁴/ 3.006 × 10⁻⁶) = 1.83 × 10⁻⁵ M

               pH = -㏒(1.83 × 10⁻⁵) = 4.74

As a result, the pH of the solution after adding 50.0 mL of 0.500 M HCl is 4.74.

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THF is a poor solvent for Friedel-Crafts reactions using aluminum chloride or boron trifluoride because THF forms stable adducts with these compounds, THF is a strong Lewis base, and THF has a very low boiling point. Here options B, C, and D are correct.

THF is a polar aprotic solvent that is commonly used for many organic reactions, but it is a poor solvent for Friedel-Crafts reactions that use aluminum chloride or boron trifluoride. This is because THF can form stable adducts with these compounds, which can reduce their reactivity and prevent them from effectively catalyzing the reaction. Therefore, option B is correct.

In addition, THF is a strong Lewis base, which means that it can coordinate with Lewis acids like aluminum chloride or boron trifluoride and form adducts. This further reduces the reactivity of the Lewis acid, making the reaction less effective. Therefore, option C is also correct.

Finally, the boiling point of THF is relatively low (66°C), which can make it difficult to maintain the reaction at the desired temperature. This can affect the reaction kinetics and the yield of the product. Therefore, option E is also correct.

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Complete question:

Why is THF (tetrahydrofuran) a poor solvent for Friedel-Crafts reactions that use aluminum chloride or boron trifluoride?

A. THF is unstable in the presence of these compounds

B. THF forms stable adducts with these compounds

C. THF is a strong Lewis base

D. THF is a strong Lewis acid

E. The boiling point of THF is too low.

The chemical equation for the reaction between sulfur and carbon to produce carbon disulfide is:

S2(g) + C(s) → CS2(g)

To determine the number of grams of CS2(g) that can be prepared, we need to use the stoichiometry of the reaction and the ideal gas law.

We need to calculate the number of moles of CS2(g) that can be produced by 9.80 mol of S2(g). From the equation, we can see that one mole of S2(g) reacts with one mole of C(s) to produce one mole of CS2(g). Therefore, the number of moles of CS2(g) produced will be equal to 9.80 mol.

We need to use the ideal gas law to calculate the volume of the CS2(g) produced. We know that the reaction vessel has a volume of 5.75 L and is held at 900 K. Assuming the pressure remains constant, we can use the ideal gas law: PV = nRT

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

We can rearrange this equation to solve for n: n = PV/RT

Substituting the values we have:

n = (1 atm)(5.75 L)/(0.08206 L·atm/mol·K)(900 K)

n = 0.318 mol

Therefore, the mass of CS2(g) produced will be:

mass = n × M

where M is the molar mass of CS2(g) which is 76.14 g/mol.

mass = 0.318 mol × 76.14 g/mol

mass = 24.2 g

Therefore, 24.2 grams of CS2(g) can be prepared by heating 9.80 mol S2(g) with excess carbon in a 5.75 L reaction vessel held at 900 K until equilibrium is attained.

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The symbol that is a measure of the favorability of a reaction, considering both the enthalpy and entropy, is Gibbs free energy (ΔG).

The Gibbs free energy change of a reaction takes into account the change in enthalpy (ΔH) and the change in entropy (ΔS) of the system, and is given by the equation:

ΔG = ΔH - TΔS

where T is the temperature in Kelvin.

A negative value of ΔG indicates that the reaction is spontaneous and favors the formation of products, while a positive value of ΔG indicates that the reaction is non-spontaneous and favors the formation of reactants. At equilibrium, ΔG is zero, indicating that the reaction is in a state of balance between the reactants and products.

Find out more about enthalpy and entropy

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Agriculture: Copper (II) sulfate is used as a fungicide to control plant diseases, while iron metal is used as a nutrient for plants.

Fe + CuSO4 Reaction.

Reactant and state: Copper (II) sulphate, solid (CuSO4.5H2O)

Product and state: Iron (II) sulphate, solid

(FeSO4) and Copper metal, solid (Cu)

Word equation: Iron metal + Copper (II) sulphate sulphate - Copper metal + Iron (II)

Balanced formula: Fe + CuSO4 Cu + FeSO4

Type of reaction: This is a single displacement or substitution reaction, where iron replaces copper in the copper sulfate compound.

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