If hydrochloric acid reacts with 13.7g of sodium sulfate, how many grams of sulfuric acid are produced?

The given carboxylic acid is reduced via reaction with excess lithium aluminum deuteride. The appropriate acidic workup is performed following this reduction. The final product(s) would best be described as an alcohol.

Lithium aluminum deuteride is a powerful reducing agent used in organic chemistry. Lithium aluminum deuteride is an odorless, white crystalline powder that is soluble in tetrahydrofuran (THF) and diethyl ether (Et2O). It is often utilized as a source of deuterium. When heated, it emits hydrogen and deuterium. Lithium aluminum deuteride (LiAlD4) is a lithium salt of aluminum hydride with deuterium. It is a strong reducing agent and is frequently utilized in organic synthesis.

The process of adding an electron or hydrogen to a substance is known as reduction, and it is the opposite of oxidation. During the reaction of a carboxylic acid with lithium aluminum deuteride, the carbonyl group (C=O) is reduced to an alcohol (R–OH). Acidic workup is used to quench the reaction and neutralize the unreacted reagent after the lithium aluminum deuteride has reduced the carbonyl group in a carboxylic acid.

Carboxylic acids are a class of organic compounds with a carboxyl functional group that consists of a carbonyl group and a hydroxyl group. Acetic acid, formic acid, and butyric acid are examples of common carboxylic acids. The formula R–COOH is used to represent them. The acidity of carboxylic acids is due to the presence of the acidic proton in the hydroxyl group. The hydrogen ion, H+, is generated when the proton is dissociated.

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No, the products obtained from the reaction of ethylbenzene with [tex]Br_2[/tex] and [tex]FeBr_3[/tex] in the presence of light or heat will be different from the products obtained from the reaction of ethylbenzene with [tex]Br_2[/tex] / light or heat.

In the first reaction, [tex]Br_2[/tex] and [tex]FeBr_3[/tex] act as a source of electrophilic bromine, which attacks the aromatic ring of ethylbenzene, leading to the formation of 1-bromoethylbenzene. The mechanism for this reaction is an electrophilic aromatic substitution, where the electrophilic [tex]Br^+[/tex] ion is generated in situ by the reaction of [tex]Br_2[/tex] with [tex]FeBr_3[/tex].

In the second reaction, [tex]Br_2[/tex] acts as a source of free radical bromine, which undergoes a free radical substitution reaction with ethylbenzene, leading to the formation of 1,2-dibromoethylbenzene. This reaction proceeds through a free radical mechanism, where the [tex]Br_2[/tex] molecule is split into two free radicals by the action of light or heat.

Therefore, the products obtained from the two reactions will be different. In the first reaction, 1-bromoethylbenzene will be formed, while in the second reaction, 1,2-dibromoethylbenzene will be formed.

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When 0.0400 mol KOH is added to 1.0 L of a solution that is 0.25 M in NH3 and 0.20 M in NH4NO3, the pH increases only slightly.

The statement that best explains this is that the weak acid (NH4+) will combine with OH- to create a weak base (NH3). Explanation: NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH–(aq)The ammonium ion (NH4+) acts as a weak acid that combines with hydroxide ion (OH–) to form ammonia (NH3) and water (H2O).

It is important to remember that ammonia is not strong enough to raise the pH significantly and that ammonium is a weak acid that won't produce a lot of hydroxides. Therefore, the pH change will be negligible. The explanation for the above reaction is as follows: NH4+ + OH– ⇌ NH3 + H2O In this equilibrium, the weak acid (NH4+) will combine with OH– to create a weak base (NH3), resulting in the pH not rising significantly.

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Answer: The aldehyde or ketone undergoes hydrolysis and forms an alkane, which is the source of octane in the product mixture of this reaction.

The source of octane in the product mixture of this reaction is the hydroboration reaction. This reaction involves the addition of a boron hydride, such as BH3, to an alkene in the presence of a hydrocarbon solvent.

The addition of boron hydride creates a boron-alkyl species, which then reacts with water and is converted into an alcohol. The alcohol then undergoes oxidation and forms an aldehyde or a ketone, depending on the conditions. The aldehyde or ketone then undergoes hydrolysis and forms an alkane, which is the source of octane.

To summarize, the hydroboration reaction of an alkene in the presence of a hydrocarbon solvent produces an alcohol. The alcohol then undergoes oxidation and forms an aldehyde or a ketone, depending on the conditions. The aldehyde or ketone then undergoes hydrolysis and forms an alkane, which is the source of octane in the product mixture of this reaction.

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The mass in grams of potassium chloride in 430 ml of a .193 m potassium chloride solution is 14.4 grams. Potassium Chloride is a compound that contains potassium and chlorine in a 1:1 ratio.

The mass in grams of potassium chloride contained in 430 ml of a .193m potassium chloride solution can be calculated by first determining the molarity of the solution.

Molarity = moles of solute / volume of solution in liters. The solution's molarity is 0.193 mol/L because it is given in the problem statement.

For the quantity of solute, compute the number of moles of solute first:Number of moles of solute = Molarity × volume of solution in liters= 0.193 mol/L × 0.43 L= 0.08299 moles of KCl

The mass of potassium chloride using the molar mass of KCl:Mass of KCl = moles of KCl × molar mass of KCl= 0.08299 moles × 74.55 g/mol (molar mass of KCl)= 6.1819 g = 6.18 g (rounded to two decimal places)

Therefore, the mass in grams of potassium chloride contained in 430 ml of a .193m potassium chloride solution is 14.4 grams.

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This ratio shows that to prepare a buffer with a pH of 4.42 using formic acid, you require the ratio of [sodium formates]/[formic acid] to be 49.23:1.

Explanation:

To prepare a buffer with a pH of 4.42 using formic acid, you need to determine the ratio of [sodium formate]/[formic acid].

A buffer solution is a solution that can resist changes in pH, even when subjected to acid or base. The buffer solution comprises a weak acid or a weak base with its conjugate base or acid, respectively.

Suppose you want to prepare a buffer with a pH of 4.42 using formic acid, with a Ka of 1.8x10^-4. Find the ratio of [sodium format]/[formic acid]. Here, we can use the Henderson-Hasselbalch equation, which is:

pH = pKa + log ([A-] / [HA])

Here, [A-] represents the conjugate base concentration, and

[HA] represents the weak acid concentration.

Rearranging the above equation gives:

log([A-]/[HA]) = pH - pKa putting values gives:

log([A-]/[HA]) = 4.42 - (-log 1.8x10^-4) lo([A-]/[HA]) = 4.42 + 3.74log([A-]/[HA]) = 8.16log([A-]/[HA]) = 1.74                                                                        

Now, taking antilog of both sides: [A-]/[HA] = 10^1.74[A-]/[HA] = 49.23:1

This ratio shows that to prepare a buffer with a pH of 4.42 using formic acid, you require the ratio of [sodium formate]/ [formic acid] to be 49.23:1.

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At standard pressure, which substance becomes less soluble in water as temperature increases from 10.°C to 80.°C is : d. NH4Cl

Solubility of substance increases with temperature, as higher temperatures allow more particles to dissolve in solvent. However, there are some exceptions, where solubility decreases with increasing temperature.

In this case, we are looking for substance that becomes less soluble in water as temperature increases from 10°C to 80°C at standard pressure.

The correct answer is d. NH4Cl.

At standard pressure, solubility of NH4Cl decreases with increasing temperature due to its endothermic dissolution process. As temperature increases, heat absorbed by the solution also increases, which makes dissolution process less favorable. Therefore, solubility of NH4Cl decreases with increasing temperature.

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CaCl2 is a salt that forms as the result of ionic bonds. An ionic bond is a bond that forms between a metal and a nonmetal when they react. One of the atoms will be electronegative, while the other will be electropositive.

When an atom is electropositive, it is more likely to give up its electrons, whereas an electronegative atom is more likely to take up an electron or electrons.

A covalent bond is formed between two nonmetal atoms when they react. Unlike an ionic bond, which occurs between a metal and a nonmetal, a covalent bond occurs between two nonmetal atoms.

The electrons are shared in a covalent bond, with each atom receiving one. As a result, both atoms have a stable number of electrons in their outermost shell.

A bond in which one atom is more electronegative than the other and thus attracts electrons more strongly is known as a polar bond.

The positive end of the molecule is the less electronegative end, and the negative end is the more electronegative end.

A hydrogen bond is a weak bond that occurs between a hydrogen atom and an electronegative atom such as nitrogen, oxygen, or fluorine.

Despite being weak, hydrogen bonds are crucial in many biological processes, such as the formation of DNA. When two atoms are identical, the bond between them is nonpolar.

In the case of a covalent bond, this occurs when the two atoms share electrons equally.

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The molality of CH3OH is 0.03077 m and the mole fraction of CH3OH is 0.1326.

To calculate the molality of CH3OH (methanol) and the mole fraction of solvent in a solution that is 7.50% by mass CH3OH in CH3CH2OH (ethanol), we can use the following steps:

1. Calculate the moles of CH3OH present in the solution:

Mass of CH3OH = 7.50% by mass × 0.100 L solution = 0.00750 L CH3OH

Moles of CH3OH = 0.00750 L ÷ 24.3 g/mol = 0.0003077 mol CH3OH

2. Calculate the molality of CH3OH:

Molality of CH3OH = moles of CH3OH ÷ 0.100 L solution

= 0.0003077 mol ÷ 0.100 L = 0.03077 m

3. Calculate the moles of CH3CH2OH present in the solution:

Mass of CH3CH2OH = 100% - 7.50% = 92.50% by mass × 0.100 L solution = 0.09250 L CH3CH2OH

Moles of CH3CH2OH = 0.09250 L ÷ 46.1 g/mol = 0.002005 mol CH3CH2OH

4. Calculate the mole fraction of CH3OH:

Mole fraction of CH3OH = moles of CH3OH ÷ total moles

= 0.0003077 mol ÷ (0.0003077 mol + 0.002005 mol) = 0.1326

Therefore, the molality of CH3OH is 0.03077 m and the mole fraction of CH3OH is 0.1326.



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A group 2A metal will form a stable ion with a charge of +2. Examples of group 2A metals include magnesium (Mg), calcium (Ca), and strontium (Sr).

A group 3A metal will form a stable ion with a charge of +3. Examples of group 3A metals include boron (B), aluminum (Al), and gallium (Ga).

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1. A metal of group 2A, plus

2. A metal from group 3A, - 3+

A What is charge?

Both positive and negative charges are possible. We are aware that a positive charge is created when a species has more protons than electrons. A negative ion, on the other hand, is one that has more electrons than protons.

We now understand that metals mostly produce positive ions. The group that the metal belongs to in the periodic table determines how much charge is on the ions.

The ions' charges are as follows:

1. A metal of group 2A, plus

2. A metal from group 3A, - 3+

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A cholesterol sample is prepared using acetyl CoA molecules in which both the methyl group and the carboxyl functional group of the acetyl are radiolabeled with 14c. In the cholesterol product, the 14C label would appear in the acetate component.

Cholesterol is a waxy substance that your liver produces and is found in animal-based foods. Cholesterol is crucial for the functioning of your body. It helps your body produce hormones, vitamin D, and bile acids, which aid in the digestion of fat. However, having too much of it in your blood raises your risk of heart disease and stroke. 14C is a radiolabeled carbon isotope. Isotopes are variants of the same element that have a different number of neutrons. Carbon-14 (14C) is an isotope of carbon that has 6 protons and 8 neutrons in its nucleus. In the cholesterol product, the 14C label would appear in the acetate component.

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GaAs (Gallium Arsenide) is a semiconductor widely used to manufacture solid-state lasers in CD and DVD players. GaAs is a compound composed of Gallium and Arsenic. Gallium is a metal, whereas Arsenic is a nonmetal and GaAs make covalent bonds.

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Readers may want to continue reading a work if they are intrigued by the characters, interested in the plot, or invested in the themes and messages presented.

In general, act sets the stage for the rest of the work, introducing key characters, establishing conflicts, and setting the tone and mood.

If a reader finds these elements compelling or engaging, they may be motivated to continue reading to see how the story unfolds and how the characters develop. Additionally, Act I may introduce questions or mysteries that pique the reader's curiosity and encourage them to keep reading to find the answers.

Thus, a reader may want to continue reading a work if they are in interested in the plot.

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When 7.00 mL of 0.100 M HCl(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH3(aq) and 0.100 M in NH4Cl(aq), the pH of the solution decreases by 0.24.

This is because the added acid increases the total concentration of H+ ions in the solution, resulting in a lower pH.

When 7.00 mL of 0.100 M HCl(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in NH3(aq) and 0.100 M in NH4Cl(aq),

the change in pH will depend on the relative amounts of acid and base present in the buffer solution.

In order to calculate the change in pH, we must consider the acid dissociation constants (Ka) for both the NH3 and NH4Cl, as well as the total amount of base and acid in the buffer solution.

The Ka value for NH3 is 1.8 x 10^-5, and the Ka value for NH4Cl is 5.6 x 10^-10.

To calculate the change in pH, we must first calculate the concentrations of the two species present in the buffer solution after 7.00 mL of 0.100 M HCl is added.

The total volume of the solution after the addition of the acid is 107.00 mL. This means that the NH3 concentration is 0.093 M and the NH4Cl concentration is 0.093 M.

Using the Ka values, we can then calculate the total amount of H+ ions present in the solution. This is equal to (1.8 x 10^-5)x(0.093) + (5.6 x 10^-10)x(0.093) = 1.71 x 10^-5.

Using the H+ concentration, we can then calculate the pH of the solution using the formula pH = -log[H+].

In this case, the pH of the solution is equal to 4.76. This means that the change in pH is equal to -0.24, as the original pH of the buffer solution was 5.00.

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The statement, "if a gas is colder than its critical temperature, less pressure is required to liquefy it," is true.

The critical temperature is the temperature at which a gas can't be condensed into a liquid through an increase in pressure alone.

If the temperature exceeds the critical temperature, the gas can only exist as a gas regardless of the pressure applied, and no amount of pressure can cause the gas to condense into a liquid at or above the critical temperature.

A gas is typically liquefied by increasing the pressure and reducing the temperature.

A gas can be condensed into a liquid by reducing the pressure or increasing the temperature if the gas is below its critical temperature.

If the gas is above the critical temperature, no amount of pressure can cause it to liquefy. When a gas is below its critical temperature, less pressure is required to liquefy it.

The relationship between pressure and temperature can be shown using a phase diagram.

A phase diagram is a graph of pressure versus temperature that shows the conditions under which different phases of a substance can exist. The critical temperature is depicted as a point on a phase diagram.

Above the critical temperature, there is no distinction between the gas and liquid phases. Below the critical temperature, the liquid and gas phases can coexist at a specific pressure known as the vapor pressure.

As a result, to liquefy a gas, the pressure must be raised above the vapor pressure at a temperature below the critical temperature. Therefore, if a gas is colder than its critical temperature, less pressure is required to liquefy it.

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The correct option is: Increasing the temperature increases the frequency and force of collisions between gas molecules and the container walls, causing the pressure to increase.

When the temperature of a gas in a closed container is increased, the gas molecules gain kinetic energy and move faster, colliding with the container walls more frequently and with greater force.

According to the kinetic theory of gases, the pressure of a gas is directly proportional to the frequency and force of collisions between gas molecules and the container walls.

Therefore, doubling the temperature of a gas in a closed container would also double the pressure of the gas.

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The pressure of the dry gas alone can be calculated using the ideal gas law: PV = nRT and the pressure is  736.2 torr.

The pressure of dry gas alone is 736.2 torr. Step-by-step explanation: Given that, the Volume of oxygen gas = 250 ml. Temperature = 25 oC Pressure = 760 torr, Vapor pressure of water at 25 oC = 23.8 torrTo find: The pressure of the dry gas alone.

Formula used,V2 = (P1 - P2) * (V1 - Vw) / P2Where,V2 = Volume of gas aloneP1 = Pressure of gas collectedP2 = Vapor pressure of water at temperature T1V1 = Volume of gas collected Vw = Volume of water vapor formedCalculation,P1 = 760 torrP2 = 23.8 torrV1 = 250 mlVw = V1 * P2 / P1= 250 * 23.8 / 760= 7.84 mlV2 = (P1 - P2) * (V1 - Vw) / P2= (760 - 23.8) * (250 - 7.84) / 760= 231.82 mlPressure of dry gas alone = P1 * V2 / V1= 760 * 231.82 / 250= 736.2 torr.

Hence, the pressure of the dry gas alone is 736.2 torr.

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The volume, in liters, of the 40% alcoholic solution needed to make a mixture that is 25% alcohol by volume is 4 L.

To find the amount of 40% alcoholic solution needed to make a mixture that is 25% alcohol by volume, we need to use the following formula:

C₁V₁ + C₂V₂ = CfVf

where C₁ is the concentration of the first solution, V₁ is the volume of the first solution, C₂ is the concentration of the second solution, V₂ is the volume of the second solution, Cf is the desired concentration of the resulting mixture, and Vf is the volume of the resulting mixture.

In this case, we know the first solution is 40% alcohol by volume and the second solution 10% alcoholic by volume, and we need to make a mixture that is 25% alcoholic by volume. We need to know the volume of the first solution, V₁.

Plugging in the values, we get:

C₁V₁ + C₂V₂ = CfVf

0.40V₁ + (0.10)(4) = (0.25)(4 + V₁ )

Solving for the value of V₁, we get:

0.40V₁ + 0.40 = 1 + 0.25V₁

0.15V₁ = 0.60

V₁ = 4

Therefore, 4 liters of the first solution is needed.

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Strong acids are substances that have a high affinity for protons, meaning that they can donate or accept protons in order to form an acid-base equilibrium. The following substances are strong acids: HF, HI, HCl, H2SO4, HNO3, HBr, HClO, HClO2, HClO3, HClO4, H2S, CH3COOH, H3PO4, NH3, NH4Cl, KOH, FeCl3, H2N2, Ca(OH)2, and CH3NH2.

HF is a hydrogen halide and is the strongest of the acids listed above. It is used in industrial applications as a strong oxidizing agent. HI is another hydrogen halide, and it is used in the production of organic compounds. HCl, also known as hydrochloric acid, is a strong acid that is commonly used in the chemical industry. H2SO4 is a strong mineral acid used in the production of fertilizers and dyes.

HNO3 is a strong oxidizing agent and is used in the production of fertilizers and explosives. HBr is a strong acid used in the production of organic compounds. HClO, HClO2, HClO3, and HClO4 are strong oxidizing agents that are used in the chemical industry.

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The percent dissociation of HF is 144%. This result may seem greater than 100%, but it is possible for the percent dissociation to exceed 100% in cases where the concentration of the dissociated species exceeds the initial concentration of the undissociated species.

What is Percent Dissociation?

Percent dissociation is a measure of the extent to which a substance dissociates in a solution. It is defined as the ratio of the concentration of the dissociated species to the initial concentration of the substance, expressed as a percentage.

The first step in solving this problem is to write the equation for the dissociation of hydrofluoric acid (HF) in water:

HF + H2O ⇌ H3O+ + F-

Ka = [H3O+][F-] / [HF]

Since the pH of the solution is given as 4, we know that:

[H3O+] = 10^-4 M

We can use the given initial concentration of HF and the expression for Ka to solve for the concentration of F- at equilibrium. Since HF is a weak acid, we can assume that the dissociation is small compared to the initial concentration, so we can use the approximation [HF] ≈ [HF]0.

Ka = [H3O+][F-] / [HF]0

[F-] = Ka [HF]0 / [H3O+]

[F-] = (7.2 × 10^-4)(0.2 mol / 0.1 L) / (10^-4 M)

[F-] ≈ 0.288 M

The percent dissociation of HF is defined as:

% dissociation = ([F-] / [HF]0) × 100%

% dissociation = (0.288 M / 0.2 mol / 0.1 L) × 100%

% dissociation = 144%

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Answer: The solution that will precipitate sulfate ions is B. BaCl2.

How do you test for sulfate ions?

The most reliable test for sulfate ions is to add a few drops of barium chloride to the test solution. If sulfate ions are present, they will combine with the barium ions to create a white precipitate of barium sulfate.

In the presence of barium ions, sulfuric acid is added to the test solution to look for the sulfate ions that are there. A white precipitate of barium sulfate is formed as a result of the reaction.

The production of a white precipitate of barium sulfate means that sulfate ions are present. In order to eliminate carbonates and other anions, the test solution should be treated with a few drops of dilute hydrochloric acid before testing.

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High-performance liquid chromatography (HPLC)  is the method used.

The process of chromatography separates mixtures into their constituents by distributing the constituents of a mixture between two phases: a stationary phase and a mobile phase.

Separation is based on the differential partitioning of analytes between these two phases.

The resolution of a chromatographic separation is a function of the differences in retention times and peak widths between two peaks of interest.

The resolution between two compounds for a liquid separation using a packed chromatography column can be improved using several methods.

Here are some of the methods that can be used to improve the resolution between two compounds for a liquid separation using a packed chromatography column:1.

Using a smaller particle size. A smaller particle size stationary phase decreases HETP and broadens the range of flow rates that can be used for a separation, providing higher resolution.2.

Increasing the length of the column. A longer column provides a larger surface area, more separation can occur, and thus higher resolution can be obtained.3. Changing the particle size distribution.

Changing the particle size distribution of the stationary phase can result in a greater variation of pore sizes, resulting in a greater variety of interactions between the analytes and the stationary phase.

This leads to an increase in resolution.4. Changing the solvent or buffer system. Altering the solvent or buffer system to optimize the separation conditions can result in an increase in resolution.

Solvent changes, pH changes, or changing the ionic strength of the buffer system can be used.5. Modifying the temperature.

Modifying the temperature can affect the degree of analyte interaction with the stationary phase, thereby affecting the separation.

It is also necessary to note that liquid chromatography, which is frequently referred to as high-performance liquid chromatography (HPLC),

has a variety of advantages over gas chromatography (GC), which are better suited for volatile or small molecular weight analytes.

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The true statement given about the bonding electrons is option b. "The bonding electrons will spend more time around the O atom as it attracts the electrons more strongly".

Carbon dioxide is a linear molecule that consists of two oxygen atoms and one carbon atom. The C-O bond in [tex]CO_2[/tex] is polar, which means that the electrons are shared unequally between the atoms. As oxygen is more electronegative than carbon, it attracts the electrons more strongly, and hence, the bonding electrons spend more time around the O atom than the C atom.

In other words, option b is the correct statement about the bonding electrons in carbon dioxide ([tex]CO_2[/tex]) molecule.

Thus bonding electron spends more time around the O atom as it attracts the electrons more strongly than the C atom.

Therefore correct option is b. The bonding electrons will spend more time around the O atom as it attracts the electrons more strongly.

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Answer:itd a bro

Explanation:dont trust just need points

To dilute 45.0 ml of a 1.20 M solution of NaBr to a 0.0400 M solution, you need to add enough water to a total volume of 226.25 ml.

The dilution formula is M1V1 = M2V2, where M1 and V1 are the initial molarity and volume of the solution and M2 and V2 are the desired molarity and volume of the dilute solution.

Calculate V2 (the desired volume) by rearranging the equation and solving for V2: V2 = (M1V1) / M2.

V2 = (1.20M * 45.0ml) / 0.0400M = 226.25ml.

Therefore, to create a 0.0400 M solution of NaBr from a 1.20 M solution of NaBr, you need to add enough water to a total volume of 226.25 ml.

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Answer:

carbon-magnesium

Explanation:

H3C - Mg - Br

Formation and dissociation constants are related in that they are inverses of each other and must have a product that equals the equilibrium constant of the reaction.

Formation and dissociation constants, also known as Kf and Kd respectively, represent the equilibrium concentrations of the reactants and products in a chemical reaction.

Kf is the constant of formation, which is the product of the concentrations of the products of the reaction, divided by the product of the concentrations of the reactants.

Kd is the dissociation constant, which is the product of the concentrations of the reactants, divided by the product of the concentrations of the products .

Kf and Kd are related in that the product of Kf and Kd must equal Kw, which is the equilibrium constant for the reaction.

The value of Kw is constant, meaning that regardless of the equilibrium concentrations of reactants and products, Kf and Kd must be inverses of each other such that the product of Kf and Kd must equal Kw.

Therefore, formation and dissociation constants are related in that they are inverses of each other and must have a product that equals the equilibrium constant of the reaction.

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The one contains the less solute at the given temperature and the pressure is the unsaturated solution.

The unsaturated solution is the solution that contains the less solute than the saturated solution at the given temperature and the pressure. The Unsaturated solutions are the solutions in which the amount of the dissolved solute is the less than the saturation point of solvent.

If the amount of the dissolved solute will be equal to the saturation point of solvent, then the solution is called the saturated solution. The solution in the which the solute can further to be dissolved at the any fixed temperature is called the unsaturated solution.

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An ionic equation shows species dissolved in solution. This equation is the most accurate representation of the chemical change occurring.

What is an ionic equation? An ionic equation is a type of chemical equation that shows the dissociated species in a when ionic compounds are involved.                                                                                               Only the ions that react or are changed during the reaction are shown in this type of equation.A chemical change is the process of converting one substance to another through chemical reactions. When one or more substances undergo a chemical reaction to create a new substance with new properties, a chemical change occurs. The reactants are transformed into new substances through a chemical change

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At Standard Temperature and Pressure (STP) the volume of the gas at is 216.1 ml.

A sample of ideal gas occupies 208ml at 36.2 degree Celsius and 704 torr what is the volume at stp

For a sample of ideal gas, the relationship between volume, pressure, and temperature is given by the Ideal Gas Law:

PV = nRT

Where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, R is the universal gas constant, and T is the temperature of the gas.

At STP (Standard Temperature and Pressure), the pressure of the gas is 1 atm and the temperature of the gas is 0°C (273 K). Therefore:

P1 = 704 torr

V1 = 208 mL

T1 = 36.2°C = 309.35 K

P2 = 1 atm

V2 = ?

T2 = 0°C = 273 K

To find V2, we can use the following equation:

V2 = V1(P2/P1)(T1/T2)

Plugging in the given values:

V2 = 208 mL (1 atm/704 torr) (309.35 K/273 K)

V2 = 208 mL (0.939) (1.132)

V2 = 216.1 mL

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