which interactions can contribute to the intrinsic binding energy during enzymatic catalysis?electrostatic interactionspermanent covalent bondingvan der waals interactionsnucleophilic attack by serinehydrogen bonding

Answer:

G, A, A, T, C, C, G, A, A, T, G, G, T

Explanation:

When a strong acid and a strong base are mixed in equal amounts, they undergo a neutralization reaction, resulting in the formation of water and a salt.

Therefore, the mixture of 1 M NaOH and 1 M HCl will not result in a neutral solution but instead will form sodium chloride and water. On the other hand, the mixture of 1 M NH3 and 1 M HCl, and the mixture of 1 M KOH and 1 M HBr, will also undergo neutralization reactions but will result in the formation of ammonium chloride and potassium bromide, respectively. The mixture of 1 M NaOH and 1 M HI will also not result in a neutral solution but will form sodium iodide and water due to the reaction between the strong base NaOH and the weak acid HI.

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The net ionic equation for the acid-base hydrolysis equilibrium that is established when ammonium nitrate is dissolved in water is given below:Answer: NH4+(aq) + H2O(l) ⇌ H3O+(aq) + NH3(aq) + NO3-(aq).

The ionic equation is a chemical equation in which the electrolytes in aqueous solution are represented by their actual ions rather than their complete formulas. It indicates that ions undergo a chemical reaction to produce a new compound.

The net ionic equation displays the actual chemical reaction taking place in an aqueous solution. It is derived by eliminating spectator ions, which do not play any active role in the chemical reaction.The acid-base hydrolysis equilibrium is as follows:NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq).

The ammonium ion hydrolyzes to form ammonium ion and hydronium ions. The nitrate ion is a spectator ion that does not participate in the reaction. Therefore, the net ionic equation is given by:NH4+(aq) + H2O(l) ⇌ H3O+(aq) + NH3(aq).

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When a stock solution is diluted, the number of moles of solute (NaNO3 in this case) remains constant. Therefore, we can use the following equation to find the final molarity of the diluted solution:

M1V1 = M2V2

where M1 is the initial molarity (0.275 M), V1 is the initial volume (15.00 mL), M2 is the final molarity (what we want to find), and V2 is the final volume (100.0 mL).

First, we need to convert the initial volume from milliliters to liters:

V1 = 15.00 mL = 0.01500 L

Next, we can substitute the given values into the equation:

(0.275 M) × (0.01500 L) = M2 × (0.1000 L)

Solving for M2, we get:

M2 = (0.275 M × 0.01500 L) ÷ 0.1000 L

M2 = 0.04125 M

Therefore, the final molarity of the NaNO3 solution is 0.04125 M.

To calculate the volume of

[tex]5.9 \times 10^23[/tex]

molecules of propane gas trapped in a container at a pressure of 253.3 kPa and a temperature, we can use the ideal gas law equation, PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature in Kelvin.

We need to convert the number of molecules to moles. The molar mass of propane is 44.1 g/mol, so the number of moles of propane is

[tex]5.9 \times 10^23[/tex]

molecules /

[tex]6.022 \times 10^23[/tex]

molecules/mol = 0.98 moles.

We need to convert the pressure to atmospheres (atm), which is the unit typically used with the ideal gas law. 253.3 kPa / 101.3 kPa/atm = 2.50 atm.

We also need to convert the temperature to Kelvin by adding 273.15 to the Celsius temperature. Let's assume the temperature is 25°C, so T = 25°C + 273.15 = 298.15 K.

We can plug in the values into the ideal gas law equation and solve for V:

[tex]V = (nRT) / P = (0.98 mol \times 0.0821 L•atm/mol•K \times 298.15 K) / 2.50 atm = 29.6 L[/tex]

The volume of

[tex]5.9 \times 10^23[/tex]

molecules of propane gas trapped in a container at a pressure of 253.3 kPa and a temperature of 25°C is 29.6 L.

The ideal gas law equation is a useful tool to calculate the volume of a gas sample when its pressure, temperature, and amount of substance are known.

It is important to convert the units to the appropriate ones and use the correct value for the gas constant depending on the units used.

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As a result, the theoretical yield of Fe3O4 in moles is 1.426 x 105 molewhile the actual yield is 2.41 mole and the percent yield is 0.107%.

The reaction's chemically balanced equation is as follows:

4H₂ + Fe₃O₄ = 3Fe + 4H₂O

Fe₃O₄has a molar mass of 231.5326 g/mol7.

We need to know the mass of Fe₃O₄ created in order to calculate the real amount of Fe₃O₄ in moles. 559 g1 of Fe₃O₄ were produced.

We must first determine the quantity of Fe utilized in the reaction in order to compute the theoretical yield of magnitude  Fe₃O₄  moles. 7.97 million g, or 7.97 x 106 g, of Fe are used1. Fe has a molar mass of 55.845 g/mol7. Hence, the amount of Fe utilized in the reaction was:

1.426 x 105 mole = (7.97 x 106 g) / (55.845 g/mole)

One mole of Fe creates one mole of Fe3O4 according to the chemical equation. As a result, 1.426 x 105 mole is the theoretical yield of magnitude Fe3O4 in moles.

In order to determine percent yield, we apply the formula:

(Actual yield / Theoretical yield) times 100% equals the percent yield.

Replacement of values

Yield is calculated as follows:

(559 g/(1.426 x 105 mole/231.5326 g/mol)) x 100% = **0.107%**6.

What common chemical formulas are there?

Listed below are some typical chemical formulas:

- H₂O, or water

- NaCl: Salt

Baking soda is NaHCO₃.

- NaClO for bleach

- Table sugar (sucrose): C12H22O11

- CO₂: Carbon dioxide

- NH₃ for ammonia

- H₂O₂, or hydrogen peroxide

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Answer:

To find the empirical formula of a compound, we need to determine the simplest whole number ratio of the atoms present in the compound. We can do this by dividing each element's mass by its molar mass to get the number of moles of each element, and then dividing each number of moles by the smallest number of moles obtained. The molar masses of carbon, hydrogen, and oxygen are 12.01 g/mol, 1.008 g/mol, and 16.00 g/mol, respectively. Number of moles of C = 324 g / 12.01 g/mol = 26.98 mol Number of moles of H = 48.5 g / 1.008 g/mol = 48.11 mol Number of moles of O = 16.0 g / 16.00 g/mol = 1.

Answer:

C27H48O

Explanation:

To determine the empirical formula of the compound, we need to find the simplest whole number ratio of atoms in the compound. We can do this by assuming that we have 100 g of the compound, and finding the number of moles of each element in this amount.

Number of moles of carbon (C): 324 g / 12.01 g/mol = 26.98 mol

Number of moles of hydrogen (H): 48.5 g / 1.01 g/mol = 48.02 mol

Number of moles of oxygen (O): 16.0 g / 16.00 g/mol = 1.00 mol

Next, we divide each of these mole values by the smallest value to get the simplest ratio:

C: 26.98 mol / 1.00 mol = 26.98

H: 48.02 mol / 1.00 mol = 48.02

O: 1.00 mol / 1.00 mol = 1.00

We can see that the simplest ratio of atoms in the compound is approximately C27H48O. However, we need to express this as a whole number ratio, so we divide each subscript by the smallest subscript (which is 1):

Empirical formula: C27H48O

Therefore, the empirical formula of the compound is C27H48O.

Tech A is correct who says that overheated catalysts cannot be restored and must be replaced.

In chemistry, a catalyst is any substance that speeds up a reaction without consuming itself. Many crucial biochemical reactions are catalysed by enzymes, which are substances that occur naturally.

The majority of solid catalysts are made of metals, or the oxides, sulphides, and halides of metals, as well as of the semimetallic elements silicon, aluminium, and boron. Solid catalysts are frequently dispersed in materials known as catalyst supports, while gaseous and liquid catalysts are typically used in their pure form or in combination with appropriate carriers or solvents.

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The concentration of the pyridinium cation at equilibrium is 3.96 x 10^-9 M.Assuming that pyridinium bromide is a weak acid, we can use the acid dissociation constant (Ka) to calculate the concentration of the pyridinium cation at equilibrium. The Ka for pyridinium bromide is 5.6 x 10^-6.

Using the expression for Ka, we have:

Ka = [H+][C5H5NH+] / [C5H5NHBr]

At equilibrium, [H+] = 10^-pH = 10^-3.10 = 7.94 x 10^-4 M

Substituting the values into the equation and solving for [C5H5NH+], we get:

[C5H5NH+] = Ka * [C5H5NHBr] / [H+] = 5.6 x 10^-6 * [C5H5NHBr] / 7.94 x 10^-4 = 3.96 x 10^-9 M

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The name given to an aqueous solution of HBr is "hydrobromic acid." Option E is correct.

Hydrobromic acid (HBr) is a strong acid that forms when hydrogen bromide gas dissolves in water. It is a clear, colorless liquid having a pungent odor as well as it is highly corrosive. Hydrobromic acid is commonly used in the laboratory and in various industrial applications, including the production of pharmaceuticals, dyes, and other chemicals. It is also used as a reagent in organic chemistry for various types of reactions.

An aqueous solution is the solution in which water will be the solvent. In other words, it is a solution where water is the substance that dissolves other substances, which are called solutes. Many substances can dissolve in water to form aqueous solutions, including salts, acids, and bases.

Hence, E. is the correct option.

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--The given question is incomplete, the complete question is

"The name given to an aqueous solution of HBr is . group of answer choices A) bromic acid B) hypobromous acid C) hydrogen bromide D) bromous acid E) hydrobromic acid."--

If Abby doubles the initial monomer concentration and quadruples the initiator concentration from the original settings, the outcome of the new polymer will likely have a higher number average molecular weight.

Abby's outcome of the new polymer will likely have a higher number average molecular weight because increasing the monomer concentration will lead to more monomer units being available for polymerization, while increasing the initiator concentration will lead to more initiation events, resulting in a higher degree of polymerization.

However, the exact molecular weight of the new polymer will depend on the efficiency of the polymerization reaction and any potential side reactions or termination events that may occur. It is possible that increasing the initiator concentration could also lead to more side reactions, such as chain transfer or termination, which could affect the final molecular weight distribution.

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While considering a good solvent alternative for diethyl ether, the best one will be ethyl acetate. It can be used for extraction due to its polarity and less toxicity.

Diethyl ether is one of the commonly used solvent in extraction process of non-polar or slightly polar organic compounds. This is because it does not have hydrogen bonding. So here methanol cannot be used as it has extensive hydrogen bonding and non-polar compounds might not dissolve.

Water also cannot be used because of its polar nature. So organic compounds does not dissolve. Tetrahydrofuran can be used as a solvent, but toxicity levels are higher compared to diethyl ether.

So the alternative that can be used is ethyl acetate, which is also widely used solvent in extraction of non-polar compounds. Also it has less toxicity compared to THF.

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Answer: 1.57g and 70.5%

Explanation:

Theoretical Yield

The theoretical yield of a reaction is the absolute maximum amount of product that could be created with the amounts of reactants.

This problem gives us the amount of hydrochloric acid, which is 6.37 grams. The molar mass of HCl is the molar mass of hydrogen plus the molar mass of chlorine, which is 36.46 g/mol.

To find the moles of HCl, we just divide the mass by the molar mass.

6.37/36.46 = 0.175 moles HCl

Since oxygen is in excess, the amount used in the reaction will be dictated by the amount of HCl used in the reaction. It does not need to be taken into consideration when determining the amount of reactant since it is in excess.

To find the theoretical yield of water, we will do stoichiometry.

Balancing the equation

Written out with the chemical symbols, this equation is

HCl + O2 ⇒H2O + Cl

This is not balanced, since there is 1 hydrogen on the left side and 2 on the right, and 2 oxygens on the left and 1 on the right.

To balance this, we can put coefficients in front of some reactants and products to make sure there are equal amounts of everything on each side.

The balanced equation will be 4HCl + O2 ⇒ 2H2O + Cl

Now, there are 4 hydrogens on the left and 4 on the right, as well as 2 oxygens on the left and 2 on the right. It is balanced.

We can see by looking at the coefficients of the balanced equation that every 4 moles of HCl consumed will produce 2 mole of H2O, so the ratio is 1:2.

To do stoichiometry, we will multiply the moles of HCl by the ratio of H2O to HCl, which is just dividing by 2.

The theoretical yield of water is then 0.175 moles HCl * [tex]\frac{1moleH2O}{2moleHCl}[/tex] = 0.1874 moles H2O.

Our theoretical yield is 0.0874 moles H2O. But the question and the actual yield are in grams, so we will convert this to grams. To convert moles to grams, just multiply the moles by the molar mass. The molar mass of water is 18.0 g/mol, so

0.0874*18.0 = 1.57 g

The theoretical yield is 1.57 g H2O

Percent Yield

Percent yield is much easier. Percent yield is

((actual yield)/(theoretical yield))*100

In this case, our actual yield is 1.11 grams and our theoretical yield is 3.15 grams, so

[tex]\frac{1.11}{1.57} =[/tex] 70.5%

70.5% is the percent yield.

The presence of hydrogen would be indicated by a pale blue flame that is nearly invisible in broad daylight, but the presence of carbon dioxide would be indicated by the flame going out.

A flame ignited by hydrogen gas emits a barely perceptible pale blue flame under normal lighting conditions. This is due to the flame that hydrogen gas produces mostly emitting light in the ultraviolet spectrum, which is invisible to the human eye.

On the other side, a flame is put out when carbon dioxide gas is added to it. This is because carbon dioxide is a gas that does not support burning and is not flammable.

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The LD50 concentration of CuSO₄ for brine shrimp is reported to be around 2.75 ppm (parts per million), which means that if 50% of the brine shrimp population were exposed to this concentration of CuSO₄, they would die as a result of the exposure

The LD50 and ED50 are both terms commonly used in toxicology to express the effectiveness or toxicity of a substance.

The LD50, which stands for "lethal dose 50," is the amount of a substance required to cause death in 50% of the test population. It is typically expressed in units of milligrams or micrograms of the substance per kilogram of body weight of the test animal.

On the other hand, the ED50 stands for "effective dose 50," which is the amount of a substance required to produce a desired effect in 50% of the test population. It is commonly used in pharmacology to measure the potency of a drug.

In the case of the brine shrimp, the LD50 concentration of CuSO4 (copper sulfate) would be the amount of CuSO4 that would cause the death of 50% of the shrimp population in a given test. It is important to note that the LD50 can vary depending on various factors such as the species being tested, the method of exposure, and the duration of exposure.

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The percentage of plants showing phototropism is the dependent variable, and the color of the light is the independent variable. Therefore, the correct answer is: "the percentage of plants showing phototropism is the dependent variable, and the color of the light is the independent variable."

The dependent variable is the variable that is being measured or observed, and its value depends on the independent variable, which is the variable that is being manipulated or changed in the experiment. In this experiment, the percentage of plants showing phototropism is being measured, which means that it is the dependent variable. The color of the light is being manipulated, which means that it is the independent variable.

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The enthalpy change or AH for the reaction is -1560.9 kJ/mol. This negative value indicates that the reaction is exothermic, meaning it releases energy in the form of heat to the surroundings.

C2H4 (g) + 3O2 (g) → 2H2O (g) + 2CO2 (g)

We can use the standard enthalpies of formation (∆Hf°) of the products and reactants to determine the overall enthalpy change. The ∆Hf° values can be found in a reference table or online database.

The balanced equation tells us that 1 mole of C2H4 reacts with 3 moles of O2 to form 2 moles of H2O and 2 moles of CO2. So, we can write:

ΔH° = [2∆Hf°(H2O) + 2∆Hf°(CO2)] - [∆Hf°(C2H4) + 3∆Hf°(O2)]

Substituting the ∆Hf° values for each substance, we get:

ΔH° = [2(-241.8 kJ/mol) + 2(-393.5 kJ/mol)] - [(52.3 kJ/mol) + 3(0 kJ/mol)] ΔH° = -1560.9 kJ/mol

Therefore, the enthalpy change or AH for the reaction is -1560.9 kJ/mol. This negative value indicates that the reaction is exothermic, meaning it releases energy in the form of heat to the surroundings.

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the new freezing point of the solution is 0°C - 6.19°C = -6.19°C.

Assuming complete dissociation of the rock salt in water, the number of particles in the solution is 0.50 moles of rock salt x 2 particles per formula unit (NaCl) = 1.0 mole of particles.

The molal concentration of the solution can be calculated as follows:

molality (m) = moles of solute / mass of solvent in kg

mass of solvent = 300 g = 0.3 kg

m = 1.0 moles / 0.3 kg = 3.33 m

The freezing point depression (ΔTf) of the solution can be calculated using the formula:

ΔTf = Kf x m

where Kf is the freezing point depression constant of water, which is 1.86 °C/m.

ΔTf = 1.86 °C/m x 3.33 m = 6.19 °C

Therefore, the new freezing point of the solution is 0°C - 6.19°C = -6.19°C.

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Performing a reaction in a vacuum chamber to control the experiment is a valid approach to prevent external variables from impacting the reaction.

However, it may not be sufficient to determine if the reaction is endothermic or exothermic and measure the amount of energy absorbed or given off. This is because the vacuum chamber only isolates the reaction from the environment, but it does not provide a way to measure the energy changes that occur during the reaction. To measure the energy changes, the scientist should use techniques such as calorimetry, which directly measures the heat absorbed or released by the reaction.

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To fix the issue, the student should carefully reapply the sample in a smaller amount, allow it to dry completely, and adjust the solvent system if necessary to achieve proper separation of the components on the TLC plate.

The student might have made the following technical mistake while running the TLC plate:

1. Overloading the sample: When spotting the TLC plate, the student may have applied too much sample, causing the components to streak rather than separate into individual spots. To resolve this, the student should apply a smaller amount of sample and ensure that it is evenly distributed.

2. Insufficient drying: If the student did not allow the spotted sample to dry properly before placing it in the solvent, it can cause the components to streak. To prevent this, the student should ensure the sample is completely dry before running the TLC plate.

3. Inappropriate solvent system: Although the student used a 10% ethyl acetate/hexane mixture, it is possible that this solvent system was not suitable for the specific sample. Adjusting the solvent ratio or trying different solvent systems could help achieve better separation.

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it would take 3.16 litres of water to make a 0.19 M solution of NaI from 0.60 moles of NaI.

To calculate the volume of water needed to make a 0.19 M solution of NaI, we need to use the formula:

Moles of solute = Molarity x Volume (in liters)

We can rearrange this formula to solve for the volume of water:

Volume (in liters) = Moles of solute / Molarity

First, let's calculate the number of moles of NaI in 0.60 moles:

Moles of NaI = 0.60 moles

Now, we can use the formula above to calculate the volume of water needed:

Volume (in litres) = Moles of NaI / Molarity

Volume (in litres) = 0.60 moles / 0.19 M

Volume (in litres) = 3.16 litres

Therefore, it would take 3.16 litres of water to make a 0.19 M solution of NaI from 0.60 moles of NaI.
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HCl after dilution: Final concentration = 0.107 g/ml,  NH₃ after dilution: Final concentration = 0.016875 g/ml

To calculate the concentration of HCl after dilution:
Step 1: Find the mass of HCl in the 10 ml concentrated solution.
Mass = Volume × Density × Concentration
Mass = 10 ml × 1.07 g/ml × 0.25
Mass = 26.75 g
Step 2: Calculate the final volume after dilution, which is the volume of the volumetric flask.
Final volume = 250 ml
Step 3: Calculate the final concentration of HCl.
Final concentration = Mass / Final volume
Final concentration = 26.75 g / 250 ml
Final concentration = 0.107 g/ml (or 10.7% by mass)
To calculate the concentration of NH₃ after dilution:
Step 1: Find the mass of NH3 in the 15 ml concentrated solution.
Mass = Volume × Density × Concentration
Mass = 15 ml × 0.75 g/ml × 0.15
Mass = 1.6875 g
Step 2: Calculate the final volume after dilution, which is the volume of the volumetric flask.
Final volume = 100 ml
Step 3: Calculate the final concentration of NH₃.
Final concentration = Mass / Final volume
Final concentration = 1.6875 g / 100 ml
Final concentration = 0.016875 g/ml (or 1.6875% by mass)

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The acid dissociation constant of hexanoic acid is 4.93 × 10^-10 mol/L.

When measuring the pH of a solution of hexanoic acid to be 4.96, the acid dissociation constant of hexanoic acid can be calculated. This can be done through the following equation:

Ka = [H3O+][A-] / [HA]whereKa

= acid dissociation constantH3O+

= hydronium ionA-

= conjugate baseHA

= acidThe pH of the solution of hexanoic acid is measured to be 4.96.

Thus, [H3O+] is equal to 10^-4.96 or 7.02 × 10^-5 M.

The initial concentration of the hexanoic acid is equal to the concentration of the undissociated acid or [HA].The acid dissociation constant of hexanoic acid can be calculated by plugging the known values into the equation:

Ka = [7.02 × 10^-5][A-] / [HA]The concentration of the conjugate base, A-, is equal to the concentration of the dissociated acid, which is equal to [H3O+].

Thus,Ka = [7.02 × 10^-5]^2 / [HA]Ka = 4.93 × 10^-10 mol/L

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The molarity of the diluted solution is 0.125M.

To find the molarity of the diluted solution after adding 25 ml of water to 125 ml of a 0.15 M sodium hydroxide solution, you can follow these steps:

1. Determine the initial volume (V1) and molarity (M1) of the sodium hydroxide solution: V1 = 125 ml and M1 = 0.15 M.
2. Determine the volume of water added (V2): V2 = 25 ml.
3. Calculate the total volume of the diluted solution (Vt):

      Vt = V1 + V2

      Vt = 125 ml + 25 ml

      Vt = 150 ml.
4. Use the dilution equation M1V1 = M2V2, where M2 is the molarity of the diluted solution.
5. Solve for M2:

      M2 = (M1V1) / Vt

      M2 = (0.15 M × 125 ml) / 150 ml

      M2 = 18.75 / 150

      M2 = 0.125 M.

So, the molarity of the diluted solution after adding 25 ml of water to 125 ml of a 0.15 M sodium hydroxide solution will be 0.125 M.

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The part 2 in the attached image will undergo reduction with sodium borohydride.

Sodium borohydride (NaBH₄) is a commonly used reducing agent in organic chemistry. It is primarily used to reduce carbonyl groups (aldehydes and ketones) to their corresponding alcohols. Sodium borohydride can also reduce some other functional groups, such as imines and acid chlorides, but its reactivity towards these groups is much lower compared to carbonyls.

In general, the carbonyl groups in a molecule are the most likely functional groups to undergo reduction with sodium borohydride. Other functional groups such as alkenes, alkynes, aromatic rings, and alcohols are generally unreactive towards sodium borohydride.

Therefore, if your compound contains any carbonyl groups (such as aldehydes or ketones), those groups would be the most likely candidates for reduction with sodium borohydride. Part 2 is a carbonyl group, hence it will react.

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To calculate the molar mass of the unknown substance, we need to use the formula:

Molar mass = (mass of solute) / (number of moles of solute)

First, let's calculate the number of moles of solute in the solution:

Number of moles = (concentration) x (volume in liters)

We know that the concentration of the solution is 0.45 M, and the volume of the solution is 425 mL, which is equivalent to 0.425 L. Substituting these values into the formula, we get:

Number of moles = 0.45 M x 0.425 L

Number of moles = 0.19125 moles

Next, we can calculate the mass of the solute (the unknown substance) by using the formula:

mass = number of moles x molar mass

Rearranging the formula, we get:

molar mass = mass / number of moles

We know that the mass of the solute is 12 grams, and we have already calculated the number of moles as 0.19125 moles. Substituting these values into the formula, we get:

molar mass = 12 g / 0.19125 moles

molar mass = 62.8 g/mol

Therefore, the molar mass of the unknown substance is 62.8 g/mol.

1. The mass (in grams) of FeBr₃ produced from 65 g of Br₂ is 80.13 g

2. The mole of CO₂ formed from 10 moles of C₅H₁₂ is 50 moles

3. The moles of MnCl₂ prepared from 52.1 grams of MnO₂ is 0.6 mole

The mass of FeBr₃ produced can be obtained as illustrated below:

2Fe + 3Br₂ -> 2FeBr₃

From the balanced equation above,

480 g of Br₂ reacted to produce 591.7 g of FeBr₃

Therefore,

65 g of Br₂ will react to produce = (65 × 591.7) / 480 = 80.13 g of FeBr₃

Thus, the mass of FeBr₃ produced is 80.13 g

The mole of CO₂ formed can be obtained as follow:

C₅H₁₂ + 8O₂ -> 5CO₂ + 6H₂O

From the balanced equation above,

1 mole of C₅H₁₂ reacted to produce 5 moles of CO₂

Therefore,

10 moles of C₅H₁₂ will react to produce = 10 × 5 = 50 moles of CO₂

Thus, the mole of CO₂ formed is 50 moles

First, we shall obtain the mole of 52.1 g of MnO₂, Details below:

Mole = mass / molar mass

Mole of MnO₂ = 52.1 / 86.94

Mole of MnO₂ = 0.6 mole

Finally, we shall determine the mole of MnCl₂ prepared can be obtained as follow:

MnO₂ + 4HCl -> MnCl₂ + Cl₂ + 2H₂O

From the balanced equation above,

1 mole of MnO₂ reacted to produce 1 mole of MnCl₂

Therefore,

0.6 mole of MnO₂ will also react to produce 0.6 mole of MnCl₂

Thus, the mole of MnCl₂ prepared is 0.6 mole

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[tex] \:\:\:\:\:\:\star [/tex]For the general solubility equilibrium [tex]\sf \underline{AB \longrightarrow A^+ + B^-} [/tex]the solubility product has the following expression-

[tex] \:\:\:\:\:\:\star\longrightarrow \sf\underline{K_{(sp)} = [A^+] \times [B^-]}\\[/tex]

As per question, we are given that-

[tex] \:\:\:\:\:\:\star [/tex] Now that we have all the required values, so we can substitute these values into the Ksp expression and solve for Ksp-

[tex] \:\:\:\:\:\:\star\longrightarrow \sf\underline{K_{(sp)} = [A^+] \times [B^-]}\\[/tex]

[tex] \:\:\:\:\:\:\longrightarrow \sf K_{(sp)} = 0.00343 \:M\times 0.00343\:M\\[/tex]

[tex] \:\:\:\:\:\:\longrightarrow \sf K_{(sp)} = 0.00343 \:molL^{-1}\times 0.00343\:molL^{-1}\\[/tex]

[tex] \:\:\:\:\:\:\longrightarrow \sf \underline{K_{(sp)} = 1.17649\times 10^{-5} \: mol^2L^{-2}}\\[/tex]