c. What predictions can you make about ΔH, ΔS, and ΔG for the general reaction A(s) + 2B(g) 3C(g) + D(g) + heat? What conclusions can you make about the spontaneity of the reaction? Explain your reasoning. (4 points)


d. The reaction 2NO2(g) N2(g) + 2O2(g) is spontaneous at all temperatures, yet it hardly happens under normal conditions. How can this be? (2 points)

Answer:

conc. fuming, 1 mol H2SO4

Dilute NaOH

Br2

Dilute H2SO4

Explanation:

The  synthesis of ortho-bromophenol follows the reaction sequence shown in the image attached.

First of all, the phenol is sulphonated using concentrated sulphuric acid at 100°C.  Carrying out the reaction at 100°C ensures that the para-isomer predominates. Lower temperatures favour the formation of the ortho isomer. Dilute sodium hydroxide is added before the addition of bromine.

Bromine molecule is then added. The incoming electrophile now attaches to the ortho position. Dilute acid is  added at 100°C to remove the -SO3H thereby obtaining the Ortho-bromophenol

Answer:

Check the explanation

Explanation:

Kindly check the attached image below to see the step by step explanation to the question above.

Answer:

pH = 8.93

Explanation:

In this case, this titration is the case of a weak acid and a strong base. Now, at the equivalence point, it's supposed that we have the same moles of each reactant in solution, and we will expect that the pH would have to be 7. However, as the acid is pretty weak, there's a little difference in the solution because of the grade of dissociation of the acid, and the pH will be higher than 7. To know this, we first need to calculate the volume of added base:

M₁V₁ = M₂V₂

With this expression, let's calculate the volume of the base:

V₂ = M₁V₁ / M₂

V₂ = 0.1917 * 220 / 0.1787 = 236 mL

So, at the equivalence point, 236 mL are needed to neutralize this reaction. As the moles are the same for each reactant, we just need to calculate the concentration of the acid in this part. This will be the sum between the initial volume of acid and the calculated volume of base:

V of solution = 236 + 220 = 456 mL or 0.456 L

Then, the new concentration of the acid is:

[C₂H₅COOH] = 0.1917 * 0.220 / 0.456 = 0.0924 M

Now, the reaction with the base is the following:

C₂H₅COOH + KOH --------> C₂H₅COOK + H₂O

This means that in the equivalence point we have the propionic potassium and water, so, if take this and dissociates into it's ions we can calculate the pH of the solution:

C₂H₅COO⁻ + H₂O <-------> C₂H₅COOH + OH⁻

With this reaction in solution in the equivalence point, we just need the Kb of propionate ion, and this can be calculated with the value of the pKa of the acid:

Ka = 10^(-pKa)

Ka = 1.29x10⁻⁵

Now the value of Kb can calculated using the following expression:;

Kb = Kw / Ka ---> replacing we have

Kb = 1x10⁻¹⁴ / 1.29x10⁻⁵

Kb = 7.75x10⁻¹⁰

Now, with this value and the above reaction we can write an ICE chart to calculate the [OH⁻] and then, the pH of solution:

     C₂H₅COO⁻ + H₂O --------> C₂H₅COOH + OH⁻   Kb = 7.75x10⁻¹⁰

i)       0.0924                                    0               0

e)     0.0924-x                                  x               x

The Kb expression:

Kb = [C₂H₅COOH] [OH⁻] / [C₂H₅COO⁻]

7.75x10⁻¹⁰ = x² / 0.0924-x ---> Kb is very small, so this substraction can be neglected to just 0.0924 assuming x will be very small too.

7.75x10⁻¹⁰ = x² / 0.0924

7.75x10⁻¹⁰ * 0.0924 = x²

x = [OH⁻] = 8.46x10⁻⁶ M

With this value, we can calculate pOH and then the pH:

pOH = -log(8.46x10⁻¹⁰) = 5.07

Finally the pH:

pH = 14 - pOH

pH = 14 - 5.07

Answer:

The new elution order expected will be the following:

Please see below for details and explanation.

Explanation:

Which compound will elute first depends on a number of factors. The compound with the lowest boiling point will elute before another compound with a higher boiling point and so on. By extension, the volatility of the compound will also be considered when predicting elution order. Thirdly, how the solutes interact with each other during the stationary phase. I've listed the boiling points below:

1-pentanol 138 °C

ethylene diamine 116 °C

diethylene glycol 245 °C

The advantage of using temperature programmed chromatogram is that it changes retention times (time needed for the solute to pass through the column). And it will be according to the respective boiling points.

Hope that answers the question, have a great day!

Answer:

0.1 M KOH

Explanation:

The thymol blue indicator will appear blue in a basic solution, particularly in a solution with a pH greater than 8.0. A diluted sodium hydroxide solution is an example of such a basic environment where thymol blue would turn blue.

The thymol blue indicator will appear blue in a basic solution. Specifically, thymol blue changes color from yellow to blue over a pH range of approximately 8.0 to 9.6. This means that for thymol blue to exhibit a blue color, it needs to be in a solution with a pH greater than 8.0, indicating a basic environment.

An example of a solution in which thymol blue would appear blue is a diluted sodium hydroxide (NaOH) solution, as NaOH is a strong base that would increase the pH of the solution.

Answer:

decrease the molar solubility of the solid relative to its molar solubility in pure water.

Explanation:

The common ion effect is defined as a decrease in the solubility of a solute because of the addition of a second solute with which it has a common ion. If a solution already contains a solute MX and another ionic solid containing BX is added to the solution, the X^- anion is common to the both species. Hence the presence of MX will decrease the solubility of BX compared to the solubility of BX in pure water.

Generally, when a soluble ionic solid is placed in a solution containing a common ion, the molar solubility of the solid decreases compared to its molar solubility in pure water.

Answer:

The mass of the solute and the volume of the solution.

Explanation:

Hello,

In this case, given the formula of molarity:

[tex]M=\frac{n_{solute}}{V_{solution}}[/tex]

In such a way, since the moles could not be directly measured, we must measure the mass of the solute and by using its molar mass, one could compute its moles. Moreover, since the solution is composed by the solvent (typically water) and the solute, we consequently must measure the volume of the solution needed for the preparation of such concentration-known solution. In such a way, we can actually prepare the required solution.

Best regards.

For the formation of 2 M solution of potassium nitrate, mass and volume of the solution has been measured.

Molarity can be defined as the mass of solute present in a liter of solution. The molarity has been used for the determination of the concentration of the compounds.

It can be expressed as mol/L. The molarity (M) has expression:

[tex]M=\rm \dfrac{solute\;mass}{solute\;molar\;mass}\;\times\;Solution\;Volume[/tex]

For the formation of 2 M potassium nitrate solution, the mass of the solute and the volume of the solution has to be measured.

For more information about molarity, refer to the link:

https://brainly.com/question/12127540

Given question is incomplete. The complete question is as follows.

When 72.8 g of benzamide ([tex]C_{7}H_{7}NO[/tex]) are dissolved in 600 g of a certain mystery liquid X, the freezing point of the solution is [tex]6.90^{o}C[/tex] less than the freezing point of pure X. Calculate the mass of ammonium chloride [tex](NH_{4}Cl)[/tex] that must be dissolved in the same mass of X to produce the same depression in freezing point. The van't Hoff factor i = 70 for ammonium chloride in X. Be sure your answer has a unit symbol, if necessary, and round your answer to significant digits.

Explanation:

The given data is as follows.

        Mass of solute (benzamide), [tex]w_{B}[/tex] = 72.8 g

        Mass of solvent (X), [tex]w_{A}[/tex] = 600 g

          [tex]\Delta T_{f} 6.90^{o}C[/tex]

  Molar mass of benzamide, [tex]M_{w_{B}}[/tex] = 121.14 g/mol

We know that,

              [tex]\Delta T_{f} = k_{f} \times X \times m[/tex]   (for non-dissociating)

   [tex]6.90 = k_{f} \times \frac{72.8 \times 1000}{121.14 \times 600}[/tex] ...... (1)

For other experiment, when [tex]NH_{4}Cl[/tex] is taken :

       Mass of [tex]NH_{4}Cl[/tex], ([tex]w_{NH_{4}Cl}[/tex]) = ?

  Molar mass of [tex]NH_{4}Cl[/tex] = 53.491 g/mol

  Mass of solvent (X) = 600 g

        [tex]\Delta T_{f} = 6.90^{o}C[/tex]

           i = Van't Hoff factor = 1.70

As,     [tex]\Delta T_{f} = i \times k_{f} \times m[/tex]

      [tex]6.90 = 1.70 \times k_{f} \times \frac{w_{NH_{4}Cl} \times 1000}{53.491 \times 600}[/tex] ........... (2)

Now, we will divide equation (1) by equation (2) as follows.

         [tex]w_{NH_{4}Cl} \times 1 = \frac{72.8 \times 53.491}{1.70 \times 121.14}[/tex]  

                = 18.90 g

Therefore, we can conclude that the mass of ammonium chloride [tex](NH_{4}Cl)[/tex] that must be dissolved in the same mass of X to produce the same depression in freezing point is 18.90 g.

Final answer:

The mass of ammonium chloride required to produce the same freezing point depression as benzamide can't be calculated without additional data. The van't Hoff factor is crucial for understanding the effects of ionic compounds in freezing point depression calculations.

Explanation:

To calculate the mass of ammonium chloride that must be dissolved to produce the same depression in freezing point as benzamide, we will need to use the freezing point depression concept, which states that the change in freezing point (ΔTf) is equal to the molal freezing point depression constant of the solvent (Kf) multiplied by the molality (m) of the solution. The given data is as follows.

       Mass of solute (benzamide), w_(B) = 72.8 g

       Mass of solvent (X), w_(A) = 600 g

         \Delta T_(f) 6.90^(o)C

 Molar mass of benzamide, M_{w_(B)} = 121.14 g/mol

We know that,

             \Delta T_(f) = k_(f) * X * m   (for non-dissociating)

  6.90 = k_(f) * (72.8 * 1000)/(121.14 * 600) ...... (1)

For other experiment, when NH_(4)Cl is taken :

      Mass of NH_(4)Cl, (w_{NH_(4)Cl}) = ?

 Molar mass of NH_(4)Cl = 53.491 g/mol

 Mass of solvent (X) = 600 g

       \Delta T_(f) = 6.90^(o)C

          i = Van't Hoff factor = 1.70

As,     \Delta T_(f) = i * k_(f) * m

     6.90 = 1.70 * k_(f) * \frac{w_{NH_(4)Cl} * 1000}{53.491 * 600} ........... (2)

Now, we will divide equation (1) by equation (2) as follows.

        w_{NH_(4)Cl} * 1 = (72.8 * 53.491)/(1.70 * 121.14)  

               = 18.90 g

Therefore, we can conclude that the mass of ammonium chloride (NH_(4)Cl) that must be dissolved in the same mass of X to produce the same depression in freezing point is 18.90 g.

Answer:

9.1 KJ

Explanation:

We must first put down the reaction equation;

H2(g) + Br2(g) ----> 2HBr(g)

Secondly we find the number of moles of HBr involved;

number of moles of HBr= mass of HBr/ molar mass of HBr

But molar mass of HBr= 80.91 g/mol

Given mass of HBr as given in the question= 20.1g

Hence;

Number of moles of HBr= 20.1 g/80.91g

Number of moles of HBr= 0.25 moles of HBr

Lastly we calculate the heat transferred from the number of moles involved;

If 2 moles of HBr has a heat of formation of 72.80KJ

Then 0.25 moles of HBr will have a heat of formation of 0.25× 72.80/2= 9.1 KJ

Then; 20.1 g of HBr will have a heat of formation of 9.1 KJ

Answer:

The Earth's bioshere consist of the parts of Earth where life exists. Ecosystems.

Explanation:

Final answer:

The reaction in question demonstrates the application of Le Chatelier's Principle, with shifts in equilibrium occurring in response to decreased pressure, increased temperature, and the removal of a reactant.

Explanation:

The reaction presented is dealing with changes in reaction conditions in a chemical equilibrium situation. This is directly related to Le Chatelier's Principle, which states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. In this scenario:

Answer:

The correct answer is 0.00582 grams.

Explanation:

In order to solve the question, let us consider the vapor pressure of H2O, as hydrogen gas is collected over water, therefore, we have to consider the vapor pressure of water in the given case. Let us assume that the pressure is 760 torr or 1 atm.

It is known that the vapor pressure of water at 40 degree C is 53.365 torr (Based on the data).

Therefore, the pressure of H2 will be,

P = 760-55.365 = 704.635 torr or 704.635/760 = 0.9272 atm

The volume of the hydrogen gas collected in the tube is 80 ml or 0.08 L

Temperature in Kelvin will be 40+273 = 313 K

To calculate the moles of hydrogen (H2) gas, there is a need to use the ideal gas equation, that is, PV= nRT, in this R is the gas constant, whose value is 0.0821 L atm/molK, and n is the moles of the gas.

By inserting the values in the equation we get:

PV = nRT

n = PV/RT = 0.9272 *0.08 / 0.0821 * 313

n = 0.00289 moles

The mass of H2 will be moles * molar mass = 0.00289 * 2.016

= 0.00582 grams.

Answer:

C) 3.40 x 10-3

Explanation:

Hello,

In this case, for the given pOH, we are able to compute the concentration of hydrixide ions by applying the following formula:

[tex]pOH=-log([OH^-])[/tex]

[tex][OH^-]=10^{-pOH}=10^{-2.468}\\[/tex]

[tex][OH^-]=0.0034M=3.4x10^{-3}M[/tex]

Therefore, answer is C).

Best regards.

Answer:

3.40 × 10⁻³ M

Explanation:

The pOH scale is used to express the acidity or basicity of a solution.

The pOH of this solution is 2.468, so it is basic. We can calculate the concentration of hydroxide ions in the solution using the following expression.

[tex]pOH = -log [OH^{-} ]\\\[[OH^{-}] = antilog-pOH = antilog-2.468 = 3.40 \times 10^{-3} M[/tex]

Answer:

The correct answer is 89.6 L

Explanation:

We have the following chemical equation and the molar masses for the reaction:

3H₂(g)   +   N₂ -->   2 NH₃

 6 g            28 g      34 g

That means that 3 moles of H₂ (6 g) reacts with 1 mol of N₂ (28 g) and gives 2 moles of NH₃ (34 g). In order to calculate how many liters of NH₃ result from the reaction of 12 grams of H₂ and 28 grams of N₂, we have to first figure out which reactant is the limiting reactant. According to the equation, if 6 grams of H₂ reacts with 28 g of N₂, and we have 12 grams:

6 g H₂------- 28 g N₂

12 g H₂-------- X = 12 g H₂ x 28 g N₂/6 g H₂ = 56 g N₂

We need 56 g of N₂ but we have 28 g of N₂, so N₂ is the limiting reactant. With the limiting reactant we can calculate the moles of product (NH₃) we will obtain:

We have 28 g N₂ -----> 28 g/14 g/mol = 2 moles N₂

1 mol N₂ ----------- 2 moles NH₃

2 mol N₂ --------- X = 2 mol N₂ x 2 moles NH₃/1 mol N₂ = 4 mol NH₃

Finally, we convert the moles of NH₃ to liters:

1 mol gas at STP = 22.4 L

Liters NH₃ obtained = 4 moles NH₃ x 22.4 L/1 mol = 89.6 L

Final answer:

When 12 grams of H2 and 28 grams of N2 react completely according to the equation 3H2(g) + N2(g) → 2NH3(g), 34.08 grams of NH3 are produced, with some H2 remaining unreacted due to it being in excess.

Explanation:

The question asks about the result when 12 grams of H2 and 28 grams of N2 react to completion according to the balanced chemical equation 3H2(g) + N2(g) -> 2NH3(g). This equation signifies that one mole of nitrogen gas (N2) reacts with three moles of hydrogen gas (H2) to produce two moles of ammonia (NH3). Given the molar masses (N2 = 28.02 g/mol, H2 = 2.02 g/mol), we can determine that the initial amounts provided are excess H2. Precisely, 28 grams of N2 is one mole, and 12 grams of H2 is six moles, which is enough to fully react according to the stoichiometry of the equation. The reaction of 28.02 g of N2 and 6.06 g of H2 produces 34.08 g of NH3, according to mass conservation principles. Since more H2 is provided than required, only 6.06 g will be consumed, leaving excess H2 unreacted. Therefore, the reaction yields 34.08 grams of NH3.

Answer:

Explanation:

Global winds are the winds that occur in the belts that are found all over the planet. Like local winds, global winds are caused by differences in heat in the atmosphere.

Polar Easterlies, from 60-90 degrees latitude.

Prevailing Westerlies, from 30-60 degrees latitude.

Tropical Easterlies, from 0-30 degrees latitude.

Answer:

A,C,D

Explanation:

I was very confused because I wasn't sure for this question so i went on brainly to find a answer and all of them said A,B,C and i was confused cause that was not right low and behold i put in this as a geuss and it was right hope this helped anyone wondering!! but correct answer is A,C,D

Answer:

2NaHCO3(aq)+H2SO4(aq)→Na2SO4(aq)+2CO2(g)+2H2O(l)

Explanation:

As we know that

acid + carbonate  →  salt + carbon dioxide + water

So, the general (un-balanced) equation would be-

NaHCO3(aq)+H2SO4(aq)→Na2SO4(aq)+CO2(g)+H2O(l)

Now we will write the net ionic reactions

[tex]HCO_3^-+ H3O^+[/tex] ----> CO2(g)↑+2H2O(l)

[tex]Na ---> Na^+ + e^-[/tex]

[tex]2H^+ + 2e^- ---> H2[/tex]

[tex]SO_4^{2-} ---- SO_4 + 2e^-[/tex]

Adding all the above equation, we get

2NaHCO3(aq)+H2SO4(aq)→Na2SO4(aq)+2CO2(g)+2H2O(l)

Answer:

See explaination

Explanation:

The electrons geometry shows the special distribution of the electrons around of the central atom of the molecule.

The molecular geometry shows the special distribution of the atoms that form the molecule.

Please kindly check attachment for further solution.

Answer:

The new volume is 1.62 L

Explanation:

Boyle's law says:

"The volume occupied by a given gas mass at constant temperature is inversely proportional to the pressure." It is expressed mathematically as:

Pressure * Volume = constant

o P * V = k

Charles's law is a law that says that when the amount of gas and pressure are kept constant, the ratio between volume and temperature will always have the same value:

[tex]\frac{V}{T}=k[/tex]

Gay-Lussac's law indicates that when there is a constant volume, as the temperature increases, the gas pressure increases. And when the temperature is decreased, the gas pressure decreases. So this law indicates that the quotient between pressure and temperature is constant.

Gay-Lussac's law can be expressed mathematically as follows:

[tex]\frac{P}{T}=k[/tex]

Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law.

[tex]\frac{P*V}{T}=k[/tex]

Having an initial state 1 and a final state 2 it is possible to say that:

[tex]\frac{P1*V1}{T1} =\frac{P2*V2}{T2}[/tex]

Standard temperature and pressure (STP) indicate pressure conditions P = 1 atm and temperature T = 0 ° C = 273 ° K. Then:

Replacing:

[tex]\frac{1 atm* 1.2 L}{273K} =\frac{0.8 atm*V2}{294K}[/tex]

Solving:

[tex]V2=\frac{1 atm*1.2 L}{273 K} *\frac{294 K}{0.8 atm}[/tex]

V2= 1.62 L

The new volume is 1.62 L

Answer: Concentration of N₂ is 4.8.[tex]10^{9}[/tex] M.

Explanation: [tex]K_{c}[/tex] is a constant of equilibrium and it is dependent of the concentrations of the reactants and the products of a balanced reaction. For

N2(g) + 2 O2(g) ⇄ 2 NO2(g)

[tex]K_{c}[/tex] = [tex]\frac{[NO2]^{2} }{[N2][O2]^{2} }[/tex]

From the question concentration of NO2 is twice of O2:

[NO2] = 2[O2]

Substituting this into [tex]K_{c}[/tex]:

[tex]K_{c}[/tex] = [tex]\frac{[2O2]^{2} }{[N2][O2]^{2} }[/tex]

8.3.[tex]10^{-10}[/tex] = [tex]\frac{4O2^{2} }{[N2].O2^{2} }[/tex]

[N2] = [tex]\frac{4O2^{2} }{8.3.10^{-10}.O2^{2} }[/tex]

[N2] = [tex]\frac{4}{8.3.10^{-10} }[/tex]

[N2] = 4.8.[tex]10^{9}[/tex]

The concentration of N2 in the equilibrium is [N2] = 4.8.[tex]10^{9}[/tex]M.

The concentration of [tex]N_2[/tex] is [tex]4.8.10^9M[/tex]

Since k_e represent the equilibrium constant and based on the concentrations of the reactants and the products of a balanced reaction.

Also, the reaction: N2(g) + 2 O2(g) ⇌ 2 NO2(g), And, the concentration of NO2 is twice the concentration of O2 gas

Kc = 8.3 × 10-10 at 25°C.

NO2 is twice of O2.

Now

[tex]8.310^{-10} = \frac{4O_2^2}{N_2O_2^2} \\\\N_2 = \frac{4O_2^2}{8.3.10^{-10}O_2^2}\\\\ = 4\div 8.3.10^{-10}\\\\= 4.8.10^9[/tex]

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Answer:

See explaination

Explanation:

H O / I MOH the OH da Anomer My nequatorial bovel SOH niematerial and CH2OH Yr axial boud - H o H o . H (Auro) 2 flip . note: 1)Fischer projection : L- Sugar ( OH group is in the left side ) at C-5 carbon (bottom most chiral center )

In the Haworth projection given above in the qsn it is Beta anomer( -CH2OH & C-OH are in same side i.e. Cis to each other ). It is a L -sugar .

I omitted C2 -H bond (axial ) in the chair conformation of alpha anomer.

See attachment for further solution

Transitioning from a Haworth to a Fischer projection, every carbon atom becomes a line end or intersection point. Correct position and orientation of the hydroxyl group determines the chair configuration. Carbohydrates, which contain carbon, hydrogen, and oxygen atoms in a 1:2:1 ratio, have monosaccharide, disaccharide, and polysaccharide subtypes.

The question is asking you to convert a Haworth projection of a carbohydrate into a Fischer projection and a chair conformation of its opposite anomer. This involves understanding several concepts in organic chemistry. The Haworth projection is a common way of representing cyclic sugars, which are typically found in ring forms in aqueous solutions. The Fischer projection is used to represent the stereochemistry of the molecule. In converting the Haworth projection to a Fischer projection, each carbon atom in the ring becomes the end of a line or a point where lines intersect in the Fischer projection. The most oxidized carbon (typically the one attached to an oxygen atom) is placed at the top.

Next, to draw the chair conformation, you must understand axial and equatorial bonds, which describe the orientation of chemical groups around the carbon atoms in the ring. Axial bonds are vertical and shunted to the side, while equatorial bonds are more horizontal and radiate outwards. For the opposite anomer, if the original has the hydroxyl group below the plane in an alpha position, the opposite anomer would have it above the plane in a beta position.

Carbohydrates bear the stoichiometric formula (CH₂O)n, indicating a carbon to hydrogen to oxygen ratio of 1:2:1. Classified into monosaccharides, disaccharides, and polysaccharides, carbohydrates are a crucial chemical group in biology.

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Answer:

The molecular weight of this compound is 386.4 g/mol

Explanation:

Step 1: Data given

MAss of the compound = 11.09 grams

Volume of diethyl ether = 180.9 mL

Osmotic pressure = 3.88 atm

Temperature = 298 K

The compound =  nonvolatile and non-electrolyte

Step 2: Calculate molar concentration

π = i*M*R*T

⇒with π = the osmotic pressure = 3.88 atm

⇒with i = the van't Hoff factor = 1

⇒with C = the molar concentration = = TO BE DETERMINED

⇒ with R = the gas constant = 0.08206 L*atm/mol*K

⇒with T = 298 K

C = 3.88 / (0.08206*298)

C = 0.1587 M

Step 3: Calculate moles compound

C = moles / volume

moles = 0.1587 M * 0.1809 L

Moles compound = 0.0287 moles

Step 4: Calculate molecular weight of the compound

Molar mass = mass / moles

Molar mass compound = 11.09 grams / 0.0287 moles

Molar mass compound = 386.4 g/mol

The molecular weight of this compound is 386.4 g/mol

Answer:

See explanation below

Explanation:

This reaction is known as Ketone hydrolisis in acid medium. This involves the formation of an hemi cetal, and then, the acetal. This is often used to convert ketones or aldehydes in ethers.

The first step involves the reaction with the acid. The carbonile reacts with the acid and forms an alcohol there. The next step is the reaction of the alcohol, in this case, the methanol to form the hemi cetal. Then in the third step, we repeat the first step, using acid to turn the OH group into a great leaving group such water. Then the water leaves the molecule, leaving the space wide open in the next step for methanol, and the acetal is formed.

See picture for the curved arrow mechanism

Protonation of 4-methylpentan-2-one creates a positive carbon center. First, methanol attacks this center, then a proton transfer occurs. After water is lost forming an oxonium ion, a second methanol attacks the intermediate and deprotonation results in an acetal.

In order to draw the curved arrow mechanism for the formation of an acetal from acidic methanol and 4-methylpentan-2-one, we proceed as follows:

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Answer:

5 kJ/g  

Explanation:

There are two energy flows in this reaction.

q₁ = heat from reaction

q₂ = heat to warm the solution

 q₁   +    q₂     = 0

m₁ΔH + m₂CΔT = 0

Data:

m₁ = 0.258 g

V₂ = 100 mL

  C = 4.184  J°C⁻¹g⁻¹

T_i = 22 °C

T_f = 25.1 °C

Calculations

(a) Mass of solution

[tex]\text{Mass} = \text{100 mL} \times \dfrac{\text{1.00 g}}{\text{1 mL}} = \text{100 g}[/tex]

(b) ΔT

ΔT = T_f - T_i = 25.1 °C - 22 °C = 3.1°C

(c) ΔH

[tex]\begin{array}{ccccl}m_{1}\Delta H & +& m_{2}C \Delta T& = &0\\\text{0.258 g}\times \Delta H& + & \text{100 g} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times 3.1 \, ^{\circ}\text{C} & = & 0\\0.258 \Delta H \text{ g} & + & \text{1300 J} & = & 0\\&&0.258 \Delta H \text{ g} & = & \text{-1300 J} & & \\& &\Delta H & = & \dfrac{\text{-1300 J}}{\text{0.258 g}}\\\\& & & = & \text{-5000 J/g}\\& & & = & \textbf{-5 kJ/g}\\\end{array}[/tex]

[tex]\text{The reaction produces $\large \boxed{\textbf{5 kJ}}$ per gram of potassium.}[/tex]

Note: The answer can have only one significant figure because you measured the initial temperature of the water only to the nearest degree.

From the calculation, the heat generated from the solution is -194.4 kJ/mol

A calorimeter is an instrument that is used to measure heat.

Now we know that number of moles of the potassium = 0.258 g /39 g/mol = 0.0066 moles

Total mass present = 0.258 g + 100 g = 100.258 g

Temperature change = 25.1°C  - 22°C = 3.1°C

Now;

H = -(100.258  * 4.128 *  3.1)/ 0.0066

= -194.4 kJ/mol

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Answer:   11.6g of estrogen are needed to generate an osmotic pressure of 4.45 atm when dissolved in 234 ml of a benzene solution at 298 K.

Explanation:

To calculate the amount of solute, we use the equation for osmotic pressure, which is:

[tex]\pi=iMRT[/tex]

Or,

[tex]\pi=i\times \frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}\times RT[/tex]

where,

[tex]\pi[/tex] = osmotic pressure of the solution = 4.45 atm

i = Van't hoff factor = 1 (for non-electrolytes)

Let Mass of solute (estrogen)  = x g  

Volume of solution = 234 mL

R = Gas constant = [tex]0.0821Latmmol^{-1}K^{-1}[/tex]

T = temperature of the solution = [tex]298K[/tex]

Putting values in above equation, we get:

[tex]4.45=1\times \frac{x\times 1000}{272.4\times234}\times 0.0821Latmmol^{-1}K^{-1}\times 298K[/tex]

[tex]x=11.6g[/tex]

Hence, 11.6g of estrogen are needed to generate an osmotic pressure of 4.45 atm when dissolved in 234 ml of a benzene solution at 298 K.

The question is about calculating the quantity of estrogen needed to generate a specific osmotic pressure in a benzene solution using the Van 't Hoff equation. The moles of estrogen needed is calculated first, then converted into grams using the molar mass of estrogen.

This question is based on the concept of osmotic pressure and solution chemistry, for a nonvolatile, nonelectrolyte compound in solution. The Van 't Hoff equation (π= nRT/V) can be used to solve the problem, where π refers to the osmotic pressure, n is the amount of solute in moles, R is the gas constant (0.0821 L·atm/K·mol for this problem), T is the temperature in Kelvin, and V is the volume in liters. Given the osmotic pressure (4.45 atm), the temperature (298 K), and the volume (0.234 L), you can find the number of moles of estrogen needed. After calculating the amount in moles, use the molar mass of the estrogen (272.4 g/mol) to find the mass in grams. Hence, the quantity of estrogen needed can be calculated.

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Answer:

0.100 m AlCl3  will have the highest boiling point

Explanation:

Step 1: Data given

The molal boiling point elevation constant for water is 0.51°C/m

Since those are all  aqueous solutions, the have the same molal boiling point elevation constant

Step 2:

0.100 m C6H12O6

ΔT = i*Kb*m

⇒with ΔT is the boiling point elevation = TO BE DETERMINED

⇒with i = the van't Hoff factr = 1

⇒with Kb = The molal boiling point elevation constant for water is 0.51°C/m

⇒with m = the molality = 0.100m

ΔT = 1 * 0.51 * 0.100

ΔT  = 0.051 °C

0.100 m NaCl

ΔT = i*Kb*m

⇒with ΔT is the boiling point elevation = TO BE DETERMINED

⇒with i = the van't Hoff factr =  Na+ + Cl- = 2

⇒with Kb = The molal boiling point elevation constant for water is 0.51°C/m

⇒with m = the molality = 0.100m

ΔT =2 * 0.51 * 0.100

ΔT = 0.102 °C

0.100 m AlCl3

ΔT = i*Kb*m

⇒with ΔT is the boiling point elevation = TO BE DETERMINED

⇒with i = the van't Hoff factr =  Al^3+ + 3Cl- = 4

⇒with Kb = The molal boiling point elevation constant for water is 0.51°C/m

⇒with m = the molality = 0.100m

ΔT =4 * 0.51 * 0.100

ΔT = 0.204 °C

0.100 m MgCl2

ΔT = i*Kb*m

⇒with ΔT is the boiling point elevation = TO BE DETERMINED

⇒with i = the van't Hoff factr =  Mg^2+ +2Cl- = 3

⇒with Kb = The molal boiling point elevation constant for water is 0.51°C/m

⇒with m = the molality = 0.100m

ΔT =3 * 0.51 * 0.100

ΔT = 0.153 °C

0.100 m AlCl3  will have the highest boiling point

Answer: The amount of heat released is 42739 Joules

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

[tex]Q=m\times c\times \Delta T[/tex]

Q = Heat released =?

c = heat capacity of granite = [tex]0.790J/g^0C[/tex]

Initial temperature  = [tex]T_i[/tex] = [tex]41.2^0C[/tex]

Final temperature of the calorimeter  = [tex]T_f[/tex]  = [tex]-12.9^0C[/tex]

Change in temperature ,[tex]\Delta T=T_f-T_i=(-12.9-41.2)^0C=-54.1^0C[/tex]

Putting in the values, we get:

[tex]Q=1000g\times 0.790J/g^0C\times -54.1^0C=-42739J[/tex]

As heat comes out to be negative, that means the heat has been released and the amount of heat released is 42739 Joules

Answer:

9.1 seconds

Explanation:

Given that for a second order reaction

1/[A]t = kt + 1/[A]o

Where [A]t= concentration at time = t= 0.340M

[A]o= initial concentration = 0.820M

k= rate constant for the reaction=0.190m-1s-1

t= time taken for the reaction (the unknown)

Hence;

(0.340)^-1 = 0.190×t + (0.820)^-1

t= (0.340)^-1 - (0.820)^-1/0.190

t= 9.1 seconds

Hence the time taken for the concentration to decrease from 0.840M to 0.340M is 9.1 seconds.

Final answer:

To calculate the time it takes for the concentration of compound A to decrease in a second-order reaction, the integrated rate law 1/[A] - 1/[A]0 = kt is used with the provided rate constant and initial and final concentrations.

Explanation:

Calculating Time for a Second-Order Reaction:

The question pertains to the time it takes for the concentration of compound A to decrease from 0.820 M to 0.340 M in a second-order reaction with a rate constant of 0.190 M-1·s-1 at 300 °C. The integrated rate law for a second-order reaction is given by:

1/[A] - 1/[A]0 = kt

Where [A]0 is the initial concentration, [A] is the concentration at time t, k is the rate constant, and t is the time. Using the provided concentrations, we can solve for t:

1/0.340 M - 1/0.820 M = (0.190 M-1·s-1)t

After calculating the left side, we can isolate t:

t = (1/0.340 M - 1/0.820 M) / (0.190 M-1·s-1)

Hence:

(0.340)^-1 = 0.190×t + (0.820)^-1

t= (0.340)^-1 - (0.820)^-1/0.190

t= 9.1 seconds

Hence the time taken for the concentration to decrease from 0.840M to 0.340M is 9.1 seconds.

Answer:

decrease

Explanation:

Answer:

[tex]m = 1\,g[/tex]

Explanation:

The molarity is the ratio of the amount of moles solvent to volume of the solute.

[tex]0.010\,M = \frac{0.010\,moles\,NaOH}{1\,L}[/tex]

The quantity of solute is determined by simple rule of three:

[tex]n = \left(\frac{2.5\,L}{1\,L} \right)\cdot (0.010\,mole)[/tex]

[tex]n = 0.025\,moles[/tex]

The molecular weight of NaOH is [tex]39.997\,\frac{g}{mole}[/tex], the mass of solute is:

[tex]m = 1\,g[/tex]