Consider this reaction occurring at 298 K: N2O(g) + NO2(g) 2 3 NO(g) a. Show that the reaction is not spontaneous under standard conditions by calculating A Grx b. If a reaction mixture contains only N20 and NO2 at partial pressures of 1.0 atm each, the reaction will be spontaneous until some NO forms in the mixture. What maximum partial pressure of NO builds up before the reaction ceases to be spontaneous? c. Can the reaction be made more spontaneous by an increase or decrease in temperature? If so, what temperature is required to make the reaction spontaneous under standard conditions?

Answer:

First structure is sophorose, second structure is turanose, third structure is trehalose, fourth structure is none of the above.

Trehalose is a non-reducing disaccharide, similar to sucrose, with two glucose units linked at their anomeric carbons. Sophorose is a reducing sugar yielding two glucose anomers upon hydrolysis. Turanose is also a reducing sugar but yields one molecule each of glucose (aldose) and fructose (ketose) upon hydrolysis.

Identifying trehalose, sophorose, and turanose among the disaccharides requires understanding their structural characteristics and behavior upon hydrolysis. Trehalose is a non-reducing sugar, which means it doesn't have a free anomeric carbon that can form an open-chain aldehyde or ketone upon hydrolysis. From the given information, we know that trehalose upon hydrolysis gives two molecules of aldoses, which means it must have two glucose units. As trehalose is a non-reducing sugar, it is similar to sucrose in having a glycosidic bond between the two anomeric carbons.

Sophorose is a reducing sugar that upon hydrolysis yields two molecules of an aldose that are anomers of each other, implying the presence of two glucose molecules. Given that it is reducing, at least one of the anomeric carbons must not be involved in the glycosidic linkage, allowing it to open and reduce other compounds.

Turanose, like sophorose, is a reducing sugar, but it gives one molecule of an aldose and one molecule of a ketose upon hydrolysis. This indicates that turanose is composed of one glucose and one fructose molecule. The glucose component would be linked through its anomeric carbon, leaving the fructose part with a potential ketose structure.

Answer:

The peak at mass 100 with a 8% relative abundance is the molecular ion peak

Explanation:

Molecular ion peak has the highest charge to mass ratio,

Mass of 100 is same as mass to charge ratio =100

Answer: The Molar mass of Cro2 (or Chromium (IV) oxide) is 83.9949 g/mol.

Explanation: The atomic mass of Chromium is 51.9961 and the atomic mass of Oxygen is 15.9994. You then add both numbers and multiply by 2. Put it in the calculator like this: 51.9961 + 15.9994 x 2 and you will get 83.9949 g/mol.

Answer:

ΔS = -114.296 J/K

Explanation:

Step 1: Data given

Temperature = 298 K

Number of moles N2 = 2.19 moles

S°(N2) = 191.6 J/K*mol

S°(O2)= 161.1 J/k*mol

S°(N2O) = 219.96 J/K*mol

Step 2: The balanced equation

2N2(g) + O2(g) ⇆ 2N2O(g)

Step 3: Calculate ΔSrxn

ΔSrxn = ∑S°products -  ∑S°reactants

ΔSrxn = (2*219.96) - (161.1 + 2*191.6) J/K*mol

ΔSrxn = -104.38 J/K

Step 4: Calculate ΔS for 2.19 moles

The reaction is for 2 moles N2

ΔS = -114.296 J/K

Final answer:

The entropy change for a chemical reaction can be determined using standard absolute entropies. In this case, the given reaction of N2(g) and O2(g) to form N2O(g) can be analyzed using these principles.

Explanation:

The entropy change for the given reaction can be calculated using the standard absolute entropies at 298K. The formula to calculate the entropy change is ΔS = ΣnS(products) - ΣnS(reactants). In this case, for the reaction 2N2(g) + O2(g) → 2N2O(g), the entropy change can be calculated using the standard absolute entropies of each component.

Step 1: Data given

Temperature = 298 K

Number of moles N2 = 2.19 moles

S°(N2) = 191.6 J/K*mol

S°(O2)= 161.1 J/k*mol

S°(N2O) = 219.96 J/K*mol

Step 2: The balanced equation

2N2(g) + O2(g) ⇆ 2N2O(g)

Step 3: Calculate ΔSrxn

ΔSrxn = ∑S°products -  ∑S°reactants

ΔSrxn = (2*219.96) - (161.1 + 2*191.6) J/K*mol

ΔSrxn = -104.38 J/K

Step 4: Calculate ΔS for 2.19 moles

The reaction is for 2 moles N2

ΔS = -114.296 J/K

Answer: Third stream’s temperature is [tex]33.2^{o}C[/tex]  and its relative humidity is 50%.

Explanation:

According to the Psychrometric chart.

For the first stream, we have the following.

  [tex]T_{1} = 40^{o}C [/tex],       R.H = 40%

      V = 3L/s ,         W = 19 gm of moisture/kg of dry air

    [tex]h_{1}[/tex] = 89 kj/kg ,     [tex]v_{1} = 0.913 m^{3}/kg[/tex]

For the second stream, we have the following.

   [tex]T_{2} = 15^{o}C[/tex]

       R.H = 80%

        V = 1 L/s

     W = 8.5 gm of moisture/kg of dry air

  [tex]h_{2}[/tex] = 36.5 kj/kg

  [tex]v_{2} = 0.828 m^{3}/kg[/tex]

Now,

     [tex]ma_{1} = \frac{V}{v_{1}}[/tex]

              = [tex]\frac{3}{0.913}[/tex]

              = 3.286 kg/s

     [tex]ma_{2} = \frac{V}{v_{2}}[/tex]

              = [tex]\frac{1}{0.828}[/tex]

              = 1.2077 kg/s

We will calculate the value of [tex]ma_{3}[/tex] as follows.

  [tex]ma_{3} = ma_{1} + ma_{2}[/tex]

            = 3.286 kg/s  + 1.2077 kg/s

             = 4.5 kg/s

Now, heat balance will be as follows.

     [tex]ma_{1}h_{1} + ma_{2}h_{2} = ma_{3}h_{3}[/tex]

        [tex]h_{3}[/tex] = 74.7855 kj/kg

Hence, the moisture balance will be calculated as follows.

        [tex]ma_{1}W_{1} + ma_{2}W_{2} = ma_{3}W_{3}[/tex]

       [tex]W_{3}[/tex] = 16.155

Now, again using the psychrometric chart we will find the value of  temperature and relative humidity as follows.

Values of [tex]T_{3}[/tex] and [tex]R.H_{3}[/tex] against [tex]W_{3}[/tex] and [tex]h_{3}[/tex] are:

     [tex]T_{3} = 33.2^{o}C[/tex]

           R.H = 50%

Thus, we can conclude that third stream’s temperature is [tex]33.2^{o}C[/tex]  and its relative humidity is 50%.

To calculate the temperature and relative humidity of the third stream formed by mixing two humid streams, we can use the concept of partial pressure and the definition of relative humidity. Using the given data, we can determine that the third stream's temperature is approximately 27.4°C and its relative humidity is 59.4%.

To calculate the temperature and relative humidity of the third stream formed by adiabatically mixing two humid streams, we can use the concept of partial pressure and the definition of relative humidity.

First, we need to calculate the mole fraction of water vapor in each stream using the given relative humidity values.

Then, we can calculate the partial pressures of water vapor in each stream using the mole fraction and the saturation vapor pressure at the respective temperatures.

Finally, the total pressure of the third stream can be calculated using the ideal gas law, and the mole fraction of water vapor in the third stream can be used to calculate its relative humidity.

Using the given data, we can determine that the third stream's temperature is approximately 27.4°C and its relative humidity is 59.4%.

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Answer:

(a)  Distillate (D) = 0.597 kg

    Bottom product (B) = 0.403 kg

(b) Compositions of distillate  product:

     Benzene = 0.995

     Toluene = 0.05

   Compositions of bottom product:

    Benzene = 0.0149

    Naphthalene = 0.1241

    Toluene = 0.8610

(c) Fraction of toluene fed that is recovered in the bottom product = 99.14%

Explanation:

See the attached file for the calculation

Answer:

a. Copper (II) oxide is the limiting reactant.

b. [tex]m_{N_2}=5.30g[/tex]

c. [tex]Y=87\%[/tex]

d. [tex]m_{NH_3}=2.60gNH_3[/tex]

Explanation:

Hello,

In this case, for the given reaction:

[tex]2 NH_3(g) + 3 CuO(s) \rightarrow N_2(g) + 3 Cu(s) + 3 H_2O(l)[/tex]

a. The limiting reactant is identified by computing the available moles of ammonia and the moles of ammonia that react with 45.2 g of copper (I) oxide as shown below:

[tex]n_{NH_3}^{Available}=9.05gNH_3*\frac{1molNH_3}{17gNH_3}=0.532molNH_3\\n_{NH_3}^{Reacted}=45.2gCuO*\frac{1molCuO}{79.545gCuO}*\frac{2molCuO}{3molCuO} =0.379molNH_3[/tex]

In such a way, as there more ammonia available than that is reacted, we say it is in excess and the copper (II) oxide the limiting reactant.

b. Here, with the reacting moles of ammonia, we compute the yielded grams of nitrogen:

[tex]m_{N_2}=0.379molNH_3*\frac{1molN_2}{2molNH_3}*\frac{28gN_2}{1molN_2}\\m_{N_2}=5.30g[/tex]

c. Now, since the 5.30 g of nitrogen are the expected grams of it, the percent yield of nitrogen is compute by dividing the real obtained mass over the theoretical previously computed mass:

[tex]Y=\frac{4.61g}{5.30g} *100\%\\Y=87\%[/tex]

d. Finally, as 0.532 moles of ammonia are available, but just 0.379 moles react, the unreacted moles are:

[tex]n_{NH_3}=0.532molNH_3-0.379molNH_3=0.153molNH_3[/tex]

That in grams are:

[tex]m_{NH_3}=0.153molNH_3*\frac{17gNH_3}{1molNH_3}=2.60gNH_3[/tex]

Best regards.

The limiting reactant is CuO, the percent yield of nitrogen gas is 29%.

The equation of the reaction is;

2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l)

Number of moles of NH3 =  9.05 g/17 g/mol = 0.53 moles

Number of moles of CuO = 45.2 g/80 g/mol = 0.565 moles

Since 2 moles of NH3 reacts with 3 moles of CuO

0.53 moles of NH3 reacts with 0.53 moles × 3 moles/ 2 moles

= 0.795 moles

We can see that there is not enough CuO in the system hence it is the limiting reactant.

Number of moles of N2 produced = 0.565 moles × 28 g/mol =15.82 g of N2

Percent yield = Actual yield/Theoretical yield × 100/1

Percent yield = 4.61 g/15.82 g  × 100/1

Percent yield = 29%

If 2 moles of NH3 reacts with 3 moles of CuO

x moles of NH3 reacts with 0.565 moles of CuO

x =  2 moles × 0.565 moles/3 moles

x = 0.38 moles

Number of moles of excess reactant left over = 0.53 moles - 0.38 moles

= 0.15 moles

Mass of excess reactant left over = 0.15 moles × 17 g/mol = 2.55 g

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Answer:

1. The balanced equation is given below:

CaCO3 —> CaO + CO2

2. The coefficients are: 1, 1, 1

Explanation:

Step 1:

The word equation is given below:

calcium carbonate decomposes to produce calcium oxide and carbon dioxide.

Step 2:

The elemental equation. This is illustrated below:

Calcium carbonate => CaCO3

calcium oxide => CaO

carbon dioxide => CO2

CaCO3 —> CaO + CO2

A careful observation of the above equation shows that the equation is balanced.

The coefficients are: 1, 1, 1

Final answer:

The balanced equation for the decomposition reaction of calcium carbonate is CaCO3(s) → CaO(s) + CO2(g), representing the formation of calcium oxide and carbon dioxide gas from calcium carbonate.

Explanation:

The decomposition reaction of calcium carbonate forming calcium oxide and carbon dioxide can be represented by the following balanced chemical equation:

CaCO3(s) → CaO(s) + CO2(g)

This equation indicates that one mole of calcium carbonate (CaCO3) decomposes to form one mole of calcium oxide (CaO) and one mole of carbon dioxide (CO2) gas upon heating. The reactants and products are represented using their smallest possible integer coefficients, reflecting the principle of conservation of mass and the stoichiometry of the decomposition reaction.

Answer:

The structure in the first image file attached below shows the arrangement of atoms in hexagonal close packing

we need to show that the ratio between the height of the unit cell divided by its edge length is 1.633

In the structure, the two atoms are shown apart. But in fact the two atoms are touching. Therefore, the edge length is the sum of the radius of two atoms. If we assume r as radius then the expression for edge length will be as follows.

a = 2r    ..............(1)

other attached files show additional solutions in steps

The ideal c/a ratio in an HCP unit cell is 1.633, depicting the most compact arrangement of spheres. However, real HCP metals, influenced by bonding characteristics and atomic sizes, display varying c/a ratios. For Mg with known atomic radius in HCP arrangement, the lattice constants, c/a ratio, and theoretical density can be calculated.

The HCP unit cell (hexagonal-close-packed) is one of the simplest lattice types in crystal structures. It includes two types of atoms, one at the corners (and center) of a hexagonal prism and the other at the top and bottom of the prism. The ideal ratio of c/a (height of the unit cell c divided by edge length a) corresponds to the most compact possible arrangement of spheres, and is theoretically calculated to be √(8/3), or approximately 1.633.

In real-world applications, various HCP metals display c/a ratios varying slightly, such as 1.58 for Be and 1.89 for Cd. This depends on bonding characteristics and varying atomic sizes in the metal crystal structure. For example, for Mg (magnesium) which has a known atomic radius of 0.1605 nm in HCP arrangement, we can calculate the lattice constants, c and a, and thus the c/a ratio. Theoretical density can be determined using the mass of atoms in a unit cell, Avogadro's number, and the volume of unit cell.

Learn more about HCP Unit Cell Geometry here:

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Answer:

Problem 1 => 8778 joules

Problem 2 => 14,630 joules

Explanation:

Reference the Heating Curge of Water Problem posted earlier. These are temperature change fragments of that type problem. As you read the problem note the 'temperature change' phrase in the problem. This should signal you to use the q = m·c·ΔT expression as opposed to the q = m·ΔH expression where no temperature change is noted; i.e., melting or evaporation/boiling.

For the listed problems...

Problem 1: Amount of heat needed to heat 150 grams water from 21.0 to 35.0 Celcius.

Note the temperature change in the problem context => use q = m·c·ΔT ...

q = (150g)(4.18j/g·°C)(35.00°C - 21.0°C) = 8778 joules (4 sig. figs. based on the 150.0g value of water)

Problem 2: Amount of heat needed to heat 250.0 grams water from 31.0 to 45.0 Celcius.

Same type problem. Note temperature change in text of problem => use q = m·c·ΔT ...

q = (250.0g)(4.18j/g·°C)(45.0°C - 31.0°C) = 14,630 joules

Hope this helps. Doc

Answer:

The correct answer is the first option. No, the structures above cannot undergo a Fischer esterification reaction to form an ester.

Explanation:

The reaction that will take place can be found in the attached file. The reaction does not require any catalyst and it cannot undergo a Fischer esterification reaction to form an ester.

The accurate answer is the first option which is No, the structures above cannot undergo a Fischer esterification response to form an ester.

The reaction that will take the spot can be found in the connected file. The reaction does not require any catalyst and it cannot experience a Fischer esterification reaction to create an ester.

Therefore the correct option is No, the structures above cannot experience a Fischer esterification reaction to form an ester.

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Answer:

Reaction Quotient, Kq = {[a-ketoglutarate]x[L-alanine]}/{[L-glutamate]x[pyruvate]}

or, Kq = {(1.6x10-2)x(6.25x10∧-3)}/{(3x10∧-5)x(3.3x10-4)} = 1.01 x 10∧4

Since Kq > Keqb ; therefore the reaction will proceed in the backward direction, in order words the reaction will not occur in forward direction. i.e formation of reactants will be favored.

Explanation:

Question: The question is incomplete. Below is the complete question and the answer;

While ethanol (CH3CH2OH is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by reacting ethylene CH2CH2) with water vapor at elevated temperatures. A chemical engineer studying this reaction fills a 50.0 L tank at 22. °C with 24. mol of ethylene gas and 24. mol of water vapor. He then raises the temperature considerably, and when the mixture has come to equilibrium determines that it contains 15.4 mol of ethylene gas and 15.4 mol of water vapor The engineer then adds another 12. mol of water, and allows the mixture to come to equilibrium again. Calculate the moles of ethanol after equilibrium is reached the second time. Round your answer to 2 significant digits.

Answer:

Number of moles of ethanol = 11 mol

Explanation:

SEE THE ATTACHED FILE FOR THE CALCULATION

Answer:

B. CH3COOH pH > 4.7 (4.8)

Explanation:

∴ mol NaOH = (5 E-3 L)*(0.10 mol/L) = 5 E-4 mol

⇒ mol CH3COOH = (0.05 L)*(0.20 mol/L) = 0.01 mol

⇒ C CH3COOH = (0.01 mol - 5 E-4 mol) / (0.105 L)

⇒ C CH3COOH = 0.0905 M

∴ mol CH3COONa = (0.05 L )*(0.20 mol/L) = 0.01 mol

⇒ C CH3COONa =  (0.01 mol + 5 E-4 mol) / (0.105 L )

⇒ C CH3COONa = 0.1 M

∴ Ka = ([H3O+]*(0.1 + [H3O+])) / (0.0905 - [H3O+]) = 1.75 E-5

⇒ 0.1[H3O+] + [H3O+]² = (1.75 E-5)*(0.0905 - [H3O+])

⇒ [H3O+]² 0.1[H3O+] = 1.584 E-6 - 1.75 E-5[H3O+]

⇒ [H3O+]² + 0.1000175[H3O+] - 1.584 E-6 = 0

⇒ [H3O+] = 1.5835 E-5 M

∴ pH = - Log [H3O+]

⇒ pH = - Log (1.5835 E-5)

⇒ pH = 4.8004 > 4.7

Final answer:

After adding NaOH to the acetate buffer, the acetate ion concentration increases and the pH of the solution will be greater than the pKa of acetic acid (4.7). Option D, with C2H3O2- and pH > 4.7, is correct.

Explanation:

The question involves calculating the pH of an acetate buffer solution after the addition of sodium hydroxide, NaOH. The buffer consists of acetic acid (HC2H3O2) and sodium acetate (NaC2H3O2). The pKa of acetic acid is 4.7, and equal concentrations of the buffer components are typically ideal. When NaOH is added, it neutralizes some of the acetic acid, converting it to acetate and increasing the pH slightly.

Before the addition of NaOH, the buffer has equal concentrations of acetic acid and acetate, so its pH is close to the pKa of acetic acid, which is 4.7. After adding NaOH, more acetic acid is converted to acetate, but the pH will still be maintained close to the pKa since the buffer action resists changes in pH. Therefore, the correct pairing is that the acetate ion (C2H3O2-) will be present in greater concentration and the pH of the solution will be greater than 4.7 after NaOH is added, making option D correct.

Answer:

35.6 g

Explanation:

Step 1: Calculate the [H⁺]

We use the following expression

pH = -log [H⁺]

[H⁺] = antilog -pH = antilog -13.90 = 1.259 × 10⁻¹⁴ M

Step 2: Calculate the [OH⁻]

We use the ionic product of water (Kw).

Kw = 10⁻¹⁴ = [H⁺] × [OH⁻]

[OH⁻] = 0.7943 M

Step 3: Calculate the moles of OH⁻

[tex]\frac{0.7943mol}{L} \times 0.800 L = 0.635 mol[/tex]

Step 4: Calculate the required moles of KOH

KOH is a strong base that dissociates according to the following equation.

KOH → K⁺ + OH⁻

The molar ratio of KOH to OH⁻ is 1:1. Then, the required moles of KOH are 0.635 moles.

Step 5: Calculate the mass of KOH

The molar mass of KOH is 56.11 g/mol.

[tex]0.635 mol \times \frac{56.11g}{mol} =35.6 g[/tex]

The question is incomplete, the table of the question is given below

Answer:

I) xA= 0.34, yA= 0.55

ii) 76.2 mole % vapor

iii) Percentage of vapor volume = 98%

Explanation:

i) xA= 0.34, yA= 0.55

 xA= 0.34, yA= 0.55

ii)      0.50 = 0.55 nv + 0.34 nL

     Therefore, nV =    0.762 mol vapor and nL = 0.238 mol liquid

This shows 76.2 mole % vapor

iii)  ρA= 0.791 g/cm3 and,  ρE = 0.789 g/cm3

Therefore, ρ = 0.790 g/cm3

Now, we have:

MA = 58.08 g/mol and ME= 46.07 g/mol

So Ml = (0.34 x 58.08)+[(1 -0.34) x 46.07] = 50.15 g/mol

1 mol liquid = (0.762 mol vapor/0.238 mol liquid) = 3.2 mol vapor

Liquid volume = Vl= [1 mol x (50.15 g/mol)] / (0.790 g/cm3) = 63.48 cm3

Vapour volume = Vv = 3.2 mol x(22400 cm3/mol) x [(65+273)/273] = 88747 cm3

Therefore, percentage of vapour volume = 88747 / (88747+63.48) = 99.9 %

Find the figure in the attachment

3.47 x [tex]10^{23}[/tex] atoms of gold have mass of 113.44 grams.

Explanation:

Data given:

number of atoms of gold = 3.47 x [tex]10^{23}[/tex]

mass of the gold in given number of atoms = ?

atomic mass of gold =196.96 grams/mole

Avagadro's number = 6.022 X [tex]10^{23}[/tex]

from the relation,

1 mole of element contains 6.022 x [tex]10^{23}[/tex] atoms.

so no of moles of gold given = [tex]\frac{3.47 X 10^{23} }{6.022 X 10^{23} }[/tex]

0.57 moles of gold.

from the relation:

number of moles = [tex]\frac{mass}{atomic mass of 1 mole}[/tex]

rearranging the equation,

mass = number of moles x atomic mass

mass = 0.57 x 196.96

mass = 113.44 grams

thus, 3.47 x [tex]10^{23}[/tex] atoms of gold have mass of 113.44 grams

The mass of 3.47 x 10^23 gold atoms calculated using Avogadro's number and the molar mass of gold equals approximately 113.52 grams. To calculate this, the given number of atoms was first converted into moles and the number of moles were then multiplied by the molar mass.

In chemistry, to calculate the mass of a certain number of atoms, we use the concept of a mole and Avogadro's number (6.022 x 1023). For gold atoms, the molar mass is 197 g/mol. Given that the number of gold atoms is 3.47 x 1023, we can use the relationship that 1 mole of gold atoms = 6.022 x 1023 gold atoms. Therefore, the mass of 3.47 x 1023 gold atoms can be calculated as follows:

First, we convert the number of atoms to moles using Avogadro's number: 3.47 x 1023 atoms / 6.022 x 1023 atoms/mol = 0.576 moles of gold.

Then, we multiply this moles by the molar mass of gold to get the mass in grams: 0.576 moles * 197 g/mol = 113.52 g.

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Answer:

[H3O+]= 2.88×10^-12M

pH= 11.54

Explanation:

pOH= -log[OH] = -log[3.5*10^-3]= 2.46

pH= 14- pOH= 14-2.46= 11.54

High pH means that a solution is basic while high pOH means that a solution is acidic,

Hence the solution is basic

Answer:

No, the copper metal weighs more

Explanation:

When the electrochemical cell is set up, zinc will function as the oxidation half cell (anode) while copper will function as the reduction half cell(cathode).

The cathode increases in size due to deposition of the anode metal on the cathode rod. Hence copper (the cathode) will be heavier in size while the anode zinc will be lighter and smaller in size than the cathode.

The half-life of the radioactive isotope is 8 days because after 16 days (which is two half-lives), the sample is reduced from 20 x 10^10 atoms to 5 x 10^10 atoms.

The question deals with the concept of the half-life of a radioactive isotope, which is the time required for half of the radioactive atoms in a sample to decay. According to the problem statement, an initial sample containing 20 x 1010 atoms decays to 5 x 1010 atoms in 16 days. To find the half-life (t1/2), we can use the fact that after one half-life, the number of atoms would be halved. Starting with 20 x 1010 atoms, after one half-life, there would be 10 x 1010 atoms, and after two half-lives, there would be 5 x 1010. This indicates that two half-lives have passed in 16 days, making the half-life 8 days.

The correct answer is 8 days, option E.

The enolate ion is formed from a carbonyl function via a strong base, leading to resonance structures between keto and enolate forms. The resonance involves shifting of electron pairs, with the enolate form being more prevalent under basic conditions.

An enolate ion is formed when a strong base, B, abstracts a proton α to a carbonyl function. The negatively charged oxygen donates a pair of electrons, pushing the electrons from the double bond into the alpha carbon to form the enolate form. The curved arrow represents the movement of a pair of electrons.

The resonance of the keto form to the enolate form involves shifting of electron pairs, with the electron pair on the carbon moving to form a C=O bond and the π bond between carbon and oxygen moving to the oxygen, forming a p-orbital with a lone pair of electrons and making the oxygen negatively charged.

By this mechanism, the keto and enolate forms are interconvertible with the enolate form being more prevalent under basic conditions. The full resonance structures for both the keto and enolate forms would consist of all their bonds, charges, and nonbonding electron pairs.

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Answer:

copper

Explanation:

These are usually copper wires with no insulation. They make the path through which the electricity flows. One piece of the wire connects the current from the power source (cell) to the load

Answer:

In a Grignard reagent, the carbon bonded to the magnesium has a partial negative charge, because carbon is more electronegative than magnesium. This makes this carbon of the Grignard nucleophilic.

Explanation:

Organometallic compounds have covalent bonds between carbon atoms and metallic atoms. Organometallic reagents are useful because they have nucleophilic carbon atoms, in contrast to the electrophilic carbon atoms of the alkyl halides. The majority of metals are more electropositive than carbon, and the bond is biased with a positive partial charge on the metal and a negative partial charge on the carbon.

Answer:

In a Grignard reagent, the carbon bonded to the magnesium has a partial negative charge, because carbon is more electronegative than magnesium. This makes this carbon of the Grignard nucleophilic In a Grignard reagent, the carbon bonded to the magnesium has a partial positive charge, because carbon is more electronegative than magnesium. This makes this carbon of the Grignard less electrophilic.

Explanation:

The Grignard reagent is a highly reactive organomagnesium compound formed by the reaction of a haloalkane with magnesium in an ether solvent. The carbon atom of a Grignard reagent has a partial negative charge. Hence a Grignard reagent will have a general formula RMgX. Where R must contain a carbanion.

Grignard reagents are versatile reagents used in many applications in organic chemistry.

Answer: The value of [tex]E^{o}[/tex] for the half-cell reaction is 0.222 V.

Explanation:

Equation for solubility equilibrium is as follows.

          [tex]AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)[/tex]

Its solubility product will be as follows.

       [tex]K_{sp} = [Ag^{+}][Cl^{-}][/tex]

Cell reaction for this equation is as follows.

     [tex]Ag(s)| AgCl(s)|Cl^{-}(0.1 M)|| Ag^{+}(1.0 M)| Ag(s)[/tex]

Reduction half-reaction: [tex]Ag^{+} + 1e^{-} \rightarrow Ag(s)[/tex],  [tex]E^{o}_{Ag^{+}/Ag} = 0.799 V[/tex]

Oxidation half-reaction: [tex]Ag(s) + Cl^{-}(aq) \rightarrow AgCl(s) + 1e^{-}[/tex],   [tex]E^{o}_{AgCl/Ag}[/tex] = ?

Cell reaction: [tex]Ag^{+}(aq) + Cl^{-}(aq) \rightarrow AgCl(s)[/tex]

So, for this cell reaction the number of moles of electrons transferred are n = 1.

    Solubility product, [tex]K_{sp} = [Ag^{+}][Cl^{-}][/tex]

                                               = [tex]1.77 \times 10^{-10}[/tex]

Therefore, according to the Nernst equation

           [tex]E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}[/tex]

At equilibrium, [tex]E_{cell}[/tex] = 0.00 V

Putting the given values into the above formula as follows.

         [tex]E_{cell} = E^{o}_{cell} - \frac{0.0592 V}{n} log \frac{[AgCl]}{[Ag^{+}][Cl^{-}]}[/tex]

        [tex]0.00 = E^{o}_{cell} - \frac{0.0592 V}{1} log \frac{1}{[Ag^{+}][Cl^{-}]}[/tex]    

       [tex]E^{o}_{cell} = \frac{0.0592}{1} log \frac{1}{K_{sp}}[/tex]

                  = [tex]0.0591 V \times log \frac{1}{1.77 \times 10^{-10}}[/tex]

                  = 0.577 V

Hence, we will calculate the standard cell potential as follows.

           [tex]E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode}[/tex]

       [tex]0.577 V = E^{o}_{Ag^{+}/Ag} - E^{o}_{AgCl/Ag}[/tex]

       [tex]0.577 V = 0.799 V - E^{o}_{AgCl/Ag}[/tex]

       [tex]E^{o}_{AgCl/Ag}[/tex] = 0.222 V

Thus, we can conclude that value of [tex]E^{o}[/tex] for the half-cell reaction is 0.222 V.

The E° for the half-reaction can be calculated by using the Nernst equation, solubility product constant (Ksp) and the standard reduction potential for Ag+(aq) + e− ⇀ ↽ − Ag(s) which is +0.799 V.

The E° for the half-reaction, AgCl (s) + e− ⇀ ↽ − Ag (s) + Cl− (aq) can be calculated using the Nernst equation, considering the relationship between standard reduction potential and solubility product constant.

First, we can rewrite the half-reaction for the dissolution of AgCl: AgCl(s) ⇀ ↽ − Ag+(aq) + Cl−(aq). The solubility product constant (Ksp) for this reaction is 1.77 × 10^−10.

We also know that the standard reduction potential of Ag+(aq) + e− ⇀ ↽ − Ag(s) is +0.799 V.

We can relate this data through the Nernst equation: E=E°−(RT/nF)lnQ. Here, we can set Q as Ksp and adjust the equation to solve for our wanted E.

Using these values in Nernst equation would yield the E for the half-reaction.

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Answer:

Check the explanation

Explanation:

kindly check the attached image below to see the answer to question A and B

(c) The energy balance equation if the pressure is constant (ΔE=0) is,

ΔE= ΔU + Δ[tex]E_{k}[/tex]

From the above relation, it is clear that some heat energy used to raises the temperature of the air.

Hence, the internal energy is not equal to zero. Therefore, the rate of transfer of heat to air is not equal to the rate of thane in kinetic energy of the air.

That is ΔE ≠ Δ[tex]E_{k}[/tex]

Answer:

a) [tex]\delta E_k[/tex] = 112.164 W

b) [tex]\delta E_k[/tex]  = 42.567 W

c) From what we've explained in part b;  increase in temperature of the system is caused by rate of heat transfer . Therefore, not all heat is used to  increase kinetic energy. Hence, since not all the heat is used to increase the kinetic energy . It is not valid and it is incorrect to say that rate of heat transfer is equal to the change in kinetic energy.

Explanation:

The given data include:

Inlet diameter [tex](d_1)[/tex] = 7 cm = 0.7 m

Inlet velocity [tex](v_1)[/tex] = 42 m/s

Inlet pressure [tex](P_1)[/tex] = 130 KPa

Inlet temoerature [tex](T_1)[/tex] = 300°C = (300 + 273.15) = 573.15 K

a)  Assuming Ideal gas behaviour

Inlet Volumetric flowrate [tex](V_1)[/tex] = Inlet velocity [tex](v_1)[/tex] × area of the tube

[tex]= v_1*(\frac{\pi}{4})0.07^2\\\\= 0.161635 \ m^3/s[/tex]

Using Ideal gas law at Inlet 1

[tex]P_1V_1 =nRT_1[/tex]

where ; n = molar flow rate of steam

making n the subject of the formula; we have:

[tex]n = \frac{P_1V_1}{RT_1}[/tex]

[tex]n = \frac{130*42}{8.314*573.15}[/tex]

[tex]n= 4.4096*10^{-3} \ Kmol/s[/tex]

Moleular weight of air = 28.84 g/mol

The mass flow rate = molar flowrate × molecular weight of air

[tex]= 4.4096*10^{-3} \ * 28.84\\= 0.12717 \ kg/s[/tex]

Finally: the kinetic  energy at Inlet [tex](\delta E_k) = \frac{1}{2}*mass \ flowrate *v_1^2[/tex]

= [tex]0.5*0.12717*42^2[/tex]

= 112.164 W

b) If the air is heated to 400°C;

Then temperature at 400°C = (400 + 273.15)K = 673.15 K

Thee pressure is also said to be constant ;

i.e [tex]P_1= P_2[/tex] = 130 KPa

Therefore; the mass flow rate is also the same ; so as the molar flow rate:

Thus; [tex]n= 4.4096*10^{-3} \ Kmol/s[/tex]

Using Ideal gas law at Inlet 2

[tex]P_2V_2 = nRT_2[/tex]

making [tex]V_2[/tex] the subject of the formula; we have:

[tex]V_2 = \frac{nRT_2}{P_2}[/tex]

[tex]V_2 = \frac{4.4906*10^{-3}*8.314*673.15}{130}[/tex]

[tex]V_2 = 0.189836 \ m^3/s[/tex]

Assuming that the diameter is constant

[tex]d_1 = d_2 = 0.07 \ cm[/tex]

Now; the velocity at outlet = [tex]\frac{V_2}{\frac{\pi}{4}d_2^2}[/tex]

= [tex]\frac{0.0189836}{\frac{\pi}{4}(0.07)^2}[/tex]

= 49.33 m/s

Change in kinetic energy [tex]\delta E_k[/tex]  = [tex]\frac{1}{2}*mass \ flowrate * \delta V[/tex]

= [tex]0.5*0.12717 *(49.33^2 -42^2)[/tex]

= 42.567 W

c).

From what we've explained in part b; increase in temperature of the system is caused by rate of heat transfer . Therefore, not all heat is used to  increase kinetic energy. Hence, since not all the heat is used to increase the kinetic energy . It is not valid and it is incorrect to say that rate of heat transfer is equal to the change in kinetic energy.

Answer: pH of the solution after the addition of 600.0 ml of LiOH is 7.

Explanation:

The given data is as follows.

    Volume of [tex]HClO_{4}[/tex] = 900.0 ml = 0.9 L,

   Molarity of [tex]HClO_{4}[/tex] = 0.18 M,

Hence, we will calculate the number of moles of [tex]HClO_{4}[/tex] as follows.

       No. of moles = Molarity × Volume

                             = 0.18 M × 0.9 L

                             = 0.162 moles

Volume of NaOH = 600.0 ml = 0.6 L

Molarity of NaOH = 0.27 M

No. of moles of NaOH = Molarity × Volume

                                     = 0.27 M × 0.6 L

                                     = 0.162 moles

This shows that the number of moles of [tex]HClO_{4}[/tex] is equal to the number of moles of NaOH.

Also we know that,

         [tex]HClO_{4} + NaOH \rightarrow NaClO_{4} + H_{2}O[/tex]

As 1 mole of [tex]HClO_{4}[/tex] reacts with 1 mole of NaOH then all the hydrogen ions and hydroxide ions will be consumed.

This means that pH = 7.

Thus, we can conclude that pH of the solution after the addition of 600.0 ml of LiOH is 7.

Answer:

The heat is  [tex]H= -8.044KJ[/tex]

Explanation:

From the question we are told that

   The pressure is  [tex]P = 1 \ atm[/tex]

    The boiling point is  [tex]B_P = 34^oC[/tex]

     The heat of vaporization at 34°C is  = [tex]26.5 kJ/mol[/tex]

       The specific heat of the liquid is [tex]c_p = 2.32 J/g^oC[/tex]

      The  mass is  [tex]m = 22.5g[/tex]

  The no of moles of the sample of [tex]C_2 H_5OC_2H_5[/tex]  is given as

               [tex]No \ mole (n) = \frac{Mass \ of \ sample }{Molar \ mass }[/tex]

    The  Molar mass for [tex]C_2 H_5OC_2H_5[/tex]  is a value = [tex]= 74.12 g/mol[/tex]

Substituting the value into the above equation

          [tex]n = \frac{22.5}{74.12}[/tex]

                           [tex]= 0.30356 \ mol[/tex]

      The heat H is mathematically as

                [tex]H =- nH_{vap}[/tex]

The negative sign show that the heat is for condensing

            [tex]H = 0.30356 * 26.5 *10^{3} J/mol[/tex]

                [tex]H= -8.044KJ[/tex]

The enthalpy change for condensing a 22.5 g sample of gaseous ether at its normal boiling point of 34.6 °C is approximately -8.045 kJ.

The question asks about the enthalpy change (H) during the condensation of a sample of ether. To calculate the enthalpy change for condensing 22.5 g of diethyl ether at its boiling point, we'll use the given enthalpy of vaporization ([tex]H_{vap}[/tex]) and the mass of the ether. Considering that the enthalpy of vaporization ([tex]H_{vap}[/tex]) is the amount of energy required to vaporize 1 mole of a substance at its boiling point, the enthalpy of condensation will be the negative of this value since condensation is the reverse process.

First, we convert the mass of ether to moles:

22.5 g * (1 mol / 74.12 g/mol) ≈ 0.3036 mol

Next, multiply the moles by the enthalpy of vaporization (Hvap) but with a negative sign for condensation:

-0.3036 mol * 26.5 kJ/mol ≈ -8.045 kJ

This is the enthalpy change for the condensation process of the ether sample.

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Hope this helped

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