You replicate the CEC analysis of a secondary alcohol you performed in the lab using a reverse phase TLC plate producing the TLC plate below. Based on this TLC plate and the mnemonic in your notes, what can you conclude about the stereochemistry of the alcohol? (1 pts)

Explanation:

Let us assume that 50 carbon atoms are available with possible 100 site bondings. It is given here that 94% are occupied by hydrogen (94 out of 100) and 6% (6 out of 100) are occupied by chlorine atom.

Therefore, moles of carbon = 50

                  moles of hydrogen = 94

                  moles of chlorine = 6

Therefore,

  Mass of 50 carbon atoms = [tex]50 \times 12.01 g/mol[/tex] = 600.5 g/mol

  Mass of 94 hydrogen atoms = [tex]94 \times 1.008 g/mol[/tex] = 94.752 g/mol

  Mass of 6 chlorine atoms = [tex]6 \times 35.45 g/mol[/tex] = 212.7 g/mol

Therefore, concentration of chlorine is as follows.

              [tex]C_{cl} = \frac{m_{cl}}{m_{c} + m_{H} + m_{cl}}[/tex]

                         = [tex]\frac{212.7}{907.952} \times 100[/tex]

                         = 23.42%

thus, we can conclude that the concentration of Cl is 23.42%.

The calculation of Chlorine concentration in high-density polyethylene after 6% Hydrogen replacement involves determining the increased mass from Chlorine substitution, considering the significant difference in atomic masses of Hydrogen and Chlorine. Chlorine's contribution to the weight of the new compound would be considerable.

High-density polyethylene has a chemical formula of (C2H4)n, where 'n' denotes the number of repeating units in the polymer chain. Each molecule of ethylene contributes two hydrogen atoms. Therefore, if we assume 6% of all hydrogen atoms are replaced by chlorine atoms, we have to add the weight of chlorine into the composition.

Under normal conditions, Chlorine (Cl) has an atomic weight of approximately 35.45 g/mol, and Hydrogen (H) has an atomic weight of approximately 1.008 g/mol. Therefore, the introduction of Chlorine substitutes a weight of 35.45 g for every 1.008 g of Hydrogen.

To calculate the weight percent of Chlorine in the composition, we determine the total weight contributed by chlorine and divide it by the total weight of the new compound, then multiply by 100%. As a result, the weight percentage of Chlorine when 6% of hydrogen atoms are replaced, following the calculations and the Principle of Atomic Substitution, is quite significant due to the considerable difference in atomic weights of Chlorine and Hydrogen.

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Answer:

radius = 156 pm

Explanation:

The relation between radius and edge length of unit cell of BCC is

r=a[tex]\sqrt{3}[/tex]/4

Given

a = 360 pm

Therefore

r = r = radius = 360[tex]\sqrt{3}[/tex]/4= 155.88 pm

Or

156 pm

The question is incomplete, complete question is :

In the Haber reaction, patented by German chemist Fritz Haber in 1908, dinitrogen gas combines with dihydrogen gas to produce gaseous ammonia. This reaction is now the first step taken to make most of the world's fertilizer. Suppose a chemical engineer studying a new catalyst for the Haber reaction finds that 348 liters per second of dinitrogen are consumed when the reaction is run at 205°C and 0.72 atm. Calculate the rate at which ammonia is being produced.

Answer:

The rate of production of ammonia is 217.08 grams per second.

Explanation:

[tex]N_2+3H_2\rightarrow 2NH_3[/tex]

Volume of dinitrogen used in a second = 348 L

Temperature of the gas = T = 205°C = 205+273 K = 478 K

Pressure of the gas = P = 0.72 atm

Moles of dinitrogen = n

[tex]n=\frac{PV}{RT}=\frac{0.72 atm\times 348 L}{0.0821 atm L/mol K\times 478 K}=6.385 mol[/tex]

According to reaction, 1 mole of dinitriogen gives 2 mole of ammonia.Then 6.385 moles of dinitrogen will give:

[tex]\frac{2}{1}\times 6.385 mol=12.769 mol[/tex]

Mass of 12.769 moles of ammonia;

12.769 mol 17 g/mol = 217.08 g

217.08 grams of ammonia is produced per second.So, the rate of production of ammonia is 217.08 grams per second.

The CO2 concentration has been increasing steadily for several decades, with an average annual increase about 1.4 ppm between 1958 and 2014, somewhat lower between 1970 and 1980 and higher between 2004 and 2014. The most recent CO2 concentration measured I found is 414.0 ppm up to July 2021.

It appears that the student's question is missing specific data, however, I can provide information on the average increase in

CO2 concentration

over certain periods. The concentration of CO2 in the atmosphere has been steadily increasing since the beginning of the industrial revolution, with more prominent increases in recent decades due to increased industrial activity and deforestation. For example, between 1958 and 2014, the CO2 concentration in the atmosphere reportedly increased by about 1.4 parts per million (ppm) per year on average. The rate was somewhat lower between 1970 and 1980 (about 1.3 ppm/year), but higher between 2004 and 2014 (about 2 ppm/year). The most recent CO2 concentration recorded I could find is 414.0 ppm up to July 2021. It's important to note that

increasing CO2 concentration

is a significant contributor to global warming and climate change.

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Answer:

328.4KJ

Explanation:

Before we move on to calculate enthalpy change, we calculate the amount of heat Q

Q= mcΔT

m = density * volume = 250 * 1.25 = 312.5g

c = 3.74J/g.k

ΔT = 7.80 + 273.15K = 280.95K

Q= 312.5 * 3.74 * 280.95 = 328,360.312 J= 328.4KJ(1000J = 1KJ, so divide by 1000)

The enthalpy change in the reaction is same as amount of heat transferred = 328.4KJ

To calculate the standard enthalpy change for the given chemical reaction, balance the equation, then apply standard enthalpies of formation for the reactants and products, and use Hess's Law. The calculated standard enthalpy change for this reaction is -433.37 kJ/mol.

To calculate the standard enthalpy change ([tex]H_orxn[/tex]) for the reaction [tex]CH_4(g) + Cl_2(g) \rightarrow CCl_4(l) + HCl(g)[/tex], we need to first balance the reaction:

[tex]CH_4(g) + 4Cl_2(g) \rightarrow CCl_4(l) + 4HCl(g)[/tex]

Then we use the standard enthalpies of formation ( Hf) for each substance to calculate [tex]H_orxn[/tex] using Hess's Law:

[tex]H_o = [ H_f[CCl_4(l)] + 4 H_f[HCl(g)]] - [ H_f[CH_4(g)] + 4 H_f[Cl_2(g)]][/tex]

[tex]H_o = [-139 kJ/mol + 4(-92.31 kJ/mol)] - [-74.87 kJ/mol + 4(0 kJ/mol)][/tex]

[tex]H_o = -139 kJ/mol - 369.24 kJ/mol - (-74.87 kJ/mol)[/tex]

[tex]H_o = -433.37 kJ/mol[/tex]

ΔH°rxn for CH₄(g) + Cl₂(g) → CCl₄(l) + HCl(g), is -433.37 kJ

To calculate the enthalpy change (ΔH°rxn) for the reaction CH₄(g) + Cl₂(g) → CCl₄(l) + HCl(g), we first need to balance the equation and use the standard enthalpies of formation (ΔH°f) provided for each compound.

Balance the chemical equation:

The balanced equation is: CH₄(g) + 4Cl₂(g) → CCl₄(l) + 4HCl(g)

Write the enthalpy of formation values:

[tex]\Delta H^\circ_f[\text{CH}_4(g)] &= -74.87 \, \text{kJ/mol} \\\\\Delta H^\circ_f[\text{CCl}_4(\ell)] &= -139 \, \text{kJ/mol} \\\\\Delta H^\circ_f[\text{HCl(g)}] &= -92.31 \, \text{kJ/mol} \\\\\Delta H^\circ_f[\text{Cl}_2(g)] &= 0 \, \text{kJ/mol} \quad (\text{by convention})[/tex]

Calculate the total ΔH°f for the products:

[tex]\text{For CCl}_4(\ell) \text{ and HCl(g):}\\\\ & {1 \, \text{mol of CCl}_4(\ell) \times (-139 \, \text{kJ/mol})} + {4 \, \text{mol of HCl(g)} \times (-92.31 \, \text{kJ/mol})} \\& = -139 \, \text{kJ} + (-369.24 \, \text{kJ}) \\& = -508.24 \, \text{kJ} \\[/tex]

Calculate the total ΔH°f for the reactants:

[tex]\text{For CH}_4(g) \text{ and Cl}_2(g): \\\\& {1 \, \text{mol of CH}_4(g) \times (-74.87 \, \text{kJ/mol})} + {4 \, \text{mol of Cl}_2(g) \times (0 \, \text{kJ/mol})} \\& = -74.87 \, \text{kJ} \\[/tex]

Determine ΔH°rxn:

[tex]\Delta H^\circ_\text{rxn}: & = \Sigma \Delta H^\circ_f(\text{products}) - \Sigma \Delta H^\circ_f(\text{reactants}) \\\\& = -508.24 \, \text{kJ} - (-74.87 \, \text{kJ}) \\& = -433.37 \, \text{kJ}[/tex]

Therefore, the enthalpy change for the reaction ΔH°rxn is -433.37 kJ.

Answer:

≅ 16.81 kJ

Explanation:

Given that;

mass of acetone = 31.5 g

molar mass of acetone = 58.08 g/mol

heat of vaporization for acetone = 31.0 kJ/molkJ/mol.

Number of moles = [tex]\frac{mass}{molar mass}[/tex]

Number of moles of acetone = [tex]\frac{31.5}{58.08}[/tex]

Number of moles  of acetone = 0.5424 mole

The heat required to vaporize 31.5 g of acetone can be determined by multiplying the number of moles of acetone with the heat of vaporization of acetone;

Hence;

The heat required to vaporize 31.5 g of acetone = 0.5424 mole × 31.0 kJ/mol

The heat required to vaporize 31.5 g of acetone = 16.8144 kJ

≅ 16.81 kJ

Answer:

No mass of hexane could be left over by the chemical reaction.

C₆H₁₄ is the limitin reagent

Explanation:

This is a combustion reaction were the oxygen is one of the reactant and it burns a compound in order to generate water and carbon dioxide.

In this case, we have liquid hexane combustion, so the reaction is:

2C₆H₁₄(l) +  19O₂(g)  →   12CO₂(g) +  14H₂O(g)

In this situation we are asked for the mass of a reactant that could be left over, this is the excess reagent.

We convert the masses of reactants to moles:

9.48 g . 1mol  / 86g = 0.110 moles of C₆H₁₄

43 g . 1 mol/32 g =  1.34 moles of O₂

The hexane may be the excess reagent but we confirm like this:

19 moles of O₂ need 2 moles of hexane to react

Then, 1.34 moles of O₂ will react with (1.34 . 2) / 19 = 0.141 moles

We need 0.141 moles, and we only have 0.110. Hexane is the limiting reagent so no mass could be left over by the chemical reaction.

In conclussion, oxygen is excess reactant. We verify:

2 moles of hexane need 19 moles of O₂ to react

Then, 0.110 moles of hexane will react with (0.110 . 19) / 2 = 1.04 moles of O₂

As I have 1.34 moles of oxygen, value is higher so in this case we are having mass that could be left over by the reaction.

At a constant temperature of 50 °C, a volume of 1.0 L would contain approximately 0.045 moles of gas, whereas a volume of 3.0 L would contain approximately 0.134 moles of gas. These values are based on the standard molar volume of 22.4 L/mol at Standard Temperature and Pressure (STP).

In order to make this comparison, we can use Avogadro's Law which states that equal volumes of gases, at the same temperature and pressure, will contain an equal number of molecules or moles. We would also use the standard molar volume of an ideal gas at Standard Temperature and Pressure (STP), which is about 22.4 L/mol.

Given that our system's temperature is constant at 50 °C and comparing volumes of 1.0 L and 3.0 L, we simply calculate the number of moles in each volume. For 1.0 L, if we stay consistent with molar volume at STP, it equates to about 0.045 moles of gas (1 L / 22.4 L per mole). For 3.0 L, it would equate to about 0.134 moles of gas (3 L / 22.4 L per mole).

Please note that this approximation is based on the standard molar volume at STP, and accuracy may vary at temperatures significantly different from STP (such as 50 °C).

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Answer:

Ca(OH)2(s) + 2HCl(aq) →  CaCl2(aq) + 2H2O(l)

Explanation:

Step 1: Data given

Mass of Ca(OH)2 = 9.49 grams

Volume of a 0.490 M HCl solution = 30.0 mL = 0.030 L

Step 2: The unbalanced equation

Ca(OH)2 + HCl →  CaCl2 + H2O

On the left side we have 2x O (in Ca(OH)2 )  and on the right side we have 1x O (in H2O). To balance the amount of O we have to multiply H2O, on the right side , by 2.

Ca(OH)2 + HCl →  CaCl2 + 2H2O

On the right side we have 4x H (in 2H2O), and on the left side we have 3x H  (2x in Ca(OH)2 and 1x in HCl). To multiply the amount of H, we have to multiply HCl on the left side by 2.

Ca(OH)2(s) + 2HCl(aq) →  CaCl2(aq) + 2H2O(l)

Answer:

Asking for and receiving a paper that someone who is in another lab section wrote is an honor violation. True

An oral discussion with a classmate regarding a paper's topics and format is cheating. False

Purchasing papers from tutoring companies is allowed. False

Sending a paper that you wrote to a student who is currently in another section is an honor violation. True

Misconduct in research or scholarship includes fabrication, falsification, or plagiarism in proposing, performing, reviewing, or reporting research. It does not include honest error or honest differences in interpretations or judgements of data. True

Discussing an exam or laboratory practical with anyone prior to the conclusion of the week that final exams and laboratory practicals are given would be considered an honor violation. True

Explanation:

The Aggie code of honour is a code of academic integrity of the Texas A&M University. It spells out the codes of academic integrity and responsible research. Instructors are to include this code in all syllabi. It targets the application of the highest degree of integrity in academic research and forbids misconducts such as cheating, plagiarism, falsification et cetera.

The statements regarding the Aggie Honor System and Academic Integrity rules for CHEM 111/112/117 are generally correct, with exceptions for discussions about paper topics, which is not innately cheating, and purchasing of papers, which is not allowed.

Based on the Aggie Honor System Rules and the Academic Integrity statement on the CHEM 111/112/117 syllabus, the following assessments can be made about the statements:

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Answer: Cyclohexene

Explanation:

Cyclohexane belongs to the Alkenes family. Alkenes react in the cold with pure liquid bromine, or with a solution of bromine in an organic solvent like tetrachloromethane. The double bond breaks, and a bromine atom get attached to each carbon. The bromine loses its original red-brown color to give a colorless liquid. In the case of the reaction with ethene, 1,2-dibromoethane is formed. When bromine is added to cyclohexane in the dark room, there won't be any reaction. If the mixture is exposed to light however, free bromine radicals are generated. In this condition, polybrominated products can be produced as well.

The chemical used to quench the residual bromine at the end of the isomerization reaction is usually an organic compound containing a reducing agent, such as sodium bisulfite (NaHSO3) or sodium hydrogen sulfite (NaHSO3).

The chemical used to quench the residual bromine at the end of the isomerization reaction is usually an organic compound containing a reducing agent, such as sodium bisulfite (NaHSO3) or sodium hydrogen sulfite (NaHSO3). These compounds react with bromine to form non-toxic salts that can be easily removed. For example, NaHSO3 reacts with bromine to form sodium bromide (NaBr) and sodium sulfate (Na2SO4). This reaction effectively removes the residual bromine and prevents it from causing further reactions or harming the desired product.

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Fusion and Fission both are nuclear reactions where the major difference between them is Fusion is combining process whereas fission is a breaking process.

Explanation:

Similarity:

Both Fission and Fusion are nuclear reactions that releases vast amount of energy.

Difference:

Answer:

The molar mass of aluminum is 26.9 grams per mole. We start by converting the grams of aluminum to moles of aluminum.

m

o

l

A

l

=

3.78

g

A

l

×

1

m

o

l

A

l

26.9

g

=

0.1405

m

o

l

A

l

We use Avogadro's number to determine the number of aluminum atoms present in 0.1405 mol Al.

A

t

o

m

s

A

l

=

0.1405

m

o

l

A

l

×

6.022

×

10

23

1

m

o

l

A

l

=

8.46

×

10

22

a

t

o

m

s

A

l

Answer:

0.846×10²³

Explanation:

molar mass of aluminum is 26.9 g/mol.  

converting the grams of aluminum to moles of aluminum.

mol Al = 3.78 g ×1 mol Al÷26.9 g = 0.1405 mol of Al

to determine the number of aluminum atoms present in 0.1405 mol Al we use the Avogrado's number which is equal to 6.022×10²³

Atoms of  Al = 0.1405 mol Al×6.022×10²³

= 0.846×10²³

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The pH changes significantly through the different stages of titration of triethylamine with HCl. After adding 11.00 mL of HCl, the pH is approximately 11.30, demonstrating a basic solution. At 20.60 mL and 25.00 mL of HCl added, the pH drops to approximately 2.83 and 1.95, respectively, indicating an acidic solution due to the excess HCl.

To find the pH during the titration of 20.00 mL of 0.1000 M triethylamine, [((CH₃CH₂)₃N)] , with 0.1000 M HCl solution, we need to follow the steps below at different volumes of HCl addition:

(a) After adding 11.00 mL of HCl

Calculate moles of triethylamine:
moles of [((CH₃CH₂)₃N)] = 0.1000 M x 0.0200 L = 0.00200 mol

Calculate moles of HCl added:
moles of HCl = 0.1000 M * 0.0110 L = 0.00110 mol

Moles of triethylamine remaining:
0.00200 mol - 0.00110 mol = 0.00090 mol

Volume of solution after adding HCl:
20.00 mL + 11.00 mL = 31.00 mL or 0.0310 L

Concentration of triethylamine:
[((CH₃CH₂)₃N)] = 0.00090 mol / 0.0310 L ≈ 0.0290 M

Using  Kᵇ, find [OH⁻]:
Kᵇ = 5.2 x 10⁻⁴, set up ICE table for equilibrium calculation, estimate [OH⁻].

Find pOH and then pH:
pOH = -log[OH⁻]; pH = 14 - pOH

pH ≈ 11.30

(b) After adding 20.60 mL of HCl

Moles of HCl added:
0.1000 M x 0.0206 L = 0.00206 mol

Moles of triethylamine remaining:
0.00200 mol - 0.00206 mol = -0.00006 mol. This indicates an excess of HCl.

Moles of excess HCl:
0.00006 mol

Volume of solution:
20.00 mL + 20.60 mL = 40.60 mL or 0.0406 L

Concentration of H⁺:
[H⁺] = 0.00006 mol / 0.0406 L ≈ 0.00148 M

Find pH:
pH = -log[H⁺]

pH ≈ 2.83

(c) After adding 25.00 mL of HCl

The pH values during the titration are (a) 10.65 after adding 11.00 mL of HCl, (b) 3.97 at equivalence point (20.60 mL), and (c) 1.95 after adding 25.00 mL of HCl.

To find the pH during the titration of 20.00 mL of 0.1000 M triethylamine, (CH₃CH₂)₃N (Kb = 5.2 × 10⁻⁴), with 0.1000 M HCl solution after different volumes of titrant have been added, follow these steps:

Initial pH Calculation (before titration)

1. Calculate the pOH from Kb and the initial concentration of triethylamine:

Kb = 5.2 × 10⁻⁴

[B] = 0.1000 M

Using the formula:

[tex]& K_b = \frac{[\text{OH}^-][\text{BH}^+]}{[B]} \\[/tex]

We assume that [OH⁻] = [BH⁺]. Thus, [tex]K_b = \frac{[\text{OH}^-]^2}{[B]} \\[/tex]

[tex]& [\text{OH}^-] = \sqrt{K_b \cdot [B]} = \sqrt{5.2 \times 10^{-4} \cdot 0.1000} = 0.0072 \, \text{M} \\\\[/tex]

[tex]& \text{pOH} = -\log[\text{OH}^-] = -\log(0.0072) \approx 2.14 \\\\ & \text{pH} = 14 - \text{pOH} = 14 - 2.14 = 11.86 \\[/tex]

After Adding 11.00 mL of HCl

2. Calculate the moles of HCl added:

n(HCl) = M * V = 0.1000 M * 0.01100 L = 0.001100 mol

Initial moles of (CH₃CH₂)₃N = 0.1000 M * 0.02000 L = 0.002000 mol

Remaining (CH₃CH₂)₃N = 0.002000 mol - 0.001100 mol = 0.000900 mol

Moles of (CH₃CH₂)₃NH⁺ formed = 0.001100 mol

Total volume = 20 mL + 11 mL = 31 mL = 0.031 L

Compute the concentrations:

[B] = 0.000900 mol / 0.031 L ≈ 0.029 M

[HB+] = 0.001100 mol / 0.031 L ≈ 0.035 M

Using the Henderson-Hasselbalch equation:

[tex]pH = pKa + \log\left(\frac{[B]}{[\text{HB}^+]}\right)\\\\pKa = 14 - \text{pKb} = 14 - 3.28 = 10.72\\\\pH = 10.72 + \log\left(\frac{0.029}{0.035}\right) = 10.72 + \log(0.829) \approx 10.65[/tex]

At Equivalence Point (20.60 mL)

3. Calculate moles at equivalence point:

Moles of  (CH₃CH₂)₃N = 0.002000 mol

Equivalence moles of HCl = 0.002060 mol

All the base has been neutralized:

(CH₃CH₂)₃NH⁺ in solution = 0.002060 / 0.04060 = 0.0507 M

Calculate pH:

[H+] from the equilibrium of (CH₃CH₂)₃NH⁺ :

[tex]K_a \text{ for } (CH_3CH_2)_3NH^+ = \frac{1}{K_b} = \frac{1}{5.2 \times 10^{-4}} = 1.923 \times 10^3[/tex]

[tex]& [\text{H}^+] = \sqrt{1.923 \times 10^3 \times 0.0507 \, \text{M}} = 0.09825 \, \text{M} \\\\& \text{pH} = -\log(0.09825) \approx 3.97 \\[/tex]

After Adding 25.00 mL of HCl

4. Calculate moles of excess HCl:

HCl added (total) = 0.002500 moles

Excess HCl = 0.002500 - 0.002000 = 0.000500 mol

Total volume = 20 mL + 25 mL = 45 mL = 0.045 L

[tex]& [\text{H}^+] = \frac{0.000500 \, \text{mol}}{0.045 \, \text{L}} \approx 0.0111 \, \text{M} \\[/tex]

[tex]& \text{pH} = -\log(0.0111) \approx 1.95 \\[/tex]

In summary, the pH values during the titration are 10.65 after adding 11.00 mL of HCl, 3.97 at equivalence point (20.60 mL), and 1.95 after adding 25.00 mL of HCl.

Answer:

   ΔrxnH  = -580.5 kJ

Explanation:

To solve this question we are going to help ourselves with Hess´s law.

Basically the strategy here is to work  in an algebraic way with the three first reactions so as to reprduce the desired equation when we add them together, paying particular attention to place the reactants and products in the order that they are in the desired equation.

Notice that in the 3rd reaction we have 2 mol Na₂O (s) which is a reactant but with a coefficient of one, so we will multiply this equation by 1/2-

The 2nd equation has Na₂SO₄ as a reactant and it is a product in our required equation, therefore we will reverse the 2nd . Note the coefficient is 1 so we do not need to multiply.

This leads to the first equation and since we need to cancel 2 NaOH, we will nedd to multiply by 2 the first one.

Taking  1/2 eq 3 + (-) eq 2 + 2 eq 1 should do it.

      Na₂O (s) + H₂ (g) ⇒ 2 Na (s) + H₂O(l)             ΔrxnHº = 259 / 2 kJ  1/2 eq3

+    2NaOH(s)  + SO₃(g) ⇒  Na₂SO₄ (s) + H₂O (l)   ΔrxnHº = -418 kJ     - eq 2

+    2Na (s) + 2 H₂O (l)  ⇒   2 NaOH (s) + H₂ (g)    ΔrxnHº = -146 x 2    2 eq 1

Na₂O (s) + SO₃ (g)  ⇒ Na₂SO₄ (s)    ΔrxnHº =  259/2 + (-418) + (-146) x 2 kJ

                                                          ΔrxnH  = -580.5 kJ

Final answer:

To find the enthalpy change for the reaction Na2O(s) + SO3(g) → Na2SO4(s), we apply Hess's Law and manipulate the given reactions to find that the enthalpy change is -288.5 kJ.

Explanation:

To determine the enthalpy change (\(\Delta H^o_{rxn}\)) for the reaction (4) Na2O(s) + SO3(g) → Na2SO4(s), we use Hess's Law which states that if a reaction is the sum of two or more other reactions, the enthalpy changes of the constituent reactions can be added to determine the enthalpy change of the overall process. We need to find a combination of reactions (1), (2), and (3) that will result in reaction (4).

Here are the given reactions with their enthalpy changes:

Na(s) + H2O(l) → NaOH(s) + 1/2 H2(g), \(\Delta H^o_{rxn} = -146 kJ\)

Na2SO4(s) + H2O(l) → 2NaOH(s) + SO3(g), \(\Delta H^o_{rxn} = +418 kJ\)

2Na2O(s) + 2H2(g) → 4Na(s) + 2H2O(l), \(\Delta H^o_{rxn} = +259 kJ\)

To solve for the enthalpy of reaction (4), we need to reverse reaction (2) and halve reaction (3) to cancel out unwanted substances and leave Na2O(s) + SO3(g) on the left-hand side and Na2SO4(s) on the right-hand side.

Reversed reaction (2):\[\text{Na2SO4(s) + H2O(l) → 2NaOH(s) + SO3(g)}\], \(\Delta H^o_{rxn} = -418 kJ\) (since we reversed it)

Halved reaction (3):\[\text{Na2O(s) + H2(g) → 2Na(s) + H2O(l)}\], \(\Delta H^o_{rxn} = +259 kJ / 2 = +129.5 kJ\) (since we halved it)

Finally, we add the modified reaction (2) and (3):

(4) Na2O(s) + SO3(g) → Na2SO4(s), the enthalpy change is \(\Delta H^o_{rxn} = (-418 kJ) + (+129.5 kJ) = -288.5 kJ\).

Answer:

0.1724 M is the concentration of magnesium ions in this solution.

Explanation:

Mass of magnesium hydroxide gas = 10.00 g

Molar mass of magnesium hydroxide = 58 g/mol

Moles of magnesium hydroxide = [tex]\frac{10.00 g}{58 g/mol}=0.1724 mol[/tex]

Volume of the solution = V = 1.00 L

Molarity or concentration of the magnesium hydroxide:M

[tex]M=\frac{\text{Moles of solute}}{\text{Volume of solution in L}}[/tex]

[tex]M=\frac{0.1724 mol}{1.00 L}=0.1724 M[/tex]

1 mole of magnesium hydroxide contains 1 mole of magnesium ion.Then 0.1724 M of magnesium hydroxide will have :

[tex][Mg^{2+}]=1\times 0.1724 M=0.1724 M[/tex]

0.1724 M is the concentration of magnesium ions in this solution.

The concentration of magnesium hydroxide in the solution is 0.1714 Molar. This is calculated by dividing the number of moles of magnesium hydroxide by the volume of the solution in the volumetric flask.

The subject of this question falls under Chemistry, and it's asking about calculating concentration, specifically for a solution of magnesium hydroxide. In chemistry, concentration is commonly expressed in moles per liter (M). It's necessary to convert the mass of magnesium hydroxide to moles by using its molar mass.

First, to solve this question, we need to know the molar mass of magnesium hydroxide (Mg(OH)2), which is about 58.3197 grams/mole. Dividing the given mass of magnesium hydroxide (10.00g) by the molar mass gives us the number of moles. Thus, 10.00 g / 58.3197 g/mol ≈ 0.1714 mol of Mg(OH)2.

The whole solution was made up in a 1.00 L volumetric flask. Concentration is defined as the amount of solute per unit volume of solvent. Hence, the concentration of Mg(OH)2 is 0.1714 mol / 1.00 L = 0.1714 M (molar).

On note, the temperature given in the question (25◦C) does not affect the calculation of the concentration in this case, nor does the pH of the water used.

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Answer: Option (b) is the correct answer.

Explanation:

A laboratory device which is used to separate fluids, gases or liquid on the basis of their density is known as a centrifuge. When a vessel containing the sample is spins at a high speed then separation takes place as the centrifugal force pushes the heavier material out of the vessel during this process.  

So, if we want the sample to safely run through the centrifuge then it is important that it should be free from impurities.

Thus, we can conclude that in order to safely run a sample in a centrifuge, it is important to make sure the sample is free from impurities.

When the safety run sample should be in a centrifuge so the sample should be free from impurities.

It is used to distinguish gases or liquids that depend upon the density is called a centrifuge. At the time When a vessel comprised of the sample should be spinning at a high speed so the separation took place since the centrifugal force pushes the heavier material that should be out of the vessel at the time of the process.  

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Answer:

1. The overall equation is ClO-(aq)+I-(aq) → Cl-(aq)+IO-(aq)

2. The intermediates include: HClO(aq), OH-(aq) and HIO(aq)

3. The rates are k[ClO-][H2O], k[I-][HClO] and k[OH-][HIO]

4. No,

The rate depends on [OH-], so it's not consistent with the actual rate law

Explanation:

1

Given

(1) ClO−(aq) + H2O(l) ⇌ HClO(aq) + OH−(aq) [fast]

(2) I−(aq) + HClO(aq) → HIO(aq) + Cl−(aq) [slow]

(3) OH−(aq) + HIO(aq) → H2O(l) + IO−(aq) [fast]

Add up the three equations

ClO−(aq) + H2O(l) ⇌ HClO(aq) + OH−(aq) [fast]

I−(aq) + HClO(aq) → HIO(aq) + Cl−(aq) [slow]

OH−(aq) + HIO(aq) → H2O(l) + IO−(aq) [fast]

Remove all common terms {H2O(l) + I-(aq) +HClO(aq) +OH-(aq) +HIO(aq) => HClO(aq) + OH-(aq) + HIO(aq)

+H2O(l)}

We're left with

ClO-(aq) + I-(aq) => Cl-(aq) + IO-(aq)

2.

There are intermediates generated but they are not visible in the overall equation.

The intermediates include: HClO(aq), OH-(aq) and HIO(aq)

3.

The three steps are bimolecular.

The rates are k[ClO-][H2O], k[I-][HClO] and k[OH-][HIO]

4. Let K represents equilibrium constant

At step 1,

K1 = [HClO][OH-]/[ClO-]

Simplify;

K1 [ClO-]= [HClO][OH-]

K1[ClO-]/[OH-] = [HClO]

Determine the rate at step 2

= k2[I-][HClO]

= K1k2[I-][ClO-]/[OH-]

= k[ClO-][I-]/[OH-]

The answer is no

Final answer:

The overall equation derived from the given mechanism is ClO−(aq) + I−(aq) → IO−(aq) + H2O(l) + Cl−(aq), with HClO, OH−, and HIO as intermediates. Molecularity for all steps is bimolecular with specific rate laws for each step, and the mechanism is consistent with the experimental rate law rate = k[ClO−][I−].

Explanation:

The student's question pertains to deriving the overall equation, identifying intermediates, determining molecularity and rate laws for each step, and verifying the consistency of a proposed mechanism with the experimental rate law in a chemical reaction series involving species such as ClO−(aq), H2O(l), I−(aq), and others.

Answers to the Student's Question

The overall equation is ClO−(aq) + I−(aq) → IO−(aq) + H2O(l) + Cl−(aq).

The intermediates in the reaction mechanism are HClO, OH−, HIO.

For the first step, the molecularity is bimolecular and the rate law is k1[ClO−][H2O]. For the second step, it's bimolecular with a rate law of k2[I−][HClO]. The third step is also bimolecular with a rate law of k3[OH−][HIO].

The mechanism is consistent with the actual rate law, which is rate = k[ClO−][I−], because the rate-determining step involves these reactants.

Explanation:

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Final answer:

The molarity of acetone in the reaction mixture is 0.8 M.

Explanation:

To find the molarity of acetone in the reaction mixture, we need to use the molarity formula: Molarity (M) = Number of moles (n) / Volume of solution (V) in liters. Given that the number of moles of acetone is 0.020 and the volume of the mixture is 25 mL (0.025 L), we can substitute these values into the formula to find the molarity of acetone:

Molarity of acetone (MA) = 0.020 moles / 0.025 L = 0.8 M

Answer:

Vapor pressure of solution is 98.01 Torr

Explanation:

Colligative property. In this case, we apply the lowering vapor pressure.

ΔP = P° . Xm

P° - P' = Vapor pressure of pure solvent - Vapor pressure of solution = ΔP

Xm is the molar fraction for solute. We try to determine it:

Moles of solute / Total moles → Total moles = Solute moles + Solvent moles

10.3 g . 1mol / 154 g = 0.0669 moles of biphenyl

27.8 g . 1mol / 78 g = 2.32 moles of benzene

Total moles = 2.32 moles + 0.0669 moles = 2.3869 moles

Xm for solute = 0.0669 / 2.3869 = 0.028

Let's replace data:

100.84 Torr - P' = 100.84 Torr . 0.028

P' = - (100.84 Torr . 0.028 - 100.84Torr) = 98.01 Torr

Answer:

The vapor pressure of the solution is 84.9 torr

Explanation:

Biphenyl is a nonvolatile, nonionizing solute

Temperature = 25.0 °C

Vapor pressure = 100.84 torr

Mass of biphenyl = 10.3 grams

Mass of benzene = 27.8 grams

Molar mass biphenyl = 154.21 g/mol

Molar mass benzene = 78.11 g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles biphenyl = 10.3 grams / 154.21 g/mol

Moles biphenyl = 0.0668 moles

Moles benzene = 27.8 grams / 78.11 g/mol

Moles benzene = 0.356 moles

Step 3: Calculate total moles

Total moles = 0.0668 moles + 0.356 moles

Total moles= 0.4228 moles

Step 4: Calculate mol fraction benzene

Mol fraction benzene = moles benzene / total moles

Mol fraction benzene = 0.356 moles / 0.4228 moles

Mol fraction benzene = 0.842

Step 5: Calculate vapor pressure of the solution

Psol = Xbenzene  * P°benzene

⇒Psol = the vapor pressure of the solution

⇒Xbenzene = mol fraction of benzene

⇒P°benzene = the vapor pressure of pure benzene

Psol = 0.842 * 100.84 torr

Psol = 84.9 torr

The vapor pressure of the solution is 84.9 torr

Explanation:

The given data is as follows.

      [HCOOH] = 0.2 M,       [NaOH] = 2.0 M,

         V = 500 ml,   [Benzoic acid] = 0.2 M

First, we will calculate the number of moles of benzoic acid as follows.

   No. of moles of benzoic acid = Molarity × Volume

                         = [tex]2 \times 0.475[/tex]

                         = 0.095 mol

And, moles of NaOH present in the solution will be as follows.

    No. of moles of NaOH = Molarity × Volume

                          = [tex]2 \times 0.025[/tex]

                          = 0.05 mol

Hence, the ICE table for the chemical equation will be as follows.

         [tex]C_{6}H_{5}COOH + NaOH \rightarrow C_{6}H_{5}COONa + H_{2}O[/tex]

Initial:        0.095           0.05            0             0

Equlbm:  (0.095 - 0.05)  0            0.05

        pH = [tex]pK_{a} + log \frac{Base}{Acid}[/tex]  

              = [tex]4.2 + log \frac{0.05}{0.045}[/tex]

              = 4.245

For,  

         [tex]HCOOH + NaOH \rightarrow HCOONa + H_{2}O[/tex]

Initial:       0.2x     2(0.5 - x)               0

Equlbm:   0.2x - 2(0.5 - x)                 0             2(0.5 - x)

As,

           pH = [tex]pK_{a} + log \frac{Base}{Acid}[/tex]  

          4.245 = 3.75 + [tex]log \frac{Base}{Acid}[/tex]

      [tex]log \frac{Base}{Acid}[/tex] = 0.5

    [tex]\frac{Base}{Acid}[/tex] = 3.162

Now,

        [tex]\frac{2(0.5 - x)}{0.2x - 2(0.5 - x)}[/tex] = 3.162

               x = 0.464 L

Volume of NaOH = (0.5 - 0.464) L

                             = 0.036 L

                             = 36 ml               (as 1 L = 1000 mL)

And, volume of formic acid is 464 mL.

36 ml of NaOh and 464 ml of HCOOH would be enough to form 500 ml of a buffer with the same pH as the buffer made with benzoic acid and NaOH.

We can arrive at this answer through the following calculation:

Amount of moles of NaOH: [tex]2 * 0.025 = 0.05 mol[/tex]

Amount of moles of benzoic acid: [tex]2*0.475=0.095mol[/tex]

[tex]pH=pK{a}+log\frac{base}{acid}[/tex]

[tex]4.2+log\frac{0.05}{0.045}=4.245[/tex]

We must repeat this calculation, with the values shown for HCOOH and NaOH. In this case, we can calculate as follows:

[tex]pH=pK{a} +log\frac{base}{acid} \\4.245=3.75+log\frac{base}{acid} \\log\frac{base}{acid}=0.5\\\frac{base}{acid} = 3.162[/tex]

[tex]\frac{2(0.5-x)}{0.2x-2(0.5-x)} =0.464L[/tex]

NaOH volume:

[tex](0.5-0.464) L\\0.036L ----- 36mL[/tex]

HCOOH volume:

[tex]500 mL-36mL = 464 mL[/tex]

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The Fischer esterification mechanism involves two main steps: the addition of methanol (MeOH) to a carboxylic acid to form a tetrahedral intermediate, and the loss of water (H2O) to create the ester.

The Fourier esterification is a fundamental mechanism in Chemistry and encompasses two main steps. Part 1 involves the Addition of Methanol (MeOH) to a carboxylic acid to form a tetrahedral intermediate. Here, the lone pair of electrons on the oxygen of the MeOH attacks the carbonyl carbon of the acid, pushing the electrons up onto the carbonyl oxygen and forming the tetrahedral intermediate. Part 2 is the Loss of water (H2O) to create the ester. The hydroxyl group of the intermediate acts as a leaving group, departing as a water molecule. The lone pair of electrons on the oxygen then comes back down and forms the pi bond, hence forming the ester.

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In the Fischer esterification mechanism, a carboxylic acid and an alcohol react, leading to the formation of an ester and water. It's called a 'key tetrahedral intermediate' and the process involves the loss of a water molecule.

The Fischer esterification mechanism is a fundamental process in organic chemistry that involves the transformation of a carboxylic acid and an alcohol into an ester. This process is initiated by the addition of an alcohol, such as MeOH (methanol), to the carboxylic acid via a nucleophilic attack, forming a tetrahedral intermediate.

The carboxylic acid group of this intermediate subsequently loses a water molecule (H2O) in a dehydration process, resulting in the formation of the desired ester.

For example, when ethanol reacts with acetic acid (ethanoic acid), the ester known as ethyl acetate (methyl ethanoate) is produced. Notably, esters are recognized for their fragrant odors, deriving from various plants and their fruits, such as honey methyl phenylacetate.

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Answer:

The retention factor, k is 2.49

Explanation:

According to the theory of High-Performance Liquid Chromatography (HPLC), the retention factor (or capacity), k, of a column is the ratio of the retention time of a retained analyte (toluene) to that of the  un-retained solute (methane).

This implies that:

k = [tex]\frac{16.35 (mins)}{ 6.39 (mins)}[/tex]  =  [tex]\frac{995 (secs)}{399 (secs)}[/tex]

 retention factor, k  = 2.49.

Note that there is no unit for retention factor, as it is a ratio.

Final answer:

The retention factor for toluene in the given gas chromatography system, calculated using the retention times of toluene and methane, is approximately 1.56.

Explanation:

The question involves calculating the retention factor (k) for toluene using gas chromatography (GC) data. The retention time is the time a compound spends in the column before being detected, which varies by compound owing to differences in interaction with the stationary phase. The retention factor is a measure of this time relative to an unretained solute, providing insight into the compound's affinity for the stationary phase versus the mobile phase.

To calculate the retention factor (k) for toluene, we use the formula: k = (t-R - t-M) / tM, where tR is the retention time of the analyte (toluene) and tM is the retention time of an unretained solute (methane, in this case). For toluene, t-R = 16.35 min, and for methane, t-M = 6.39 min. Substituting these values into the equation, we find k for toluene as follows: k = (16.35 - 6.39) / 6.39 = 9.96 / 6.39 ≈ 1.56.

Answer:

a. Yes, there was a difference in the number of times your substrate underwent addition in the two Friedel-Crafts reactions.

b. There was a difference- due to the fact that the  first added alkyl group increaseD the electron density on benzene ring.

Explanation:

Yes, there was a difference in the number of times the substrate underwent addition in the two Friedel-Crafts reactions .

There was a difference - There was an increase in the  rate of second Friedel-Craft reaction due to the fact that the  first added alkyl group increaseD the electron density on benzene ring.

Therefore the rate of the addition reaction increases.

Friedel-Crafts reactions exhibit differences in substrate addition frequency due to the presence of electron-donating alkyl groups, which increase reactivity through polyalkylation, and activating groups that stabilize intermediates and affect reactivity.

The differences in the number of times your substrate underwent addition in the two Friedel-Crafts reactions can be attributed to the nature of the substituents added during the process. In the Friedel-Crafts alkylation, alkyl groups are introduced onto an aromatic ring which are electron-donating substituents. These alkyl groups render the product more reactive than the starting material, leading to the possibility of polyalkylation, where multiple alkyl groups can be added if additional reaction takes place. This is due to the alkyl groups' ability to activate the ring towards further electrophilic attack. Polyalkylation often results in a mixture of products and for synthetic purposes, this could be considered a disadvantage as it complicates purification and affects yield.

On the other hand, the presence of an activating group, such as a methoxy substituent, in Friedel-Crafts alkylation reactions can stabilize the ring carbocation intermediate and increase the rate of reaction, leading to a difference in reactivity compared to reactions without such group. Thus, the control of reactivity and selectivity in Friedel-Crafts reactions is critical and differences in the substrate or reaction conditions can have a significant impact on the outcome of the reaction.

Answer:

b. atomizes liquids into small solid particles

Explanation:

Spray drying -

It refers to the process of getting a dry powder from the liquid or the slurry ,with the help of hot gas , is referred to as spray drying .

The method is used in the field of pharmacy .

It is a type of atomizer or spray nozzle which helps to disperse the liquid into the controlled drop size spray .

Small solid particles are generated by this method .

Hence , from the given scenario of the question ,

The correct answer is b. atomizes liquids into small solid particles  .

Explanation:

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Answer:

The concentration of the KBr solution is 6.27 mol/L

Explanation:

Step 1: Data given

Mass of KBr = 224 grams

Molar mass KBr = 119.0 g/mol

Volume = 300 mL = 0.300 L

Step 2: Calculate moles KBr

Moles KBr = mass KBr / molar mass KBr

Moles KBr = 224 grams / 119.0 g/mol

Moles KBr = 1.88 moles

Step 3: Calculate concentration KBr solution

Concentration solution = moles KBr / volume

Concentration solution = 1.88 moles / 0.300 L

Concentration solution = 6.27 mol / L

The concentration of the KBr solution is 6.27 mol/L

Answer: The nuclear equations for the given process is written below.

Explanation:

The chemical equation for the bombardment of neutron to U-238 isotope follows:

[tex]_{92}^{238}\textrm{U}+n\rightarrow _{92}^{239}\textrm{U}[/tex]

Beta decay is defined as the process in which neutrons get converted into an electron and a proton. The released electron is known as the beta particle.

[tex]_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta[/tex]

The chemical equation for the first beta decay process of [tex]_{92}^{239}\textrm{U}[/tex] follows:

[tex]_{92}^{239}\textrm{U}\rightarrow _{93}^{239}\textrm{Np}+_{-1}^0\beta[/tex]

The chemical equation for the second beta decay process of [tex]_{93}^{239}\textrm{Np}[/tex] follows:

[tex]_{93}^{239}\textrm{Np}\rightarrow _{94}^{239}\textrm{Pu}+_{-1}^0\beta[/tex]

Hence, the nuclear equations for the given process is written above.

Explanation:

It is known that formula for the ionization energy of hydrogen atom is as follows.

               E = [tex]\frac{13.6 eV}{n^{2}}[/tex]

or,          n = [tex]\sqrt{\frac{13.6}{E}}[/tex]

The value of energy is given as 0.544 eV. Therefore, we will calculate the value of n as follows.

                   n = [tex]\sqrt{\frac{13.6}{E}}[/tex]

                      = [tex]\sqrt{\frac{13.6}{0.544 eV}}[/tex]

                      = 5

Thus, we can conclude that n equals to 5 for a hydrogen atom if 0.544 eV of energy can ionize it.

Final answer:

The principal quantum number n of a hydrogen atom that requires 0.850 eV of energy to ionize from an excited state is 4. This is calculated using the formula for the energy levels of hydrogen, En = -13.6 eV/n², and solving for n.

Explanation:

The question relates to the energy levels of a hydrogen atom and how much energy is required to ionize it when the electron is in an excited state. The ionization energy of hydrogen in the ground state (n = 1) is 13.6 eV. To find the principal quantum number n for a particular amount of energy required to ionize the hydrogen atom from an excited state, we use the formula: En = -13.6 eV/n². We can set this equal to the energy needed to ionize the atom, in this case, 0.850 eV, and solve for n.

First, we rearrange the formula so it looks like this:

-13.6 eV/n² = -0.850 eV
Then we solve for n:

n² = 13.6 eV / 0.850 eV
n² = 16
n = √16
n = 4

Thus, the hydrogen atom is in the n = 4 excited state when 0.850 eV of energy can ionize it.

The question is incomplete, here is the complete question:

Ethylene is the starting point for a wide array of industrial chemical syntheses. For example, worldwide about 8.0 x 10¹⁰ of polyethylene are made from ethylene each year, for use in everything from household plumbing to artificial joints. Natural sources of ethylene are entirely inadequate to meet world demand, so ethane from natural gas is "cracked" in refineries at high temperature in a kinetically complex reaction that produces ethylene gas and hydrogen gas.

Suppose an engineer studying ethane cracking fills a 30.0 L reaction tank with 38.0 atm of ethane gas and raises the temperature to 400°C. He believes Kp = 0.4 at this temperature. Calculate the percent by mass of ethylene the engineer expects to find in the equilibrium gas mixture.

Answer: The mass percent of ethylene gas is 9.20 %

Explanation:

We are given:

Initial partial pressure of ethane gas = 38.0 atm

The chemical equation for the dehydrogenation of ethane follows:

                  [tex]C_2H_6\rightleftharpoons C_2H_4+H_2[/tex]

Initial:             38

At eqllm:      38-x         x       x

The expression of [tex]K_p[/tex] for above equation follows:

[tex]K_p=\frac{p_{C_2H_4}\times p_{H_2}}{p_{C_2H_6}}[/tex]

We are given:

[tex]K_p=0.40[/tex]

Putting values in above equation, we get:

[tex]0.40=\frac{x\times x}{38-x}\\\\x=-4.10,3.70[/tex]

Neglecting the negative value of 'x' because partial pressure cannot be negative

So, equilibrium partial pressure of ethane = 38 - x = 38 - 3.70 = 34.30 atm

Equilibrium partial pressure of ethene = x = 3.70 atm

To calculate the number of moles, we use the equation given by ideal gas which follows:

[tex]PV=nRT[/tex]         ..........(1)

To calculate the number of moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]     .....(2)

We are given:

[tex]P=34.3atm\\V=30.0L\\R=0.0821\text{ L atm }mol^{-1}K^{-1}\\T=400^oC=[400+273]=673K[/tex]

Putting values in equation 1, we get:

[tex]34.3atm\times 30.0L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 673K\\\\n=\frac{34.3\times 30.0}{0.0821\times 673}=18.62mol[/tex]

Molar mass of ethane gas = 30 g/mol

Moles of ethane gas = 18.62 mol

Putting values in equation 2, we get:

[tex]18.62mol=\frac{\text{Mass of ethane}}{30g/mol}\\\\\text{Mass of ethane gas}=(18.62mol\times 30g/mol)=558.6g[/tex]

We are given:

[tex]P=3.70atm\\V=30.0L\\R=0.0821\text{ L atm }mol^{-1}K^{-1}\\T=400^oC=[400+273]=673K[/tex]

Putting values in equation 1, we get:

[tex]3.70atm\times 30.0L=n\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 673K\\\\n=\frac{3.70\times 30.0}{0.0821\times 673}=2.01mol[/tex]

Molar mass of ethylene gas = 28 g/mol

Moles of ethylene gas = 2.01 mol

Putting values in equation 2, we get:

[tex]2.01mol=\frac{\text{Mass of ethylene}}{28g/mol}\\\\\text{Mass of ethylene gas}=(2.01mol\times 28g/mol)=56.28g[/tex]

[tex]\text{Mass percent of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100[/tex]

Mass of ethylene = 56.28 g

Mass of mixture = [558.6 + 56.28] g = 641.88 g

Putting values in above equation, we get:

[tex]\text{Mass percent of ethylene}=\frac{56.28g}{641.88g}\times 100=9.20\%[/tex]

Hence, the mass percent of ethylene gas is 9.20 %