Hydrogen gas and nitrogen gas react to form ammonia gas. What volume of ammonia would be produced by this reaction if 6.9 m3 of hydrogen were consumed? Also, be sure your answer has a unit symbol, and is rounded to the correct number of significant digits.

Answer: The volume of the container is [tex]2.8497m^3[/tex]

Explanation:

To calculate the volume of water, we use the equation given by ideal gas, which is:

[tex]PV=nRT[/tex]

or,

[tex]PV=\frac{m}{M}RT[/tex]

where,

P = pressure of container = 200 kPa

V = volume of container = ? L

m = Given mass of water = 2.61 kg = 2610 g   (Conversion factor: 1kg = 1000 g)

M = Molar mass of water = 18 g/mol

R = Gas constant = [tex]8.31\text{L kPa }mol^{-1}K^{-1}[/tex]

T = temperature of container = [tex]200^oC=[200+273]K=473K[/tex]

Putting values in above equation, we get:

[tex]200kPa\times V=\frac{2610g}{18g/mol}\times 8.31\text{L kPa }\times 473K\\\\V=2849.7L[/tex]

Converting this into cubic meter, we use the conversion factor:

[tex]1m^3=1000L[/tex]

So, [tex]\Rightarrow \frac{1m^3}{1000L}\times 2849.7L[/tex]

[tex]\Rightarrow 2.8497m^3[/tex]

Hence, the volume of the container is [tex]2.8497m^3[/tex]

Answer: 85 minutes

Explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

Half life for second order kinetics is given by:

[tex]t_{\frac{1}{2}=\frac{1}{k\times a_0}[/tex]

[tex]t_{\frac{1}{2}[/tex] = half life = 15 min

k = rate constant =?

[tex]a_0[/tex] = initial concentration = 100 (say)

[tex]15min=\frac{1}{k\times 100}[/tex]

[tex]k=\frac{1}{1500}[/tex]

Integrated rate law for second order kinetics is given by:

[tex]\frac{1}{a}=kt+\frac{1}{a_0}[/tex]

a= concentration left after time t = [tex]100-\farc{85}{100}\times 100=15[/tex]

[tex]\frac{1}{15}=\frac{1}{1500}\times t+\frac{1}{100}[/tex]

[tex]t=85min[/tex]

Thus after 85 minutes after the start of the reaction, it will be 85% complete.

Answer: The given statement is false.

Explanation:

Precipitation reaction is defined as the chemical reaction in which two aqueous solution upon mixing together results in the formation of an insoluble solid.

For example, [tex]NaCl(aq) + AgNO_{3} \rightarrow NaNO_{3}(aq) + AgCl(s)[/tex]

Here AgCl is present in solid state so, it is the precipitate.

But it is not necessarily true that two aqueous solutions will always result in the formation of a precipitate.

For example, [tex]NaCl(aq) + KNO_{3}(aq) \rightarrow KCl(aq) + NaNO_{3}(aq)[/tex]

Hence, we can conclude that the statement precipitation reactions always occur when two aqueous solutions are mixed, is false.

The statement is false; precipitation reactions occur when two aqueous solutions form an insoluble product. Whether a reaction occurs depends on the solubility rules of the compounds formed.

The statement that precipitation reactions always occur when two aqueous solutions are mixed is false. A precipitation reaction occurs when two solutions are mixed and an insoluble product, the precipitate, forms. Whether a precipitation reaction occurs depends on the solubility rules of the ionic compounds formed during the mixing of the two solutions. If none of the possible combinations result in an insoluble product, then no precipitation will occur. This is described by solubility rules, which predict the solubility of different ionic compounds in water.

Answer:

2s²2p⁴

Explanation:

Oxygen is an element on the periodic table with a total of 8 electrons. It's electronic configuration is given as 2,6.

Using the orbital notation we write as 1s²2s²2p⁴

Also, the valence electrons are the electrons in the outermost shell of an atom. These electrons mostly determine the chemical properties of an atom.

Oxygen has a total of 6 electrons in its outermost shell and it is given as 2s²2p⁴

Answer:

2s²2p⁴

Explanation:

You start with 100 kg raw potatoes.

8% of mass is lost by peeling:

100 - ( 8 / 100) × 100 = 92 kg of peeled raw potatoes

By drying the water contents is reduces from 75% to 7%, that means a 68% loss by weight:

92 - ( 68 / 100) × 92 = 92 - 62.56 = 29.44 kg of dried potatoes

The mass of dried potatoes produced from each 100 kg of raw potatoes is 24.7312 kg.

To solve this problem, we will follow the steps outlined in the question:

1. Start with 100 kg of raw potatoes.

2. Calculate the loss of mass due to peeling. Since 8% by weight is lost, we have:

 Mass lost due to peeling = 8% of 100 kg = 0.08 * 100 kg = 8 kg.

3. The remaining mass after peeling is:

 Mass after peeling = Initial mass - Mass lost due to peeling = 100 kg - 8 kg = 92 kg.

4. Now, we need to adjust the water content from 75% to 7%. Let's denote the mass of the dried potatoes as m. The amount of water to be removed from the potatoes is the difference between the initial water content and the final water content:

 Water to be removed = (75% - 7%) * m.

5. Since the initial water content is 75%, the amount of water in the potatoes after peeling is:

 Initial water mass = 75% of 92 kg = 0.75 * 92 kg = 69 kg.

6. The final water content should be 7%, so the mass of water in the dried potatoes will be:

 Final water mass = 7% of m = 0.07 * m.

7. The difference in water content is the amount of water that needs to be removed:

 Water to be removed = Initial water mass - Final water mass.

8. Since the mass of the dried potatoes \( m \) consists of 7% water and 93% solids (including starch), we can express the mass of the solids as:

Solids mass = 93% of m = 0.93 * m.

9. The solids mass after drying must be equal to the solids mass after peeling, which is the total mass after peeling minus the initial water mass:

 Solids mass after peeling = Mass after peeling - Initial water mass = 92 kg - 69 kg = 23 kg.

10. Now we can set up the equation:

Solids mass after drying = Solids mass after peeling,

0.93 * m = 23 kg.

11. Solving for m, we get:

[tex]\( m = \frac{23 \text{ kg}}{0.93} \).[/tex]

12. Calculate the mass of the dried potatoes:

[tex]\( m = \frac{23}{0.93} \approx 24.7312 \) kg.[/tex]

Therefore, the mass of dried potatoes produced from each 100 kg of raw potatoes is approximately 24.7312 kg.

The answer is: 24.7312.

Answer: The empirical formula for the given compound is [tex]FeC_{47}H_{66}O_{26}[/tex]

Explanation:

[tex]Fe_wC_xH_yO_z+O_2\rightarrow CO_2+H_2O[/tex]

where, 'w', 'x', 'y' and 'z' are the subscripts of Iron, carbon, hydrogen and oxygen respectively.

We are given:

Mass of [tex]CO_2=3.1737g[/tex]

Mass of [tex]H_2O=0.90829g[/tex]

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 3.1737 g of carbon dioxide, [tex]\frac{12}{44}\times 3.1737=0.865g[/tex] of carbon will be contained.

In 18g of water, 2 g of hydrogen is contained.

So, in 0.90829 g of water, [tex]\frac{2}{18}\times 0.90829=0.101g[/tex] of hydrogen will be contained.

Percent of Fe in [tex]Fe_2O_3[/tex] = [tex]\frac{(2\times \text{molar mass of Fe}}{\text{molar mass of }Fe_2O_3}\times 100[/tex]

Molar mass of iron = 55.85 g/mol

Molar mass of iron (III) oxide = 159.69 g/mol

Putting values in above equation, we get:

[tex]\%\text{ mass of iron in }Fe_2O_3=\frac{2\times 55.85}{159.69}\times 100=69.94\%[/tex]

So, the amount of iron present in 0.1230 g of [tex]Fe_2O_3=\frac{69.94}{100}\times 0.1230=0.0860g[/tex] of iron.

To formulate the empirical formula, we need to follow some steps:

Moles of Carbon =[tex]\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.865g}{12g/mole}=0.072moles[/tex]

Moles of Hydrogen = [tex]\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.101g}{1g/mole}=0.101moles[/tex]

Moles of Oxygen = [tex]\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.6448g}{16g/mole}=0.0403moles[/tex]

Moles of Iron = [tex]\frac{\text{Given mass of iron}}{\text{Molar mass of iron}}=\frac{0.0860g}{55.85g/mole}=0.00153moles[/tex]

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.00153 moles.

For Carbon = [tex]\frac{0.072}{0.00153}=47.05\approx 47[/tex]

For Hydrogen  = [tex]\frac{0.101}{0.00153}=66.01\approx 66[/tex]

For Oxygen  = [tex]\frac{0.0403}{0.00153}=26.33\approx 26[/tex]

For Iron  = [tex]\frac{0.00153}{0.00153}=1[/tex]

The ratio of Fe : C : H : O = 1 : 47 : 66  : 26

Hence, the empirical formula for the given compound is [tex]Fe_1C_{47}H_{66}O_{26}=FeC_{47}H_{66}O_{26}[/tex]

Answer:

0.82V

Explanation:

0.48V+0.34V=0.82V

Answer:

-.14

Explanation:

Answer:

The frequencies of the two lines are:

a) [tex]3.79\times 10^{14} s^{-1}[/tex]

b)[tex]7.14\times 10^{14} s^{-1}[/tex]

When we heat rubidium compound we will see red color.

Explanation:

[tex]\nu=\frac{c}{\lambda }[/tex]

c = speed of light

[tex]\lambda [/tex] = wavelength of light

a) Frequency of the light when wavelength is equal to [tex]7.9\times 10^{-7} m[/tex]

[tex]\nu=\frac{c}{\lambda }[/tex]

[tex]\nu=\frac{3\times 10^8m/s)}{7.9\times 10^{-7}}[/tex]

[tex]\nu=3.79\times 10^{14} s^{-1}[/tex]

This frequency corresponds to red light

b) Frequency of the light when wavelength is equal to [tex]4.2\times 10^{-7} m[/tex]

[tex]\nu=\frac{c}{\lambda }[/tex]

[tex]\nu=\frac{3\times 10^8m/s)}{4.2\times 10^{-7}}[/tex]

[tex]\nu=7.14\times 10^{14} s^{-1}[/tex]

This frequency corresponds to violet light

When we heat rubidium compound we will see red color.

When rubidium ions are heated to a high temperature, two lines are observed in its line spectrum at wavelengths (a) 7.9 × 10⁻⁷ m and (b) 4.2 × 10⁻⁷ m. The frequencies of the two lines are 3.8 × 10¹⁴ Hz and 7.1 × 10¹⁴ Hz. The color observed when we heat a rubidium compound is red and blue.

When rubidium ions are heated to a high temperature, two lines are observed in its line spectrum at wavelengths (a) 7.9 × 10⁻⁷ m and (b)  4.2 × 10⁻⁷ m. To find the frequencies of these lines, we can use the equation v = c/λ, where v is the frequency, c is the speed of light, and λ is the wavelength. By plugging in the given wavelengths, we can calculate the frequencies.

The frequency of line (a) is 3.8 × 10¹⁴ Hz, and the frequency of line (b) is 7.1 × 10¹⁴ Hz.

When a rubidium compound is heated, we observe two lines in the line spectrum. The first line has a wavelength of 7.9 × 10⁻⁷ m, which corresponds to a red color. The second line has a wavelength of  4.2 × 10⁻⁷ m, which corresponds to a blue color.

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Answer:

pH of Bfr = 4.83

Explanation:

The pH of the buffer solution prepared from acetic acid and calcium acetate in sufficient water is 4.84.

H(Ac) = 0.4/1.4L = 0.2857 M

Ca(OH)₂ = 0.25/1.4L = 0.17857 M

= 0.2857 M H(Ac)  + 0.17857 M Ca(OH)₂

= 0.2857 M H(Ac) + 2(0.17857) OH⁻

=  0.2857 M H(Ac) + 0.3571 M OH⁻

H(Ac) ⇄ H⁺   OH⁻

0.2857 M   ⇄ 0.3571 M

[tex]H^+ = \frac{[K_a][H(_{AC}]}{OH^-} \\\\H^+ = \frac{(1.8 \times 10^{-5}) (0.2857)}{(0.3571)} \\\\H^+ = 1.44\times 10^{-5}[/tex]

pH = -Log(H⁺)

pH = - Log(1.44 x 10⁻⁵)

pH = 4.84

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Answer:

Here's what I get  

Explanation:

In Step 1, the hydroxide ion removes a proton from the OH group (acid-base equilibrium).

In Step 2, the alkoxide non undergoes an internal SN2 attack on the back side of the carbon bearing the bromine.

The bromide ion is ejected, and the final product is the epoxide.

Answer: The empirical formula of compound is [tex]C_4H_{12}Si[/tex].

Explanation:

Mass of Sample= 0.702 g

Mass of [tex]CO_2[/tex] = 1.4 g

Mass of [tex]H_2O[/tex] = 0.86 g

Mass of [tex]SiO_2[/tex] = 0.478 g  

First we have to calculate moles of[tex]CO_2[/tex], [tex]H_2O[/tex] and [tex]SiO_2[/tex] formed.

1. Moles of [tex]CO_2=\frac{1.4g}{44g/mol}=0.032mol[/tex]

Now , Moles of carbon == Moles of [tex]CO_2[/tex] = 0.032

2.  Moles of [tex]H_2O=\frac{0.86g}{18g/mol}[/tex]=0.048mol​​​

Now , Moles of hydrogen = [tex]2\times[/tex] Moles of [tex]H_2O[/tex] =[tex]2\times 0.048=0.096mol[/tex]

3.  Moles of [tex]SiO_2=\frac{0.478g}{60g/mol}=0.008[/tex] mol

Now , Moles of silicon = Moles of [tex]SiO_2[/tex] = 0.008 moles

Therefore, the ratio of number of moles of C : H : Si is  = 0.032 : 0.096 : 0.008

For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C= [tex]\frac{0.032}{0.008}=4[/tex]

For H =[tex]\frac{0.096}{0.008}=12[/tex]

For Si=[tex]\frac{0.008}{0.008}=1[/tex]

Thus, C: H: Si = 4 : 12 : 1

The simplest ratio represent empirical formula.

Hence, the empirical formula of compound is [tex]C_4H_{12}Si[/tex].

Final answer:

To find the empirical formula of the compound, calculate the moles of each element from the given masses of CO2, H2O, and SiO2. The empirical formula for this compound is CH4Si.

Explanation:

To determine the empirical formula of the compound, first, calculate the moles of each element using the given masses of CO2, H2O, and SiO2. Next, find the ratio of the moles of each element to each other and simplify if necessary to get the empirical formula. In this case, the empirical formula of the compound is **CH4Si**.

Answer:

The answer is B

[The extracting solvent should be non-toxic and readily available]

Explanation:

The extracting solvent should not pose a risk to life (or the risk have to be minimum) and be available in large quantity.

A. Most of the extracting solvents are volatile, and is a good think because you can remove them by distillation.

C. The extraction solvent should not react with the solute, because you lose the extracted compound.

D. The extraction solvent should not be miscible with water. Usually you extract compounds organic compounds from water using an appropriate extracting solvent.

The correct attribute for an extracting solvent is that it should be non-toxic and readily available, making (b) the right answer. Volatility, reactivity with the solute, and miscibility with water are undesirable features for an extraction solvent.

An extracting solvent is used in liquid-liquid extraction to separate compounds based on their solubility properties. The ideal characteristics of an extracting solvent include being non-toxic, readily available, not miscible with water, and having a different density than water to facilitate easy separation. For example, diethyl ether is an effective extracting solvent because of its ability to dissolve non-ionic organic compounds, immiscibility with water, and the fact that ionic compounds are generally insoluble in it. In contrast, Toluene (C₆H₅-CH₃) is a nonpolar solvent which is ideal for dissolving nonpolar substances like octane , but not polar substances such as water (H₂O) or ionic compounds like sodium sulfate .

In selecting the correct answer to the student's question, the attribute that an extracting solvent should have is that it should be non-toxic and readily available. Solvents that are volatile, reactive with the solute, or miscible with water are not ideal for extraction purposes. Therefore, the correct choice is (b).

Answer : The vapor pressure of solution is 23.67 mmHg.

Solution:

As the relative lowering of vapor pressure is directly proportional to the amount of dissolved solute.

The formula for relative lowering of vapor pressure will be,

[tex]\frac{p^o-p_s}{p^o}=\frac{w_2M_1}{w_1M_2}[/tex]

where,

[tex]p^o[/tex] = vapor pressure of pure solvent  (water) = 23.76 mmHg

[tex]p_s[/tex] = vapor pressure of solution= ?

[tex]w_2[/tex] = mass of solute  (sucrose) = 12.25 g

[tex]w_1[/tex] = mass of solvent  (water) = 176.3 g

[tex]M_1[/tex] = molar mass of solvent (water) = 18.02 g/mole

[tex]M_2[/tex] = molar mass of solute (sucrose) = 342.3 g/mole

Now put all the given values in this formula ,we get the vapor pressure of the solution.

[tex]\frac{23.76-p_s}{23.76}=\frac{12.25\times 18.02}{176.3\times 342.3}[/tex]

[tex]p_s=23.67mmHg[/tex]

Therefore, the vapor pressure of solution is 23.67 mmHg.

Final answer:

The vapor pressure of a solution made by dissolving 12.25 grams of sucrose in 176.3 grams of water at 25 °C is calculated as 23.68 mm Hg, using Raoult's Law and the vapor pressure of pure water (23.76 mm Hg) as the basis.

Explanation:

Calculation of Vapor Pressure of a Sucrose Solution

To calculate the vapor pressure of a solution made by dissolving 12.25 grams of sucrose in 176.3 grams of water at 25 °C, where the vapor pressure of pure water is 23.76 mm Hg, we use Raoult's Law. Raoult's Law states that the vapor pressure of a solvent in a solution (Πsolvent) is equal to the vapor pressure of the pure solvent (Πpure solvent) times the mole fraction of the solvent (χsolvent) in the solution.

First, calculate the moles of sucrose (C12H22O11): Moles = 12.25 g / 342.3 g/mol.

Then, calculate the moles of water (H2O): Moles = 176.3 g / 18.02 g/mol.

Next, compute the mole fraction of water: χH2O = moles of H2O / (moles of H2O + moles of C12H22O11).

Finally, calculate vapor pressure of the solution: Πsolution = Πpure water * χH2O.

Working through the calculations:

Moles of sucrose = 12.25 / 342.3 = 0.0358 mol.
Moles of water = 176.3 / 18.02 = 9.785 mol.
Mole fraction of water = 9.785 / (9.785 + 0.0358) = 0.99636.
Vapor pressure of the solution = 23.76 mm Hg * 0.99636 = 23.68 mm Hg.

Thus, the vapor pressure of the sucrose solution at 25 °C is 23.68 mm Hg.

Hey there!:

Δt =  m * Kf

Boiling point of a solution =>  Boiling point + Δt ( Kc * m )

Boiling point of benzene solution => 80.1 + 2.53 m

Boiling point of CCl₄ solution => 76.8 +5.03 m

Since the boiling points are the same 80.1 + 2.53 m = 76.8 +5.03 m

3.3 = 2.5 m

m =  3.3 / 2.5 => 1.32 moles of solute per kg of solvent

Bp =   80.1 + 2.53 * 1.32 => 83.4396 ºC

Hope this helps!

Complete question:

Imagine two solutions with the same concentration and the same boiling point, but one has benzene as the solvent and the other has carbon tetrachloride as the solvent. Determine that molal concentration, m (or b), and boiling point, Tb.

benzene boiling point=80.1 Kb=2.53

carbon tetrachloride boiling point=76.8 Kb=5.03

Answer:

m = 1.32 mol/kg

Boiling point: 83.4°C

Explanation:

When a nonvolatile solute is added to a pure solvent, the boiling point of the solvent increases, a phenomenon called ebullioscopy. This happens because of the interactions between the solute and the solvent. The temperature variation (new boiling point - normal boiling point) can be calculated by:

ΔT = m*Kb*i

Where m is the molal concentration (moles o solute/mass of solvent in kg), Kb is the ebullioscopy constant of the solvent, and i is the van't Hoff factor, which indicates how much of the solute dissociates. Let's assume that i is equal in both solvents and equal to 1 (the solvent dissociates completely)

Calling the new boiling point as Tb, for benzene:

Tb - 80.1 = m*2.53*1

Tb = 2.53m + 80.1

For carbon tetrachloride:

Tb - 76.8 = m*5.03*1

Tb = 5.03m + 76.8

Because Tb and m are equal for both:

5.03m + 76.8 = 2.53m + 80.1

2.5m = 3.3

m = 1.32 mol/kg

So, substituting m in any of the equations (choosing the first):

Tb = 2.53 * 1.32 + 80.1

Tb = 83.4°C

Answer:

[tex]y_{O2} =4.3[/tex]%

Explanation:

The ethanol combustion reaction is:

[tex]C_{2}H_{5} OH+3O_{2}[/tex]→[tex]2CO_{2}+3H_{2}O[/tex]

If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:

[tex]x*1.10(excess)*\frac{3 O_{2}moles }{etOHmole}[/tex]

Dividing the previous equation by x:

[tex]1.10(excess)*\frac{3 O_{2}moles}{etOHmole}=3.30\frac{O_{2}moles}{etOHmole}[/tex]

We would need 3.30 oxygen moles per ethanol mole.

Then we apply the composition relation between O2 and N2 in the feed air:

[tex]3.30(O_{2} moles)*\frac{0.79(N_{2} moles)}{0.21(O_{2} moles)}=121.414 (N_{2} moles )[/tex]

Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:

[tex]3.30(O_{2} moles)-0.85(etOHmoles)*\frac{3(O_{2} moles)}{1(etOHmoles)} =0.75O_{2} moles[/tex]

Calculate the number of moles of CO2 and water considering the same:

[tex]0.85(etOHmoles)*\frac{3(H_{2}Omoles)}{1(etOHmoles)}=2.55(H_{2}Omoles)[/tex]

[tex]0.85(etOHmoles)*\frac{2(CO_{2}moles)}{1(etOHmoles)}=1.7(CO_{2}moles)[/tex]

The total number of moles at the reactor output would be:

[tex]N=1.7(CO2)+12.414(N2)+2.55(H2O)+0.75(O2)\\ N=17.414(Dry-air-moles)[/tex]

So, the oxygen mole fraction would be:

[tex]y_{O_{2}}=\frac{0.75}{17.414}=0.0430=4.3[/tex]%

Answer:The oxygen present at the tertiary carbon would be eliminated.The suggested mechanism of the reaction can be found in attachment

Explanation:

The Oxygen atom at the tertiary carbon atom would be eliminated because the removal of this oxygen in form of water after the protonation by sulphuric acid would lead to the formation of a stable tertiary carbocation which is vary stable.

The tertiary carbocation is  stable on account of inductive effect of the methy groups.

The oxygen atom at the primary carbon would not be eliminated as its elimination would result in a primary carbocation which is unstable in nature,.

The mechanism of the overall reaction is following:

1. In the first step the OH group present at the tertiary carbocation is protonated  by sulphuric acid and on account of this protonation the OH group turns into a good leaving group and leaves as (water) H₂O.

2. Once the H₂O molecule is eliminated it leads to the formation of a stable tertiary carbocation.

3. The tertiary carbocation so formed is electrophilic in nature and as there is one more OH group present at the primary carbon which is 3 carbons away . The OH group is weakly nucleophilic in nature and can appreciably attack the carbocation . The attack of OH at the carbocation leads to the formation of a 5-membered ring containing oxygen as heteroatom.

4.The 5-membered ring so formed has Oxygen as hetero atom which is protonated so the protonated oxygen atom is deprotonated using H₂O.

This further leads to the product formation.

Kindly refer the attachment for the complete reaction mechanism:

Answer:Acid catalyst is needed to increase the electrophilicity of Carbonyl group of Carboxylic acid as alcohol is a weak nucleophile.

Alternatively esters can be synthesised by converting carboxylic acid into acyl chloride using thionyl chloride(SOCl_{2} and then further treating acyl chloride with alcohol.

Carboxylic acid and esters can be easily distinguished on the basis of IR as carboxylic acid would contain a broad intense peak in 2500-3200cm_{-1} corresponding to OH stretching frequency whereas esters would not contain any such broad intense peak.

Alcohol and esters can also be distinguished using IR as alcohols would contain a broad intense peak at around 3200-3600cm_{-1}

Explanation: For the synthesis of esters using alcohol and carboxylic acid we need to add a little amount of acid in the reaction . The acid used here increases the electrophilicity of carbonyl carbon and hence makes it easier for a weaker nucleophile like alcohol to attack the carbonyl carbon of acid.

The oxygen of the carbonyl group is protonated using the acidic proton which  leads to the generation of positive charge on the oxygen. The positive charge generated is delocalised over the whole acid molecule and hence the electrophilicity of carbonyl group is increased. Kindly refer attachment for the structures.

If we simply mix the acid and alcohol then no appreciable reaction would take place between them and ester formation would not take place because the carboxylic acid in that case is not a good electrophile whereas alcohol is also not a very strong nucleophile which can attack the carbonyl group.

Alternatively we can use thionyl chloride or any other reagent which can convert the carboxylic acid into acyl chloride. Acyl chloride is very elctrophilic and alcohol can very easily attack the acyl chloride and esters could be synthesized.

The carboxylic acid and ester can very easily be distinguished on the basis of broad intense OH stretching frequency peak at around 2500-3200cm_{-1} . The broad intense OH stretching frequency peak is present in carboxylic acids as they contain OH groups and absent in case of esters .

Likewise esters and alcohols can also be distinguished on the basis IR spectra as alcohols will have broad intense spectra  at around 3200-3600cm_{-1}corresponding to OH stretching frequency whereas esters will not have any such peak. Rather esters would be having a Carbonyl stretching frequency at around 1720-1760

Final answer:

An acid catalyst is required for ester formation from carboxylic acids and alcohols due to the poor leaving group ability of -OH. Alternative ester synthesis can avoid acids by using acyl chlorides. IR spectroscopy differentiates carboxylic acids, esters, and alcohols by their unique O-H and C=O stretching vibrations.

Explanation:

An acid catalyst is required to make an ester from a carboxylic acid and an alcohol due to the poor leaving group ability of -OH in carboxylic acids. The acid catalyst speeds up the reaction by protonating the carbonyl oxygen, which allows the alcohol to act as a nucleophile and attack more readily. An alternative method to create esters, that avoids using an acid catalyst, involves converting the carboxylic acid into an acid chloride (using thionyl chloride), followed by a reaction with an alcohol.

In IR spectroscopy, distinguishing between a carboxylic acid and an ester involves looking for the presence of a broad O-H stretch around 2500-3000 cm-1 in acids that are absent in esters. Esters, however, will show a strong C=O stretch around 1735-1750 cm-1. To differentiate an alcohol from an ester, look for the O-H stretch in alcohols around 3200-3600 cm-1 which is sharper and more defined compared to the broad O-H stretch observed in carboxylic acids and absent in esters.

Answer:

There's no reducing sugar.

Explanation:

The Benedict test is another of the oxidation reactions, which, as we know, helps us to recognize reducing sugars, that is, those compounds that have their free anomeric OH, such as glucose, lactose or maltose or cellobiose, in the Benedict reaction can reduce the Cu2 + that presents a blue color, in an alkaline medium, the cupric ion (given by cupric sulfate) is able to be reduced by the effect of the aldehyde group of sugar (CHO) to its Cu + form. This new ion is observed as a red brick precipitate corresponding to cuprous oxide (Cu2O), which precipitates from the alkaline solution with a red-orange color, this precipitate is considered as evidence that there is a reducing sugar.

Benedict's reagent is composed of:

*Cupric sulfate.

*Sodium citrate.

*Anhydrous sodium carbonate.

The evidence of Benedict's reaction is the formation of the precipitate Ion Cuprous (Cu2O).

It means:

Cu+2 ------> Cu+1

Then, if we don't have a reducing sugar, the reactive is not going to react with our sample.

Answer: The molarity of Iron (III) chloride is 0.622 M.

Explanation:

Molarity is defined as the number of moles present in one liter of solution.  The equation used to calculate molarity of the solution is:

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]

Or,

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

We are given:

Mass of iron (III) chloride = 1.01 g

Molar mass of iron (III) chloride = 162.2 g/mol

Volume of the solution = 10 mL

Putting values in above equation, we get:

[tex]\text{Molarity of Iron (III) chloride}=\frac{1.01g\times 1000}{162.2g/mol\times 10mL}\\\\\text{Molarity of Iron (III) chloride}=0.622M[/tex]

Hence, the molarity of Iron (III) chloride is 0.622 M.

Answer:

TRUE

Explanation:

Isozymes are enzymes that differ in amino acid sequence but catalyze the same chemical reaction.  

These enzymes usually have different enzyme kinetics, that is, they have different Km and Vmax values,

For example, three isozymes of lactate dehydrogenase (A, B, and C) have Km values of 0.260, 0.172, and 0.052 mmol/L, respectively.

Answer:

The fraction of the copper’s electrons removed is [tex]9.076\times 10^{-13}[/tex].

Explanation:

Mass of copper ball = 50.0 g

Moles of copper = [tex]\frac{50.0 g}{63.5 g/mol}[/tex]

1 mole = [tex]N_A=6.022\times 10^{23}[/tex]

Number of copper atoms =[tex]\frac{50.0 g}{63.5 g/mol}\times 6.022\times 10^{23} [/tex]

1 atom of copper has 29 protons

Total number of protons in 50.0 g of copper =[tex]\frac{50.0 g}{63.5 g/mol}\times 6.022\times 10^{23} \times 29=1.3751\times 10^{25}[/tex]

Since an atom is a neutral specie which means number of protons are equal to number of electrons.

Total number of electrons = [tex]1.3751\times 10^{25}[/tex]....(1)

Net charge on the copper ball = [tex]2.00/mu C=2.00\times 10^{-6} C[/tex]

Q=Ne

Q = Total charge

N = Number of electrons

e = charge on an electron = [tex]1.602\times 10^{-19} C[/tex]

[tex]2.00\times 10^{-6} C=N\times 1.602\times 10^{-19} C[/tex]

[tex]N =1.248\times 10^{13} [/tex]

Total number of electrons removed = N = [tex]1.248\times 10^{13} [/tex]

Fraction of the copper’s electrons has been removed:

[tex]\frac{\text{Number of electrons removed}}{\text{Total electrons}}[/tex]

[tex]\frac{1.248\times 10^{13}}{1.3751\times 10^{25}}=9.076\times 10^{-13}[/tex]

The fraction of the copper’s electrons removed is [tex]9.076\times 10^{-13}[/tex].

To calculate the fraction of electrons removed from a 50.0 g ball of copper with a 2.00 µC charge, we determine the total number of electrons in the ball and then find the number corresponding to the charge. We conclude that approximately 9.083 x 10−11% of the copper's electrons have been removed.

To determine the fraction of copper's electrons that have been removed, we'll need to calculate the total number of electrons in the 50.0 g ball of copper and then see how many electrons correspond to a charge of 2.00 µC.

First, we calculate the number of moles of copper in the 50.0 g ball. With an atomic mass of 63.5 g/mol for copper, we have:

Number of moles = 50.0 g / 63.5 g/mol = 0.7874 moles

Next, using Avogadro's number (6.022 x 1023 atoms/mol), we find the number of copper atoms:

Number of copper atoms = 0.7874 moles x (6.022 x 1023 atoms/mol) = 4.739 x 1022 atoms

Since each copper atom contributes 29 electrons, the total number of electrons in the ball is:

Total number of electrons = 29 x 4.739 x 1022 atoms = 1.374 x 1024 electrons

To find the number of electrons that corresponds to a charge of 2.00 µC, we use the electron charge (1.602 x 10−19 C/electron):

Number of electrons removed = 2.00 µC / (1.602 x 10−19 C/electron) = 2.00 x 10−6 C / (1.602 x 10−19 C/electron) = 1.248 x 1013 electrons

Now, we find the fraction of electrons removed:

Fraction = Number of electrons removed / Total number of electrons = 1.248 x 1013 / 1.374 x 1024 ≈ 9.083 x 10−11%

So, approximately 9.083 x 10−11% of the copper's electrons have been removed.

Answer: n-Butanol are converted using SN2 and tert-butanol is converted using SN1

Explanation: For the conversion of n-butanol into butyl chloride using Hydrogen Chloride the reaction would follow SN2 mechanism.

SN2 reaction mechanism occurs only in the case of primary substrates as it is a one step mechanism that happens in a concerted manner. It involves backside attack of nucleophile on the substrate such that the nucleophile attacks from the back side and leaving group leaves from the front side.

In this reaction since hydroxy group (OH) is not a good leaving group hence firstly we need to convert it into a good leaving group. When we treat n-butanol with HCl hydroxy group is protonated and now it turns into a good leaving group as it can leave as H₂O.

Cl⁻  here acts as nucleophile and now attacks the primary carbon center from the back side which contains the protonated hydroxy group as a leaving group.

In the case of tertiary butanol the reaction follows SN1  mechanism and it is 2 step mechanism.

In the first step hydroxy group is protonated and as it becomes a good leaving group it leaves and leads to the formation of a stable tertiary carbocation as an intermediate.

In the second step this intermediate carbocation is attacked by the Cl⁻  nucleophile which leads to the formation of tertiary butyl chloride.

Kindly find in attachment the reaction mechanism for both the reactions.

Answer:

The answer would be Lead.

Explanation:

Answer:

(R) - hexyl acetate

Explanation:

Hello,

This reacción is a nucleophilic substitution SN2.

The configuration (s), means that the groups around the chiral carbon are organized appose to the clock hands movement. But when the reaction happens, these configurations become an (r) configuration, it means the groups around the chiral carbon organize according to the clock hands movement.

Generally, these reactions are related to nucleophilic species, an example is the ion acetate, a conjugated acid which is a weak nucleophilic, for this reason, the transition state is more energetic, it means, less stable than if the reaction occurs with a strong nucleophilic.

Look the image to compare the two configurations of the reactant and product.

An SN2 reaction between (S)-2-bromohexane and acetate ion leads to the formation of (R)-2-acetoxypentane due to the backside attack characteristic of such reactions, resulting in an inversion of the original stereochemistry around the chiral center.

The student's question involves a nucleophilic substitution (specifically an SN2 reaction) where the nucleophile, acetate ion (CH3CO2-), replaces the bromide ion in (S)-2-bromohexane. Under the assumption of inversion of configuration, the product will be (R)-2-acetoxypentane. The initial (S)-2-bromohexane has a specific three-dimensional arrangement with the bromine atom attached to the second carbon in a configuration that is opposite to that of the other substituents when viewed in a certain manner.

In an SN2 reaction, the nucleophile attacks from the opposite side of the leaving group, leading to an inversion of the stereochemistry at the carbon center undergoing the substitution. Therefore, the product will be the R-enantiomer of 2-acetoxypentane since the attacking acetate ion approaches from the side opposite to the leaving bromide ion, flipping the configuration.

The stereochemical outcome of an SN2 reaction is critical in organic synthesis as it controls the formation of specific enantiomers of chiral molecules.

Answer:

True

Explanation:

When we consume food, our body converts the excess amount of food in our body into triglycerides and stores it. These triacylglycerols or triglycerides, commonly known as fats, come under the category of lipids.

Triglycerides are the stored form of energy, that our body uses at the time of need.

Adipose tissues are known as fat tissues, as their main function is the storage of triglycerides in our body.

Answer:

Concentration of sodium ions in final solution is 0.1428 mol/L.

Concentration of nitrate ions in final solution is 0.2857 mol/L.

Explanation:

Concentration =[tex]\frac{\text{Moles of compound}}{\text{Volume of the solution (L)}}[/tex]

[tex]Na_2CrO_4(aq)+2AgNO_3(aq)\rightarrow Ag_2CrO_4(s)+2NaNO_3(aq)[/tex]

Moles of silver nitrate in 10 mL of 1 M silver nitrate.

Volume of the  silver nitrate solution = 10 mL = 0.010 L

[tex]1 M =\frac{\text{Moles of}Ag_2CrO_4}{\text{Volume of the solution in L}}[/tex]

Moles of silver nitrate =[tex]1 M\times 0.010 L=0.010 mol[/tex]

Moles of sodium chromate in 25 mL of 0.1 M silver nitrate.

Volume of the  silver chromate solution = 25 mL = 0.025 L

[tex]0.1 M =\frac{\text{Moles of}Na_2CrO_4}{\text{Volume of the solution in L}}[/tex]

Moles of sodium nitrate =[tex]0.1 M\times 0.025 L=0.0025 mol[/tex]

According to reaction, 1 mole of sodium chromate reacts with 2 mole of silver nitrate.

Then , 0.0025 mol of sodium chromate will react with :

[tex]\frac{1}{2}\times 0.0025 mol=0.00125 mol[/tex] of silver nitrate.

1 mol of sodium chromate gives 2 mol of sodium ions and 1 mol of chromate ions.

Then 0.0025 mol of sodium chromate will give 0.0050 mol of sodium ions.

Volume of the solution after mixing =

10 mL + 25 mL = 35 mL =0.035 L

Concentration of sodium ions in final solution:

[tex][Na^+]=\frac{0.0050 mol}{0.035 mL}=0.1428 mol/L[/tex]

Concentration of sodium ions = 0.1428 mol/L

1 mole of silver nitrate gives 1 mol of silver ion and 1 mole nitrate ion

Then 0.010 moles of silver nitrate will give 0.010 moles of nitrate ions.

Concentration of nitrate ion in the final solution:

[tex][NO_3^{-}]=\frac{0.010 mol}{0.035 L}=0.2857 mol/L[/tex]

Concentration of nitrate ions = 0.2857 mol/L

The concentration of sodium ion after the two solutions are combined is 0.1428M.

Concentration will be calculated in terms of molarity as:

M = n/V, where

n = no. of moles

V = volume

Given chemical reaction is:
2AgNO₃ + Na₂CrO₄ → Ag₂CrO₄ + 2NaNO₃

Given volume of AgNO₃ = 10mL = 0.010L

Molarity of AgNO₃ = 1M

Moles of AgNO₃ = 0.010×1 = 0.010 moles

Given volume of Na₂CrO₄ = 25mL = 0.025L

Molarity of Na₂CrO₄ = 0.1M

Moles of Na₂CrO₄ = 0.025×0.1 = 0.0025 moles

From the stoichiometry of the reaction it is clear that:

1 mole of Na₂CrO₄ = produce 2 moles of sodium ions

0.0025 moles of Na₂CrO₄ = produce 2×0.0025=0.0050 moles of sodium ions

Total volume of final solution = 25mL + 10mL = 35mL = 0.035L

Now we calculate the concentration in terms of molarity of sodium ions as:
M = 0.0050/0.035 = 0.1428M

Hence, the concentration of sodium ions is 0.1428M.

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Answer: The molarity of [tex]Pb(NO_3)-2[/tex] solution is 0.34 M.

Explanation:

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

For lead chloride:

Given mass of lead chloride = 18.86 g

Molar mass of lead chloride = 278.1 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of lead chloride}=\frac{18.86g}{278.1g/mol}=0.068mol[/tex]

[tex]Pb(NO_3)_2+2NaCl\rightarrow PbCl_2+2NaNO_3[/tex]

By Stoichiometry of the reaction:

1 mole of lead chloride is formed by 1 mole of lead nitrate

So, 0.068 moles of lead chloride will be formed from = [tex]\frac{1}{1}\times 0.068=0.068mol[/tex] of lead nitrate

[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]

We are given:

Volume of solution = 200 mL = 0.200 L   (Conversion factor: 1 L = 1000 mL)

Moles of lead nitrate = 0.068 moles

Putting values in above equation, we get:

[tex]\text{Molarity of }Pb(NO_3)_2\text{ solution}=\frac{0.068mol}{0.065L}\\\\\text{Molarity of }Pb(NO_3)_2\text{ solution}=0.34M[/tex]

Hence, the molarity of [tex]Pb(NO_3)-2[/tex] solution is 0.34 M.

The molarity of the Pb(NO3)2(aq) solution can be calculated by first finding the number of moles of PbCl2 formed in the reaction and then dividing it by the original volume of Pb(NO3)2 solution. The calculated molarity of the Pb(NO3)2(aq) solution is 0.339 M.

The calculation of molarity of the Pb(NO3)2(aq) solution is necessary in this case. In this precipitation reaction, lead nitrate reacts with sodium chloride to form lead chloride, which precipitates out, and sodium nitrate. The molar mass of lead chloride (PbCl2) is 278.1 g/mol.

Step 1: Find the number of moles of PbCl2. For this, you can divide the mass of the precipitate (PbCl2) obtained by the molar mass of PbCl2. So, the number of moles are 18.86 g / 278.1 g/mol = 0.0678 mol.

Step 2: Calculate the amount in liters of original solution of lead nitrate. Here, since the given volume is in milliliters, you need to convert it to liters. Therefore, 200.0 mL = 0.2000 L.

Step 3: Calculate the molarity. Molarity is the number of moles of solute divided by volume of the solution in liters. Hence, M = 0.0678 mol / 0.2000 L = 0.339 M. Therefore, the molarity of Pb(NO3)2(aq) solution is 0.339 M.

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Answer: The value of equilibrium constant, [tex]K_c[/tex] for the given reaction is 12.85.

Explanation:

For the given chemical equation:

                         [tex]A+2B\rightleftharpoons C[/tex]

At t = 0      0.350M     0.650M      0.300M

At [tex]t=t_{eq}[/tex]  (0.350 - x)    (0.650 - 2x)     (0.300 + x)

We are given:

Equilibrium concentration of A = 0.220 M

Forming an equation for concentration of A at equilibrium:

[tex]0.350-x=0.220\\x=0.130[/tex]

Thus, the concentration of B at equilibrium becomes = [tex]0.650-(2\times 0.130)=0.390M[/tex]

Equilibrium concentration of C = 0.430 M

The expression of [tex]K_c[/tex] for the given chemical equation is:

[tex]K_c=\frac{[C]}{[A][B]^2}[/tex]

Putting values in above equation:

[tex]K_c=\frac{0.430}{0.220\times (0.390)^2}\\\\K_c=12.85[/tex]

Hence, the value of equilibrium constant, [tex]K_c[/tex] for the given reaction is 12.85.

The equilibrium constant, Kc, can be calculated using the concentrations of the reactants and products at equilibrium.

The equilibrium constant, denoted as Kc, is the mathematical expression that relates the concentrations of the reactants and products at equilibrium. For the reaction A + 2B ⇌ C, the equilibrium constant expression is given by:

Kc = [C] / ([A] * [B]^2)

Using the given concentrations at equilibrium ([A] = 0.220 M and [C] = 0.430 M), we can substitute these values into the expression to calculate the value of Kc.

Kc = 0.430 / (0.220 * (0.650)^2)

Calculating this expression will give you the value of the equilibrium constant, Kc.

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Explanation:

[tex]HBr(aq) + LiOH(aq) \rightarrow LiBr(aq)+H_2O(l)[/tex]

a)Molarity of LiOH,[tex]M_1[/tex] = 0.0024 M

Volume of LiOH = [tex]V_1[/tex]

Molarity of HBr,[tex]M_2[/tex] = 0.0026 M

Volume of HBr = [tex]V_2=22mL=[/tex]

According to reaction , 1 mol of LiOH neutralizes 1 mol of HBr.

[tex]M_1V_1=M_2V_2[/tex]

[tex]V_1=\frac{0.0026 M\times 22mL}{0.0024 M}=23.8333 mL[/tex]

[tex]M_1=\frac{\text{Molesof LiOH}}{V_1}[/tex]

Moles of LiOH = [tex]0.0024 mol/L\times 23.8333 ml=0.0571 mol[/tex]

Mass of 0.0571 mol of LiOH:

[tex]0.0571 mol\times 2 g/mol=1.3704 g[/tex]

1.3704 g of 0.0024 M LiOH that would neutralize 22 mL of 0.0026 M HBr.

b) According to reaction , 1 mol LiOH gives 1 mol of LiBr.

Then ,0.0571 mol of LiOH will give:

[tex]\frac{1}{1}\times 0.0571 mol=0.0571 mol[/tex] of LiBr

0.0571 moles of salt are produced in the reaction

c) Moles of salt = 0.0571 mol

Volume of the solution = 22 ml+ 23.8333 mL= 45.8333 mL = 0.0458333 L

Molar concentration of the salt:

[tex]{LiBr}=\frac{0.0571 mol}{0.0458333 L}=1.2458 mol/L[/tex]

1.2458 mol/L is the molar concentration of the salt after the reaction is complete.

Hydrobromic acid and lithium hydroxide react to form lithium bromide and water. 23.8 mL of 0.0024 M LiOH would neutralize 22 mL of 0.0026 M HBr, producing 0.0000572 mol of LiBr. The molar concentration of the salt solution after the reaction = 0.00125 M.

The balanced chemical equation for the reaction between hydrobromic acid (HBr) and lithium hydroxide (LiOH) is given by: HBr(aq) + LiOH(aq) → LiBr(aq) + H2O(l).

Considering the balanced equation, we observe a ratio of 1:1 between HBr and LiOH. In (a), to calculate the volume of 0.0024 M LiOH necessary to neutralize 22 mL of 0.0026 M HBr, we use the concept of molarity (M) which is equal to mol/L. Therefore, the solution of 0.0024 M LiOH has 0.0024 mol of LiOH per L. As the ratio is 1:1, we also need 0.0026 mol/L * 0.022L = 0.0000572 mol of LiOH, which means we need 0.0000572 mol / 0.0024 mol/L = 0.0238L or 23.8 mL of LiOH.

In (b), as the reaction produces 1 mol of LiBr(salt) for every mole of HBr or LiOH, there will be the same amount of salt produced, so 0.0000572 mol of salt is produced.

In (c), the molar concentration of salt solution will be the number of moles of the salt divided by total volume of the solution. Assuming volumes of reactants add up, total volume = 0.022 L + 0.0238 L = 0.0458 L. Therefore, the molar concentration = 0.0000572mol / 0.0458L = 0.00125 M.

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Answer:

Partial pressure of methane: 1.18 atm

Partial pressure of ethane: 1.45 atm

Partial pressure of propane: 2.35 atm

Explanation:

Let the total moles of gases in a container be n.

Total pressure of the gases in a container =P = 5.0 atm

Temperature of the gases in a container =T = 23°C = 296.15 K

Volume of the container = V = 10.0 L

[tex]PV=nRT[/tex] (Ideal gas equation)

[tex]n=\frac{PV}{RT}=\frac{5.0 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.0564 mol[/tex]

Moles of methane gas =[tex]n_1=\frac{8.00 g}{16.04 g/mol}=0.4878 mol[/tex]

Moles of ethane gas =[tex]n_2=\frac{18.00 g}{30.07 g/mol}=0.5986 mol[/tex]

Moles of propane gas =[tex]n_3=?[/tex]

[tex]n=n_1+n_2+n_3[/tex]

[tex]n_3=n-n_1-n_2=2.0564 mol-0.4878 mol-0.5986 mol= 0.9700 mol[/tex]

Partial pressure of all the gases can be calculated by using Raoult's law:

[tex]p_i=P\times \chi_i[/tex]

[tex]p_i[/tex] = partial pressure of 'i' component.

[tex]\chi_1[/tex] = mole fraction of 'i' component in mixture

P = total pressure of the mixture

Partial pressure of methane:

[tex]p_1=P\times \chi_1=P\times \frac{n_1}{n_1+n+2+n_3}=P\times \frac{n_1}{n}[/tex]

[tex]p_1=5.00 atm\times \frac{0.4878 mol}{2.0564 mol}=1.18 atm[/tex]

Partial pressure of ethane:

[tex]p_2=P\times \chi_2=P\times \frac{n_2}{n_1+n+2+n_3}=P\times \frac{n_2}{n}[/tex]

[tex]p_2=5.00 atm\times \frac{0.5986 mol}{2.0564 mol}=1.45 atm[/tex]

Partial pressure of propane:

[tex]p_3=P\times \chi_3=P\times \frac{n_3}{n_1+n+2+n_3}=P\times \frac{n_3}{n}[/tex]

[tex]p_3=5.00 atm\times \frac{0.9700 mol}{2.0564 mol}=2.35 atm[/tex]

After calculating moles for each gas, the partial pressure for methane, ethane, and propane was calculated as 1.21 atm, 1.45 atm, and 2.34 atm respectively.

To calculate the partial pressure of each gas, we will first need to calculate the moles of each gas. For methane, the molar mass is 16.04 g/mol, so 8.00 g / 16.04 g/mol = 0.499 mol. The molar mass of ethane is 30.07 g/mol, so 18.0 g / 30.07 g/mol = 0.598 mol. The total moles of gas can be calculated by the total pressure and volume using the Ideal gas law, PV=nRT. The total moles of gases are 5.0 atm *10.0L/(0.0821*296.15K) = 2.06 mol. Hence, the moles of propane are 2.06 - 0.499 - 0.598 = 0.963 mol.

Using Dalton's law of partial pressures, the partial pressure of each gas can be calculated by (moles of gas/ total moles) * total pressure. Hence, the partial pressures of methane, ethane, and propane are  0.499 / 2.06 * 5.00 atm = 1.21 atm, 0.598 / 2.06 * 5.00 atm = 1.45 atm, and 0.963 / 2.06 * 5.00 atm = 2.34 atm, respectively.

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