(a) Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water. (b) A commercial bleaching solution contains 3.62 mass % sodium hypochlorite, NaOCI. What is the mass of NaOCI in a bottle containing 2.50 kg of bleaching solution?

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 : The time taken for the reaction is, 28 s.

Explanation :

Expression for rate law for first order kinetics is given by :

[tex]k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}[/tex]

where,

k = rate constant  = 0.0632

t = time taken for the process  = ?

[tex][A_o][/tex] = initial amount or concentration of the reactant  = 1.28 M

[tex][A][/tex] = amount or concentration left time 't' = [tex]1.28\times \frac{17}{100}=0.2176M[/tex]

Now put all the given values in above equation, we get:

[tex]0.0632=\frac{2.303}{t}\log\frac{1.28}{0.2176}[/tex]

[tex]t=28s[/tex]

Therefore, the time taken for the reaction is, 28 s.

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: 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:

The number of atoms in the gemstones of that tiara is [tex]8.9125\times 10^{24} atoms[/tex].

Explanation:

Number of diamonds ion tiara = 888

Mass of  each diamond = 1.0 carat = 0.200 g (given)

Mass of 888 diamonds in tiara:

[tex]888\times 0.200 g=177.600 g[/tex]

Given that diamond is a form of carbon.

Atomic mass of  carbon atom = 12 g/mol

Moles of 77.600 g of carbon =[tex]\frac{177.600 g}{12 g/mol}=14.800 mol[/tex]

Number of atoms of carbon 14.800 moles:

[tex]14.800 mol\times 6.022\times 10^{23} atoms =8.9125\times 10^{24} atoms[/tex]

The number of atoms in the gemstones of that tiara is [tex]8.9125\times 10^{24} atoms[/tex].

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 new rms speed is 2 times the original rms speed.

Explanation:

The root‑mean‑square speed, rms, is related to temperature, , by the formula

[tex]_{rms}[/tex]= √3 / ℳ

For a given gas,  

[tex]_{rms}[/tex] ∝ √

or

[tex]_{rms,2}[/tex] / [tex]_{rms,1}[/tex] = √[tex]_{2}[/tex] / [tex]_{1}[/tex]

In this case, is quadrupled.

√4 = 2

The new rms speed is 2 times the original rms speed.

The root-mean-square (rms) speed of gas molecules is proportional to the square root of the temperature. If the absolute temperature is quadrupled, the new rms speed is 2 times the original rms speed.

Effect of Temperature on Root Mean Square Speed of Gas Molecules

The root-mean-square (rms) speed, [tex]V_{rms[/tex] , of the molecules in a gas is related to the absolute temperature, T, by the equation:

[tex]V_{rms[/tex] = [tex]\sqrt((3kBT) / m)[/tex]

where kB is the Boltzmann constant and m is the mass of a molecule. This shows that Vrms is proportional to the square root of the temperature. If the temperature T is quadrupled, the new temperature T' is 4T.

Therefore, the new rms speed [tex]V_{rms[/tex]' becomes:

[tex]V_{rms[/tex]' = [tex]\sqrt((3kB * 4T) / m)[/tex]= [tex]\sqrt(4) * \sqrt((3kBT) / m)[/tex] = 2 * [tex]V_{rms[/tex]

This means that the new rms speed is 2 times the original rms speed.

Explanation:

According to Faraday's law, the amount of a substance deposited or liberated in electrolysis process is proportional to the quantity of electric charge passed and to the equivalent weight of the substance.

Formula to calculate the mass of substance liberated according to Faraday's law is as follows.

             m = [tex](\frac{Q}{F})(\frac{M}{Z})[/tex]

where,          m = mass of substance liberated at electrode

                     Q = electric charge passing through the substance

                     F = Faraday constant = 96,487 C [tex]mol^{-1}[/tex]

                     M = molar mass of the substance

                     Z = valency number of ions of the substance

Since, it is given that mass is 3 kg or 3000 g (as 1 kg = 1000 g), molar mass of Al is 27, Z is 3.

Therefore, putting the values in the above formula as follows.

                      m = [tex](\frac{Q}{F})(\frac{M}{Z})[/tex]

                     3000 g = [tex](\frac{Q}{96,487 C mol^{-1}})(\frac{27}{3})[/tex]

                        Q = 32162333.33 C

As it is given that V = 4.50 Volt. Also, it is known that

                       Energy = [tex]V \times Q[/tex]

Therefore, calculate the energy as follows.

                      Energy = [tex]V \times Q[/tex]

                                   = [tex]4.50 V \times 32162333.33 C[/tex]

                                   = 144730500 J

As it is known that [tex]3.6 \times 10^{6}[/tex] J = 1 KW Hr

So, convert 144730500 J into KW Hr as follows.

                   [tex]\frac{144730500 J \times 1 KW Hr}{3600000 J}[/tex]

                          = 40.202 KW Hr

Thus, we can conclude that the number of kilowatt-hours of electricity required to produce 3.00 kg of aluminum from electrolysis of compounds from bauxite is 40.202 KW Hr when the applied emf is 4.50V.

The production of aluminum from bauxite involves several chemical reactions and an electrolysis process in a Hall-Héroult cell. Aluminum oxide is reduced to aluminum metal during electrolysis. The exact amount of electricity required cannot be stated without additional industrial data.

The process of making aluminum from bauxite involves several stages of chemical reactions. Initially, bauxite, AlO(OH), reacts with hot sodium hydroxide to form soluble sodium aluminate, leaving behind impurities. Aluminum hydroxide is then precipitated out and heated to form aluminum oxide, Al2O3, which is dissolved in a molten mixture of cryolite and calcium fluoride. An electrolytic cell, specifically called the Hall-Héroult cell, is used for the electrolysis process. During electrolysis, the reduction of aluminum ions to aluminum metal takes place at the cathode, while oxygen, carbon monoxide, and carbon dioxide form at the anode. The amount of electricity required to produce 3 kg of aluminum using a 4.5V emf cannot be directly calculated without additional data such as the exact chemical conversion efficiencies and energy losses in the process, which are typically proprietary information for aluminum production companies.

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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.

The answer would be Lead.

Answer:

The correct answer is :It is a molecule composed of carbon and hydrogen only.

Explanation:

Hydrocarbons are defined as compounds which majorly made up carbon and hydrogen.There three types of hydrocarbons:

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:

-68.4 kJ

Explanation:

The standard enthalpy of vaporization = 23.3 kJ/mol

which means the energy required to vaporize 1 mole of ammonia at its boiling point (-33 °C).

To calculate heat released when 50.0 g of ammonia is condensed at -33 °C.

This is the opposite of enthalpy of vaporization which means that same magnitude of heat is released.

Thus,  Q = -23.3 kJ/mol

Where negative sign signifies release of heat

Given: mass of 50.0 g

Molar mass of ammonia = 17.034 g/mol

Moles of ammonia = 50.0 /17.034 moles = 2.9353 moles

Also,

1 mole of ammonia when condenses at -33 °C releases 23.3 kJ

2.9412 moles of ammonia when condenses at -33 °C releases 23.3×2.9353 kJ

Thus, amount of heat released when 50 g of ammonia condensed at -33 °C= -68.4 kJ, where negative sign signifies release of heat.

The heat released when 50.0 g of ammonia condenses at its boiling point is -68.4 kJ. This is calculated by multiplying the moles of ammonia by the enthalpy of vaporization and recognizing that heat is released in condensation.

To solve this problem, we need to understand the concept of enthalpy of vaporization, which is the heat needed to convert 1 mole of a substance from a liquid to a gas at constant pressure and temperature. For ammonia (NH3), which boils at -33 °C, the enthalpy of vaporization is 23.3 kJ/mol. However, we want the heat released when 50.0 g (around 2.94 moles) of ammonia condenses, which is the reverse process of vaporization. Thus, the energy would be released rather than absorbed.

Now, let's calculate this value. We multiply the number of moles of ammonia by the enthalpy of vaporization:

2.94 moles x 23.3 kJ/mol = 68.4 kJ

Since this is the reverse of the process of vaporization, heat is released, so the enthalpy change is negative (-68.4 kJ). Therefore, the correct answer is -68.4 kJ.

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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: 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:

Explanation:

1) Data:

a) Tb₁ = 100.000°C

b) Kb = 0.512 °C/m

c) mass of solute = 13.62 g

d) mass of solvent = 217.5 g

e) Tb₂ = 100.094°C

f) Solute: nonvolatile and nonelectrolyte

g) MM = ?

2) Chemical principles and formulae:

a) The boiling point elevation of a non-volatile solute is a colligative property, which follows this equation:

Where:

b) Molality, m:

c) Molar mass, MM:

3) Solution:

i) ΔTb = Tb₂ - Tb₁ = 100.094°C - 100.000°C = 0.094°C

ii) ΔTb = Kb × m ⇒ m = ΔTb / Kb = 0.094°C / (0.512°C/m) = 0.1836 m

iii) m = number of moles of solute / kg of solvent ⇒

    number of moles of solute = m × kg of solvent = 0.1836 m × 0.2175 kg

    number of moles of solute = 0.03993 mol

iv) MM = mass in grams / number of moles = 13.62 g / 0.03993 mol = 341.1 g/mol

Final answer:

The molecular weight of the synthesized compound is calculated using the change in boiling point and the molal boiling point elevation constant to first determine molality and then the number of moles of the compound. The molecular weight is then found by dividing the mass of the compound by the moles of the compound, yielding a value of 341.1 g/mol.

Explanation:

To calculate the molecular weight of the synthesized compound, we apply the boiling point elevation formula ΔT = Kb × m, where ΔT is the change in boiling point, Kb is the molal boiling point elevation constant of water, and m is the molality of the solution. Given Kb for water is 0.512°C/m and ΔT is 0.094°C (100.094°C - 100.000°C), we can calculate the molality:

m = ΔT / Kb = 0.094°C / 0.512°C/m = 0.1836 m

Next, we calculate the moles of solute using molality and the mass of the solvent (water):

moles of solute = molality × mass of solvent in kg = 0.1836 m × 0.2175 kg = 0.0399 mol

The molecular weight (MW) is then found by dividing the mass of the compound by the moles of the compound:

MW = mass of compound / moles of compound = 13.62 g / 0.0399 mol = 341.1 g/mol

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:

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: The given statement is false.

Explanation:

Heat flux is defined as the flow of heat or energy per unit time in per unit area. S.I. unit of heat flux is watts per square meter.

Heat flux is represented by the symbol [tex]\phi _{q}[/tex].

So, it means heat flux is not measured in watts only as it includes per unit area also.

Therefore, we can conclude that the given statement heat flux, or the rate of heat transfer per unit area, is measured in watts (Watts) only, is false.

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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Answer : The normal boiling point of ethanol will be, [tex]348.67K[/tex] or [tex]75.67^oC[/tex]

Explanation :

The Clausius- Clapeyron equation is :

[tex]\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})[/tex]

where,

[tex]P_1[/tex] = vapor pressure of ethanol at [tex]30^oC[/tex] = 98.5 mmHg

[tex]P_2[/tex] = vapor pressure of ethanol at normal boiling point = 1 atm = 760 mmHg

[tex]T_1[/tex] = temperature of ethanol = [tex]30^oC=273+30=303K[/tex]

[tex]T_2[/tex] = normal boiling point of ethanol = ?

[tex]\Delta H_{vap}[/tex] = heat of vaporization = 39.3 kJ/mole = 39300 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

[tex]\ln (\frac{760mmHg}{98.5mmHg})=\frac{39300J/mole}{8.314J/K.mole}\times (\frac{1}{303K}-\frac{1}{T_2})[/tex]

[tex]T_2=348.67K=348.67-273=75.67^oC[/tex]

Hence, the normal boiling point of ethanol will be, [tex]348.67K[/tex] or [tex]75.67^oC[/tex]

Answer: A group 1 alkali metal bonded to fluoride, such as LiF.

Explanation:

Electronegativity is defined as the property of an element to attract a shared pair of electron towards itself. The size of an atom increases as we move down the group because a new shell is added and electron gets added up.

1. A strong acid made of hydrogen and a halogen, such as HCl : A polar covalent bond is defined as the bond which is formed when there is a difference of electronegativities between the atoms. Electronegativity difference = electronegativity of chlorine - electronegativity of hydrogen = 3-2.1= 0.9

2. A group 1 alkali metal bonded to fluoride, such as LiF: Ionic bond is formed when there is complete transfer of electron from a highly electropositive metal to a highly electronegative non metal.

Electronegativity difference = electronegativity of fluorine - electronegativity of lithium= 4-1= 3

3. Carbon bonded to a group 6A (16) nonmetal chalcogen, such as in CO: A polar covalent bond is defined as the bond which is formed when there is a difference of electronegativities between the atoms.

Electronegativity difference = electronegativity of oxygen - electronegativity of carbon= 3.5-2.5= 1.0

4. A diatomic gas, such as nitrogen [tex](N_2)[/tex]: Non-polar covalent bond is defined as the bond which is formed when there is no difference of electronegativities between the atoms.

Electronegativity difference = 0

Thus the greatest electronegativity difference between the bonded atoms is in LiF.

Among the given compounds, a group 1 alkali metal bonded to fluoride, such as LiF, has the greatest electronegativity difference because fluoride is highly electronegative while group 1 alkali metals like lithium have low electronegativity.

To determine which of the given compounds has the greatest electronegativity difference between the bonded atoms, we need to consider the electronegativities of the individual atoms involved. Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons.

A strong acid made of hydrogen and a halogen, such as HCl, has a significant electronegativity difference due to chlorine being much more electronegative than hydrogen. However, a group 1 alkali metal bonded to fluoride, such as LiF, will typically have an even greater electronegativity difference. This is because fluoride is one of the most electronegative elements, and lithium is a metal with a much lower electronegativity. Carbon bonded to a group 6A (16) nonmetal chalcogen, such as in CO, will have a moderate electronegativity difference, but less than that between lithium and fluoride. A diatomic gas, such as nitrogen (N₂), will have no electronegativity difference because both atoms are the same and thus have the same electronegativity.

Therefore, a group 1 alkali metal bonded to fluoride, such as LiF, has the greatest electronegativity difference of the options provided.

Answer:

The partial pressure of component A and B is 75 kPa and 25kPa respectively.

Explanation:

Total pressure of ideal gas mixture = 100kPa

Mole fraction of component B, [tex]\chi_B= 0.25[/tex]

Mole fraction of component A ,[tex]\chi_A[/tex]

As we know sum of all mole fraction in a mixture is equal to zero.

[tex]\chi_A+\chi+B=1[/tex]

[tex]\chi_A=1-\chi_B=1-0.25=0.75[/tex]

For partial pressure of each component we will apply Dalton's law of partial pressure:

[tex]p^{o}_i=p_{total}\times \chi_i[/tex]

Partial pressure of component a in a mixture:

[tex]p^{o}_A=p\times \chi_A=100kPa\times 0.75=75 kPa[/tex]

Partial pressure of component a in a mixture:

[tex]p^{o}_B=p\times \chi_B=100kPa\times 0.25=25 kPa[/tex]

Hey there!:

In Kjeldahl's method  estimation of amount of nitrogen in a sample, the sample is first digested by using sulphuric acid , which converts the nitrogen in the sample to ammonium sulphate.

The ammonium ions in the digested product are converted to ammonia gas by adding excess of base like NaOH. The ammonia gas thus produced is collected by condensation.

The amount of ammonia produced is estimated by titration with standard solution of acid. We are using 0.1 N HCl in this case.

Volume of 0.1 N HCl used = 11.95 mL

Volume of of 0.1 N HCl used in blank titration =0.1 mL

Therefore,actual volume us of 0.1 N HCl

used by analyte( ammonia solution) = 11.95 mL - 0.1 mL  

= 11.85 mL

0.1 N HCl = 0.1 M HCl , as HCl is a monobasic acid and therefore its molar mass will be equal to its equivalent mass.

Therefore, concentration of HCl solution used = 0.1 M = 0.1 mol L-1

Volume of HCl solution used =11.85 mL =0.01185 L

Nº.of moles of  HCl used =  0.1 mol L-1  * 0.01185 L

= 0.001185 moles

The equation for reaction between NH3 (aq) and HCl during titration is  :

NH3 (aq) + HCl (aq) <= >  NH4Cl

From the above equation, we see that each mole of HCl reacts completely with 1 mole of NH3 during titration.

Therefore, no. of moles of NH3 that would have been neutralized by 0.001185 moles of HCl = 0.001185 moles

From the formula of  NH3 , 1 mole of  NH3 contains 1 mole of N atoms.

Therefore, nº of moles of :

N - atoms present = 0.001185 moles

Moar mass of N= 14 g/mol

Mass of 0.001185 moles of N =0.001185 moles * 14 g/ mol

=  0.01659 g

% of N in protein is given as 16%.

16%.of mass of protein  = mass of nitrogen in protein sample  = 0.01659 g

(16/100) *mass of protein  =  0.01659 g

mass of protein  =  0.1036875 g

% of protein  in potato flour = [mass of protein in  potato flour / mass of  potato flour] * 100%

( 0.1036875 g / 5.724 ) * 100%

=  1.812 %

Hope this helps!

Final answer:

The percent protein was found to be 1.8113 %.

Explanation:

To calculate the percent protein in the potato flour, we first need to correct the titration volume for the blank, then figure out the amount of Nitrogen (N) in the potato flour and finally convert this to percent protein using the factor that proteins in potato flour contain 16% nitrogen. Since each gram of nitrogen corresponds to 6.25 grams of protein (since 100/16 = 6.25), we can use this conversion factor to get the final percent protein.

Volume of HCl used for the sample is 11.95 mL

Volume of HCl used for the blank is 0.1 mL

Net volume of HCl used for the sample is 11.85 mL (11.95 mL - 0.1 mL)

Since the concentration of HCl is 0.1 N, we multiply the net volume by the normality to find the milliequivalents (meq) of nitrogen:
11.85 mL * 0.1 N = 1.185 meq

To find the grams of nitrogen, we use the fact that 1 meq of N equals 0.014 g:
1.185 meq * 0.014 g/meq = 0.01659 g of N

To convert grams of N to percent nitrogen in the sample, we divide by the sample mass and multiply by 100:
(0.01659 g / 5.724 g) * 100 = 0.2898 % N

Multiply the percent N by the conversion factor to get percent protein:
 0.2898 % N * 6.25 = 1.8113 % Protein

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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Answer:

low supply of oxaloacetate in the citric acid cycle

Explanation:

When there is low supply of oxalo acetate the acetyl CoA gets converted to ketone bodies to enter the TCA cycle.

In the condition of low level of glucose, the supply of oxaloacetate decreases. Thus making it unavailable to react with acetyl CoA, in this condition ketogenesis occur i.e. acetyl CoA gets converted to ketone bodies.

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:

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:

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 : 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.