When 1.6968 g of an organic iron compound containing Fe, C, H, and O was burned in O2, 3.1737 g of CO2 and 0.90829 g of H2O were produced. In a separate experiment to determine the mass percent of iron, 0.5446 g of the compound yielded 0.1230 g of Fe2O3. What is the empirical formula of the compound?

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: 2800 g

Explanation:

[tex]CaCO_3(s)\rightarrow CaO(s)+CO_2(g)[/tex]

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number [tex]6.023\times 10^{23}[/tex] of particles.

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

Given mass = 5 kg = 5000 g

[tex]\text{Number of moles}=\frac{5000g}{100g/mol}=50moles[/tex]

1 mole of [tex]CaCO_3[/tex] produces = 1 mole of [tex]CaO[/tex]

50 moles of [tex]CaCO_3[/tex] produces =[tex]\frac{1}{1}\times 50=50moles[/tex] of [tex]CaO[/tex]

Mass of [tex]CaO=moles\times {\text{Molar mass}}=50moles\times 56g/mole=2800g[/tex]

2800 g of [tex]CaO[/tex] is produced from 5.0 kg of limestone.

A decomposition reaction splits the reactants into two or more products. The mass of the quicklime produced from 5 kg of limestone is 2800 gm.

Mass is the amount or the weight of the substance occupied in the system. It can be calculated in grams or kilograms.

The decomposition reaction of the Limestone can be shown as:

[tex]\rm CaCO_{3} \rightarrow CaO + CO_{2}[/tex]

The number of the mole of limestone is given as:

[tex]\rm Moles = \rm \dfrac {Mass}{Molar \;mass}[/tex]

Here, mass is 5000 gm and the molar mass is 100 g/mol

Substituting values in the equation above:

[tex]\begin{aligned}\rm n &= \dfrac{5000}{100}\\\\&= 50\;\rm mol\end{aligned}[/tex]

The stoichiometry coefficient of the reaction gives:

1 mole of limestone = 1 mole quicklime

So, 50 moles of limestone = x moles of quicklime

Solving for x:

[tex]\begin{aligned}\rm x &= \dfrac{50 \times 1}{1}\\\\&= 50 \;\rm mole\end{aligned}[/tex]

Mass of quicklime is calculated as:

[tex]\begin{aligned}\rm mass &= \rm moles \times \rm molar \; mass\\\\&= 50 \times 56\\\\&= 2800\;\rm gm\end{aligned}[/tex]

Therefore, 2800 gm of quicklime is produced.

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Answer:

It will take 6.6 hours for 75% of the lead to decay.

Explanation:

The radioactive decay follows first order rate law

The half life and rate constant are related as

[tex]k=rate constant=\frac{0.693}{halflife}=\frac{0.693}{3.3}=0.21h^{-1}[/tex]

The rate law for first order reaction is

[tex]time=\frac{1}{k}(ln[\frac{A_{0}}{A_{t}}][/tex]

Where

A0 = initial concentration = 1 g

At= final concentration = 0.25 g (as 75% undergoes decay so 25% left]

[tex]time=\frac{1}{0.21}(ln(\frac{1}{0.25})=6.6hours[/tex]

The equilibrium pressure of H2 is 0.96 atm and the impossible solution of the quadratic equation is -1.379.

The equilibrium pressure of H2 is calculated by creating ICE table as follows;

            2 N H3 ( g ) ⟷ N2( g ) + 3H2

I:           1                         1              1

C:         -2x                      x             3x

E:        1 - 2x                    1 + x         1 + 3x

[tex]KP = \frac{(N_2)(H_2)^3}{(NH_3)^2} \\\\0.83 = \frac{(1 + x)(1 + 3x)^3}{(1 - 2x)^2}[/tex]

0.83(1 - 2x)² = (1 + x)(1 + 3x)³

0.83(1 - 4x + 4x²) = (1 + x)((1 + 3x)³)

0.83 - 3.32x + 3.32x² = (1 + x)((1 + 3x)³)

0.83 - 3.32x + 3.32x² = 1 + 10x + 36x² + 54x³ + 27x⁴

27x⁴ + 54x³ + 32.68x² + 13.32x + 0.17 = 0

x = -1.379 or - 0.013

H2 = 1 + 3(-1.379)

H2 = -3.13 atm

H2 = 1 + 3(-0.013)

H2 = 0.96 atm

Thus, the equilibrium pressure of H2 is 0.96 atm and the impossible solution of the quadratic equation is -1.379.

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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 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:

27.77 kJ/mol is the activation energy for the reaction.

Explanation:

Rate of the reaction at 323 K =[tex]k_1=k[/tex]

Rate of the reaction at 373 K =[tex]k_2=4k[/tex]

Activation energy for the reaction is calculated by formula:

[tex]\log \frac{k_2}{k_1}=\frac{E_a}{2.303\times R}[\frac{T_2-T_1}{T_2\times T_1}][/tex]

[tex]E_a[/tex] = Activation energy

[tex]T_1[/tex] = Temperature when rate of the reaction was [tex]k_1[/tex]

[tex]T_2[/tex] = Temperature when rate of the reaction was [tex]k_2[/tex]

Substituting the values:

[tex]\log \frac{4k}{k}=\frac{E_a}{2.303\times 8.314 J /mol K}[\frac{373 K-323K}{373 K\times 323 K}][/tex]

[tex]E_a=27,776.98 J/mol=27.77 kJ/mol[/tex]

27.77 kJ/mol is the activation energy for the reaction.

Answer: The correct answer is [tex]4molH_2\times \frac{2molH_2O}{2molH_2}=4 molH_2O[/tex]

Explanation:

We are given:

Moles of hydrogen gas = 4 moles

As, oxygen is given in excess. Thus, is considered as an excess reagent and hydrogen is considered as a limiting reagent because it limits the formation of products.

For the given chemical equation:

[tex]2H_2(g)+O_2(g)\rightarrow 2H_2O(l)[/tex]

By Stoichiometry of the reaction:

2 moles of hydrogen produces 2 moles of water molecule.

So, 4 moles of hydrogen will produce = [tex]\frac{2molH_2O}{2molH_2}\times 4molH_2=4mol[/tex] of water.

Hence, the correct answer is [tex]4molH_2\times \frac{2molH_2O}{2molH_2}=4 molH_2O[/tex]

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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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: 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:

Rate = 1.09*10^-3 (mol H2/L)/s

Explanation:

Given:

Initial concentration of H2, C1 = 0 M

Final concentration of H2, C2 = 0.101 M

Time taken, t = 93.0 s

To determine:

The rate of the given reaction

Calculation:

The decomposition of PH3 is represented by the following chemical reaction

[tex]4 PH3(g)\rightarrow P4(g) + 6 H2(g)[/tex]

Reaction rate in terms of the appearance of H2 is given as:

[tex]Rate = +\frac{1}{6}*\frac{\Delta [H2]]}{\Delta t}[/tex]

[tex]Rate = +\frac{1}{6}*\frac{C2[H2]-C1[H2]}{\Delta t}[/tex]

Here C1(H2) = 0 M and C2(H2) = 0.101 M

Δt = 93.0 s

[tex]Rate = \frac{(0.101-0.0)M}{93.0 s} =1.09*10^{-3} M/s[/tex]

Since molarity M = mole/L

rate = 1.09*10^-3 (mol H2/L)/s

Final answer:

Without specific numeric values for the individual atomic heat capacities of the elements in sodium carbonate, directly answering the heat capacity multiple-choice question using Kopp's rule is not feasible. Additional data is required to estimate the heat capacity of solid sodium carbonate.

Explanation:

The question regarding the heat capacity of solid sodium carbonate (Na2CO3) can be answered by applying Kopp's rule, which is a method used to estimate the heat capacities of solids. According to Kopp's rule, the heat capacity of a compound in the solid state is the sum of the atomic heat capacities of the individual elements that compose the compound. Atomic heat capacities can be estimated using the Law of Dulong and Petit, which suggests that at room temperature, the molar heat capacity of many solid elements roughly equates to 3R, where R is the gas constant with a value approximately equal to 8.314 J/(mol·K).

Applying Kopp's rule would involve calculating the total molar heat capacity for sodium, carbon, and oxygen in sodium carbonate, based on their atomic weights and expected contributions. However, since specific numeric values are not provided in the question or reference information to apply Kopp's rule directly, directly answering the multiple-choice question is not feasible without additional data on the individual atomic heat capacities of the elements involved in sodium carbonate.

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:

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:

4600 Liters NH₃(g)

Explanation:

Applying Avogadro's law to the reaction of hydrogen and nitrogen gas forming ammonia, we find that 6.9 m3 of hydrogen will form about 4.6 m3 of ammonia, assuming that temperature and pressure remain constant.

The subject in question relates to the application of Avogadro's law to chemical reactions involving gases. More specifically, we're investigating the reaction between hydrogen gas and nitrogen gas to produce ammonia gas. As per Avogadro's law, gases react in definite and simple proportions by volume, if all gas volumes are measured at the same temperature and pressure.

Focusing on the hydrogen to ammonia conversion, the reaction equation N₂(g) + 3H₂(g) turns into 2NH3(g). We can surmise that three volumes of hydrogen gas (H2) react to form two volumes of ammonia gas (NH3). Considering that 6.9 m3 of hydrogen gas is consumed, by the rule of three, we can infer that the reaction would result in approximately 4.6 m3 of ammonia gas.

Bear in mind, these calculations are assuming that the temperature and pressure remain constant during the reaction and that it goes to completion.

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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 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: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 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.

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:

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: 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 final volume of oxygen gas will be, 1.292 ml

Explanation :

Charles' Law : It is defined as the volume of gas is directly proportional to the temperature of the gas at constant pressure and number of moles.

[tex]V\propto T[/tex]

or,

[tex]\frac{V_1}{V_2}=\frac{T_1}{T_2}[/tex]

where,

[tex]V_1[/tex] = initial volume of oxygen gas = 1.75 ml

[tex]V_2[/tex] = final volume of oxygen gas = ?

[tex]T_1[/tex] = initial temperature of gas = [tex]97.8^oC=273+97.8=370.8K[/tex]

[tex]T_2[/tex] = final temperature of gas = [tex]0.7^oC=273+0.7=273.7K[/tex]

Now put all the given values in the above formula, we get the final volume of the oxygen gas.

[tex]\frac{1.75ml}{V_2}=\frac{370.8K}{273.7K}[/tex]

[tex]V_2=1.292ml[/tex]

Therefore, the final volume of oxygen gas will be, 1.292 ml

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:

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:

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]

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.