When solid (NH4)(NH2CO2)(NH4)(NH2CO2) is introduced into and evacuated flask at 25∘C25∘C, the total pressure of gas at equilibrium is 0.30.3 atm. What is the value of Kpat25∘CKpat25∘C?

Answer: The value of [tex]K_p[/tex] for the reaction is 0.169

Explanation:

We are given:

Initial partial pressure of A = 1.00 atm

Initial partial pressure of B = 1.00 atm

The given chemical equation follows:

                  [tex]A(g)+2B(g)\rightleftharpoons C(g)+D(g)[/tex]

Initial:         1.00     1.00

At eqllm:     1-x      1-2x           x        x

We are given:

Equilibrium partial pressure of C = 0.211 atm = x

So, equilibrium partial pressure of A = (1.00 - x) = (1.00 - 0.211) = 0.789 atm

Equilibrium partial pressure of B = (1.00 - 2x) = (1.00 - 2(0.211)) = 0.578 atm

Equilibrium partial pressure of D = x = 0.211 atm

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

[tex]K_p=\frac{p_C\times p_D}{p_A\times (p_B)^2}[/tex]

Putting values in above equation, we get:

[tex]K_p=\frac{0.211\times 0.211}{0.789\times (0.578)^2}\\\\K_p=0.169[/tex]

Hence, the value of [tex]K_p[/tex] for the reaction is 0.169

Final answer:

To find the number of moles of gas in the bicycle tire, we can use the ideal gas law equation: PV = nRT. Given the pressure and volume of the gas, we can solve for n using the equation n = PV / RT. The number of moles of gas in the bicycle tire is approximately 0.185 moles.

Explanation:

To find the number of moles of gas in the bicycle tire, we can use the ideal gas law equation:

PV = nRT

Where:

Given that the pressure is 765 torr (which is equivalent to 1.01 atm) and the volume is 5.65 L, we can rearrange the equation to solve for n:

n = PV / RT

Substituting the given values, we get:

n = (1.01 atm) x (5.65 L) / (0.0821 L·atm/mol·K x (25 + 273) K)

Simplifying, we find that the number of moles of gas in the bicycle tire is approximately 0.185 moles.

Answer: The equilibrium concentration of [tex]H_3O^+[/tex] ion is [tex]8.3064\times 10^{-2}M[/tex]

Explanation:

We are given:

Molarity of oxalic acid solution = 0.20 M

Oxalic acid [tex](H_2C_2O_4)[/tex] is a weak acid and will dissociate 2 hydrogen ions.

               [tex]H_2C_2O_4(aq.)+H_2O\rightleftharpoons H_3O^+(aq.)+HC_2O_4^-(aq.)[/tex]

Initial:        0.20

At eqllm:    0.20-x                            x                 x

The expression of first equilibrium constant equation follows:

[tex]Ka_1=\frac{[H_3O^+][HC_2O_4^{-}]}{[H_2C_2O_4]}[/tex]

We know that:

[tex]Ka_1\text{ for }H_2C_2O_4=0.059[/tex]

Putting values in above equation, we get:

[tex]0.059=\frac{x\times x}{(0.20-x)}\\\\x=-0.142,0.083[/tex]

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of hydronium ion = x = 0.083 M

                 [tex]HC_2O_4^-(aq.)+H_2O\rightarrow H_3O^+(aq.)+C_2O_4^{2-}(aq.)[/tex]

Initial:         0.083  

At eqllm:    0.083-y                      0.083+y               y

The expression of second equilibrium constant equation follows:

[tex]Ka_2=\frac{[H_3O^+][C_2O_4^{2-}]}{[HC_2O_4^-]}[/tex]

We know that:

[tex]Ka_2\text{ for }H_2C_2O_4=6.4\times 10^{-5}[/tex]

Putting values in above equation, we get:

[tex]6.4\times 10^{-5}=\frac{(0.083+y)\times y}{(0.083-y)}\\\\y=-0.083,0.0000639[/tex]

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of hydronium ion = y = 0.0000639 M

Total concentration of hydronium ion = [x + y] = [0.083 + 0.0000639] = 0.0830639 M

Hence, the equilibrium concentration of [tex]H_3O^+[/tex] ion is [tex]8.3064\times 10^{-2}M[/tex]

The equilibrium concentration of H3O+ in a solution of oxalic acid is linked to the acid dissociation constant Ka. Calculation of this requires specific values related to the nature of oxalic acid, a weak acid, and its behaviour in solution.

In order to find the equilibrium concentration of H3O+ in a solution of oxalic acid, we must first understand that oxalic acid is a weak acid, and thus does not fully ionize in solution. In the ionization process of oxalic acid H2C2O4, it would donate a proton (H+) to water (H2O), forming a hydronium ion (H3O+) and a mono hydrogen oxalate ion. The equilibrium concentration of H3O+ (hydronium ions) thus corresponds to the acid dissociation constant Ka, with the equation Ka=[H3O+][HC2O4-]/[H2C2O4]. Considering the initial concentration of the oxalic acid, specific calculations will be required to find an accurate equilibrium concentration which is not possible without the value of Ka. So, the specific answer depends on the Ka value of oxalic acid.

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Answer:

c) Boiling point of the reaction mixture (reflux temperature)

Explanation:

Hello,

At first, it is important to remember that esterification is an organic chemical reaction related with the neutralization of organic acids and alcohols to form esters, in this case from 1-butanol and acetic acid to n-butyl acetate as shown below:

[tex]CH_3COOH+CH3(CH_2)2CH_2OH \rightleftharpoons CH_3COOCH_2(CH_2)2CH3+H_2O[/tex]

It is shown that is a reaction which equilibrium condition is present since the n-butyl acetate is likely to come back to the acetic acid and the 1-butanol. Moreover, it is necessary to catalyze esterification with sulfuric acid and including constant heating and stirring, nonetheless, such heating induces boiling of the reacting mixture containing the acetic acid and the 1-butanol which are likely to boil. Therefore, reflux must be implemented as shown on the attached picture to prevent reactant lost which shift the reaction leftwards, diminishing n-butyl acetate yield, thus, the most accurately way to describe the actual temperature is c) boiling point of the reaction mixture (reflux temperature) since acetic acid and 1-butanol have a composition which modifies their boiling point into an only one that is the mixture's boiling point which is also related with the temperature at which the reflux is performed.

Best regards.

The option that best describes most accurately the actual, internal reaction temperature is Boiling point of the reaction mixture (reflux temperature).

It is said to be a very hard task when one is trying to monitor and control the temperature of a reaction chamber when a person do not have expensive and well equipped laboratory tools.

People often uses phase when the above is not in place as it is Phase said to be a form of melting or boiling occur at particular temperatures, and all heat exchanged are said to be done in a phase transition goes into the phase transition.

The main reason of refluxing a solution is done so as to heat a solution in a manner where one can control the outcome at a constant temperature and this is the option that is best for the Fischer esterification experiment of 1-butanol with acetic acid to create n-butyl acetate.

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Answer:

[tex]S_{N}i[/tex] is the major step in forming acid chloride from carboxylic acid and thionyl chloride

Explanation:

Answer:

We need 69 grams of NaHCO3

Explanation:

Step 1: Data given

Volume = 550 mL = 0.550 L

Molarity H2SO4 = 0.75 M

Step 2: The balanced equation

H2SO4(aq) + 2NaHCO3(aq) → Na2SO4(aq) + 2H2O(l) + 2CO2(g)

Step 3: Calculate moles of H2SO4

Moles H2SO4 = molarity * volume

Moles H2SO4 = 0.75 M * 0.550 L

Moles H2SO4 =  0.4125 moles H2SO4

Step 4: Calculate moles NaHCO3

For 1 mol H2SO4 we need 2 moles NaHCO3 to produce 1 mol Na2SO4 and 2 moles H2O and 2 Moles CO2

For 0.4125 moles H2SO4 we need 2*0.4125 = 0.825 moles NaHCO3

Step 5: Calculate mass NaHCO3

Mass NaHCO3 = moles * molar mass

Mass NaHCO3 = 0.825 moles * 84.0 g/mol

Mass NaHCO3 = 69.3 grams ≈ 69 grams

We need 69 grams of NaHCO3

Answer: The value of [tex]\Delta S^o[/tex] for the surrounding when given amount of CO gas is reacted is 197.77 J/K

Explanation:

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate entropy change is of a reaction is:

[tex]\Delta S^o_{rxn}=\sum [n\times \Delta S^o_{(product)}]-\sum [n\times \Delta S^o_{(reactant)}][/tex]

For the given chemical reaction:

[tex]2CO(g)+2NO(g)\rightarrow 2CO_2(g)+N_2(g)[/tex]

The equation for the entropy change of the above reaction is:

[tex]\Delta S^o_{rxn}=[(2\times \Delta S^o_{(CO_2(g))})+(1\times \Delta S^o_{(N_2(g))})]-[(2\times \Delta S^o_{(CO(g))})+(2\times \Delta S^o_{(NO(g))})][/tex]

We are given:

[tex]\Delta S^o_{(CO_2(g))}=213.74J/K.mol\\\Delta S^o_{(N_2(g))}=191.61J/K.mol\\\Delta S^o_{(CO(g))}=197.67J/K.mol\\\Delta S^o_{(NO(g))}=210.76J/K.mol[/tex]

Putting values in above equation, we get:

[tex]\Delta S^o_{rxn}=[(2\times (213.74))+(1\times (191.61))]-[(2\times (197.67))+(2\times (210.76))]\\\\\Delta S^o_{rxn}=-197.77/K[/tex]

Entropy change of the surrounding = - (Entropy change of the system) = -(-197.77) J/K = 197.77 J/K

We are given:

Moles of CO gas reacted = 2.00 moles

By Stoichiometry of the reaction:

When 2 mole of CO gas is reacted, the entropy change of the surrounding will be 197.77 J/K

So, when 2.00 moles of CO gas is reacted, the entropy change of the surrounding will be = [tex]\frac{197.77}{2}\times 2.00=197.77J/K[/tex]

Hence, the value of [tex]\Delta S^o[/tex] for the surrounding when given amount of CO gas is reacted is 197.77 J/K

The entropy change for the surroundings when 2.00 moles of CO(g) react at standard conditions is -96.94 J/K.

To calculate the entropy change for the surroundings, we can use the following equation:

ΔSsurroundings = -ΔSsystem - ΔSuniverse

where ΔSsystem is the entropy change of the system and ΔSuniverse is the entropy change of the universe.

The entropy change of the system can be calculated from the standard molar entropies of the reactants and products:

ΔSsystem = ΣS°products - ΣS°reactants

The standard molar entropies of the reactants and products can be found in a standard thermodynamics data table. For the reaction given in the question, the standard molar entropies are as follows:

| Species | S° (J/mol·K) |

|---|---|---|

| CO(g) | 197.69 |

| NO(g) | 210.76 |

| CO2(g) | 213.64 |

| N2(g) | 191.50 |

Substituting these values into the equation for ΔSsystem, we get:

ΔSsystem = (2 mol × 213.64 J/mol·K) + (1 mol × 191.50 J/mol·K) - (2 mol × 197.69 J/mol·K) - (2 mol × 210.76 J/mol·K)

ΔSsystem = -96.94 J/K

The entropy change of the universe is always positive for a spontaneous process. Since the reaction given in the question is spontaneous, the entropy change of the universe is positive. Therefore, the entropy change for the surroundings is negative:

ΔSsurroundings = -ΔSsystem - ΔSuniverse

ΔSsurroundings = -(-96.94 J/K) - (+)

ΔSsurroundings = -96.94 J/K

Therefore, the entropy change for the surroundings when 2.00 moles of CO(g) react at standard conditions is -96.94 J/K.

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D. Dalton's Law

Explanation:

As the pressure of the gas is related to the sum of the partial pressures of the individual gases present in the mixture was explained by Dalton's law, the given system is an example of Dalton's law.

Boyle's law relates the inverse proportionality of volume and pressure of an ideal gas.

Charles's Law reveals the direct relationship of temperature and volume of an ideal gas.

Avogadro's Law states the relationship between the volume of gas and number of molecules at same pressure as well as temperature.

(D) Dalton's Law states that the pressure of the mixture is found by adding the pressures of the two individual gases.

Dalton’s Law also known as the Law of Partial Pressures states that in a mixture of non-reacting gases the total pressure exerted is equal to the sum of the partial pressures of the gases in the mixture.

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Answer: The volume of NaOH required is 22.39 mL

Explanation:

We are given:

Mass of sample = 0.6543 g

Mass percent of oxalic acid = 53.66 %

This means that 53.66 grams of oxalic acid is present in 100 grams of sample

Mass of oxalic acid in the given amount of sample = [tex]\frac{53.66}{100}\times 0.6543=0.351g[/tex]

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

Given mass of oxalic acid = 0.351 g

Molar mass of oxalic acid = 90 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of oxalic acid}=\frac{0.351g}{90g/mol}=0.0039mol[/tex]

The chemical equation for the reaction of oxalic acid and NaOH follows:

[tex]C_2H_2O_4+2NaOH\rightarrow Na_2C_2O_4+2H_2O[/tex]

By Stoichiometry of the reaction:

1 mole of oxalic acid reacts with 2 moles of NaOH

So, 0.0039 moles of oxalic acid will react with = [tex]\frac{2}{1}\times 0.0039=0.0078mol[/tex] of NaOH

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

Moles of NaOH = 0.0078 moles

Molarity of solution = 0.3483 M

Putting values in above equation, we get:

[tex]0.3483M=\frac{0.0078\times 1000}{V}\\\\V=\frac{0.0078\times 1000}{0.3483}=22.39mL[/tex]

Hence, the volume of NaOH required is 22.39 mL

To determine the volume of NaOH needed to neutralize a 0.6543-g sample of a solid containing 53.66% oxalic acid, we calculate the moles of oxalic acid in the sample and use stoichiometry to determine the moles of NaOH required for neutralization. Finally, we use the formula for molarity to find the volume of NaOH needed.

Oxalic acid, H2C2O4, is a diprotic acid, meaning it can donate two protons (H+) in an acid-base reaction. To determine the volume of 0.3483 M NaOH needed for neutralization of a 0.6543-g sample of the solid containing 53.66% oxalic acid by mass, we need to first calculate the moles of oxalic acid in the sample.

First, we calculate the moles of H2C2O4 in the sample:

Moles of H2C2O4 = mass of H2C2O4 / molar mass of H2C2O4

Next, we use stoichiometry to calculate the moles of NaOH required for neutralization:

Moles of NaOH = 2 * moles of H2C2O4

Finally, we use the formula for molarity to determine the volume of NaOH needed:

The volume of NaOH (L) = moles of NaOH / molarity of NaOH

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Answer:

The molar solubility of the metal thiocyanate is [tex]1.127\times 10^{-4} M[/tex].

Explanation:

Concentration of potassium thiocyanate = 0.421 M

Concentration of thiocyanate ion =[tex][SCN^-]= 0.421 M[/tex]

Concentration of metal ion =[tex][M^{2+}]= ?[/tex]

The solubility product of metal thiocyanate = [tex]K_{sp}=2.00\times 10^{-5}[/tex]

[tex]M(SCN)_2\rightleftharpoons M^{2+}+2SCN^-[/tex]

                          S          2S

At equilbrium

                         S       (2S+0.421)

The expression of solubility product is given by :

[tex]K_{sp}=[M^{2+}]\times [SCN^-]^2[/tex]

[tex]2.00\times 10^{-5}=S\times (2S+0.421)^2[/tex]

Solving for S:

[tex]S = 1.128\times 10^{-4} M[/tex]

[tex][M^{2+}]=\frac{2.00\times 10^{-5}}{(0.421 M)^2}=1.127\times 10^{-4} M[/tex]

The molar solubility of the metal thiocyanate is [tex]1.127\times 10^{-4} M[/tex].

Answer:

98.6 g/mol.

Explanation:

Equation of the reaction

HX + NaOH--> NaX + H2O

Number of moles = molar concentration × volume

= 0.095 × 0.03

= 0.00285 moles

By stoichiometry, 1 mole of HX reacted with 1 mole of NaOH. Therefore, number of moles of HX = 0.00285 moles.

Molar mass = mass ÷ number of moles

= 0.281 ÷ 0.00285

= 98.6 g/mol.

Answer: The complete ionic equation is written below.

Explanation:

Complete ionic equation is defined as the equation in which all the substances that are strong electrolyte are present in an aqueous are represented in the form of ions.

The balanced molecular equation for the reaction of lead (II) nitrate and potassium sulfate follows:

[tex]Pb(NO_3)_2(aq.)+K_2SO_4(aq.)\rightarrow 2KNO_3(aq.)+PbSO_4(s)[/tex]

The complete ionic equation for the above equation is:

[tex]2Pb^{2+}(aq.)+2NO_3^{-}(aq.)+2K^{+}(aq.)+SO_4^{2-}(aq.)\rightarrow 2K^+(aq.)+2NO^{3-}(aq.)+PbSO_4(s)[/tex]

Hence, the complete ionic equation is written above.

Answer:

15.95 g

Explanation:

Calculation of the moles of sulfur as:-

Mass = 3.5 g

Molar mass of sulfur = 32.065 g/mol

The formula for the calculation of moles is shown below:

[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]

Thus,

[tex]Moles= \frac{3.5\ g}{32.065\ g/mol}[/tex]

[tex]Moles= 0.1092\ mol[/tex]

From the reaction,

[tex]S+3F_2\rightarrow SF_6[/tex]

1 mole of sulfur on reaction forms 1 mole of sulfur hexafluoride

0.1092 mole of sulfur on reaction forms 0.1092 mole of sulfur hexafluoride

Molar mass of sulfur hexafluoride = 146.06 g/mol

Mass= Moles*Molar mass = 0.1092*146.06 g = 15.95 g

15.95 g is the maximum amount of [tex]SF_6[/tex] that can be produced from the reaction of 3.5 g of sulfur with of fluorine.

The maximum amount of SF6 that can be produced from the reaction of 3.5 g of sulfur with fluorine is 47.7 g.

The question is asking about the maximum amount of sulfur hexafluoride (SF6) produced from the reaction of 3.5 g of sulfur with an unknown amount of fluorine. To determine the maximum amount, we need to calculate the limiting reactant and use its stoichiometry to find the amount of SF6 produced. The balanced equation for the reaction is:

S (s) + 3F2 (g) → SF6 (g)

First, we need to find the molar mass of sulfur (S) and calculate the moles of sulfur:

Molar mass of sulfur (S) = 32.06 g/mol

Moles of sulfur = 3.5 g / 32.06 g/mol = 0.109 mol

Next, we need to find the molar mass of fluorine (F2) and calculate the moles of fluorine:

Molar mass of fluorine (F2) = 38.00 g/mol

Using the stoichiometry of the balanced equation, we can see that 1 mole of sulfur reacts with 3 moles of fluorine to produce 1 mole of SF6. Therefore, the moles of fluorine needed to react with 0.109 mol of sulfur is:

Moles of fluorine = 3 moles of fluorine/mol of sulfur × 0.109 mol of sulfur = 0.327 mol

Finally, we can use the moles of fluorine and the molar mass of SF6 to calculate the mass of SF6 produced:

Molar mass of SF6 = 146.06 g/mol

Mass of SF6 produced = moles of SF6 × molar mass of SF6 = 0.327 mol × 146.06 g/mol = 47.7 g (rounded to two decimal places)

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Answer:

The volume CO2 produced is 65.8 L

Explanation:

Step 1: Data given

Mass of octane = 41.9 grams

Molar mass octane = 114.23 g/mol

Step 2: The balanced equation

2C8H18 + 25O2 → 16CO2 + 18H2O

Step 3: Calculate moles octane

Moles octane = mass octane / molar mass octane

Moles octane = 41.9 grams / 114.23 g/mol

Moles octane = 0.367 moles

Step 4: Calculate moles CO2

For 2 moles octane we need 25 moles O2 to produce 16 moles CO2 and 18 moles H2O

For 0.367 moles octane we need 8*0.367 = 2.936 moles

Step 5: Calculate volume of CO2

1 mol = 22.4 L

2.936 moles = 22.4 * 2.936 = 65.8 L

The volume CO2 produced is 65.8 L

Amino acids that can donate hydrogen bonds include -

With its -NH group, Tryptophan can serve as a hydrogen-bond donor, but its aromatic ring can also serve as an acceptor.

The side chains of three amino acids—arginine, lysine, and tryptophan—contain hydrogen donor atoms.

The side chains of 2 amino acids (aspartic acid and glutamic acid) include hydrogen acceptor atoms.

The side chains of six amino acids—asparagine, glutamine, histidine, serine, threonine, and tyrosine—contain both hydrogen donor and acceptor atoms.

The side chain of threonine has the capacity to serve as a hydrogen bond donor and acceptor. An oxygen atom that is part of the side chain of the amino acid threonine has two possible hydrogen bonding roles: acceptor and donor.

Therefore, from the given list of amino acids, three amino acids can act as hydrogen donor, these include – a. threonine, e. arginine, f. tryptophan.

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At pH 7, the amino acids that can form a hydrogen bond with progesterone are threonine, aspartate, and arginine. Threonine has a polar uncharged group while aspartate and arginine carry a charge, allowing them to participate in hydrogen bonding.

The amino acids that have side chains capable of forming a hydrogen bond with progesterone at pH 7 include threonine, aspartate, and arginine. These amino acids have side chains with polar, uncharged groups or charged groups. To be specific, threonine has a polar, uncharged hydroxyl group which can act as a hydrogen donor or acceptor. Aspartate carries a negative charge at pH 7 and can act as a hydrogen bond acceptor.

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Answer:

113 mL

Explanation:

Let's apply the Ideal Gases Law for this propose:

P . V = n . R .T → (n. R . T) / P = V

We need to convert the T° to Absolute T° and pressure from mmHg to atm

T° K = 30°C + 273 = 303K

485 mmHg . 1atm / 760 mmHg = 0.638 atm

Let's replace the information obtained:

V = (n . 0.082 . 303K) / 0.638 atm

n = number of moles → 0.128 g . 1mol / 44g = 0.00291 moles

V = (0.00291 moles . 0.082 . 303K) / 0.638 atm → 0.113 L

The value can be written as 113 mL

Answer: The symbol of the ion formed is [tex]Ca^{2+}[/tex]

Explanation:

An ion is formed when a neutral atom looses or gains electrons.

Electronic configuration is defined as the representation of electrons around the nucleus of an atom.

Number of electrons in an atom is determined by the atomic number of that atom.

The element present in Group 2-A and in period 4 is Calcium (Ca)

Electronic configuration of Ca atom:  [tex]1s^22s^22p^63s^23p^64s^2[/tex]

This atom will loose 2 electrons to attain stable electronic configuration similar to Argon element (noble gas)

The electronic configration of [tex]Ca^{2+}\text{ ion}=1s^22s^22p^63s^23p^6[/tex]

Hence, the symbol of the ion formed is [tex]Ca^{2+}[/tex]

Answer: The octet rule says that in forming ions, main-group atoms gain or lose electrons in order to attain a noble gas electron configuration. Except for the He  configuration, this means that atoms gain or lose electrons to have an octet of electrons in the outermost shell.

Explanation: If x is in period 4, the ion formed has the same electron configuration as the noble ga

Answer:

D) the energy associated with the random motion of atoms and molecules.

Explanation:

Thermal energy is the manifestation of energy in the form of heat. In all materials, the atoms that make up their molecules are in continuous movement, either moving or vibrating, which implies that the atoms have a certain kinetic energy that we call heat or thermal energy. In a way, thermal energy is the internal energy of a body.

Final answer:

Thermal energy is the energy associated with the random motion of atoms and molecules, a type of kinetic energy that increases with object's temperature, making Option D the correct choice.

Explanation:

The question poses multiple options for defining thermal energy. Option A suggests solar energy, which isn't correct as solar energy is a form of radiant energy. Option B refers to the energy within chemical substances, known as chemical energy. Option C discusses energy by virtue of an object's position, commonly known as potential energy. The correct answer is Option D, which states that thermal energy is the energy associated with the random motion of atoms and molecules.

More specifically, thermal energy is related to the kinetic energy resulting from the random translational motions of particles and also includes vibrational and rotational energies within molecules. An increase in this motion, as measured by temperature, corresponds to an increase in thermal energy. Hence, thermal energy is a form of internal energy of an object, manifesting as the microscopic motion of its constituent particles.

Answer:

Δ S = 93.8 J/mol-K

Explanation:

Given,

Boiling point of chloroform = 61.7 °C

                                             = 273 + 61.7 = 334.7 K.

Enthalapy of vapourization = 31.4 kJ/mol.

Using Gibbs free   energy equation

Δ G = Δ H - T (ΔS)

at equilibrium (when the liquid is boiling), Δ G = 0

so,  0 = ΔH - T (Δ S)

      T (Δ S) = Δ H

and ΔS = ΔH / T

Δ S = (31400 J/mol.) / 334.7 K

Δ S = 93.8 J/mol-K

Final answer:

The molar entropy of vaporization (ΔvapS) of acetone can be calculated using the formula ΔvapS = ΔvapH / T. By converting the boiling point to Kelvin and substituting the values, we find ΔvapS to be 88.4 J/mol·K. The positive sign of the entropy change is expected as vaporization increases disorder.

Explanation:

The student's question pertains to the calculation of the molar entropy of vaporization (ΔvapS) of acetone at its normal boiling point. Given the normal boiling point of acetone is 56°C, and its molar enthalpy of vaporization (ΔvapH) is 29.1 kJ/mol, we can calculate the molar entropy of vaporization using the equation:

ΔvapS = ΔvapH / T

Where T is the absolute temperature in Kelvin (K). To obtain T, convert the boiling point from Celsius to Kelvin:

T = 56°C + 273.15 = 329.15 K

Now substituting the values into the equation:

ΔvapS = 29.1 kJ/mol / 329.15 K

ΔvapS = 0.0884 kJ/mol·K or 88.4 J/mol·K

The sign for entropy of vaporization is positive, which makes sense since the entropy of the system is expected to increase when a liquid turns into a gas due to the increase in disorder.

Answer:

(a) The bellow flow chart shows that the system has 0° of freedom.

(b) i - Fractional conversion of menthol: 0.960 mol CH₃OH reacted/mol fed

    ii - The percentage excess air fed: 28.5%

   iii - molecular fraction of water in the product gas: 0.178 mol H₂O/mol

(c)  Potential Problems: Remedies

    Conflagration: The gas should be vented outside to prevent fire outbreak.

    Toxicity: Put a CO detection alarm in the room.

Explanation:

See picture for the flow chat and calculation of other answers.

A space heater fed with liquid methanol and burned with excess air is analyzed to determine its fractional conversion, percentage excess air fed, and mole fraction of water. The system has zero degrees of freedom. Potential problems when the combustion products are released directly into a room include the release of harmful gases and increased humidity.

To determine if a system has zero degrees of freedom, we need to analyze the material balance and the component balance. In this case, the flow chart shows that the only input is the liquid methanol and the only outputs are the product gases. Since there are no unknown variables or degrees of freedom in the system, we can conclude that the system has zero degrees of freedom.

To calculate the fractional conversion of methanol, we can use the mole percentages of CO2 and CO in the product gas. The fractional conversion is the difference between the initial mole percentage of methanol and the mole percentage of CO and CO2. The percentage excess air fed can be calculated by comparing the mole percentage of O2 in the product gas to the stoichiometric requirement. Finally, the mole fraction of water in the product gas can be found by subtracting the sum of the mole percentages of other components from 100%.

When the combustion products are released directly into a room, potential problems include the release of harmful gases such as CO and the increase in humidity due to the presence of water vapor. To remedy these problems, it is important to ensure proper ventilation and monitoring of indoor air quality. It is also advisable to use a flue or chimney to exhaust the combustion products safely.

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Explanation:

The cross-sectional area of the specimen is calculated as follows.

         [tex]A_{o} = \frac{pi}{4} d^{2}[/tex]

                     = [tex]\frac{3.14}{4} \times (\frac{4.4}{1000})^{2}[/tex]

                     = [tex]1.5197 \times 10^{-5} m^{2}[/tex]

Equation of stress is as follows.

              [tex]\sigma = \frac{F}{A_{o}}[/tex]

And, the equation of strain is as follows.

             [tex]\epsilon = \frac{\Delta l}{l_{o}}[/tex]

Hence, the Hook's law is as follows.

              E = [tex]\frac{\sigma}{\epsilon}[/tex]

       E = [tex]\frac{\frac{F}{A_{o}}}{\frac{\Delta l}{l_{o}}}[/tex]

          = [tex]\frac{F \times l_{o}}{A_{o} \times \Delta l}[/tex]

or,    [tex]l_{o} = \frac{E \times \Delta l \times A_{o}}{F}[/tex]          

                   = [tex]\frac{129 \times 10^{9} \times \frac{0.48}{1000} \times 1.662 \times 10^{-5}}{1570}[/tex]

                  = 0.6554 m

or,         [tex]l_{o}[/tex] = 655.4 mm

Thus, we can conclude that the maximum length of the specimen before deformation if the maximum allowable elongation is 0.48 mm is 655.4 mm.

Answer:2py + 2py -----> Ï2py + Ï*2py

2px + 2px -----> Ï2px + Ï*2px

Explanation:

Molecular orbitals are constructed from atomic orbitals by linear combination of atomic orbitals. The 1s, 2s and 2pz orbitals overlap in an end to end manner hence they only form sigma bonding and anti bonding orbitals. The 2px and 2py orbitals overlap side by side and form pi bonding and anti bonding orbitals. Hence the answer.

Answer: The value of equilibrium constant for the given reaction at 517 K is 1.30

Explanation:

The chemical equation for the dissociation of [tex]CH_2Cl_2[/tex] follows:

[tex]2CH_2Cl_2(g)\rightleftharpoons CH_4(g)+CCl_4(g)[/tex]

The expression of [tex]K_{eq}[/tex] for above equation follows:

[tex]K_{eq}=\frac{[CH_4][CCl_4]}{[CH_2Cl_2]^2}[/tex]

We are given:

[tex][CH_4]_{eq}=3.69\times 10^{-2}M[/tex]

[tex][CCl_4]_{eq}=4.12\times 10^{-2}M[/tex]

[tex][CH_2Cl_2]_{eq}=3.42\times 10^{-2}M[/tex]

Putting values in above expression, we get:

[tex]K_{eq}=\frac{(3.69\times 10^{-2})\times (4.12\times 10^{-2})}{(3.42\times 10^{-2})^2}\\\\K_{eq}=1.30[/tex]

Hence, the value of equilibrium constant for the given reaction at 517 K is 1.30

Answer:

HC₂H₃O₂/KC₂H₃O₂

Explanation:

Considering the Henderson- Hasselbalch equation for the calculation of the pH of the basic buffer solution as:

[tex] pH=pK_b+log\frac{[salt]}{[acid]} [/tex]

For a best pair, the pKa value must be equal to pH.

NH₃/NH₄Cl forms a basic buffer and cannot account for pH = 5

out of the acidic buffer given,

So, HF , Ka = 3.5 × 10⁻⁴ , So pKa = 3.46

HC₂H₃O₂ , Ka = 1.8 × 10⁻⁵ , So pKa = 4.77

The best pair to show pH = 5 is HC₂H₃O₂/KC₂H₃O₂

The pair for the best choice for a pH 5 buffer is:

HC₂H₃O₂/KC₂H₃O₂

The equation that is used for calculation of the pH of the basic buffer solution as:

pH= pkb + log [salt]/ [acid]

For a best pair, the pKa value must be equal to pH.

NH₃/NH₄Cl forms a basic buffer and cannot account for pH = 5

Out of the acidic buffer given,

So, HF , Ka = 3.5 × 10⁻⁴ , So pKa = 3.46

HC₂H₃O₂ , Ka = 1.8 × 10⁻⁵ , So pKa = 4.77

The best pair to show pH = 5 is HC₂H₃O₂/KC₂H₃O₂

Find more information about pH here:

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Answer:

[tex]P_2=350\ kPa[/tex]

[tex]T_2=59^{\circ}\ C[/tex]

Explanation:

Given that

mass , m = 4 kg

Initial pressure ,[tex]P_1=700\ kPa[/tex]

Initial temperature ,[tex]T_1=59^{\circ}\ C[/tex]

The volume of rigid tanks are same

[tex]V_1=V[/tex]

[tex]V_2=2 V[/tex]

Let's take final temperature[tex] =T_2[/tex]

Given that tank is insulated that is why heat transfer in the tank will be zero.

By using energy balance

[tex]E_{in}-E_{out}=\Delta U[/tex]

[tex]\Delta U[/tex]= Change in the internal energy of the gas

[tex]0 = m C_V(T_2-T_1[/tex])           ( Cv=Specific heat capacity at constant volume)

[tex]0 = T_2-T_1[/tex]

Therefore [tex]T_1=T_2[/tex]

[tex]T_2=59^{\circ}\ C[/tex]

We know that ideal gas equation for gas

P V = m R T

P=pressure ,V=Volume ,m=mass ,R= gas constant ,T=temperature

By using mass conservation

[tex]m=\dfrac{P_1V_1}{RT_1}=\dfrac{P_2V_2}{RT_2}[/tex]

Now by putting the values in the above equation

[tex]\dfrac{700\times V}{RT_1}=\dfrac{P_2\times 2V}{RT_1}[/tex]

[tex]P_2=\dfrac{700}{2}\ kPa[/tex]

[tex]P_2=350\ kPa[/tex]

Therefore the final volume will be 350 kPa and temperature will be 59°C.

The final temperature in the tank is approximately 332.15K. When the volume doubles, the pressure is halved, yielding a final pressure of approximately 350 kPa.

In this problem, your task is to determine the final temperature and pressure in a tank after an ideal gas is allowed to expand. Given the system in question is both insulated (adiabatic) and rigid, we can infer that neither heat (Q) nor work (W) is done, as indicated by the equation AEint=Q-W = 0.

Furthermore, as the internal energy does not change, this implies that the temperature will remain constant. So, the final temperature in the tank is the same as the initial, 59°C. We must convert this into Kelvin, as the equation of state of the ideal gas requires temperature to be in Kelvin. The conversion is as follows: T(K) = T(°C) + 273.15. Therefore, the final temperature is approximately 332.15K.

On the other hand, according to the ideal gas law parameters in this problem, when the volume doubles (since the partition is removed), the pressure is halved. This is reflected by the formula P = nRT/V, where 'n' is the amount of the gas, 'R' is the ideal gas constant, 'T' is the temperature and 'V' is the volume. Therefore, the final pressure is the initial pressure divided by 2, yielding a final pressure of approximately 350 kPa.

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Answer:

substances with a higher boiling point are returning back to the flask which allows another substances with the specific context temperature (lower boiling point) to boil over and be purified.

Explanation:

The reason it happens because the lower boiling point substance vaporizes and crosses over while the other substance is waiting for its boiling point to reach

In fractional distillation, the returning condensate from the bottom of the distillation column to the flask enhances the efficiency of the process. The returned condensate serves as a mixing agent, increasing temperature gradients and refining the separation of components. More 'reflux' or returned condensate means more stages of distillation and better separation.

In the process of fractional distillation, when condensate returns to the distilling flask from the bottom of the distillation column, it serves as a mixing agent. It enhances the efficiency of the fractional distillation process. This returning condensate mixes with the rising vapor which leads to a thorough exchange of heat. This increases the temperature gradient in the column, making the distillation more effective in separating the chemical components according to their boiling points. Each 'reflux' or return of condensate causes more 'theoretical plates' or stages of distillation, refining the separation.

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Answer:

In the above reaction, the oxidation state of tin changes from 2+ to 4+.

10 moles of electrons are transferred in the reaction

Explanation:

Redox reaction is:

2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻

SnO₂²⁻ → SnO₃²⁻

Tin changes the oxidation state from +2 to +4. It has increased it so this is the oxidation from the redox (it released 2 e⁻). We are in basic medium, so we add water in the side of the reaction where we have the highest amount of oxygen. We have 2 O on left side and 3 O on right side so we add 1 water on the right and we complete with OH⁻ in the opposite side to balance the H.  

SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O Oxidation

BrO₃⁻ →  Br₂

First of all, we have unbalance the bromine, so we add 2 on the BrO₃⁻. We have 6 O in left side and there are no O on the right, so we add 6 H₂O on the left. To balance the H, we must complete with 12OH⁻. Bromate reduces to bromine at ground state, so it gained 5e⁻. We have 2 atoms of Br, so finally it gaines 10 e⁻.

6H₂O + 10 e⁻ + 2BrO₃⁻ →  Br₂ + 12OH⁻ Reduction

In order to balance the main reaction and balance the electrons we multiply  (x5) the oxidation and (x1) the reduciton

(SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O) . 5

(6H₂O + 10 e⁻ + 2BrO₃⁻ →  Br₂ + 12OH⁻) . 1

5SnO₂²⁻ + 10OH⁻ + 6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻ + 5SnO₃²⁻ + 10e⁻ + 5H₂O

We can cancel the e⁻ and we substract:

12OH⁻ - 10OH⁻ = 2OH⁻ (on the right side)

6H₂O - 5H₂O = H₂O (on the left side)

2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻

Answer:

The volume of a 0.0015 [tex]\frac{moles}{liters}[/tex] calcium sulfate solution that contains 25.0 g of calcium sulfate CaSO₄ is 122.53 liters.

Explanation:

Molarity (M) is the number of moles of solute that are dissolved in a given volume. Molarity is expressed by:

[tex]Molarity=\frac{number of moles of solute}{Dissolution volume}[/tex]

Molarity is expressed in units [tex]\frac{moles}{liter}[/tex]

In this case you have 25.0g of calcium sulfate CaSO₄. First of all you need to know the amount of moles that mass represents. For that you must first know the atomic masses of each element:

Then the molar mass of the compound calcium sulfate is:

molar mass= 40 g/mol + 32 g/mol + 4*16 g/mol= 136 g/mol

It is then possible to apply a rule of three as follows: if 136 g represents 1 mol of the compound calcium sulfate, 25 g how many moles are they?

[tex]moles=\frac{25 g*1 mole}{136 g}[/tex]

moles≅0.1838

Now you can apply a rule of three knowing the molarity of 0.0015 [tex]\frac{moles}{liters}[/tex]: if 0.0015 moles represents 1 liter of solution, 0.1838 moles how many liters are they?

[tex]volume=\frac{0.1838moles*1 liter}{0.0015moles}[/tex]

volume=122.53 liters

The volume of a 0.0015 [tex]\frac{moles}{liters}[/tex] calcium sulfate solution that contains 25.0 g of calcium sulfate CaSO₄ is 122.53 liters.

To find the volume of a 0.0015 mol/L calcium sulfate solution that contains 25.0g of calcium sulfate, first convert the mass to moles, then use the molarity to find the volume. The approximate volume is 122L.

To calculate the volume of the solution, we first need to convert the mass of calcium sulfate in grams to moles. We do this by dividing by the molar mass of calcium sulfate, which is about 136.14 g/mol.

25.0g CaSO4 * (1 mol / 136.14 g) = approx 0.183 mol CaSO4

Then, we use the molarity of the solution to find the volume. Remember that the definition of molarity is moles of solute per liter of solution.

Volume = moles / molarity = 0.183 mol / 0.0015 mol/L = approx 122 L

So, the volume of the calcium sulfate solution is approximately 122 liters.

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Explanation:

[tex]Molarity=\frac{\text{Moles of solute}}{\text{Volume of solution Liter}}[/tex]

A. 2.00 mL of 0.00250 M [tex]Fe(NO_3)_3[/tex]

Moles of ferric nitrate = n

Volume of ferric nitrate = 2.00 ml = 0.002 L ( 1 mL=0.001 L)

Molarity of ferric nitrate = 0.00250 M

[tex]n=0.00250 M\times 0.002 L=0.000005 mol[/tex]

B. 5.00 mL of 0.00250 M [tex]KSCN[/tex]

Moles of KSCN  = n'

Volume of KSCN  = 5.00 ml = 0.005 L ( 1 mL=0.001 L)

Molarity of KSCN = 0.00250 M

[tex]n'=0.00250 M\times 0.005 L=0.0000125 mol[/tex]

C. 3.00 mL of 0.050 M [tex]HNO_3[/tex]

Moles of nitric acid = n''

Volume of nitric acid = 3.00 ml = 0.003 L ( 1 mL=0.001 L)

Molarity of nitric acid = 0.050 M

[tex]n=0.050 M\times 0.003 L=0.00015 mol[/tex]

After mixing A, B and C together and their respective initial concentration before reaction.

After mixing A, B and C together the volume of the solution becomes = V

V = 0.002 L=0.005 L+0.003 L= 0.010 L

Concentration of ferric nitrate :

[tex][Fe(NO_3)_3]=\frac{0.000005 mol}{0.010 L}=0.0005 M[/tex]

Concentration of ferric ions :

[tex][Fe^{3+}]=1\times [Fe(NO_3)_3]=0.0005 M[/tex]

Concentration of nitrate ions from ferric nitrate:

[tex][NO_3^{-}]=3\times [Fe(NO_3)_3]=0.0015 M[/tex]

Concentration of KSCN :

[tex][KSCN]=\frac{0.0000125 mol}{0.010 L}=0.00125 M[/tex]

Concentration of [tex]SCN^-[/tex] ions:

[tex][SCN^-]=1\times [KSCN]=0.00125 M[/tex]

Concentration of potassium ions:

[tex][K^+]=1\times [KSCN]=0.00125 M[/tex]

Concentration of nitric acid :

[tex][HNO_3]=\frac{0.00015 mol}{0.010 L}=0.015 M[/tex]

Concentration of hydrogen ion :

[tex][H^+]=1\times [HNO_3]=0.015 M[/tex]

Concentration of nitrate ions from nitric acid  :

[tex][NO_3^{-}]=1\times [HNO_3]=0.0015 M[/tex]

Concentration of nitrate ion in mixture = 0.0015 M + 0.0015 M = 0.0030 M

[tex]Fe^{3+}+SCN^-\rightleftharpoons Fe(NCS)^{2+}[/tex]

given concentration of [tex] Fe(NCS)^{2+}[/tex] at equilbrium = [tex]3.6\times 10^{-5} M = 0.000036 M[/tex]

initially :

0.0005 M     0.00125 M        0

At equilibrium

(0.0005-0.000036) M   (0.00125-0.000036) M      0.000036 M

0.000464 M     0.001214 M               0.000036 M

The expression of an equilibrium constant will be given as;

[tex]K_c=\frac{[Fe(NCS)^{2+}]}{[Fe^{3+}][SCN^{-}]}[/tex]

[tex]=\frac{0.000036 M}{0.000464 M\times 0.001214 M}=63.91 [/tex]

The value for the equilibrium constant is 63.91.

Answer:

Chain reaction

Explanation:

A chain reaction is a reaction that sustains itself. It has the ability to continue for a very long time without adding any more materials to the reaction system. It may be succinctly described as a self propagating reaction.

In a nuclear fission, uranium-235 is bombarded with neutrons to produce unstable uranium-236 which disintegrates to form daughter nuclei and produce more neutrons that bombard more uranium-235 and the reaction continues indefinitely.