An electrochemical cell based on the following reaction has a standard cell voltage (E°cell) of 0.48 V: Sn(s) + Cu2+(aq) → Sn2+(aq) + Cu(s) What is the standard reduction potential of tin(II)? (E°(Cu2+/Cu) = 0.34 V)

Answer: The [tex]\Delta H_f[/tex] for [tex]H_2O(l)[/tex] in the reaction is 297.9 kJ/mol.

Explanation:

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

The equation that is used to calculate enthalpy change of a reaction is:

[tex]\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)][/tex]

For the given chemical reaction:

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

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

[tex]\Delta H_{rxn}=[(2\times \Delta H_f_{(H_2)})+(1\times \Delta H_f_{(O_2)})]-[(2\times \Delta H_f_{(H_2O)})][/tex]

We are given:

Enthalpy of substances present in their standard form are taken to be 0.

[tex]\Delta H_f_{(H_2)}=0kJ/mol\\\Delta H_f_{(O_2)}=0kJ/mol\\\Delta H_{rxn}=595.8kJ[/tex]

Putting values in above equation, we get:

[tex]595.8=[(2\times (0))+(1\times (0))]-[2\times (\Delta H_f_{H_2O})]\\\\\Delta H_f_{H_2O}=297.9kJ/mol[/tex]

Hence, the [tex]\Delta H_f[/tex] for [tex]H_2O(l)[/tex] in the reaction is 297.9 kJ/mol.

Answer : The pH of 0.001 M [tex]H_2SO_4[/tex] is, 2.69 and the pH after mixing the solution is, 3.30.

Explanation :

First we have to calculate the concentration of hydrogen ion.

The balanced dissociation reaction will be,

[tex]H_2SO_4\rightarrow 2H^++SO_4^{2-}[/tex]

The concentration of [tex]H_2SO_4[/tex] = x = 0.001 M

The concentration of [tex]H^+[/tex] ion = 2x = 2 × 0.001 M = 0.002 M

The concentration of [tex]SO_4^{2-}[/tex] = x = 0.001 M

Now we have to calculate the pH of 0.001 M [tex]H_2SO_4[/tex].

[tex]pH=-\log [H^+][/tex]

[tex]pH=-\log (0.002)[/tex]

[tex]pH=2.69[/tex]

Now we have to calculate the molarity after mixing the solution.

[tex]M_1V_1=M_2V_2[/tex]

where,

[tex]M_1[/tex] = molarity of [tex]H_2SO_4[/tex] solution = 0.001 M

[tex]V_1[/tex] = volume of [tex]H_2SO_4[/tex] solution = 250 ml

[tex]M_2[/tex] = molarity of after mixing = ?

[tex]V_2[/tex] = volume of after mixing = 250 + 750 = 1000 ml

Now put all the given values in the above formula, we get the molarity after mixing the solution.

[tex](0.001M)\times 250ml=M_2\times (1000ml)[/tex]

[tex]M_2=2.5\times 10^{-4}M[/tex]

The concentration of [tex]H^+[/tex] ion = [tex]2\times (2.5\times 10^{-4}M)=5\times 10^{-4}M[/tex]

Now we have to calculate the pH after mixing the solution.

[tex]pH=-\log [H^+][/tex]

[tex]pH=-\log (5\times 10^{-4})[/tex]

[tex]pH=3.30[/tex]

Therefore, the pH of 0.001 M [tex]H_2SO_4[/tex] is, 2.69 and the pH after mixing the solution is, 3.30.

Answer: 6.7 g

Explanation:

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) = 55.32 mmHg

[tex]p_s[/tex] = vapor pressure of solution = 54.21 mmHg

[tex]w_2[/tex] = mass of solute  (urea) = ? g

[tex]w_1[/tex] = mass of solvent  (water) = 100 g

[tex]M_1[/tex] = molar mass of solvent (water) = 18 g/mole

[tex]M_2[/tex] = molar mass of solute (urea) = 60 g/mole

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

[tex]\frac{55.32-54.21}{55.32}=\frac{x\times 18}{100\times 60}[/tex]

[tex]x=6.7g[/tex]

Therefore, 6.7 g of urea is needed in 100.0 g of water is needed to decrease the vapor pressure of water from 55.32 mmHg of pure water to 54.21 mmHg for the solution.

Final answer:

6.70 g of urea is required to adjust the vapor pressure of water as specified, using calculations based on Raoult's law and the relations of vapor pressures, mole fractions, and molar masses.

Explanation:

To solve this problem, we can use Raoult's law, which states that the vapor pressure of a solution is directly proportional to the mole fraction of the solvent. The formula for Raoult's law is P1 = X1P10, where P1 is the vapor pressure of the solvent in the solution, X1 is the mole fraction of the solvent, and P10 is the vapor pressure of the pure solvent.

First, we calculate the change in vapor pressure: ΔP = P10 - P1 = 55.32 mmHg - 54.21 mmHg = 1.11 mmHg. We use the given vapor pressures to find the mole fraction of water in the solution.

Using the equation for Raoult's law rearranged to find the mole fraction of the solvent (water), we get: X1 = P1 / P10 = 54.21 mmHg / 55.32 mmHg = 0.9799.

Since the mole fraction of the solute (urea) and solvent (water) together is 1, the mole fraction of urea is 1 - X1 = 0.0201. To find the moles of urea, we need the moles of water, which can be calculated from its mass (100.0 g) and molar mass (18.015 g/mol).

The moles of water are 100.0 g / 18.015 g/mol = 5.55 mol. Therefore, the moles of urea required are 5.55 mol × 0.0201 = 0.1115 mol. Finally, calculating the mass of urea required: 0.1115 mol × 60.056 g/mol (molar mass of urea) = 6.70 g.

In summary, 6.70 g of urea is needed to decrease the vapor pressure of water from 55.32 mmHg to 54.21 mmHg when dissolved in 100.0 g of water, keeping the temperature constant at 40°C.

The equilibrium constant expression for this reaction is [tex]K_{eq} = \frac{[HOCl]^{2}}{[Cl_{2}]^{2}}[/tex]. This is an example of a heterogeneous equilibrium wherein the states of the reactants and products are different. In such a case, only the concentration of the gaseous and aqueous substances are included in the equilibrium constant expression.

Further Explanation:

The equilibrium constant expression is the ratio of the concentration of the products and the concentration of reactants.  

The guidelines in writing the equilibrium constant expressions are as follows:

Note: Pure substances (i.e. solids and liquids) are not included in the equilibrium constant expression as their concentrations are constant.

The numerical equivalent of the Keq expression is called the equilibrium constant. It has a specific value for a given temperature. The equilibrium constant provides information about the spontaneity and progress of an equilibrium reaction.

Learn More

Keywords: equilibrium constant expression, equilibrium

Answer:Temperature increases

Explanation: As the gas in the container is an ideal gas so it should follow the ideal gas equation, the equation of state.

We know ideal gas equation to be PV=nRT where

   P=pressure

   V=Volume

   T=Temperature

   R=Real gas constant

   n=Number of moles

since the gas is insulated such that no heat goes into or out of the system .

When we compress the ideal gas using a piston, Thermodynamically it means that work is done on the system by the surroundings.

Now as the ideal gas is been compressed so the volume of the gas would decrease and slowly a time will reach when no more gas can be compressed that is there cannot be any further decrease in volume of the gas.

From the equation PV=nRT

Once there is no further compression is possible hence volume becomes constant so pressure of the ideal gas becomes directly proportional to the temperature as n and R are constants. Also as the pressure and volume are inversely related so an decrease in volume would lead to an increase in pressure.

As the ideal  gas is compressed so the pressure of the gas would increase since the gas molecules have smaller volume available after compression hence the gas molecules would quite frequently have collisions with other gas molecules or piston and this collision would lead to increase in speed of the gas molecules and  so the pressure would increase .

The increase in pressure would lead to an increase in temperature as show by the above ideal gas equation because the pressure and temperature are directly related.

So here we can say that work done on the system by surroundings leads to increase in temperature of the system.

Final answer:

When an ideal gas in a piston is adiabatically compressed, its temperature increases due to the increased internal energy resulting from work done on the gas.

Explanation:

When an ideal gas contained in a piston is compressed and the process is adiabatic, meaning no heat flows into or out of the system, the temperature of the gas increases. This is because the work done on the gas during the compression decreases its volume but increases its internal energy.

As the particles of the gas are forced closer together, they collide more frequently and with greater energy, leading to a rise in temperature. The change in temperature during an adiabatic process can be described by the adiabatic condition, which in the case of an ideal gas relates pressure, volume, and temperature in a specific way.

Answer: The amount of sulfur present in the original material is [tex]1.34\times 10^7tons[/tex]

Explanation:

Converting given amount of mass in tons to grams, we use the conversion factor:

1 ton = 907185 g            .......(1)

So, [tex]2.68\times 10^7=2.431\times 10^{13}g[/tex]

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

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

Given mass of sulfur dioxide = [tex]2.431\times 10^{13}g[/tex]

Molar mass of sulfur dioxide = 64 g/mol

Putting values in above equation, we get:

[tex]\text{Moles of sulfur dioxide}=\frac{2.431\times 10^{13}g}{64g/mol}=3.79\times 10^9mol[/tex]

For the given chemical reaction:

[tex]S(s)+O_2(g)\rightarrow SO_2(g)[/tex]

By Stoichiometry of the reaction:

1 mole of sulfur dioxide is produced from 1 mole of sulfur

So, [tex]3.79\times 10^9[/tex] moles of sulfur dioxide will be produced from = [tex]\frac{1}{1}\times 3.79\times 10^9=3.79\times 10^9[/tex] moles of sulfur.

Now, calculating the mass of sulfur using equation 2:

Moles of sulfur = [tex]3.79\times 10^9mol[/tex]

Molar mass of sulfur = 32 g/mol

Putting values in equation 2, we get:

[tex]3.79\times 10^9mol=\frac{\text{Mass of sulfur}}{32g/mol}\\\\\text{Moles of sulfur}=121.54\times 10^{11}g[/tex]

Converting this value in tons using conversion factor 1, we get:

[tex]\Rightarrow (\frac{1ton}{907185g})\times 121.54\times 10^{11}g\\\\\Rightarrow 13397491.6tons=1.34\times 10^7tons[/tex]

Hence, the amount of sulfur present in the original material is [tex]1.34\times 10^7tons[/tex]

Final answer:

To precipitate all of the Ca2+ ions in the CaBr2 solution, we need 0.280618 mL of 0.472 M Ba(OH)2.

Explanation:

To find the number of milliliters of 0.472 M Ba(OH)2 required to precipitate all of the Ca2+ ions, we first need to determine the number of moles of Ca2+ ions in the 197 mL of 0.671 M CaBr2 solution. Using the equation CaBr2(aq) + Ba(OH)2(aq) -> Ca(OH)2(s) + BaBr2(aq), we can see that the mole ratio between Ca2+ ions and Ba(OH)2 is 1:1. So if there are x moles of Ca2+ ions, we would need x moles of Ba(OH)2. Using the formula Molarity = Moles/Volume, we can calculate the moles of Ca2+ ions:

Moles of Ca2+ ions = (0.671 M CaBr2)(197 mL) = 0.132537 mol Ca2+

Since the mole ratio is 1:1, we would need 0.132537 mol of Ba(OH)2. Using the formula Moles = Molarity x Volume, we can calculate the volume of 0.472 M Ba(OH)2:

Volume of Ba(OH)2 = (0.132537 mol)/(0.472 M) = 0.280618 mL

Answer :

Lewis-Randall rule : Lewis-Randall rule states that, the fugacity of a component is directly proportional to the mole fraction of the component in the solution in an ideal solution.

The equation used for the Lewis-Randall rule for fugacity of species in an ideal solution is :

[tex]f_i=X_i\times f_i^o[/tex]

where,

[tex]f_i[/tex] = fugacity in the solution

[tex]f_i^o[/tex] = fugacity of a pure component

[tex]X_1[/tex] = mole fraction of component

Final answer:

The Lewis-Randall rule indicates that the fugacity of a component in an ideal solution equals the product of the component's mole fraction and the fugacity of the pure component, central to thermodynamics and mixture behavior understanding.

Explanation:

The Lewis-Randall rule provides a foundation for understanding the fugacity of species in an ideal solution. It states that the fugacity of a component in an ideal mix is equal to the product of the mole fraction of that component in the mixture and the fugacity of the pure component at the same temperature and pressure. In mathematical terms, for a component i in an ideal solution, this can be expressed as fi = xi × fi°, where fi is the fugacity of the component in the mixture, xi is the mole fraction of the component, and fi° is the fugacity of the pure component.

This rule is pivotal in the field of thermodynamics and is particularly relevant when dealing with ideal solutions where the interactions between different species are similar to those present in pure substances. It provides a basis for calculating the fugacity coefficient, which is used to assess how the real behavior of a gas differs from the ideal predicted by Raoult's law in mixtures. The Lewis-Randall rule, alongside Raoult's and Henry's laws, forms a fundamental part of the theoretical framework for understanding phase equilibria and the behavior of mixtures.

Answer:

25.275 moles of oxygen gas will be required to completely react with all the samarium metal.

Explanation:

[tex]4Sm+3O_2\rightarrow 2Sm_2O_3[/tex]

Number of moles samarium metal = 33.7 moles

According to reaction, 4 moles of  samarium reacts with 3 moles of oxygen gas.

Then 33.7 moles of samarium will react with:

[tex]\frac{3}{4}\times 33.7 mol=25.275 mol[/tex]of oxygen gas.

25.275 moles of oxygen gas will be required to completely react with all the samarium metal.

Answer:

Moles of oxygen gas required to react completely with 33.7 moles of samarium metal is [tex]\fbox{25.3 \text{ mol}}[/tex].

Explanation:

A chemical equation in which the number of atoms of each element is the same on the reactant and product side is called a balanced chemical equation.

The balanced chemical equation can be used to determine the stoichiometric ratio between the reactant and the product. The stoichiometric ratio thus enables us to calculate:

1. Amount of one reactant required to react completely with the other reactant.

2. Amount of the product that can be produced from the given amount of the reactant.

Step 1: Write the chemical equation for the reaction between samarium metal and oxygen gas.

The chemical formula for oxygen gas is [tex]\text{O}_{2}[/tex].

Samarium has +3 oxidation state within the samarium/oxygen compound. So, the chemical formula of the samarium oxygen compound is [tex]\text{Sm}_{2}\text{O}_{3}[/tex].

The chemical equation is as follows:

[tex]\fbox{\text{Sm}+\text{O}_{2} \rightarrow \text{Sm}_{2}\text{O}_{3}\\\end{minipage}}[/tex]

Step 2: Balance the chemical equation for the reaction between oxygen and samarium metal.

The number of oxygen atoms in the product side is 3 and in the reactant side is 2. Put coefficient 2 in front of [tex]\text{Sm}_{2}\text{O}_{3}[/tex] and 3 in front of [tex]\text{O}_{2}[/tex] to balance the oxygen atoms.

[tex]\fbox{\text{Sm}+3\text{O}_{2} \rightarrow 2 \text{Sm}_{2}\text{O}_{3}\\\end{minipage}}[/tex]

The number of samarium atoms in the product side is 4 and in the reactant side is 1. Put coefficient 4 in front of Sm in the reactant side.

[tex]\fbox{\\4\text{Sm}+3\text{O}_{2} \rightarrow 2 \text{Sm}_{2}\text{O}_{3}\\\end{minipage}}[/tex]

Step 3: Determine the stoichiometric ratio between samarium and oxygen from the above balanced chemical equation.

According to the balanced chemical equation, we can see that the stoichiometric ratio between samarium and oxygen is 4:3.

Step 4: Use unitary method and calculate the moles of oxygen required to completely react with the given moles of samarium metal as follows:

[tex]\text{moles of O}_{2} = \left( \text{moles of Sm} \right)\left( \frac{3 \text{ mol of O}_{2}}{4 \text{ mol of Sm}} \right)[/tex]                               ...... (1)

Step 5: Substitute 33.7 mol for moles of Sm in equation (1).

[tex]\text{Moles of O}_{2} = \left( \text{33.7 mol} \right) \left( \frac{3 \text{ mol of O}_{2}}{4 \text{ mol of Sm}} \right)\\\text{Moles of O}_{2}= 25.275 \text{ mol}\\\text{Moles of O}_{2}= 25.3 \text{ mol}[/tex]

Note:

Do not forgot to balance the reaction. The reaction must be balanced in order to calculate the amount (mol) of oxygen required to completely react with the given amount of samarium.

Learn more:

1. Balanced chemical equation https://brainly.com/question/1405182

2. Learn more about how to calculate moles of the base in given volume https://brainly.com/question/4283309

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Some basic concept of chemistry

Keywords: samarium, oxygen gas, samarium/oxygen compound, 33.7 moles, 25.3 mol, balanced equation, stoichiometric ratio, coefficient, balance, moles, completely react.

Answer:

14.9802 grams of estrogen must be added to 216.7 grams of benzene.

Explanation:

The relative lowering of vapor pressure of solution containing non volatile solute is equal to mole fraction of solute.

[tex]\frac{p_o-p_s}{p_o}=\frac{n_2}{n_1+n_2}[/tex]

Where:

[tex]p_o[/tex] = Vapor pressure of pure solvent

[tex]p_s[/tex] = Vapor pressure of the solution

[tex]n_1[/tex] = Number of moles of solvent

[tex]n_2[/tex] = Number of moles of solute

[tex]p_o = 73.03 mmHg[/tex]

[tex]p_s= 71.61 mmHg[/tex]

[tex]n_1=\frac{216.7 g}{78.12 g/mol}=2.7739 mol[/tex]

[tex]\frac{73.03 mmHg-71.61 mmHg}{73.03 mmHg}=\frac{n_2}{2.7739 mol+n_2}[/tex]

[tex]n_2=0.05499 mol[/tex]

Mass of 0.05499 moles of estrogen :

= 0.05499 mol × 272.4 g/mol = 14.9802 g

14.9802 grams of estrogen must be added to 216.7 grams of benzene.

 Final answer:

The answer involves using Raoult's law to calculate the mass of estradiol needed to lower the vapor pressure of benzene by a specified amount. The calculation uses the initial and final vapor pressures of benzene and the molecular weights of both compounds to solve for the mass of estradiol.

The question relates to the calculation of the amount of a nonvolatile solute (estradiol) needed to be added to a solvent (benzene) to produce a desired lowering of the vapor pressure, via Raoult's law. Given the initial vapor pressure of benzene and the final vapor pressure after the addition of estradiol, we can use Raoult's law and the molecular weights of benzene and estradiol to solve for the mass of estradiol required.

To find the mass of estradiol that must be added to 216.7 grams of benzene to reduce its vapor pressure from 73.03 mm Hg to 71.61 mm Hg, we first calculate the mole fraction of benzene in the solution (XBenzene) after the addition of estradiol. Using the initial and final vapor pressures (Pinitial and Pfinal), and recognizing that the vapor pressure of the solution is directly proportional to the mole fraction of benzene, we get:

Pfinal = XBenzene × Pinitial

Then, we can calculate the moles of benzene (nBenzene) and use this value to find the moles of estradiol (nEstradiol) that correspond to the desired vapor pressure change. The mass of estradiol can then be computed by multiplying nEstradiol by its molecular weight (MW). Since estradiol is a nonvolatile nonelectrolyte, its addition does not alter the benzene vapor pressure other than through the change in mole fraction.

Answer:

Diethyl ether

Explanation:

vapor pressure - the pressure exerted by the gaseous molecules on the walls of the container , is called its vapor pressure.

The compound with higher boiling point , will have lower vapor pressure ,

and the compound with lower boiling point , will have higher vapor pressure.

Hence, Boiling point and vapor pressure have inverse relation.

The vapor pressure and boiling point both, depends on the inter molecular interactions , i,e, the interaction between the molecules.  

Since, the compound with stronger inter molecular interactions, will not easily convert to gas, hence will have higher boiling point between , therefore , its vapor pressure would be less.

But the compound with less inter molecular interactions , can easily vaporize to convert to gaseous state and hence will have lower boiling point, therefore, its vapor pressure would be higher .

Among all the options , diethyl ether have lowest boiling point , hence, will have highest vapor pressure , at room temperature.

Answer:

The correct answer is option B.

Explanation:

Concentration of a  steady-state plasma ,= 25 mg/L

Weight of the volunteer =75 kg

Rate of infusion = 7.5 mg/kg hr

Concentration of steady-state plasma in 75 kg weight body:

[tex]7.5 mg/kg hr\times 75 kg=562.5 mg/hr[/tex]

Total body clearance of this drug:

[tex]\frac{562.5 mg/hr}{25 mg/L}=22.5 L/hr[/tex]

The total body clearance of this drug is 22.5 L/hr.

Answer:

True

Explanation:

In an uncompetitive inhibition, initially the substrate [S] binds to the active site of the enzyme [E] and forms an enzyme-substrate activated complex [ES].

The inhibitor molecule then binds to the enzyme- substrate complex [ES], resulting in the formation of [ESI] complex, thereby inhibiting the reaction.

This inhibition is called uncompetitive because the inhibitor does not compete with the substrate to bind on the active site of the enzyme.

Therefore, in an uncompetitive inhibition, the inhibitor molecule can not bind on the active site of the enzyme directly. The inhibitor can only bind to the enzyme-substrate complex formed.

Answer:The product formed on reaction with hydroxide ion as nucleophile is 2R-hexane-2-ol.

The product formed on reaction with water would be a 50:50 mixture of

2S-hexane-2-ol. and 2R-hexane-2-ol.

Explanation:

2S-iodohexane on reactiong with hydroxide ion would undergo SN² substitution reaction that is substitution bimolecular. Hydroxide ion has a negative charge and hence it is a quite good  nucleophile .

The rate of a SN² reaction depends on both the substrate and nucleophile . Here the substrate is a secondary carbon center having Iodine as a leaving group.SN² reaction  takes place here as hydroxide ion is a good nucleophile and it can attack the secondary carbon center from the back side leading to the formation of 2R-hexane-2-ol.

In a SN² reaction since the the nucleophile attacks from the back-side so the product formation takes place with the inversion of configuration.

When the same substrate S-2-iodohexane undergoes a substitution reaction with water as a nucleophile then the reaction occurs through (SN¹) substitution nucleophilic unimolecular mechanism .

The rate of a SN¹ reaction depends only on the nature of substrate and is independent of the nature of nucleophile.

The SN¹ reaction is a 2 step reaction , in  the first step leaving group leaves leading to the formation of a carbocation and once the carbocation is formed then any weaker nucleophile or even solvent molecules can attack leading the formation of products.

In this case a secondary carbocation would be generated in the first step and then water will attack this carbocation to form the product in the second step.

The product formed on using water as a nucleophile would be a racemic mixture of R and S isomers of hexane -2-ol in 50:50 ratio. The two products formed would be 2R-hexane-2-ol and 2S-hexane-2-ol.

Kindly refer the attachment for reaction mechanism and structure of products.

Final answer:

The value of the equilibrium constant, Kp, for the given reaction can be calculated based on the partial pressures of the reactants and products at equilibrium. In this case, the values are 1.425 atm for X₂ and 5.7 atm for X. Using the formula Kp = (p(X)²) / p(X₂), we find that Kp = 22.8.

Explanation:

The value of the equilibrium constant, Kp, for the reaction X₂(g) ⇌ 2X(g), can be calculated using the partial pressures of the reactants and products at equilibrium. In this case, the initial pressure of X₂ is 1.55 atm, and the total pressure at equilibrium is 2.85 atm.

Let's assume that the partial pressure of X₂ at equilibrium is p(X₂), and the partial pressure of X at equilibrium is p(X).

According to the equation, the number of moles of X₂ decreases by 1 (from 2 to 1) and the number of moles of X increases by 2 (from 0 to 2). This means that at equilibrium, the equilibrium partial pressure of X₂ is half of the total pressure, and the equilibrium partial pressure of X is twice the total pressure.

Therefore, p(X₂) = 2.85/2 = 1.425 atm and p(X) = 2 * 2.85 = 5.7 atm.

To calculate the value of Kp, we use the formula:

Kp = (p(X)²) / p(X₂)

Substituting the values we obtained, we get:

Kp = (5.7²) / 1.425 = 22.8

Answer:

0.2886 atm is the osmotic pressure of a solution.

Explanation:

Osmotic pressure of solution =[tex]\pi[/tex]

Concentration of the solution = c

Mass of the ammonium sulfate = 1.502 g

Moles of ammonium sulfate = [tex]\frac{1.502 g}{132.16 g/mol}=0.01136 mol[/tex]

Volume of the solution = 1 L

Concentration of the solution:

[tex]=\frac{\text{Moles of ammonium sulfate}}{\text{Volume of the solution}}[/tex]

[tex]c=\frac{0.01136 mol}{1 L}=0.01136 mol/L[/tex]

Temperature of the solution ,T= 36.54°C = 309.69 K

R = universal gas constant = 0.08206 L atm/mol K

[tex]\pi=cRT[/tex]

[tex]\pi=0.01136 mol/L\times 0.08206 L atm/mol K\times 309.69 K[/tex]

[tex]\pi=0.2886 atm[/tex]

0.2886 atm is the osmotic pressure of a solution.

To find the osmotic pressure, the temperature was converted to Kelvin, the moles of (NH₄)₂SO₄ were calculated using its molar mass, and the van't Hoff factor for the dissociation of (NH₄)₂SO₄ was determined. The gas constant and these values were then used in the osmotic pressure formula to calculate a pressure of approximately 0.08452 atm.

To calculate the osmotic pressure of a solution containing 1.502 g of (NH₄)₂SO₄ in 1 L at 36.54 Degrees Celsius, we use the formula:

π = i × M × R × T
where:

π is the osmotic pressure in atm,

i is the van't Hoff factor (number of particles the solute dissociates into),

M is the molarity of the solution,

R is the gas constant (0.08206 L·atm/mol·K), and

T is the temperature in Kelvin.

First, convert the temperature to Kelvin:

T(K) = 36.54 + 273.15 = 309.69 K
Then, calculate the moles of (NH4)2SO4 using its molar mass:

Moles = {1.502 g}÷{132.16 g/mol} ≈ 0.01136 mol
Since the solution's volume is 1 L, the molarity (M) is 0.01136 M. Ammonium sulfate ((NH₄)₂SO₄) dissociates into 2NH₄⁺ and SO₄²⁻. Therefore, the van't Hoff factor (i) is 3.

Finally, substitute the values into the formula to find the osmotic pressure:

π = 3 × 0.01136 M × 0.08206 L·atm/mol·K × 309.69 K
π ≈ 0.08452 atm
The osmotic pressure of the solution is approximately 0.08452 atm.

To count the number of valence electrons we look at the electronic configuration and add the electrons form the electronic shell with the highest principal quantum number.

Rb: [Kr] 5s¹ - 1 valence electron

Xe: [Kr] 5s² 4d¹⁰ 5p⁶ - 8 valence electrons

Sb: [Kr] 5s² 4d¹⁰ 5p³ - 5 valence electrons

I:    [Kr] 5s² 4d¹⁰ 5p⁵ - 7 valence electrons

In:  [Kr] 5s² 4d¹⁰ 5p¹ - 3 valence electrons

Rank from most to fewest valence electrons:

Xe > I > Sb > In > Rb

Rank from most to fewest valence electrons:

Xe > I > Sb > In > Rb. Each atom in the bond contributes one valence electron.

A valence electron is a kind of electron that is part of an atom's outer shell in chemistry and physics. If the outer shell is open, the valence electron can take part in the creation of a chemical bond, forming a shared pair within a single covalent bond.We examine the electronic configuration then add the electrons from the electronic shells with the largest main quantum number to determine the amount of valence electrons.

Rb: [Kr] 5s¹ - 1 valence electron

Xe: [Kr] 5s² 4d¹⁰ 5p⁶ - 8 valence electrons

Sb: [Kr] 5s² 4d¹⁰ 5p³ - 5 valence electrons

I:    [Kr] 5s² 4d¹⁰ 5p⁵ - 7 valence electrons

In:  [Kr] 5s² 4d¹⁰ 5p¹ - 3 valence electrons

Rank from most to fewest valence electrons:

Xe > I > Sb > In > Rb

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Answer: The molar mass of the gas is 35.87 g/mol.

Explanation:

To calculate the mass of gas, we use the equation given by ideal gas:

PV = nRT

or,

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

where,

P = Pressure of gas = 945 mmHg

V = Volume of the gas = 0.35 L

m = Mass of gas = 0.527 g

M = Molar mass of gas = ? g/mo

R = Gas constant = [tex]62.3637\text{ L.mmHg }mol^{-1}K^{-1}[/tex]

T = Temperature of gas =  [tex]88^oC=[88+273]=361K[/tex]

Putting values in above equation, we get:

[tex]945mmHg\times 0.35L=\frac{0.527g}{M}\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 361K\\\\M=35.87g/mol[/tex]

Hence, the molar mass of the gas is 35.87 g/mol.

Answer : The number of glucose molecule are, [tex]42.55\times 10^{23}[/tex]

Explanation : Given,

Mass of [tex]C_6H_{12}O_6[/tex] = 1273 g

Molar mass of [tex]C_6H_{12}O_6[/tex] = 180.156 g/mole

First we have to calculate the moles of [tex]C_6H_{12}O_6[/tex].

[tex]\text{Moles of }C_6H_{12}O_6=\frac{\text{Mass of }C_6H_{12}O_6}{\text{Molar mass of }C_6H_{12}O_6}=\frac{1273g}{180.156g/mole}=7.066moles[/tex]

Now we have to calculate the number of molecules of glucose.

As, 1 mole of glucose contains [tex]6.022\times 10^{23}[/tex] number of glucose molecules

So, 7.066 mole of glucose contains [tex]7.066\times 6.022\times 10^{23}=42.55\times 10^{23}[/tex] number of glucose molecules

Hence, the number of glucose molecule are, [tex]42.55\times 10^{23}[/tex]

Answer:

The dissociation constant of phenol from given information is [tex]9.34\times 10^{-11}[/tex].

Explanation:

The measured pH of the solution = 5.153

[tex]C_6H_5OH\rightarrow C_6H_5O^-+H^+[/tex]

Initially      c

At eq'm   c-x       x  x

The expression of dissociation constant is given as:

[tex]K_a=\frac{[C_6H_5O^-][H^+]}{[C_6H_5OOH]}[/tex]

Concentration of phenoxide ions and hydrogen ions are equal to x.

[tex]pH=-\log[x][/tex]

[tex]5.153=-\log[x][/tex]

[tex]x=7.03\times 10^{-6} M[/tex]

[tex]K_a=\frac{x\times x}{(c-x)}=\frac{x^2}{(c-x)}=\frac{(7.03\times 10^{-6} M)^2}{ 0.529 M-7.03\times 10^{-6} M}[/tex]

[tex]K_a=9.34\times 10^{-11}[/tex]

The dissociation constant of phenol from given information is [tex]9.34\times 10^{-11}[/tex].

Final answer:

The acid dissociation constant (Ka) for phenol, given its pH of 5.153 in a 0.529 M solution, is calculated to be 9.47 x 10⁻¹² M by converting pH to hydrogen ion concentration and applying the equilibrium expression for dissociation.

Explanation:

To determine the acid dissociation constant (Ka) for phenol from its pH, we use the formula pH = -log[H⁺], where [H⁺] is the concentration of hydrogen ions in the solution. First, convert the pH given (5.153) to [H+] concentration: [H⁺] = 10^-pH = 10^-5.153 = 7.08 x 10⁻⁶ M. The ionization of phenol in water can be represented as: C₆H₅OH (aq) → C₆H₅O⁻ (aq) + H⁺ (aq). At equilibrium, the concentration of C6H5O- and H+ will be equal and be 7.08 x 10-6 M, as the weak acid only partially dissociates. The initial concentration of phenol is 0.529 M, so the concentration of undissociated phenol at equilibrium is approximately 0.529 M (assuming the dissociation is minimal due to the weak acid nature). The Ka for phenol can be calculated using the expression Ka = [H⁺][C₆H₅O⁻]/[C₆H₅OH], substituting the equilibrium concentrations yields Ka = (7.08 x 10⁻⁶)2/0.529 = 9.47 x 10⁻¹² M.

Answer:

The pH of the final solution is 0.16 .

Explanation:

The pH of the solution is defined as negative logarithm of hydrogen ion concentration in a solution.

[tex]pH=-\log[H^+][/tex]

Concentration of HCl = 0.16 M

[tex]HCl(aq)\rightarrow H^+(aq)+Cl^-(aq)[/tex]

HCl is a string acid .1 molar of HCl gives 1 molar of of hydrogen ions.

[tex][H^+]=0.16 M[/tex]

Concentration of [tex]HNO_3[/tex] = 0.52 M

[tex]HNO_3(aq)\rightarrow H^+(aq)+NO_{3}^-(aq)[/tex]

Nitric is a string acid .1 molar of nitric acid gives 1 molar of of hydrogen ions.

[tex][H^+]'=0.52 M[/tex]

Total hydrogen ion concentration:

[tex][H^+]''=[H^+]+[H^+]'[/tex]

=0.16 M+0.52 M=0.68 M

The pH of the solution:

[tex]\pH=-\log[H^+]''=-\log[0.68 M][/tex]

pH = 0.16

The pH of the final solution is 0.16 .

Answer:

  Because it uses the residual energy of the fluid used by the first engine.

Explanation:

  A combined cycle power generation counts with two heat engines that work in tandem from the same source of heat. The engines turn the energy into mechanical energy.

  The cycle is much more efficient than the other, almost 60% more.

  I hope this answer helps you.

Answer:

HCl < CH₃COOH < NH₃ < NaOH

Explanation:

Given compounds:

Acetic acid: CH₃COOH

Ammonia; NH₃

Hydrochloric acid: HCl

Sodium hydroxide: NaOH

All the solutions are of the same molarity which is 0.1M. We need to see how these compounds dissociate to form solutions in order to establish their pH value:

For Acetic acid;

            CH₃COOH + H₂O ⇄ H₃O⁺ + CH₃COO⁻

   Acetic acid is a weak acid and it ionizes slightly in solutions. It would have a pH close to 7

For Ammonia;

       NH₃ + H₂O ⇄ NH₄⁺ + OH⁻

   Ammonia is a weak base and it ionizes slightly in solutions. It sets up an equilibrium in the process. It's would be slightly above 7

For HCl:

               HCl + H₂O → H₃O⁺ + Cl⁻

HCl is a strong acid and ionizes completely in solutions. It has a very low pH

For NaOH:

          NaOH → Na⁺ + OH⁻

  NaOH ionizes also completely in solutions and it breaks down into sodium and hydroxide ions. It is a strong base and it would have a high PH value.

                 HCl < CH₃COOH < NH₃ < NaOH

This is the trend of increasing pH

Answer:

Explanation:hi

Answer:4-nitrobenzoic acid is more acidic than Benzoic acid

4-chlorobenzoic acid is more acididc than 4-methyl benzoic acid

P-Nitrophenol is more acidic than meta-nitrophenol

Explanation:Acidity of an acid can be explained in terms of the stability of conjugate base formed.

1. 4-nitrobenzoic acid is more acidic as compared to benzoic acid because of the presence of nitro group at 4-position that is para position of the benzene ring. Nitro group is an electron withdrawing group and it withdraws the electron density through resonance effect.

Here the conjugate base would be benzoate anion which has a carboxylate anion attached with the benzene ring . So any group which can withdraw the electron density from benzoate anion will stabilise the benzoate anion and subsequently it would increase the acidity .

In case of  benzoic acid there is no extra withdrawl of electron density whereas in case of 4-nitrobenzoic acid the nitro group stabilises the benzoate anion by withdrawing electron density thereby stabilising the benzoate anion and increasing the acidity.

2. 4-chlorobenzoic acid is more acidic than 4-methyl benzoic acid because 4-chlorobenzoic acid has Cl group which is a good electron withdrawing group through inductive effect so the benzoate anion formed can be stabilised by the electron withdrawing Cl atom  which would increase the acidity of 4-chlorobenzoic acid.

4-methylbenzoic acid has an electron donating Methyl group which donates electron density through inductive effect hence a methyl group would intensify the negative charge on the benzoate anion through electron donation and subsequently it would destabilise the benzoate anion thereby decreasing its acidity.

3. In case of phenols the conjugate base formed is phenoxide anion and the negative charge that is its electron density is delocalised over the whole phenol ring. The negative charge electron density is more prominent at ortho and para position rather than the meta position. p-nitrophenol is more acidic than m-nitrophenol because p-nitrophenol has the nitro group at para position where it can stabilise the phenoxide anion through electron withdrawl via  resonance whereas in case of m-nitrophenol as the nitro group is present at meta position so it can not stabilize prominently through electron withdrawl via resonance.

In each pair, 4-nitrobenzoic acid, 4-chlorobenzoic acid, and p-nitrophenol are more acidic due to their electron withdrawing groups that can better stabilize the negative charge after ionization.

The acidity of a compound can be determined by its ability to donate a proton (hydrogen ion), and this ability is often influenced by other groups present in the compound. In each pair, the compound with the group that is more electron withdrawing will generally be more acidic because they can better stabilize the negative charge of the carboxylate ion after ionization occurs.

Benzoic acid vs. 4-nitrobenzoic acid: The presence of the nitro group in the 4-nitrobenzoic acid makes it more electron withdrawing compared to benzoic acid, making it more acidic.

4-Methylbenzoic acid vs. 4-chlorobenzoic acid: The chlorine atom in 4-chlorobenzoic acid is more electron withdrawing than the methyl group in 4-methylbenzoic acid, making 4-chlorobenzoic acid more acidic.

p-Nitrophenol vs. m-nitrophenol: In this case, p-nitrophenol would be more acidic. Although the nitro group is meta to the phenol group in m-nitrophenol, and para in p-nitrophenol, the para position allows for more effective resonance stabilization post-ionization, enhancing its acidity.

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Answer:

Dimer of two peptide chains with 1 mole of molybdenum metal each.

Explanation:

Percentage of molybdenum in protein = 0.08%

Molecular mass of nitrate reductase = 240,000 g

Mass of molybdenum = x

[tex]0.08\%=\frac{x}{240,000 g}\times 100=192 g[/tex]

Moles of molybdenum =[tex]\frac{192 g}{95.95 g/mol}=2.00 mol[/tex]

Each peptide chain of nitrate reductase contain 1 mole of molybdenum.

This means that nitrate reductase is composed of to two peptide chains. And in each peptide there is a single mole of molybdenum metal.

Answer:

As we climb a mountain to a higher altitude, we experience a pressure increase. -True

Answer:

the 4th choice.

Explanation:

the fish can't swim through the aquarium

the fish can't swim in the air

Answer:

B.Example 1 represents a liquid and example 2 represents a gas

Explanation:

Solid : It is that states of matter in which atom or molecules very close to each other. The distance between atoms or molecules is negligible. The atom or molecules can't escape out of solids.

Liquid: It is that states of matter in which atoms or molecules are close to each other but not very close as in solid.The distance between atoms or molecules is small.

Gas: It is that states of matter in which atoms or molecules are very far to each other .The distances between atoms or molecules are very large. The two two atoms can not come closer to each other at normal condition.

We are given that Zoe used two examples to represent two different states of matter

1.Example :Fish swimming around one another in an aquarium

It means fish can not escape out of aquarium. Hence, it represent the liquid states of matter.

2.Example: Fish swimming away one another in the ocean .

When fish swimming away one another ,Then they can'not come close to each other .Hence, it represents gas states of matter.

Therefore, option D is true.

Answer:Example 1 represents a solid and example 2 represents a gas

Answer:

Average molecular weight of air is 28.84 g/mol.

Explanation:

The average molecular weight of a mixture is determined from their molar composition and molecular weight.

Average molecular weight :[tex]\sum (\chi_i\times m_i)[/tex]

[tex]\chi_1[/tex] : mole fraction of the 'i' component.

[tex]m_i[/tex] = Molecular weight of i component

Average molecular weight of air with approximate molar composition of 79% nitrogen gas and 21% of oxygen gas can be calculated as:

Average molecular weight of air:

[tex]79\%\times 28 g/mol+21\%\times 32 g/mol[/tex]

[tex]=0.79\times 28 g/mol+0.21\times 32 g/mol=28.84 g/mol[/tex]

Final answer:

The average molecular weight of air can be calculated using its molar composition of 79% N2 and 21% O2.

Explanation:

The average molecular weight of air can be calculated using its molar composition of 79% N2 and 21% O2. The molar weight of N2 is 28.01 g/mol and the molar weight of O2 is 32.00 g/mol. We can calculate the average molecular weight using the formula:

Average Molecular Weight = (0.79 * 28.01 g/mol) + (0.21 * 32.00 g/mol)

Average Molecular Weight = 28.89 g/mol

Answer: [tex]3.5\times 10^{-5}mol^{-1}Lsec^{-1}[/tex]

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

[tex]C_2H_5Br(alc)+OH^-(alc)\rightarrow C_2H_5OH(l)+Br^-(alc)[/tex]

Given: Order with respect to [tex]C_2H_5Br[/tex] = 1

Order with respect to [tex]OH^-[/tex] = 1

Thus rate law is:

[tex]Rate=k[C_2H_5Br]^1[OH^-]^1[/tex]

k= rate constant

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

[tex]Rate=-\frac{1d[C_2H_5Br]}{dt}=k[C_2H_5Br]^1[OH^-]^1[/tex]

Given: [tex]\frac{d[C_2H_5]}{dt}]=1.7\times 10^{-7}[/tex]

Putting in the values we get:

[tex]Rate=1.7\times 10^{-7}=k[0.0477]^1[0.100]^1[/tex]

[tex]k=3.5\times 10^{-5}mol^{-1}Lsec^{-1}[/tex]

Thus the value of the rate constant is [tex]3.5\times 10^{-5}mol^{-1}Lsec^{-1}[/tex]

The rate constant of the reaction is 3.6 * 10^-5.

We define the rate of reaction as how quickly or slowly that a reaction is occuring. That is, the rate of disappearance of C2H5Br and OH- or rate of appearance of  C2H5OH(l)  and Br-.

Now;

Rate of disappearance of ethyl bromide = 1.7 x 10^-7 M/s.

Concentration of  ethyl bromide =   0.0477 M

Concentration of OH- = 0.100 M

Thus;

-d[C2H5Br]/dt = k  [C2H5Br] [OH-]

1.7 x 10^-7  = k [0.0477] [0.100]

k = 1.7 x 10^-7/ [0.0477] [0.100]

k = 3.6 * 10^-5

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Answer:

The enthalpy of the reaction is -123.5 kJ.

Explanation:

[tex]P4 (s) + 6 Cl_2 (g)\rightarrow 4 PCl_3 (l) ,\Delta H_1 =-1280 kJ[/tex]..(1)

[tex]P4 (s) + 10 Cl_2 (g)\rightarrow 4 PCl_5 (l) ,\Delta H_2 =-1774 kJ[/tex]..(2)

[tex]PCl_3 (l) + Cl_2(g)\rightarrow PCl_5(s),\Delta H_{rxn}=x[/tex]...(3)

(2) - (1)

[tex]4PCl_3 (l) + 4Cl_2(g)\rightarrow 4PCl_5(s),\Delta H_{rxn}=y[/tex]

Dividing equation by 4 we get (3)

[tex]PCl_3 (l) + Cl_2(g)\rightarrow PCl_5(s),\Delta H_{rxn}=\frac{y}{4}[/tex]...(3)

[tex]\Delta H_{rxn}=y=(-1774 kJ)-(-1280 kJ)=-494 kJ[/tex]

[tex]\Delta H_{rxn}=x=\frac{y}{4}={-494 kJ}{4}=-123.5 kJ[/tex]

The enthalpy of the reaction is -123.5 kJ.