Liquid hexane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 9.48 g of hexane is mixed with 43. g of oxygen. Calculate the minimum mass of hexane that could be left over by the chemical reaction. Be sure your answer has the correct number of significant digits.

Answers

Answer 1

Answer:

No mass of hexane could be left over by the chemical reaction.

C₆H₁₄ is the limitin reagent

Explanation:

This is a combustion reaction were the oxygen is one of the reactant and it burns a compound in order to generate water and carbon dioxide.

In this case, we have liquid hexane combustion, so the reaction is:

2C₆H₁₄(l) +  19O₂(g)  →   12CO₂(g) +  14H₂O(g)

In this situation we are asked for the mass of a reactant that could be left over, this is the excess reagent.

We convert the masses of reactants to moles:

9.48 g . 1mol  / 86g = 0.110 moles of C₆H₁₄

43 g . 1 mol/32 g =  1.34 moles of O₂

The hexane may be the excess reagent but we confirm like this:

19 moles of O₂ need 2 moles of hexane to react

Then, 1.34 moles of O₂ will react with (1.34 . 2) / 19 = 0.141 moles

We need 0.141 moles, and we only have 0.110. Hexane is the limiting reagent so no mass could be left over by the chemical reaction.

In conclussion, oxygen is excess reactant. We verify:

2 moles of hexane need 19 moles of O₂ to react

Then, 0.110 moles of hexane will react with (0.110 . 19) / 2 = 1.04 moles of O₂

As I have 1.34 moles of oxygen, value is higher so in this case we are having mass that could be left over by the reaction.


Related Questions

In the organic combustion reaction of 41.9 g of octane (C8H18) with excess oxygen, what volume (in L) of carbon dioxide is produced if the reaction is performed at STP?

Answers

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

If 73.5 mL of 0.200 M KI(aq) was required to precipitate all of the lead(II) ion from an aqueous solution of lead(II) nitrate, how many moles of Pb2+ were originally in the solution?

Answers

Answer:

There were 0.00735 moles Pb^2+ in the solution

Explanation:

Step 1: Data given

Volume of the KI solution = 73.5 mL = 0.0735 L

Molarity of the KI solution = 0.200 M

Step 2: The balanced equation

2KI + Pb2+ → PbI2 + 2K+

Step 3: Calculate moles KI

moles = Molarity * volume

moles KI = 0.200M * 0.0735L = 0.0147 moles KI

Ste p 4: Calculate moles Pb^2+

For 2 moles KI we need 1 mol Pb^2+ to produce 1 mol PbI2 and 2 moles K+

For 0.0147 moles KI we need 0.0147 / 2 = 0.00735 moles Pb^2+

There were 0.00735 moles Pb^2+ in the solution

Please, Please help

A flask containing 550 mL of 0.75 M H2SO4 was accidentally knocked to the floor.
How many grams of NaHCO3 do you need to put on the spill to neutralize the acid according to the following equation?
H2SO4(aq)+2NaHCO3(aq)→Na2SO4(aq)+2H2O(l)+2CO2(g)

Express your answer using two significant figures.
m= g

Answers

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

In the laboratory you dissolve 22.8 g of chromium(III) acetate in a volumetric flask and add water to a total volume of 250 mL. What is the molarity of the solution? M. What is the concentration of the chromium(III) cation? M. What is the concentration of the acetate anion?

Answers

Final answer:

The molarity of the chromium(III) acetate solution is 0.398 M. The concentration of chromium(III) cation is also 0.398 M, and the concentration of the acetate anion is 1.194 M.

Explanation:

First, we need to recall that the molar mass of chromium(III) acetate is approximately 229.13 g/mol. We can then find the moles of the compound by the equation: moles = mass (g) / molar mass (g/mol). So, moles of chromium(III) acetate = 22.8 g / 229.13 g/mol = 0.0995 mol. Then, using the formula for molarity, M = moles / volume (L). Volume must be in liters, so 250 mL is converted to 0.25 L. This gives us M = 0.0995 mol / 0.25 L = 0.398 M.

As for the concentrations of the chromium(III) cation and the acetate anion, we have to consider the formula of chromium(III) acetate, which is Cr(C2H3O2)3. This indicates that for every 1 mol of compound, there is 1 mol of Cr3+ and 3 mol of C2H3O2-. So the molar concentration of Cr3+ is the same as that of the substance, 0.398 M. The molar concentration of C2H3O2- is three times as much, 0.398 M x 3 = 1.194 M.

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Liquid methanol is fed to a space heater at a rate of 12.0 L/h and burned with excess air. The product gas is analyzed and the following dry-basis mole percentages are determined: CH3OH = 0.45%, CO2 = 9.03%, and CO = 1.81%. (a) After drawing and labeling a flowchart, verify that the system has zero degrees of freedom. (b) Calculate the fractional conversion of methanol, the percentage excess air fed, and the mole fraction of water in the product gas. (c) Suppose the combustion products are released directly into a room. What potential problems do you see and what remedies can you suggest?

Answers

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.

Final answer:

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.

Explanation:

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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A 500.0-mL buffer solution is 0.100 M in HNO2 and 0.150 M in KNO2. Determine whether each addition would exceed the capacity of the buffer to neutralize it.250.0 mg NaOH350.0 mg KOH1.25 g HBr1.35 g HI

Answers

Answer:

None of the additions will exceed the capacity of the buffer.

Explanation:

As we know a buffer has the ability to resist pH changes when small amounts of strong acid or base are added.

The pH of the buffer is given by the Henderson-Hasselbach equation:

pH = pKa + log [A⁻] / [HA]

where A⁻ is the conjugate base of the weak acid HA.

Now we can see that what is important is the ratio [A⁻] / [HA] to resist a pH change brought about by the addition of acid or base.

It follows then that once we have consumed by neutralization reaction either the acid or conjugate base in the buffer, this will lose its ability to act as such and the pH will increase or decrease dramatically by any added acid or base.

Therefore to solve this question we must determine the number of moles of acid HNO₂ and NO₂⁻ we have in the buffer and compare it with the added acid or base to see if it will deplete one of these species.

Volume buffer = 500.0 mL = 0.5 L

# mol HNO₂ = 0.5 L x 0.100 mol/L = 0.05 mol HNO₂

# mol NO₂⁻ = 0.5 L x 0.150 mol/L = 0.075 mol NO₂⁻

a. If we add 250 mg NaOH (0.250 g)

molar mass NaOH =40 g/mol

# mol NaOH =0.250 g/ 40g/mol = 0.0063 mol

0.0063 mol NaOH will be neutralized by 0.0063 mol HNO₂ and we have plenty of it, so it would not exceed the capacity of the buffer.

b. If we add 350 mg KOH (0.350 g)

molar mass KOH =56.10 g

# mol KOH = 0.350 g/56.10 g/mol = 0.0062 mol

Again the capacity of the buffer will not be exceeded since we have 0.05 mol HNO₂ in the buffer.

c. If we add 1.25 g HBr

molar mass HBr = 80.91 g/mol

# mol HBr = 1.25 g / 80.91 g/mol = 0.015 mol

0.015 mol Hbr will neutralize 0.015 mol NO₂⁻ and we have to start with 0.075 mol in the buffer, therefore the capacity will not be exceeded.

d. If we add 1.35 g HI

molar mass HI = 127.91 g/mol

# mol HI = 1.35 g / 127.91 g/mol = 0.011 mol

Again the capacity of the buffer will not be exceed since we have plenty of it in the buffer after the neutralization reaction.

Final answer:

Based on the buffer's molarity and the amount of substances added, it appears that none of the additions would exceed the buffer's capacity to neutralize them. Yet, they would consume a significant portion of the buffering agents, thereby potentially affecting the buffer's effectiveness in further reactions.

Explanation:

The subject question is asking about the buffer capacity and if certain substances when added would exceed this capacity. The buffer in this case is a solution of HNO2 (nitrous acid) and KNO2 (potassium nitrite). The buffer can neutralize added acids or bases because HNO2 can donate a proton (H+) to neutralize OH- ions from a base, and KNO2 can donate OH- ions to neutralize excess H+ ions from an acid.

Now, for each addition, let's convert the mass of the substance to moles. For NaOH, 250.0 mg is about 0.00625 moles. For KOH, 350.0 mg is roughly 0.00624 moles. For HBr, 1.25 g is about 0.00736 moles. Finally, for HI, 1.35 g is quite close to 0.00736 moles.

Comparing these measures to the buffer's concentration in the solution, all of the added substances are less than 0.100 M or 0.150 M. Thus, these added amounts will not exceed the buffer's capacity to neutralize them, although it should be noted that they would extensively consume the buffering agents in the solution.

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A titration of 0.1 M NaOH into 0.8 L of HCl was stopped once the pH reached 7 (at 25C). If 0.2 L of NaOH needed to be added to achieve this pH, what was the original concentration of the sample of HCl

Answers

Answer:

0.03 M

Explanation:

Let's consider the neutralization reaction between HCl and NaOH.

HCl + NaOH → NaCl + H₂O

0.2 L of 0.1 M NaOH were used. The moles of NaOH that reacted are:

0.2 L × 0.1 mol/L = 0.02 mol

The molar ratio of HCl to NaOH is 1:1. The moles of HCl that reacted are 0.02 mol.

0.02 moles of HCl are in 0.8 L of solution. The molarity of HCl is:

0.02 mol / 0.8 L = 0.03 M

A(g) + 2B(g) → C(g) + D(g)If you initially start with 1.00 atm of both A and B and find that at equilibrium 0.211 atm of C is present, what is the value of Kp for the reaction at the temperature the reaction was run?

Answers

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

Consider the reaction 2CO(g) + 2NO(g)2CO2(g) + N2(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 2.00 moles of CO(g) react at standard conditions. S°surroundings = J/K Are You Sure? Please check your answer for mistakes. Submit Answer

Answers

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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In fractional distillation, liquid can be seen running from the bottom of the distillation column back into the distilling flask. What effect does this returning condensate have on the fractional distillation?

Answers

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

Final answer:

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.

Explanation:

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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Predict the type of bond that would be formed between each of the following pairs of atoms(ionic, polar covalent or nonpolar covalent)

a. H and Cl
b. Mg and F
c. Li and N
d. N and S

Answers

Answer:

In between H and Cl the bond will be covalent.

In between Mg and F the bond will be ionic.

In between Li and N the bond will be ionic .

In between N and S the bond will be polar covalent.

Explanation:

Ionic bonds are defined as the bonds which are formed by the complete transfer of electrons from cation (positively charged ions) to anion (negatively charged ions). For Example: NaCl, [tex]MgF_2[/tex] etc.

Covalent bond is defined as the bond which is formed by the sharing of electrons between the atoms. For Example: HCl, [tex]CH_4[/tex] etc.

Its of two type:

Polar covalent compound: This compound is formed when difference in electronegativity between the atoms is present. When atoms of different elements combine, it results in the formation of polar covalent bond. For Example: [tex]CO_2,NO_2[/tex] etc..Non-polar covalent compound: This compound is formed when there is no difference in electronegativity between the atoms. When atoms of the same element combine, it results in the formation of non-polar covalent bond. For Example: [tex]N_2,O_2[/tex] etc.

In between H and Cl the bond will be covalent.

In between Mg and F the bond will be ionic.

In between Li and N the bond will be ionic .

In between N and S the bond will be polar covalent.

The type of bond that would be formed between each of the following pairs of atoms will be:

H and Cl - Covalent bondMg and F - Ionic bondLi and N - Ionic bondN and S - Polar covalent

Ionic bonds are formed by the transfer of electrons from cation to anion. An example is NaCl.

Covalent bond is a bond that's formed by the sharing of electrons between atoms. e.g. HCl.

A polar covalent bond is formed when there is the presence of difference in electronegativity between the atoms.

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The reaction of a carboxylic acid and thionyl chloride produces an acid chloride plus the gases SO2 and HCl. In the boxes, draw the mechanism arrows for the reaction. Be sure to add lone pars of electrons and nonzero formal charges on all species.

Answers

Answer:

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

Explanation:

In the first step, -OH group in carboxylic acid gives nucleophilic substitution reaction at S center in thionyl chloride and substitutes -Cl atomIn the second step, deprotonation takes place by chloride ion.In the third step, an intramolecular nucleophilic substitution reaction ([tex]S_{N}i[/tex]) takes place where bond electrons rearranges to produce [tex]SO_{2}[/tex], HCl and thionyl chloride.This rearrangement is highly favorable due to formation of gaseous species [tex]SO_{2}[/tex]Reaction mechanism has been shown below.

The air in a bicycle tire is bubbled through water and collected at 25 ∘C. If the total volume of gas collected is 5.65 L at a temperature of 25 ∘C and a pressure of 765 torr , how many moles of gas was in the bicycle tire?

Answers

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:

P is the pressure of the gas in atmV is the volume of the gas in Ln is the number of moles of gasR is the ideal gas constant (0.0821 L·atm/mol·K)T is the temperature of the gas in Kelvin

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.

A 25.0-g sample of ice at -6.5oC is removed from the freezer and allowed to warm until it melts. Given the data below, select all the options that correctly reflect the calculations needed to determine the total heat change for this process.

Melting point at 1 atm = 0,0°C; -2.09 J/g.°C; Cloud -4.21J/g °C, AH -6,02 kJ/mol
Check all that apply.

A. q for the temperature change from -6,5°C to 0.0°C is glven by 25.0 x 6.02 = 151 kJ
B. q for the phase change is given by 1.39 x 6.02 = 8.37 kJ
C. There are 3 separate heat change stages in this process.
D. The total heat change for the process is equal to +8.71 kJ
E. The total heat change for the process is equal to +350 kJ

Answers

Final answer:

The total heat change for the process of ice melting at -6.5oC can be calculated by considering the temperature change and the phase change. The correct calculations are: temperature change is 337.25 J and phase change is 150.5 J. The total heat change is 487.75 J or 0.48775 kJ.

Explanation:

The total heat change for the process of ice melting at -6.5oC can be calculated by considering the temperature change from -6.5oC to 0.0oC and the phase change from solid to liquid.

For the temperature change, we use the equation q = m * C * ΔT, where q is the heat change, m is the mass of ice, C is the specific heat capacity of ice, and ΔT is the temperature change. Plugging in the values, we get q = 25.0 g * 2.09 J/g.°C * (0.0-(-6.5)°C) = 337.25 J. This calculation is incorrectly represented as option A.

For the phase change, we use the equation q = m * ΔH, where q is the heat change, m is the mass of ice, and ΔH is the enthalpy of fusion. Plugging in the values, we get q = 25.0 g * 6.02 kJ/mol = 150.5 J. This calculation is incorrectly represented as option B.

Since there are only two stages in this process (temperature change and phase change), option C is false.

Therefore, the correct calculations to determine the total heat change for this process are: the temperature change is 337.25 J and the phase change is 150.5 J. Adding these up, the total heat change is 487.75 J, which is equivalent to 0.48775 kJ. This calculation is incorrectly represented as option D but not as option E.

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What volume will o.128 g of propane, c3h8 occupy at a pressure golf 485 mm Hg and a temperature of 30.0 c

Answers

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

A water sample has a pH of 8.2 and a bicarbonate concentration of 97 mg/L. What is the alkalinity of the sample in moles/L and in mg CaCO3/L?

Answers

Answer:6.94

Explanation:

Molar mass of CaCO3=40+12+16×3

=40+12+48=100g/mol

Moles=mass of substance/molar mass

=97mg/100g=0.097/100=0.00097moles/L.

PH=-log[CaCo3]=-log(0.00097)=6.94

P.s it's log to base e

Final answer:

The alkalinity of the water sample with a pH of 8.2 and bicarbonate concentration of 97 mg/L is 0.00159 moles/L when calculated as bicarbonate, and 159.14 mg CaCO3/L when expressed as equivalent calcium carbonate.

Explanation:

A student has asked about calculating the alkalinity of a water sample with a pH of 8.2 and a bicarbonate concentration of 97 mg/L, both in moles/L and in mg CaCO3/L. To find the alkalinity in moles/L, we need to consider that the molecular weight of HCO3- (bicarbonate) is approximately 61.01 g/mol. Therefore, 97 mg/L can be converted to moles/L by dividing by the molecular weight:

Alkalinity (as HCO3-) = 97 mg/L / (61.01 g/mol * 1000 mg/g) = 0.00159 mol/L.

To convert this alkalinity to mg CaCO3/L, we use the equivalent weight of CaCO3 (100.087 g/mol) since 1 mol of HCO3- is chemically equivalent to 1 mol of CaCO3 for neutralization purposes:

Alkalinity (as CaCO3) = 0.00159 mol/L * 100.087 g/mol * 1000 mg/g = 159.14 mg CaCO3/L.

Progesterone is a hormone that contains two ketone groups. The oxygen in the ketone group can function as a hydrogen bond acceptor. Select the amino acids that have side chains that can form a hydrogen bond with progesterone at pH 7.

a. threonine.

b. cysteine.

c. alanine.

d. aspartate.

e. arginine.

f. tryptophan.

Answers

Amino acids that can donate hydrogen bonds include -

a. threoninee. argininef. tryptophan

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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Final answer:

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.

Explanation:

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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For the decomposition reaction AB → A + B, the experimentally determined rate law was found to be: rate = k[AB]2 , and k = 0.20 L/mol•s. How long will it take for AB to reach one third of its initial concentration of 1.50 M? (

Answers

Answer:

It would take 20 seconds

Explanation:

Let the final concentration of AB be C

Rate = k[C]^2 = change in concentration of AB/time

k is the rate constant = 0.2 L/mol.s

Initial concentration of AB = 1.5 M

Final concentration of AB = 1/3 × 1.5 = 0.5 M

Change in concentration of AB = 1.5 - 0.5 = 1 M

0.2 × 0.5^2 = 1/time

time = 1/0.05 = 20 s

What is the change in entropy when 0.7 m3 of CO2 and 0.3 m3 of N2, each at 1 bar and 25°C, blend to form a gas mixture at the same conditions? Assume ideal gases

Answers

Explanation:

The given data is as follows.

        P = 1 bar,     T = 298 K

For [tex]CO_{2}[/tex];  [tex]V_{1C} = 0.7 m^{3}[/tex]

For [tex]N_{2}[/tex];    [tex]V_{1N} = 0.3 m^{3}[/tex]

Therefore, final volume will be as follows.

           [tex]V_{1C} + V_{1N} = 1 m^{3} = V_{2}[/tex]

Hence,            [tex]V_{2} = 1 m^{3}[/tex]

Formula for change in entropy is as follows.

          dS = [tex]nR ln (\frac{V_{2}}{V_{1}})[/tex]

For [tex]CO_{2}[/tex],  [tex]PV_{1C} = nRT[/tex]

or,         [tex]nR = \frac{PV_{1C}}{T}[/tex]            

                       = [tex]\frac{10^{5} \times 0.7}{298}[/tex]        

                       = 234.89

      [tex]dS_{CO_{2}} = 234.89 \times \frac{1}{0.7}[/tex]  

                       = 83.77 J/K

For [tex]N_{2}[/tex],       nR = [tex]\frac{P_{1}V_{1}N}{T}[/tex]

                  = [tex]\frac{10^{5} \times 0.3}{298}[/tex]

                  = 100.67

       [tex]dS_{N_{2}} = 100.67 ln (\frac{1}{0.3})[/tex]

                   = 121.20 J/K

Now, total change in the entropy is calculated as follows.

               dS = [tex]dS_{CO_{2}} + dS_{N_{2}}[/tex]

                    = (83.77 + 121.20) J/K

                    = 204.97 J/K

Thus, we can conclude that the change in entropy is 204.97 J/K.

Entropy change will be 204.97 J/K.

Firstly, let's write the given values:

Since, it's an ideal gas therefore;

P=1 barT=298 KFor CO₂= [tex]V_{CO_2} = 0.7m^3[/tex]For N₂= [tex]V_{N_2}=0.3 m^3[/tex]

Thus, the final volume will be V₂= [tex]0.7+0.3=1m^3[/tex]

What does change in entropy mean?

Entropy change can be defined as the change in the state of disorder of a thermodynamic system that is associated with the conversion of heat or enthalpy into work.

Entropy change can be calculated by using the given formula:

[tex]\text{dS}=n R ln \frac{V_2}{V_1}[/tex]

⇒For CO₂,

From ideal gas equation it is known that: pV=nRT

∴ nR=pV/T

[tex]nR=\frac{10^5 *0.7}{298} =234.89[/tex]

So, entropy change for CO₂ will be:

[tex]dS_{CO_2}=234.89ln (\frac{1}{0.7} )=83.77J/K[/tex]

⇒For N₂,

[tex]nR=\frac{10^5*0.3}{298} =100.67[/tex]

So, entropy change for N₂ will be:

[tex]dS_{N_2}=100.67 ln(\frac{1}{0.3}) =121.20 J/K[/tex]

Now, total entropy change can be calculated as:

[tex]dS_{Total}=dS_{CO_2}+dS_{N_2}\\\\=83.77+121.20\\\\=204.97 J/K[/tex]

Thus, entropy change will be 204.97 J/K.

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Draw a resonance structure, complete with all formal charges and lone (unshared) electron pairs, that shows the resonance interaction of the carboxy with the para position in benzoic acid.

Answers

Answer:see the picture attached

Explanation:

Final answer:

Resonance structures for benzoic acid highlight electron delocalization, particularly the interaction between the carboxyl group and the para position on the benzene ring. Valence electrons are shown to resonate, creating multiple valid forms which collectively define the resonance hybrid of the molecule.

Explanation:

Resonance Structures of Benzoic Acid

When drawing resonance structures for benzoic acid, especially illustrating the resonance interaction with the para position (position opposite to the carboxyl group), we focus on the delocalization of π-electrons within the benzene ring and the adjacent carboxyl group. For the carboxyl group, one of the oxygen atoms will have a double bond with the carbon to fulfill the octet rule. However, due to the equivalent nature of the oxygen atoms, the double bond can resonate between the two oxygens, creating two resonance forms. At the para position, another resonance form is created by the movement of π-electrons from the benzene ring towards the carboxyl group, thus extending the conjugation and delocalization of electrons throughout the molecule.

Each resonance structure will be carefully drawn ensuring that all atoms obey the octet rule and formal charges are correctly assigned. It's important to note that the actual molecule is better represented by a resonance hybrid, which is an average of all valid resonance forms. This concept signifies that electrons are not strictly localized in one structure, but are distributed across the molecule in a delocalized π-electron cloud.

Which of the pairs below would be the best choice for a pH 5 buffer? HF/NaF, K a (HF) = 3.5 × 10-4 HC2H3O2/KC2H3O2, K a (HC2H3O2) = 1.8 × 10-5 NH3/NH4Cl, K b (NH3) = 1.8 × 10-5

Answers

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

Henderson- Hasselbalch equation:

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

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Carbon tetrachloride can be produced by this reaction: Suppose 1.1 mol and 3.3 mol are placed in a 1.00-L flask and the flask is sealed. After equilibrium has been achieved, the mixture contains 0.82 mol . Calculate .

Answers

The question is incomplete, complete question is:

Carbon tetrachloride can be produced by this reaction:

[tex]CS_2(g) + 3Cl_2(g)[/tex] ⇌ [tex]S_2Cl_2(g) + CCl_4 (g)[/tex]

Suppose 1.1 mol  and 3.3 mol  are placed in a 1.00-L flask and the flask is sealed. After equilibrium has been achieved, the mixture contains 0.82 mol  .

Calculate [tex]K_c[/tex].

Answer:

The value of the [tex]K_c[/tex] of the reaction is 4.05.

Explanation:

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

Initial concentration of [tex]CS_2[/tex]:

[tex][CS_2]=\frac{1.1 mol}{1 L}=1.1 M[/tex]

Initial concentration of [tex]Cl_2[/tex]:

[tex][Cl_2]=\frac{3.3mol}{1 L}=3.3M[/tex]

Equilibrium concentration of [tex]CCl_4[/tex]:

[tex][CCl_4]=\frac{0.82 mol}{1 L}=0.82 M[/tex]

[tex]CS_2(g) + 3Cl_2(g)[/tex] ⇌ [tex]S_2Cl_2(g) + CCl_4 (g)[/tex]

initially :

1.1 M     3.3 M           0          0

At equilibrium

(1.1-0.82) M     (3.3-3 × 0.82) M                      0.82 M    0.82 M

0.28 M      0.84                           0.82     0.82

The expression of equilibrium constant [tex]K_c[/tex] is given by :

[tex]K_c=\frac{[S_2Cl_2][CCl_4]}{[CS_2][Cl_2]^3}[/tex]

[tex]=\frac{0.82 M\times 0.82 M}{0.28M\times (0.84 M)^3}=4.05[/tex]

The value of the [tex]K_c[/tex] of the reaction is 4.05.

A farmer uses triazine herbicide to control pigweed in his field. For the first few years, the triazine works well and almost all the pigweed dies; but after several years, the farmer sees more and more pigweed. Which of these explanations best explains what happened?

A. The herbicide company lost its triazine formula and started selling poor-quality triazine.
B. Triazine-resistant weeds were more likely to survive and reproduce
C. Natural selection caused the pigweed to mutate, creating a new triazine-resistant species
D. Triazine-resistant pigweed has less efficient photosynthesis metabolism.

Answers

Answer:

B. Triazine-resistant weeds were more likely to survive and reproduce

Explanation:

According to darwin's theory of evolution, variation is already present in some members of a population and this variation lets them survive in the adverse condition and those who do not have that variation which helps in survival are lost with time.

So as before using triazine herbicide the major population of the weed were not resistant to this herbicide so in the first few years the nonresistant weeds were lost and only resistant weed which was very less in number survived.

So after several years these resistant weeds reproduced and transferred their gene to their offsprings and became predominant in the field. Therefore the correct answer is B.

Triazine-resistant weeds survived and reproduced due to natural selection, rendering the herbicide less effective over time.

Option (B) is correct.

Option B, the development of triazine-resistant weeds, is the most likely explanation for the increasing pigweed problem. Over time, the repeated use of the same herbicide, like triazine, exerts selective pressure on weed populations.

Some pigweed plants may carry genetic mutations that make them naturally resistant to the herbicide. When the herbicide kills most pigweed, these resistant individuals survive and pass on their resistant traits to their offspring. Eventually, the population becomes dominated by triazine-resistant pigweed, making the herbicide less effective.

This is a classic example of natural selection and the evolution of herbicide resistance in weed populations, a common issue in agriculture when the same herbicide is used repeatedly. The other options are less plausible or unrelated to herbicide resistance.

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Pricing the product low in order to stimulate demand and increase the installed base, then trying to make high profits on the sale of complements that are relatively high in price, is known as the:

Answers

Answer:

razor and blade strategy

Explanation:

Razor and blade strategy -

It refers to the method of pricing , where the price of one of the item is reduced in order to increase the sale of another item , is referred to as razor and blade strategy .

It is a type of pricing tactics used by the company to indirectly earn profit for their goods and services .

Sometimes , fews goods are given free along with certain products , in order to earn profit .

Hence , from the given information of the question ,

The correct answer is razor and blade strategy .

At equilibrium, the concentrations in this system were found to be [ N 2 ] = [ O 2 ] = 0.200 M and [ NO ] = 0.400 M . N 2 ( g ) + O 2 ( g ) − ⇀ ↽ − 2 NO ( g ) If more NO is added, bringing its concentration to 0.700 M, what will the final concentration of NO be after equilibrium is re‑established?

Answers

Answer: The equilibrium concentration of NO after it is re-established is 0.55 M

Explanation:

For the given chemical equation:

[tex]N_2(g)+O_(g)\rightleftharpoons 2NO(g)[/tex]

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

[tex]K_{eq}=\frac{[NO]^2}{[N_2][O_2]}[/tex]     .....(1)

We are given:

[tex][NO]_{eq}=0.400M[/tex]

[tex][N_2]_{eq}=0.200M[/tex]

[tex][O_2]_{eq}=0.200M[/tex]

Putting values in expression 1, we get:

[tex]K_{eq}=\frac{(0.400)^2}{0.200\times 0.200}\\\\K_{eq}=4[/tex]

Now, the concentration of NO is added and is made to 0.700 M

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle. This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

The equilibrium will shift in backward direction.

                           [tex]N_2(g)+O_(g)\rightleftharpoons 2NO(g)[/tex]

Initial:               0.200    0.200        0.700

At eqllm:      0.200+x   0.200+x     0.700-2x

Putting values in expression 1, we get:

[tex]4=\frac{(0.700-2x)^2}{(0.200+x)\times (0.200+x)}\\\\x=0.075[/tex]

So, equilibrium concentration of NO after it is re-established = (0.700 - 2x) = [0.700 - 2(0.075)] = 0.55 M

Hence, the equilibrium concentration of NO after it is re-established is 0.55 M

Sulfur and fluorine react in a combination reaction to produce sulfur hexafluoride: S (s) 3F 2(g) SF 6 (g) The maximum amount of SF 6 that can be produced from the reaction of 3.5 g of sulfur with of fluorine is ________ g.

Answers

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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g 2BrO3- + 5SnO22-+ H2O5SnO32- + Br2+ 2OH- In the above reaction, the oxidation state of tin changes from to . How many electrons are transferred in the reaction

Answers

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⁻

Consider the reaction 3Fe2O3(s) + H2(g)2Fe3O4(s) + H2O(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.68 moles of Fe2O3(s) react at standard conditions. S°surroundings = J/K An error has been detected in your answer. Check

Answers

Answer:

the entropy change for the surroundings when 1.68 moles of Fe2O3(s) react at standard conditions = 49.73 J/K.

Explanation:

3Fe2O3(s) + H2(g)-----------2Fe3O4(s) + H2O(g)

∆S°rxn = n x sum of ∆S° products - n x sum of ∆S° reactants

∆S°rxn = [2x∆S°Fe3O4(s) + ∆S°H2O(g)] - [3x∆S°Fe2O3(s) + ∆S°H2(g)]

∆S°rxn = [(2x146.44)+(188.72)] - [(3x87.40)+(130.59)] J/K

∆S°rxn = (481.6 - 392.79) J/K =88.81J/K.

For 3 moles of Fe2O3 react, ∆S° =88.81 J/K,

then for 1.68 moles Fe2O3 react, ∆S° = (1.68 mol x 88.81 J/K)/(3 mol) = 49.73 J/K the entropy change for the surroundings when 1.68 moles of Fe2O3(s) react at standard conditions.

A chemist designs a galvanic cell that uses these two half-reactions:

Half-reaction Standard reduction potential
O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l) E⁰ red = +1.23 V
Fe³⁺(aq) + e⁻ → Fe²⁺(aq) E⁰ red = +0.771 V

(a) Write a balanced equation for the half-reaction that happens at the cathode.
(b) Write a balanced equation for the half-reaction that happens at the anode.
(c) Write a balanced equation for the overall reaction that powers the cell. Be sure the reaction is spontaneous as written.
(d) Do you have enough information to calculate the cell voltage under standard conditions?

Answers

Answer :

(a) Reaction at anode (oxidation) : [tex]4Fe^{2+}\rightarrow 4Fe^{3+}+4e^-[/tex]  

(b) Reaction at cathode (reduction) : [tex]O_2+4H^++4e^-\rightarrow 2H_2O[/tex]  

(c) [tex]O_2+4H^++4Fe^{2+}\rightarrow 2H_2O+4Fe^{3+}[/tex]

(d) Yes, we have have enough information to calculate the cell voltage under standard conditions.

Explanation :

The half reaction will be:

Reaction at anode (oxidation) : [tex]Fe^{2+}\rightarrow Fe^{3+}+e^-[/tex]     [tex]E^0_{anode}=+0.771V[/tex]

Reaction at cathode (reduction) : [tex]O_2+4H^++4e^-\rightarrow 2H_2O[/tex]     [tex]E^0_{cathode}=+1.23V[/tex]

To balance the electrons we are multiplying oxidation reaction by 4 and then adding both the reaction, we get:

Part (a):

Reaction at anode (oxidation) : [tex]4Fe^{2+}\rightarrow 4Fe^{3+}+4e^-[/tex]     [tex]E^0_{anode}=+0.771V[/tex]

Part (b):

Reaction at cathode (reduction) : [tex]O_2+4H^++4e^-\rightarrow 2H_2O[/tex]     [tex]E^0_{cathode}=+1.23V[/tex]

Part (c):

The balanced cell reaction will be,

[tex]O_2+4H^++4Fe^{2+}\rightarrow 2H_2O+4Fe^{3+}[/tex]

Part (d):

Now we have to calculate the standard electrode potential of the cell.

[tex]E^o=E^o_{cathode}-E^o_{anode}[/tex]

[tex]E^o=(1.23V)-(0.771V)=+0.459V[/tex]

For a reaction to be spontaneous, the standard electrode potential must be positive.

So, we have have enough information to calculate the cell voltage under standard conditions.

Integrated Problem 17.52 Get help answering Molecular Drawing questions. Draw the structure of a compound with the molecular formula C9H10O2 that exhibits the following spectral data. (a) IR: 3005 cm-1, 1676 cm-1, 1603 cm-1 (b) 1H NMR: 2.6 ppm (singlet, I

Answers

Answer:

The answer is attached and other details about the answer is also attached.

Explanation:

The molecular formula (C9H10O2) indicates five

degrees of unsaturation (see Section 14.16), which is

strongly suggestive of an aromatic ring, as well as one

additional double bond or ring. The signal just above

3000 cm-1 in the IR spectrum confirms the aromatic ring,

as does the signal just above 1600 cm-1. The 1

H NMR

spectrum exhibits two doublets between 6.9 and 7.9

ppm, each with an integration of 2. This is the

characteristic pattern of a disubstituted aromatic ring, in

which the two substituents are different from each other:

The singlet at 3.9 ppm (with an integration of 3)

represents a methyl group. The chemical shift is

downfield from the expected benchmark value of 0.9

ppm for a methyl group, indicating that it is likely next to

an oxygen atom:

The singlet at 2.6 ppm (with an integration of 3)

represents an isolated methyl group. The chemical shift

of this signal suggests that the methyl group is

neighboring a carbonyl group:

The carbonyl group accounts for one degree of

unsaturation, and together with the aromatic ring, this

would account for all five degrees of unsaturation. The

presence of a carbonyl group is also confirmed by the

signal at 196.6 ppm in the 13C NMR spectrum.

We have uncovered three pieces, which can only be

connected in one way, as shown:

This structure is consistent with the 13C NMR data: four

signals for the sp2 hybridized carbon atoms of the

aromatic ring, and two signals for the sp3 hybridized

carbon atoms (one of which is above 50 ppm because it

is next to an oxygen atom).

Also notice that the carbonyl group is conjugated to the

aromatic ring, which explains why the signal for the

C=O bond in the IR spectrum appears at 1676 cm-1,

rather than 1720 cm-1.

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