Calculate the number of pounds of CO2 released into the atmosphere when a 22.0 gallon tank of gasoline is burned in an automobile engine. Assume that gasoline is primarily octane, C8H18, and that the density of gasoline is 0.692 g⋅mL^−1. This assumption ignores additives. Also, assume complete combustion. CO2 released:

Answer:

Explanation:

The relevant fact here is:

That means that the precipitate that was thrown away, before evaporating the remaining liquid solution under vacuum, does not count; you must only use the amount of solute that was dissolved after cooling the solution to 21°C.

Then, the amount of solute dissolved in the 600 ml solution at 21°C is the weighed precipitate: 0.084 kg = 84 g.

With that, the solubility can be calculated from the followiing proportion:

      ⇒ y = 84. g solute × 100 ml solution / 600 ml solution = 14. g.

The correct number of significant figures is 2, since the mass 0.084 kg contains two significant figures.

The answer is 14. g of solute per 100 ml of solution.

To calculate the solubility of compound X at 21.° C, you would need to know the number of moles of X in the precipitate, which can be calculated if the molar mass of X is known. The solubility can then be obtained by dividing the number of moles by the volume of solution in liters. Without the molar mass, a precise value can't be obtained.

In the given scenario, we can find the solubility of compound X by determining the number of moles in the precipitate and dividing it by the volume of the solution. The mass of the precipitate given is 0.084 kg or 84 g. Assuming compound X is a simple ionic compound that follows a 1:1 ratio like NaCl, the number of moles of compound X in the solution equals the number of moles of the precipitate. Therefore, the solubility of Compound X at 21.° C can be calculated by dividing the number of moles by the volume of solution in liters. A precise value can't be obtained without the molar mass of the compound. However, if we know the molar mass of the compound, we could obtain the molarity. This molarity would represent the solubility of the compound X at 21.° C.

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Due to the conjugate base of the hydrogen atom is aromatic, Hb is regarded as the most acidic. Because the conjugate base of the hydrogen atom Hc is anti-aromatic, it is the least acidic.

The correct options are:

(A) - (a)

(B) - (d)

The hydrogen connected at the heptatriene's tertiary position (at the 7-methyl) would be particularly acidic, as its removal would leave a positive charge that could be transported around the ring via resonance.

The hydrogen connected to the pentadiene (5-methyl) at the tertiary position would not be acidic, as removing it would result in an anti-aromatic structure.

Thus, the least acidic H atom is Hc and the most acidic H atom is Hb.

Learn more about hydrogen atom, here:

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Answer:

n = 5

Explanation:

methane => 1 Carbon => CH₃

ethane => 2 Carbons => C₂H₆

propane => 3 Carbons => C₃H₈

butane => 4 Carbons  => C₄H₁₀

pentane => 5 Carbons => C₅H₁₂

hexane => 6 Carbons => C₆H₁₄

heptane => 7 Carbons => C₇H₁₆ => CH₃(CH₂)₅CH₃

Octante => 8 Carbons => C₈H₁₈

Nonane => 9 Carbons => C₉H₂₀

Decane => 10 Carbons => C₁₀H₂₂

Answer:

6) (a) 0.499; (b) 31.7 %

7) 0.15

Explanation:

6) (a) Absorbance

Beer's Law is

[tex]A = \epsilon cl\\A = \text{35.9 L&\cdot$mol$^{-1}$cm$^{-1}$} $\times$ 0.0278 mol$\cdot$L$^{-1} \times $ 0.5 cm = \mathbf{0.499}[/tex]

(b) Percent transmission

[tex]A = \log {\left (\dfrac{1}{T}}\right)}\\\\\%T = 100T\\\\T = \dfrac{\%T}{100}\\\\\dfrac{1}{T} = \dfrac{100 }{\%T}\\\\A = \log \left(\dfrac{100 }{\%T} \right ) = 2 - \log \%T\\\\0.499 = 2 - \log \%T\\\\\log \%T = 2 - 0.499 = 1.501\\\\\%T = 10^{1.501} = \mathbf{31.7}[/tex]

7) Absorbance

[tex]A = \log \left (\dfrac{I_{0}}{I} \right ) = \log \left (\dfrac{I_{0}}{0.70I_{0}} \right ) = \log \left (\dfrac{1}{0.70} \right ) = -\log(0.70) = \mathbf{0.15}}[/tex]

The absorbance of the solution with the given molar extinction coefficient, concentration, and path length is approximately 0.5, and the percent transmittance is approximately 31.6%. When the transmitted light is 70% of the initial light beam intensity, the absorbance is approximately 0.523.

The subject matter is related to the concept of absorbance in Chemistry. In this specific case, you are required to know and apply the Beer-Lambert law, which states that the absorbance (A) of a solution is directly proportional to its concentration (c) and the path length (l). The formula used is A = εcl, where ε is the molar extinction coefficient.

For part (a), plug the given values into the above formula to get: A = 35.9 L/(mol cm) * 0.0278 mol/L * 0.5 cm = 0.4987. So, the absorbance of the solution is approximately 0.5.

Now, for part (b), the percent transmittance (%T) can be calculated using the relationship A = 2 - log(%T). Solving for %T gives: %T = 10^(2-A) = 10^(2-0.4987) = 31.6%. So the %T is approximately 31.6%.

For part (7), the decrease in the intensity of light by 70% means the transmitted light is 30% of the initial intensity. The absorbance in this case can be calculated directly from the formula A = -log(I/I0) where I/I0 = 0.30. Hence, A = -log(0.30) = 0.523. So the absorbance, when the transmitted light intensity is 70% of the initial light beam intensity, is approximately 0.523.

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Answer:

Explanation:

1) Chemical equation (given):

2) Mole ratio:

3) C₈H₁₈ (l) moles:

Molar mass: 114.2285 g/mol (taken from a table or internet)

4) Proportion:

5) Ideal gas equation:

Substitute with:

Solve for V:

Approximately 30.28 liters of CO2 are produced for every gallon of octane burned.

To find the number of liters of CO2 produced for every gallon of octane burned, we need to use the balanced equation and conversion factors. From the equation, we can see that for 2 moles of octane burned, we get 16 moles of CO2. Using the molar volume of gases at STP (22.4 L/mol), we can convert the moles of CO2 to liters. Finally, we can use the given conversion factor to convert from gallons to liters.

First, let's calculate the moles of CO2 produced from 1 gallon of octane burned:

16 moles CO2 / 2 moles octane x 3.785 L / 1 gallon = 30.28 L CO2

Therefore, for every gallon of octane burned, approximately 30.28 liters of CO2 are produced.

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Answer:

Ka = 4.9 x 10ˉ⁴

Explanation:

HOAc    ⇄    H⁺    +    OAcˉ

Ka = [H⁺][OAcˉ]/[HOAc]

Given pH = 2.58 => [H⁺] = [OAcˉ] = 10ˉ²∙⁵⁸ = 2.63 x 10ˉ³M

Ka = (2.63 x 10ˉ³)²/(0.0142) = 4.9 x 10ˉ⁴

Answer:

the 2nd part

Explanation:

could conversion is called cellular respiration

Glycolysis results in the conversion of one glucose molecule into two pyruvate molecules, a net gain of two ATP molecules, and the reduction of NAD+ to NADH. It does not result in the production of CO2 or the conversion of NADH back to NAD+.

Glycolysis is a metabolic pathway occurring in the cytosol of cells where one glucose molecule is converted into two, three-carbon compounds, specifically, Pyruvate. This process is anaerobic, meaning it doesn't require oxygen. There's also a net production of two ATP molecules per glucose molecule, so the option indicating a net loss is incorrect. Furthermore, in the process of glycolysis, the compound NAD+ is actually reduced to NADH, not the other way around. CO2 production doesn't occur in glycolysis itself but in subsequent processes of cellular respiration namely the Krebs cycle and oxidative phosphorylation.

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Answer : The percent yield of the reaction is, 91.32 %

Explanation :  Given,

Mass of [tex]C_6H_5CH_3[/tex] = 1 Kg = 1000 g

Molar mass of [tex]C_6H_5CH_3[/tex] = 92.14 g/mole

Molar mass of [tex]C_6H_5COOH[/tex] = 122.12 g/mole

First we have to calculate the moles of [tex]C_6H_5CH_3[/tex].

[tex]\text{Moles of }C_6H_5CH_3=\frac{\text{Mass of }C_6H_5CH_3}{\text{Molar mass of }C_6H_5CH_3}=\frac{1000g}{92.14g/mole}=10.85mole[/tex]

Now we have to calculate the moles of [tex]C_6H_5COOH[/tex].

The balanced chemical reaction will be,

[tex]2C_6H_5CH_3+3O_2\rightarrow 2C_6H_5COOH+2H_2O[/tex]

From the balanced reaction, we conclude that

As, 2 moles of [tex]C_6H_5CH_3[/tex] react to give 2 moles of [tex]C_6H_5COOH[/tex]

So, 10.85 moles of [tex]C_6H_5CH_3[/tex] react to give 10.85 moles of [tex]C_6H_5COOH[/tex]

Now we have to calculate the mass of [tex]C_6H_5COOH[/tex]

[tex]\text{Mass of }C_6H_5COOH=\text{Moles of }C_6H_5COOH\times \text{Molar mass of }C_6H_5COOH[/tex]

[tex]\text{Mass of }C_6H_5COOH=(10.85mole)\times (122.12g/mole)=1325.002g[/tex]

The theoretical yield of [tex]C_6H_5COOH[/tex]  = 1325.002 g

The actual yield of [tex]C_6H_5COOH[/tex]  = 1.21 Kg = 1210 g

Now we have to calculate the percent yield of [tex]C_6H_5COOH[/tex]

[tex]\%\text{ yield of }C_6H_5COOH=\frac{\text{Actual yield of }C_6H_5COOH}{\text{Theoretical yield of }C_6H_5COOH}\times 100=\frac{1210g}{1325.002g}\times 100=91.32\%[/tex]

Therefore, the percent yield of the reaction is, 91.32 %

The molarity of the KOH solution is calculated by first determining the number of moles of H2SO4 using its volume and molarity. Then, the number of moles of KOH is calculated based on the stoichiometry of the reaction. Finally, the molarity of KOH is computed using the definition of molarity (moles/volume), which results in a molarity of 0.84 M for the KOH solution.

To solve the student's question regarding the molarity of a KOH solution, we'll refer to the balanced chemical equation which is given as 2KOH(aq) + H2SO4(aq) → K2SO4(aq) + 2H2O(l). This equation tells us that the ratio of moles of KOH to H2SO4 is 2:1. This means 1.00 mole of H2SO4 reacts with 2.00 moles of KOH.

First, calculate the number of moles of H2SO4 using its volume and molarity: moles H2SO4 = volume (L) × molarity (M) = 25.2 mL (which is 0.0252 L) × 1.50 M = 0.0378 mol H2SO4.

Then, calculate the number of moles of KOH, knowing that 2 moles of KOH react with each mole of H2SO4: moles KOH = 2 × moles H2SO4 = 2 × 0.0378 mol = 0.0756 mol KOH.

Finally, calculate the molarity of KOH using the definition of molarity = moles/volume(L): Molarity KOH = moles KOH / volume (L) = 0.0756 mol / 0.090 L = 0.84 M.

So, the molarity of the KOH solution is 0.84 M.

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The molarity of the KOH solution, you can use the balanced chemical equation and stoichiometry to calculate the solution concentration as 0.840 M.

Molarity of KOH Solution:

In the given titration, 25.2 ml of 1.50 M H₂SO₄  neutralized 90.0 ml of KOH solution. By using the balanced chemical equation provided, you can determine that 1 mole of H₂SO₄ reacts with 2 moles of KOH. Using this information, you can calculate the molarity of the KOH solution to be 0.840 M.

Answer: The net ionic equation is written below.

Explanation:

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which do not get involved in the chemical equation. It is also defined as the ions which are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of ammonium perchlorate and water is given as:

[tex]NH_4ClO_4(aq.)+H_2O(l)\rightarrow NH_4OH(aq.)+HClO_4(aq.)[/tex]

Ionic form of the above equation follows:

[tex]NH_4^+(aq.)+ClO_4^-(aq.)+H_2O(l)\rightarrow NH_4OH(aq.)+H^+(aq.)+ClO_4^-(aq.)[/tex]

Ammonium hydroxide will not dissociate into its ions because it is a weak base.

As, chlorate ions are present on both the sides of the reaction, thus, it will not be present in the net ionic equation.

The net ionic equation for the above reaction follows:

[tex]NH_4^+(aq.)+H_2O(l)\rightarrow NH_3^+(aq.)+H_3O^+(aq.)[/tex]

Hence, the net ionic equation is given above.

The net ionic equation for the acid-base hydrolysis equilibrium of ammonium perchlorate in water involves the dissociation of ions and the formation of hydronium ions.

The net ionic equation for the acid-base hydrolysis equilibrium when ammonium perchlorate is dissolved in water can be written as follows:

NH₄ClO₄ + H₂O → NH₄⁺ + ClO₄⁻ + H₃O⁺

This equation represents the dissociation of ammonium perchlorate into its ions in water, with the presence of hydronium ions due to the protonation of water.

Answer : The temperature in kelvins is, [tex]T>1108.695K[/tex]

Explanation : Given,

[tex]\Delta H[/tex] = 178.5 KJ/mole = 178500 J/mole

[tex]\Delta S[/tex] = 161.0 J/mole.K

Gibbs–Helmholtz equation is :

[tex]\Delta G=\Delta H-T\Delta S[/tex]

As per question the reaction is spontaneous that means the value of [tex]\Delta G[/tex] is negative or we can say that the value of [tex]\Delta G[/tex] is less than zero.

[tex]\Delta <0[/tex]

The above expression will be:

[tex]0>\Delta H-T\Delta S[/tex]

[tex]T\Delta S>\Delta H[/tex]

[tex]T>\frac{\Delta H}{\Delta S}[/tex]

Now put all the given values in this expression, we get :

[tex]T>\frac{178500J/mole}{161.0J/mole.K}[/tex]

[tex]T>1108.695K[/tex]

Therefore, the temperature in kelvins is, [tex]T>1108.695K[/tex]

Answer:

-720.2 kJ

Explanation:

We have two equations:

(I)   X + ½O₂ ⟶ XO;       ΔH = −549.3 kJ

(II) XCO₃ ⟶ XO + CO₂; ΔH =  +170.9 kJ  

From these, we must devise the target equation:

(III) X +½O₂ +CO₂ ⟶ XCO₃; ΔH = ?

The target equation has X on the left, so you rewrite Equation(I).

(I) X + ½O₂ ⟶ XO; ΔH = −549.3 kJ

Equation (IV) has XO on the right, and that is not in the target equation.

You need an equation with XO on the left, so you reverse Equation (II).  

When you reverse an equation, you reverse the sign of its ΔH.  

(IV)  XO + CO₂ ⟶ XCO₃ ;  ΔH = -170.9 kJ  

Now, you add equations (I) and (IV), cancelling species that appear on opposite sides of the reaction arrows.

When you add equations, you add their ΔH values.

You get the target equation (III):

(I)   X + ½O₂ ⟶ XO;                ΔH = -549.3 kJ

(IV) XO + CO₂ ⟶ XCO₃ ;       ΔH =  -170.9 kJ  

(III) X + ½O₂ + CO₂ ⟶ XCO₃; ΔH = -720.2 kJ

The enthalpy change ΔH for the reaction X(s)+1/2O2(g)+CO2(g)⟶XCO3(s) is +720.2 kJ, according to Hess's law.

The enthalpy change ΔH for the reaction X(s)+1/2O2(g)+CO2(g)⟶XCO3(s) can be deduced using Hess's law. According to Hess's law, the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps into which a reaction can be divided.

The two provided reactions can be manipulated to yield the desired reaction:

1. Reverse the first reaction and change the sign of ΔH:

XO(s)⟶X(s)+1/2O2(g) with ΔH=+549.3 kJ

2. Keep the second reaction as is:

XCO3(s)⟶XO(s)+CO2(g) with ΔH=+170.9 kJ

Adding these reactions gives the desired reaction:

X(s)+1/2O2(g)+CO2(g)⟶XCO3(s)

The enthalpy change ΔH for the overall reaction is the sum of the enthalpy changes for the individual reactions, or ΔH= (+549.3 kJ) + (+170.9 kJ) = +720.2 kJ.

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The equilibrium constant, Kc, of the given reaction is calculated by substitifying the equilibrium concentrations into the Kc expression. Considering the chemical equation 2 HI(g) + Cl2(g) ⇌ 2 HCl(g) + I2(s), the Kc expression is [HCl]^2 / ([HI]^2 * [Cl2]). Substituting the given equilibrium concentrations in, Kc equates to 4.67 × 10^33.

The equilibrium constant (Kc) for a chemical reaction is a number that expresses the ratio of the concentrations of the products to the concentrations of the reactants, each raised to the power of its stoichiometric coefficient in the balanced chemical equation. In the equation given: 2 HI(g) + Cl2(g) ⇌ 2 HCl(g) + I2(s), I2 (s) is a pure solid and according to the rules of equilibrium, pure solids and liquids are not included in the equilibrium expression. Therefore, the equilibrium constant expression does not include I2.

For this reaction, the Kc expression is: Kc = [HCl]^2 / ([HI]^2 * [Cl2])

Substituting the equilibrium concentrations into this expression gives: Kc = (0.13 M)^2 / ((5.6 × 10^-16 M)^2 * 0.0019 M) = 4.67 × 10^33

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Answer:

Wt Avg Atomic Wt = 28.0577 amu

Explanation:

Atomic Mass = ∑Wt Avg contributions of isotopes of an element

Si-28 => 92.21% => 0.9221 decimal fraction => Wt Avg Si-28 = 0.9221(27.9769) = 25.7695 amu

Si-29 => 4.69% => 0.0469 decimal fraction => Wt Avg Si-29 = 0.0469(28.9765) = 1.3590 amu

Si-30 => 3.10% => 0.0310 decimal fraction => Wt Avg Si-30 = 0.0310(29.9737) = 0.9292

Wt Avg Atomic Wt = ∑Wt Avg Contributions = (25.7695 + 1.3590 + 0.9292)amu = 28.0577 amu

Answer :  The average atomic mass of silicon is, 28.08 amu

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

[tex]\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i[/tex]

For isotope Si-28 :

Mass of isotope Si-28  = 27.9769 amu

Fractional abundance of isotope Si-28  = 92.21 % = 0.9221

For isotope Si-29 :

Mass of isotope Si-29 = 28.9765 amu

Fractional abundance of isotope Si-29 = 4.69 % = 0.0469

For isotope Si-30 :

Mass of isotope Si-30 = 29.9737 amu

Fractional abundance of isotope Si-30 = 3.10 % = 0.0310

Putting values in equation 1, we get:

[tex]\text{Average atomic mass }=[(27.9769\times 0.9221)+(28.9765\times 0.0469)+(29.9737\times 0.0310)][/tex]

[tex]\text{Average atomic mass }=28.08amu[/tex]

Thus, the average atomic mass of silicon is, 28.08 amu