The complete combustion of octane, a component of gasoline, is represented by the equation: 2 C8H18(l) + 25 O2(g) →16 CO2(g) + 18 H2O(l) How many liters of CO2(g), measured at 63.1°C and 688 mmHg, are produced for every gallon of octane burned? (1 gal = 3.785 L; density of C8H18(l) = 0.703 g/mL)

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 %

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:

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 : 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:

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