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1 month ago

A driver with a nearly empty fuel tank may say she is "running on fumes."

1 answer:

6 0

If the vehicle runs at 20.0 miles per gallon using liquid gasoline, the maximum distance it can cover is 1.62 miles. The fuel tank of the car holds 15 gallons, which translates to 56.781 liters using the conversion of 15 × 3.7854. To calculate the number of moles of gasoline, we apply the ideal gas equation: PV = nRT. Rearranging for n gives us n = PV/RT, where: P = 747 mmHg V = 56.781 liters R (the universal gas constant) = 0.0821 liter·atm/mol·K T = 25 ∘C, which converts to 298 K (273 + 25). 1 atm equals 760 mmHg. Thus, we find n = 747 × 56.781 / (0.0821 × 760 × 298) which equals 2.281 mol. Given that the molar mass of gasoline is 101 g/mol, we calculate the mass of the gasoline as n multiplied by the molar mass: 2.281 mol × 101 g/mol results in 230.38 g. The density of liquid gasoline is 0.75 g/mL. To find the volume of gasoline available, we divide the mass of the gasoline by its density, resulting in 230.38 g ÷ 0.75 g/mL = 307.17 mL, which converts to 0.3071 liters, or 0.0811 gallons when divided by 3.7854. With the car achieving 20.0 miles per gallon, the distance it will cover equals gallons available multiplied by miles per gallon, amounting to 0.0811 × 20, totaling 1.62 miles.

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66.667 mL of 3.000 M H2SO4 (aq) solution was neutralized by the stoichiometric amount of 4.000 M Al(OH)3 solution in a coffee cu

eduard [2782]

Answer:

$\large \boxed{\Delta_{\textbf{r}}H =\text{-4600 J$\cdot$ mol}^{-1}}$

Explanation:

This scenario is unrealistic since Al(OH)₃ is not soluble in water.

The question consists of two parts:

A. Stoichiometry — where we determine volumes, masses, and moles for the products

B. Calorimetry — where we assess the enthalpy of the reaction.

A. Stoichiometry

1. Determine the volume of Al(OH)₃

(a) The balanced chemical equation:

2Al(OH)₃ + 3H₂SO₄ ⟶ Al₂(SO₄)₃ + 6H₂O

M/V: 66.667

c/mol·L⁻¹: 4.000 3.000

(b) Moles of H₂SO₄

$\rm \text{66.667 mL H$_{2}$}SO_{4} \times \dfrac{\text{3.000 mmol H$_{2}$SO}_{4}}{\text{1 mL H$_{2}$SO}_{4}} = \text{200.00 mmol H$_{2}$SO}_{4}$

(c) Moles of Al(OH)₃

The molar ratio stands at 2 mmol Al(OH)₃: 3 mmol H₂SO₄

$\text{Moles of Al(OH)}_{3} = \text{200.00 mmol of H$_{2}$SO}_{4} \times \dfrac{\text{2 mmol Al(OH)}_{3}}{\text{3 mmol H$_{2}$SO}_{4}}\\\\= \text{133.33 mmol Al(OH)}_{3}$

(d) Volume of Al(OH)₃

$\text{Moles of Al(OH)}_{3} = \text{200.00 mmol of H$_{2}$SO}_{4} \times \dfrac{\text{1 mL Al(OH)}_{3}}{\text{4 mmol H$_{2}$SO}_{4}} = \text{50.000 mL Al(OH)}_{3}$

B. Calorimetry

This reaction has two energy exchanges.

q₁ = heat from the reaction

q₂ = heat used to heat the calorimeter

q₁ + q₂ = 0

nΔH + mCΔT = 0

Data:

Moles of Al₂(SO₄)₃ = 0.066 667 mol

C = 1.10 J°C⁻¹g⁻¹

T_initial = 22.3 °C

T_final = 24.7 °C

Calculations

(a) Mass of solution

Assume solutions are as dense as water (though not realistic).

Mass of sulfuric acid solution = 66.667 g

Mass of aluminium hydroxide solution = 50.000

TOTAL = 116.667 g

(b) ΔT

ΔT = T_final - T_initial = 24.7 °C - 22.3 °C = 2.4°C

(c) ΔH

$\begin{array}{ccccl}n\Delta H & +& mC \Delta T& = &0\\\text{0.066 667 mol }\times \Delta H& + & \text{116.667 g} \times 1.10 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times 2.4 \, ^{\circ}\text{C} & = & 0\\0.066667 \Delta H \text{ mol} & + & \text{310 J} & = & 0\\&&0.066667 \Delta H \text{ mol} & = & \text{-310 J} & & \\\end{array}\\$

$\begin{array}{ccccl}& &\Delta H & = & \dfrac{\text{-310 J}}{\text{0.066667 mol}}\\\\& &\Delta H & = & \textbf{-4600 kJ/mol}\\\end{array}\\\large \boxed{\mathbf{\Delta_{\textbf{r}}H} =\textbf{-4600 J$\cdot$ mol}^{\mathbf{-1}}}$

This result appears nonsensical, but it is derived from your given figures.

6 0

2 months ago

A piece of iron metal is heated to 155 degrees C and placed into a calorimeter that contains 50.0 mL of water at 18.7 degrees C.

VMariaS [2998]

Jsjsjsdjdjkdskkeekdks probably answer a

3 0

3 months ago

Read 2 more answers

What is the total number of atoms in 0.20 mol of propanone, CH3COCH3?

KiRa [2933]

Answer:

1.2×10²³ atoms.

Explanation:

In the problem, we see the data:

Mole of propanone = 0.20 mole

Calculating the number of atoms in propanone =?

According to Avogadro's principle, one mole of a substance contains 6.022×10²³ atoms.

This means that one mole of propanone also holds 6.022×10²³ atoms.

Thus, we can determine the atom count in 0.20 mole of propanone as:

1 mole of propanone contains 6.022×10²³ atoms.

Accordingly, 0.20 mole of propanone will have = 0.2 × 6.022×10²³ = 1.2×10²³ atoms.

Therefore, 0.20 mole of propanone contains

1.2×10²³ atoms.

6 0

1 month ago

Suppose you wanted to make a buffer of exactly ph 7.00 using kh2po4 and na2hpo4. if the final solution was 0.10 m in kh2po4, wha

Tems11 [2777]

The answer is - 0.138 M. The buffer pH can be determined using the Henderson equation. Here, $KH_2PO_4$ acts as a weak acid and $Na_2HPO_4$ serves as its corresponding conjugate base. The weak acid has two protons, while the base contains one. The equation can therefore be expressed in terms of protons transferred. Phosphoric acid can donate protons in three stages; the equation we’ve referenced pertains to the second stage, as the acid then has only two protons available and the base only one. Given the concentration of the acid as 0.10 M, we need to calculate the concentration of the base necessary to form a buffer with a pH of exactly 7.0. Substituting the values into the equation leads us to the solution. Cross-multiplying, we find that [base] = 1.38(0.10), yielding [base] = 0.138. Therefore, the concentration of the base needed for the buffer is 0.138 M.

5 0

1 month ago