What volume of a 0.452 m naoh solution is needed to neutralize 85.0 ml of a 0.176 m solution of h2so4?

Solid aluminum hydroxide reacts with a solution of hydrobromic acid. write a balanced molecular equation and a balanced net ioni

Tems11 [2777]

To determine the balanced chemical equation, it is necessary to identify the symbols and charges for each element and then balance the equation. This results in: Al(OH)3(aq) + 3HBr(aq) ---> AlBr3(aq) + 3H2O(l) The net ionic equation involves summing the charges and breaking them down into fundamental parts. Al3+ + 3OH- + 3H+ + 3Br- --> Al3+ + 3Br + H3O+ After eliminating aluminum and bromine, we have: 3OH-(aq) + 3H+(aq) --> 3H2O(l)

4 0

1 month ago

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Classify each of these soluble solutes as a strong electrolyte, a weak electrolyte, or a nonelectrolyte. Solutes Formula Hydroch

VMariaS [2998]

Answer:

The categorization of strong, weak, and non-electrolytes is detailed below, based on the examples presented in the question.

Explanation:

A strong electrolyte fully dissociates or nearly so in an aqueous environment; typically, strong acids, bases, and salts fall under this category. Examples of strong electrolytes include:

Hydrochloric acid, HCl
Calcium hydroxide, Ca(OH)2
Potassium chloride, KCl

A weak electrolytepartially ionizes in solution; weak acids and bases are primary instances. Examples consist of:

Methylamine, CH3NH2
Hydrofluoric acid, HF

A non-electrolytedoes not dissociate in an aqueous medium. Examples of non-electrolytes are:

Sucrose, C12H22O11
Methanol, CH3OH

5 0

2 months ago

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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

How does vegetation slow and prevent sediment loss?

Alekssandra [3086]

Vegetation cover serves as the most efficient and effective method to curb sediment loss. The roots of plants like grass interlink soil particles, aiding in erosion resistance, particularly against runoff water. Vegetation absorbs the force of raindrops, preventing soil particle detachment. Additionally, plants can lie flat resembling shingles on a roof, enabling runoff to travel over the soil rather than disturbing it. Tall, erect vegetation functions as a barrier against wind, diminishing its force so that it cannot dislodge soil particles from the ground surface.

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2 months ago

Identify the reaction type for each generic chemical equation

VMariaS [2998]

A + B → AB: ✔ combination

AB → A + B: ✔ breakdown

Hydrocarbon + O2 → CO2 + H2O: ✔ oxidation

AB + CD → AD + CB: ✔ exchange

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1 month ago

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