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2 months ago

BRAINLIESTTT ASAP!!!

Chemistry

2 answers:

lorasvet [2.7K]2 months ago

8 0

Although I may not be the smartest, I can definitely answer.

This represents a chemical change because the substances' chemical identities were altered. The fizzing was a clear sign, and the temperature increase was another indicator of the reaction.

Anarel [2.9K]2 months ago

4 0

Explanation:

A physical change does not alter the chemical nature of the substance involved.

Examples include changes to the shape, size, phase, mass, or density of an object.

Conversely, a chemical change results in a modification to the chemical composition of the material.

For instance, when yeast granules are added to hydrogen peroxide, the appearance of fizzing and bubbling indicates a chemical reaction producing gas.

Additionally, heat is released causing the temperature to rise.

Therefore, based on these observations, it is clear that a chemical change has occurred.

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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

1 month ago

A hardware salesman measures the mass of a box containing

lorasvet [2795]

Step $1$ : To find the mass of a single washer, divide the total measured mass by the number of washers $\frac{0.4168}{1000} =0.0004168\ Kg$ $0.0004168kg=4.168*10^{-4} kg$ . Step $2$ : Convert kilograms to milligrams knowing $1 kg=1*10^{6} mg$ ; thus $4.168*10^{-4}*10^{6} =4.168*10^{2}mg=416.8 mg$ resulting in the final answer $416.8 mg$ .

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lorasvet [2795]

Response: In transverse waves, the movement occurs perpendicular to the vibration source.

In contrast, longitudinal waves oscillate parallel to the source of vibration.

Both types share a common aspect: they facilitate energy transfer within the respective wave forms.

Clarification:

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In a chemical reaction, 300 grams of reactant A are combined with 100 grams of reactant B. Both A and

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I expect the product's mass to be 400 grams. This belief stems from the law of conservation of mass, which states that mass can neither be created nor destroyed. In a sealed system, the mass of the reactants equals the mass of the products. Therefore, since the total mass of the reactants is 400 grams, the resulting mass of the products must also be 400 grams

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The solubility of glucose at 30°C is 125 g/100 g water. Classify a solution made by adding 550 g of glucose to 400 mL of water a

VMariaS [2998]

At 30°C, glucose has a solubility of 1.25 g per gram of water. Given that the density of water at this temperature is 1 g/mL, the mass corresponding to 400 mL of water is also 400 g. Therefore, the concentration of the solution is calculated as 550 g divided by 400 g of water, which gives 1.375 g of glucose per gram of water. Since this concentration exceeds the solubility limit for glucose at this temperature, the solution can be classified as SATURATED.

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29 days ago

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