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14 days ago

BRAINLIESTTT ASAP!!!

Chemistry

2 answers:

lorasvet [956]14 days 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 [852]14 days 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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No amino acid molecule by itself can speed up or catalyze reactions between other molecules; however, when amino acids are joine

Anarel [852]

Answer:

This indicates that the enzyme is a type of protein.

Explanation:

It is important to remember that proteins are composed of vast numbers of amino acids. Because these amino acids are tiny units, they cannot function as a catalyst on their own.

However, when they form a polymer, the protein enzyme will possess varying shapes, sizes, and both physical and chemical attributes differing from a single monomer.

Additionally, for proteins to function actively, a specific number of amino acids must combine to create a distinct shape suited to interact with another molecule, thus accelerating the chemical reaction and functioning as an enzyme.

8 0

4 days ago

Read 2 more answers

Consider the following system at equilibrium:

VMariaS [1037]

Answer:

A - Increase (R), Decrease (P), Decrease(q), Triple both (Q) and (R)

B - Increase(P), Increase(q), Decrease (R)

C - Triple (P) and cut (q) down to a third

Explanation:

According to the principle of Le Chatelier, when a system reaches equilibrium and a change is introduced, the system will respond to counteract that change.

Since P and Q are reactants, raising the amount of either one or both without a proportional rise in R (which is a product) will cause the equilibrium to move towards the right. Similarly, if R decreases while P and Q remain constant, this too will push the equilibrium to the right. Thus, Increase(P), Increase(q), and Decrease(R) will lead to a rightward shift in the equilibrium.

Conversely, raising R without increasing P and Q will draw the equilibrium to the left. Likewise, cutting down P and/or Q without a similar reduction in R will shift the equilibrium leftward. Therefore, Increase(R), Decrease(P), Decrease(q), and triple both (Q) and (R) will shift the equilibrium to the left.

If there are equivalent changes in P and Q, with R remaining unchanged, then the equilibrium remains stationary. So, tripling (P) while reducing (q) to one third will not alter the equilibrium.

6 0

3 days ago

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 [944]

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}}}$