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3 years ago

A substance that can be separated into two or more substances only by a chemical change is

1 answer:

5 0

A substance that can be separated into two or more substances only by a chemical change is known as a heterogeneous mixture

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What is the process of using one or more of your five senses to measure or collect data?

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Answer:

B no observation pls mark me branilest

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How many atoms in Alka-Seltzer

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Calculate the molar solubility of copper(II) arsenate (Cu3(AsO4)2) in water. Use 7.6 x 10^-36 as the solubility product constant

Molodets [167]

Molar solubility is the number of moles of a substance (the solute) that can be dissolved per liter of solution before the solution becomes saturated. We calculate as follows:

3Cu2+ + 2(AsO4)3- = Cu3(AsO4)2

7.6 x 10^-36 = (3x^3)(2x^2)
x = 6.62 x 10^-8 M

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3 years ago

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The standard free energy of formation, ΔG∘f, of a substance is the free energy change for the formation of one mole of the subst

OLEGan [10]

Answer:

B. 2 Na(s) + O₂(g) → Na₂O₂(s); ΔG∘f=−451.0 kJ/mol

D. 2 SO(g) + O₂(g) → 2 SO₂(g); ΔG°f=−600.4 kJ/mol

Explanation:

The spontaneity of a reaction is given by the value of the standard Gibbs free energy of the reaction (ΔG°rxn). The more negative is the ΔG°rxn, the more spontaneous is a reaction.

The ΔG°rxn can be calculated using the following expression:

ΔG°rxn = ∑np × ΔG°f(products) − ∑nr × ΔG°f(reactants)

By definition, the standard Gibbs free energy of formation of simple substances in their most stable state is zero. That is why, in the reaction of formation of a compound ΔG°rxn = ΔG°f(product).

Based on the standard free energies of formation, which of the following reactions represent a feasible way to synthesize the product?

A. N₂(g) + H₂(g) → N₂H₄(g); ΔG°f=159.3 kJ/mol.

Not feasible. ΔG°rxn = ΔG°f(product) > 0.

B. 2 Na(s) + O₂(g) → Na₂O₂(s); ΔG°f=−451.0 kJ/mol

Feasible. ΔG°rxn = ΔG°f(product) < 0.

C. 2 C(s) + 2 H₂(g) → C₂H₄(g); ΔG°f=68.20 kJ/mol

Not feasible. ΔG°rxn = ΔG°f(product) > 0.

D. 2 SO(g) + O₂(g) → 2 SO₂(g); ΔG°f=−600.4 kJ/mol

Feasible. ΔG°rxn = ΔG°f(product) < 0.

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3 years ago

What would be the freezing point of a solution that has a molality of 1.324 m which was prepared by dissolving biphenyl (C12H10)

lbvjy [14]

The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.

A solution is prepared by dissolving biphenyl into naphthalene. We can calculate the freezing point depression (ΔT) for naphthalene using the following expression.

$\Delta T = i \times Kf \times m = 1 \times 6.90 \°C/m \times 1.324m = 9.14 \°C$

where,

i: van 't Hoff factor (1 for non-electrolytes)
Kf: cryoscopic constant
m: molality

The normal freezing point of naphthalene is 80.26 °C. The freezing point of the solution is:

$T = 80.26 \° C - 9.14 \° C = 71.12 \° C$

The freezing point of a 1.324 m solution, prepared by dissolving biphenyl into naphthalene, is 71.12 ° C.

Learn more: brainly.com/question/2292439

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2 years ago