Which of the following molecules is nonpolar? O A. CO2 O B. NO ос. со O D. SO2

What can you determine about the feasibility of a reaction if the enthalpy is positive and

kvasek [131]

If the enthalpy is positive and the entropy is positive, the Gibbs energy will always be positive, and the reaction will never be feasible.

<h3>What is the Gibbs Free Energy?</h3>

The Gibb Free Energy is used to obtain the feasibility of a reaction. If the Gibbs free energy is positive the reaction is not spontaneous. If the value is negative, the reaction is spontaneous while a zero values indicates equilibrium.

From the equation;

ΔG = ΔH - TΔS, it follows that if the enthalpy is positive and the entropy is positive, the Gibbs energy will always be positive, and the reaction will never be feasible.

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7 0

2 years ago

What is the empirical formula of a compound composed of 30.5 g potassium (k) and 6.24 g oxygen (o)?

VLD [36.1K]

K5O2

convert grams to moles, divide both by the smallest mole mass, multiply that until hole.

30.5 g K ÷ 39.10 = .78 mol
6.24 g O ÷ 16 = .39 mol
.78 mol ÷ .39 mol = 2.5
.39 mol ÷ .39 mol = 1
2.5 x 2 = 5
1 x 2 = 2
K5O2

5 0

3 years ago

Potasyum-40 küçük atom numaralı doğal radyoaktif birkaç element izotopundan biridir ve doğada K izotopları içerisindeki bolluk y

enot [183]

Answer:

6.707 × 10¹⁷

Explanation:

From the information given:

40 ^ K kütlesi = sütteki K kütlesi × 40 ^ K / 100 kütle yüzdesi

nerede;

sütteki K kütlesi = 1.65 mg of K/mL × 225 mL = 371.25 mg of K

∴

40 ^ K kütlesi = 371.25 × 0.012/100

40 ^ K kütlesi = 0,04455 mg = 4.455 × 10⁻⁵ grams

40 ^ K mol sayısı = 40 ^ K kütlesi / molar kütle

= 4.455 × 10⁻⁵/40

= 1.11375 × 10⁻⁶

Son olarak, 40 ^ K = mol × Avogadro sayısı atomları

= 1.11375 × 10⁻⁶ × 6.022 × 10^23

= 6.707 × 10¹⁷

3 0

3 years ago

A mixture 21.2 g of P and 79.0 g Cl2 reacts completely to form PCl3 and PCl5 as the only products. find the mass of PCl3

IRISSAK [1]

Considering the definition of reaction stoichiometry and limiting reagent, the mass of PCl₃ that could be formed is 38.26 grams.

In first place, you must know that the balanced reaction is:

4 Cl₂ + 2 P → PCl₃ + PCl₅

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Cl₂: 4 moles
P: 2 moles
PCl₃: 1 mole
PCl₅: 1 mole

The molar mass of the compounds is:

Cl₂: 70.9 g/mole
P: 31 g/mole
PCl₃: 137.35 g/mole
PCl₅: 208.25 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Cl₂: 4 moles× 70.9 g/mole = 283.6 grams
P: 2 moles× 31 g/mole= 62 grams
PCl₃: 1 mole× 137.35 g/mole= 137.35 grams
PCl₅: 1mole× 208.25 g/mole= 208.25 grams

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, it is possible to use a simple rule of three as follows: if by stoichiometry 62 grams of P reacts with 283.6 grams of Cl₂, 21.2 grams of P reacts with how much mass of Cl₂?

$mass of Cl_{2}=\frac{21.2 grams of Px283.6 grams of Cl_{2} }{62 grams of P}$

mass of Cl₂= 96.97 grams

But 96.97 grams of Cl₂ are not available, 79 grams are available. Since you have less mass than you need to react with 21.2 grams of P, Cl₂ will be the limiting reagent.

Then, it is possible to determine the mass of PCl₃ produced by another rule of three, using the limiting reagent: if by stoichiometry 283.6 grams of Cl₂ produce 137.35 grams of PCl₃, 79 grams of Cl₂ how much mass of PCl₃ will be formed?

$mass of PCl_{3}=\frac{137.35 grams of PCl_{3}x79 grams of Cl_{2} }{283.6grams of Cl_{2}}$

mass of PCl₃= 38.26 grams

In summary, the mass of PCl₃ that could be formed is 38.26 grams.

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4 0

2 years ago

A scientist prepared an aqueous solution of a 0.45 M weak acid. The pH of the solution was 2.72. What is the percentage ionizati

aleksandr82 [10.1K]

Answer:

0.42%

Explanation:

∵ pH = - log[H⁺].

2.72 = - log[H⁺]

∴ [H⁺] = 1.905 x 10⁻³.

∵ [H⁺] = √Ka.C

∴ [H⁺]² = Ka.C

∴ ka = [H⁺]²/C = (1.905 x 10⁻³)²/(0.45) = 8.068 x 10⁻⁶.

∵ Ka = α²C.

Where, α is the degree of dissociation.

∴ α = √(Ka/C) = √(8.065 x 10⁻⁶/0.45) = 4.234 x 10⁻³.

∴ percentage ionization of the acid = α x 100 = (4.233 x 10⁻³)(100) = 0.4233% ≅ 0.42%.

4 0

3 years ago

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