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3 years ago

The equilibrium constant is the ratio of the concentrations of the reactants over the concentrations of the products.

Chemistry

1 answer:

AfilCa [17]3 years ago

4 0

Answer:

The answer is false. See the explanation below, please

Explanation:

The equilibrium constant is the ratio of the concentrations of the products over the concentrations of the reactants.

See the example below:

A + B --> C + D

We calculate the Keq:

Keq = (C) x (D) / (A) x (B)

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How man grams of cl2 are consumed to produce 12.0 g of KCl

Korvikt [17]

Answer:

5.71 g

Explanation:

Step 1: Write the balanced equation

2 K + Cl₂ ⇒ 2 KCl

Step 2: Calculate the moles corresponding to 12.0 g of KCl

The molar mass of KCl is 74.55 g/mol.

12.0 g × 1 mol/74.55 g = 0.161 mol

Step 3: Calculate the moles of Cl₂ needed to produce 0.161 moles of KCl

The molar ratio of Cl₂ to KCl is 1:2. The moles of Cl₂ needed are 1/2 × 0.161 mol = 0.0805 mol

Step 4: Calculate the mass corresponding to 0.0805 moles of Cl₂

The molar mass of Cl₂ is 70.91 g/mol.

0.0805 mol × 70.91 g/mol = 5.71 g

8 0

3 years ago

Determine the freezing point and boiling point of a solution that has 68.4 g of sucrose

Ymorist [56]

Answer:

Freezing T° of solution = - 3.72°C

Boiling T° of solution = 101.02°C

Explanation:

To solve this we apply colligative properties. Firstly, freezing point depression:

ΔT = Kf . m . i

ΔT = Freezing T° of pure solvent - Freezing T° of solution

Kf = Cryoscopic constant, for water is 1.86 °C/m

m = molality (moles of solute in 1kg of solvent)

i = Ions dissolved in solution

Our solute is sucrose, an organic compound so no ions are defined. i = 1.

Let's determine the moles: 68.4 g . 1mol/ 342g = 0.2 moles

molality = 0.2 mol / 0.1kg of water = 2 m

We replace data: ΔT = 1.86°C/m . 2m . 1

Freezing T° of solution = - 3.72°C

Now, we apply elevation of boiling point: ΔT = Kb . m . i

ΔT = Boiling T° of solution - Boiling T° of pure solvent

Kf = Ebulloscopic constant, for water is 0.512 °C/m

We replace:

Boiling T° of solution - Boiling T° of pure solvent = 0.512 °C/m . 2 . 1

Boiling T° of solution = 0.512 °C/m . 2 . 1 + 100°C → 101.02°C

6 0

2 years ago

Which one would be expected to happen? If one of the buffers that contribute to pH stability in human blood is carbonic acid (H2

tino4ka555 [31]

Answer:

Option (D) is definitely the answer.

Explanation:

Before going further, it is important to know what buffers and pH represent, which are keywords to answering this question.

Buffers is a special solution that can withstand or resist changes due to pH levels which may be as a result of an introduction of acidic or basic components into the blood. In other words, they maintain the stability of pH level in the human blood.

pH blood levels on the other hand, can be grouped into three: acidity, neutrality and alkalinity. Using a pH scale, one can determine its current level. In the human blood the pH level is near neutral and needs to be on a level near 7.4 in order to avoid a high rise or a drastic fall even if acidic or basic components come in or departs the blood stream.

Therefore, if one of the buffers that contributes to pH stability in human blood is carbonic acid, which is as a result of a combination of carbon dioxide and water in the blood stream. On getting to the lungs it is converted to water and subsequently released as waste. Maintaining this stability will definitely be to decrease the concentration of carbonic acid and increase that of water instead.

5 0

3 years ago

1. A 67 kg water-skier is being pulled by a speedboat. The force causes her to accelerate at 3.8

Alborosie

Answer:

254.6N

Explanation:

F=ma

F=67×3.8

F=254.6N

4 0

3 years ago

Light energy is a form of A. neither potential nor kinetic energy. B. kinetic energy only. C. both potential and kinetic energy.

GarryVolchara [31]

Light energy is a form of electromagnetic radiation. Light consists of photons, which are produced when an object's atoms heat up. Light travels in waves and is the only form of energy visible to the human eye.

3 0

3 years ago

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