Redox reactions stand for oxidation/reduction reactions. True or false: an oxidation reaction is always paired with a reduction reaction.

Answers

Answer 1

Answer:

Answer:

The answer is: true

Explanation:

In redox reactions, the half-reactions of oxidation and reduction always occur simultaneously in pair.

The oxidation half-reaction involved the lost of electrons from a reduced substance (A) to form a oxidized substance (A⁺):

A ⇒ A⁺ + e-

In contrapossition, during the reduction half-reaction the oxidized substance (B⁺) gains electrons to form the reduced subtance (B):

B⁺ + e- ⇒ B

The overall redox reaction is obtained by the addition of the two half-reactions:

A ⇒ A⁺ + e-

B⁺ + e- ⇒ B

-----------------

A + B⁺⇒ A⁺ + B

The electrons gained by B are provided by A, which lost the same number of electrons. Thus, the oxidation/reduction reactions are paired.

Answer 2

Answer:

Final answer:

Yes, it's true that an oxidation reaction always pairs with a reduction reaction, thereby making up a redox reaction where one substance is oxidized (loses electrons) and another is reduced (gains electrons). The oxidized species is the reducing agent while the reduced one is the oxidizing agent.

Explanation:

The statement is true: an oxidation reaction is indeed always paired with a reduction reaction. This can be exemplified in the redox reactions where one substance is oxidized (loses electrons) while another is reduced (gains electrons). These reactions always occur together. The species that is oxidized is called the reducing agent, while the species that is reduced is called the oxidizing agent. Therefore, in every redox reaction, there will always be an oxidation process coupled with a reduction process.

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A balloon filled with 0.500 L of air at sea level is submerged in the water to a depth that produces a pressure of 3.25 atm. What is the volume of the balloon at this depth? a. 0.154 L b. 6.50 L c. 0.615 L d. 1.63 L d. None of the above

Answers

"0.154 L" is the volume of the balloon.

Given:

Pressure,

$P_1 = 1 \ atm$
$P_2 = 3.25 \ atm$

Volume,

$V_1 = 0.5 \ L$
$V_2 = ?$

As we know,

→ $P_1. V_1 = P_2 .V_2$

or,

→ $V_2 = (P_1. V_1)/(P_1)$

By substituting the values, we get

$= (0.5* 1)/(3.25)$

$= (0.5)/(3.25)$

$= 0.154 \ L$

Thus the above answer i.e., "option a" is correct.

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Answer:

Option a . 0.154L

Explanation:

P₁ . V₁ = P₂ . V₂

when we have constant temperature and constant moles for a certain gas.

At sea level, pressure is 1 atm so:

0.5 L . 1atm = V₂ . 3.25 atm

(0.5L . 1atm) / 3.25 atm = 0.154 L

Draw the Lewis structure for the polyatomic hydronium H3O cation. Be sure to include all resonance structures that

Answers

Answer:

Lewis structure of Hydronium ion is shown below :

Explanation:

Lewis structure : It is a representation of valence electrons on the atoms in a molecule

Here , Hydronium ion is given , which contains 1 atom of oxygen and 3 atoms of hydrogen .

Oxygen has a total of 6 valence electrons and hydrogen contains 1 valence electron .

Oxygen share its 3 valence electrons with 3 hydrogen atoms and left with 3 valence electrons. From these three valence electrons of oxygen atom two electrons will be shown as a pair of electrons on oxygen atom but a single electron can not be shown . So , to simplify this, one positive charge is shown overall .

Resonance structure will be same as the hybrid structure because all three atoms are same , that is hydrogen .

Upon adding solid potassium hydroxide pellets to water the following reaction takes place: KOH(s) → KOH(aq) + 43 kJ/mol Answer the following questions regarding the addition of 14.0 g of KOH to water: Does the beaker get warmer or colder? Is the reaction endothermic or exothermic? What is the enthalpy change for the dissolution of the 14.0 grams of KOH?

Answers

Answer:

a) Warmer

b) Exothermic

c) -10.71 kJ

Explanation:

The reaction:

KOH(s) → KOH(aq) + 43 kJ/mol

It is an exothermic reaction since the reaction liberates 43 kJ per mol of KOH dissolved.

Hence, the dissolution of potassium hydroxide pellets to water provokes that the beaker gets warmer for being an exothermic reaction.

The enthalpy change for the dissolution of 14 g of KOH is:

$n = (m)/(M)$

Where:

m: is the mass of KOH = 14 g

M: is the molar mass = 56.1056 g/mol

$n = (m)/(M) = (14 g)/(56.1056 g/mol) = 0.249 mol$

The enthalpy change is:

$\Delta H = -43 (kJ)/(mol)*0.249 mol = -10.71 kJ$

The minus sign of 43 is because the reaction is exothermic.

I hope it helps you!

14) Describe the Cloud Model.

Answers

Electron cloud is an informal term in physics. It is used to describe where electrons are when they go around the nucleus of an atom. The electron cloud model is different from the older Bohr atomic model by Niels Bohr. Bohr talked about electrons orbiting the nucleus.

A certain first-order reaction has a rate constant of 2.75 10-2 s−1 at 20.°c. what is the value of k at 45°c if ea = 75.5 kj/mol? webassign will check your answer for the correct number of significant figures. 0.0352 incorrect: your answer is incorrect.

Answers

With an activation energy $($E_a$)$ of 75.5 kJ/mol, the rate constant k for a first-order reaction at 20°C is 2.75 × 10⁻² s⁻¹. At 45°C, k is approximately 0.095 s⁻¹, determined using the Arrhenius equation.

The Arrhenius equation relates the rate constant k, temperature T, activation energy $($E_a$)$ , and the gas constant R:

$k = Ae^{-(E_a)/(RT)}$

Given that $k_1 = 2.75 * 10^(-2) \, \text{s}^(-1)$ at $T_1 = 20^\circ \text{C} = 293.15 \, \text{K}$ and $E_a = 75.5 \, \text{kJ/mol}$ , we want to find $k_2$ at $T_2 = 45^\circ \text{C} = 318.15 \, \text{K}$ .

First, let's find the value of A using the Arrhenius equation at T_1:

$2.75 * 10^(-2) = A e^{-((75.5 * 10^3))/((8.314)(293.15))}$

Solving for A:

$A \approx 3.65 \, \text{s}^(-1)$

Now, use the Arrhenius equation at $T_2$ to find $k_2$ :

$k_2 = (3.65) e^{-((75.5 * 10^3))/((8.314)(318.15))}$

Calculate $k_2$ .

$k_2 \approx 0.095 \, \text{s}^(-1)$

Therefore, the value of k at $45^\circ \text{C}$ is approximately $0.095 \, \text{s}^(-1)$ .

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Final answer:

To find the new rate constant of a first-order reaction under different temperature conditions, we can use the Arrhenius equation, which relates the rate constant, activation energy, and temperature of a reaction.

Explanation:

The student is interested in finding the value of the rate constant (k) at a different temperature for a first-order reaction. The answer can be found using the Arrhenius equation, which defines the relationship between the rate constant (k) of a reaction and the temperature at which the reaction occurs. The activation energy (Ea) is also necessary.

The Arrhenius equation is: k = A * exp(-Ea/(R*T)), where A is the pre-exponential factor, R is the universal gas constant (the value of R should be 8.314 J/mol.K to match the Ea units), and T is the temperature in Kelvin.

At the first condition, you have the value of k and the corresponding T (convert Celsius to Kelvin by adding 273.15). With these values and the known Ea, you can solve for A. Then, using the value of A, Ea, and the second T (also converted to Kelvin), you can solve for the new k.

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Using a density of 1.0 g/mL for the water added and adding in the mass of the lithium nitrate, what is the total mass of the solution?

Answers

The total mass of the solution of lithiumnitrate solution has been 99.7 grams.

Density can be defined as the mass of the solute per unit volume. The density can be expressed as g/ml or kg/L.

The mass of given Lithium nitrate = 2.5 grams.

The mass of water can be given as:

Density = $\rm (Mass)/(Volume)$

Volume of water = 97.2 ml.

The total mass of solution:

Mass of water = Density $*$ Volume

Mass of water = 1 $*$ 97.2 grams

Mass of water = 97.2 grams

The total mass = Mass of lithium nitrate + mass of water

= 2.5 + 97.2 grams

= 99.7 grams.

The total mass of the solution of lithiumnitrate solution has been 99.7 grams.

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Answer:

The total mass of the solution = 99.7 g

Note: The question is incomplete. The complete question is given below:

A "coffee-cup" calorimetry experiment is run for the dissolution of 2.5 g of lithium nitrate placed into 97.2 mL of water. The temperature of the solution is initially at 23.5oC. After the reaction takes place, the temperature of the solution is 28.3 oC.

1. Using a density of 1.0 g/mL for the water added and adding in the mass of the lithium nitrate, what is the total mass of the solution and solid?

Explanation:

mass = density * volume

density of water = 1.0 g/mL; volume of water = 97.2 mL

mass of water = 1.0 g/L * 97.2 mL

mass of water = 97.2 g

mass of lithium nitrate = 2.5 g

A solution is made by dissolving a solute (usually solid) in a solvent (usually a liquid). The solute in this reaction is lithium nitrate and the solvent is water.

Total mass of solution = mass of water + mass of lithium nitrate

Total mass of solution = 97.2 g + 2.5 g = 99.7 g

Therefore, total mass of the solution = 99.7 g

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