Five solutions on how to prevent air pollution

Use the changes in oxidation numbers to identify which atom is oxidized, reduced, the oxidizing agent, and the reducing agent. 5

Vinil7 [7]

Answer:

Reaction A:

Hydrogen atoms in H₂ are oxidized.
Oxygen atoms in O₂ are reduced.
Hydrogen gas H₂ is the reducing agent.
Oxygen gas O₂ is the oxidizing agent.

Reaction B:

Oxygen atoms in KNO₃ are oxidized.
Nitrogen atoms in KNO₃ are reduced.
Potassium nitrate (V) KNO₃ is both the oxidizing agent and the reducing agent.

Explanation:

When an atom is oxidized, its oxidation number increases.
When an atom is reduced, its oxidation number decreases.
The oxidizing agent contains atoms that are reduced.
The reducing agent contains atoms that are oxidized.

Here are some common rules for assigning oxidation states.

Oxidation states on all atoms in a neutral compound shall add up to 0.
The average oxidation state on an atom is zero if the compound contains only atoms of that element. (E.g., the oxidation state on O in O₂ is zero.)
The oxidation state on oxygen atoms in compounds is typically -2. (Exceptions: oxygen bonded to fluorine, and peroxides.)
The oxidation state on group one metals (Li, Na, K) in compounds is typically +1.
The oxidation state on group two metals (Mg, Ca, Ba) in compounds is typically +2.
The oxidation state on H in compounds is typically +1. (Exceptions: metal hydrides where the oxidation state on H can be -1.)

For this question, only the rule about neutral compounds, oxygen, and group one metals (K in this case) are needed.

<h3>Reaction B</h3>

Oxidation states in KNO₃:

K is a group one metal. The oxidation state on K in the compound KNO₃ shall be +1.
The oxidation state on N tend to vary a lot, from -3 all the way to +5. Leave that as $x$ for now.
There's no fluorine in KNO₃. The ion NO₃⁻ stands for nitrate. There's no peroxide in that ion. The oxidation state on O in this compound shall be -2.
Let the oxidation state on N be $x$ . The oxidation state of all five atoms in the formula KNO₃ shall add up to zero. $1\times (+1) + 1 \times (x) + {\bf 3} \times (-2) = 0\\x = +5$ . As a result, the oxidation state on N in KNO₃ will be +5.

Similarly, for KNO₂:

The oxidation state on the group one metal K in KNO₂ will still be +1.
Let the oxidation state on N be $y$ .
There's no peroxide in the nitrite ion, NO₂⁻, either. The oxidation state on O in KNO₂ will still be -2.
The oxidation state on all atoms in this formula shall add up to 0. Solve for the oxidation state on N: $1\times (+1) + 1 \times (y) + {\bf 2}\times (-2) = 0\\y = +3$ . The oxidation state on N in KNO₂ will be +3.

Oxygen is the only element in O₂. As a result,

The oxidation state on O in O₂ will be 0.

$\rm\stackrel{+1}{K}\stackrel{\bf +5}{N}\stackrel{\bf -2}{O}_3 \to \stackrel{+1}{K}\stackrel{\bf+3}{N}\stackrel{\bf -2}{O}_2 + \stackrel{\bf 0}{O}_2$ .

The oxidation state on two oxygen atoms in KNO₃ increases from -2 to 0. These oxygen atoms are oxidized. KNO₃ is also the reducing agent.

The oxidation state on the nitrogen atom in KNO₃ decreases from +5 to +3. That nitrogen atom is reduced. As a result, KNO₃ is also the oxidizing agent.

<h3>Reaction A</h3>

Apply these steps to reaction A.

H₂:

Oxidation state on H: 0.

O₂:

Oxidation state on O: 0.

H₂O:

Oxidation state on H: +1.
Oxidation state on O: -2.
Double check: ${\bf 2} \times (+1) + (-2) = 0$ .

$\rm \stackrel{}{2}\; \stackrel{\bf 0}{H}_2 + \stackrel{\bf 0}{O}_2\stackrel{}{\to} \stackrel{}{2}\;\stackrel{\bf +1}{H}_2\stackrel{\bf -2}{O}$ .

The oxidation state on oxygen atoms decreases from 0 to -2. Those oxygen atoms are reduced. O₂ is thus the oxidizing agent.

The oxidation state on hydrogen atoms increases from 0 to +1. Those hydrogen atoms are oxidized. H₂ is thus the reducing agent.

4 0

3 years ago

BRAINLIESTTT ASAP!! PLEASE HELP ME :)

lord [1]

There ya go.. i hope this helps! sorry if it doesn't!

3 0

3 years ago

Consider the reaction: A(aq) + 2B (aq) === C (aq). Initially 1.00 mol A and 1.80 mol B

liraira [26]

Answer:

17

Explanation:

Step 1: Calculate the needed concentrations

[A]i = 1.00 mol/5.00 L = 0.200 M

[B]i = 1.80 mol/5.00 L = 0.360 M

[B]e = 1.00 mol/5.00 L = 0.200 M

Step 2: Make an ICE chart

A(aq) + 2 B(aq) ⇄ C(aq)

I 0.200 0.360 0

C -x -2x +x

E 0.200-x 0.360-2x x

Then,

[B]e = 0.360-2x = 0.200

x = 0.0800

The concentrations at equilibrium are:

[A]e = 0.200-0.0800 = 0.120 M

[B]e = 0.200 M

[C]e = 0.0800 M

Step 3: Calculate the concentration equilibrium constant (K)

K = [C] / [A] × [B]²

K = 0.0800 / 0.120 × 0.200² = 16.6 ≈ 17

6 0

3 years ago

Please help me with this homework I really need help with this homework

Rainbow [258]

Answer:

For the first question its C, Gas

For the second one table

Explanation:

The faster particles move, the more kinetic energy they have. Within a given substance, such as water, particles in the solid state have the least amount of kinetic energy. Particles in the liquid state move faster than particles in the solid state. Therefore, they have more kinetic energy.

3 0

3 years ago

What is the predominant intermolecular force in the liquid state of each of these compounds: hydrogen fluoride (HF), carbon tetr

telo118 [61]

Answer:

HF - hydrogen bonding

CBr4 - Dispersion

NF3 - Dipole-dipole

Explanation:

Hydrogen bonding occurs when hydrogen is covalently bonded to a highly electronegative atom such as fluorine, chlorine nitrogen, oxygen etc. Hence the dominant intermolecular force in HF is hydrogen bonding.

CBr4 is nonpolar because the molecule is tetrahedral and the individual C-Br dipole moments cancel out leaving the molecule with a zero dipole moment hence the dominant intermolecular force are the dispersion forces.

NF3 has a resultant dipole moment hence the molecules are held together by dipole-dipole interaction.

4 0

2 years ago

Five solutions on how to prevent air pollution ​

Five solutions on how to prevent air pollution