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malfutka [58]
4 years ago
8

Envirnmental pollution caused by production of iron

Chemistry
1 answer:
vivado [14]4 years ago
7 0
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HOW CAN I DO AN EXPERIMENT WITH THIS THINGGGG I-
Vladimir79 [104]

Answer:

omg i love grass hopperszszszs

Explanation:

you should give it a mate

6 0
3 years ago
dentify the oxidation state of Cl in Cl 2 O 7 . +2 Identify the oxidation state of Cl in AlCl − 4 . +1 Identify the oxidation st
konstantin123 [22]

Answer:

  • <u>1. +7</u>
  • <u>2. -1</u>
  • <u>3. +3</u>
  • <u>4. +5</u>

Explanation:

<em><u>1. Identify the oxidation state of Cl in Cl₂O₇ . </u></em>

Cl₂O₇ is a neutral compound.

In a neutral compound the oxidation states add up 0.

The oxidation state of oxygen in all the binary compounds, except the peroxides (where two oxygen atoms are bonded together and with other elements) is -2.

Since there are 7 atoms of O, the total oxidation number from it is (-2) × 7 = -14.

Then, the oxidation number from the Cl atoms must add up +14 (because the compound is neutral).

Since there are two atoms of Cl each one must contribute with +14/2 = +7 to the oxidation number.

That is the oxidation state of Cl in the compound. The answer is +7.

<u><em>2 Identify the oxidation state of Cl in AlCl ₄⁻ (the formula was corrected)</em></u>

The most common oxidation state of Al is +3.

Since the net charge of the ion is -1, the oxidation number of the other atom (call it x) must be such that four times it and +3 add up -1:

  • 4x + 3 = - 1

  • 4x = -1 - 3

  • 4x = - 4

  • x = - 1

In conclusion, the oxidation state of Cl in the ion is -1

<u><em>3. Identify the oxidation state of Cl in Ba (ClO₂)₂ . </em></u>

This is a neutral compound. Thus the overall oxidation number is 0.

The oxidation state of Barium, Ba, such as of the other members of the group 2, alkaline earth metals, is +2.

The oxydation state of oxygen, O, in ternary compounds is -2

Thus, calling x the oxidation state of Cl and taking in account the number of atoms in the compound:

  • +2 + 2x + 4(-2) = 0
  • 2x + 2 - 8 = 0
  • 2x = 6
  • x = +3

Therefore, the oxidation state of Cl in this compound is +3

<u><em>4.  Identify the oxidation state of Cl in ClF₄⁺</em></u>

Fluor has oxidation state -1. Since there are four atoms, it contributes with 4 × (-1) = - 4

The overall charge of the ion is +1.

Thus, the oxidation state of Cl, x, is:

  • x - 4 = 1
  • x = 1 + 4
  • x = +5

Hence, the oxidation state of Cl in this ion is +5.

6 0
3 years ago
Read 2 more answers
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
4 years ago
A LA TEMPERATURA DE 35°C, UNA MUESTRA DE DIOXIDO DE CARBONO OCUPA UN VOLUMEN DE 350 ML. ¿Qué CAMBIO DE VOLUMEN SE PRODUCIRA SI L
Kisachek [45]

Answer:

New volume = 150 mL

Explanation:

Initial temperature, T₁ = 35°C

Initial volume, V₁ = 350 mL

We need to find the change in volume when the temperature drops to 15°C.

The relation between the temperature and the volume is given by Charle's law. Let new volume is V₂. It can be given by :

\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}\\\\V_2=\dfrac{V_1T_2}{T_1}\\\\V_2=\dfrac{350\times 15}{35}\\\\V_2=150\ mL

So, the new volume is 150 mL.

8 0
3 years ago
Which natural polymer makes up paper? A. starch B. nylon C. wool D. rubber
eduard

Answer:

starch

Explanation:

7 0
3 years ago
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