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goldfiish [28.3K]
2 years ago
11

Rusting is a chemical process that changed the strength and integrity of objects made of iron or iron alloys. Which of the follo

wing statements correctly describes the rusting process below?
4 Fe(s) + 3O2 (g) 2 Fe2O3 (s)

Iron is reduced to iron ions in this reaction.

The oxidizing agent in this reaction is iron metal.

Iron oxide is the reducing agent in this reaction.

Oxygen was reduced over the course of this reaction.
Chemistry
2 answers:
pishuonlain [190]2 years ago
6 0
Among the choices provided, the statement that correctly describes the rusting process below is that "Oxygen was reduced over the course of this reaction" as <span> iron can't be the oxidizing agent, because it is the one being oxidized.

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.
</span>
s2008m [1.1K]2 years ago
4 0

Answer : The correct option is, Oxygen was reduced over the course of this reaction.

Explanation :

Iron rusting : A chemical process in which the an iron nail react with water and oxygen to give iron oxide as a product. Rusting of iron is an oxidation-reduction reaction in which iron losses electrons to oxygen atom.

Oxidation reaction : A substance looses its electrons. In this oxidation state increases.

Reduction reaction : A substance gains electrons. In this oxidation state decreases.

The balanced chemical reaction for rusting of irons is,

4Fe(s)+3O_2(g)\rightarrow 2Fe_2O_3(s)

Half reactions of oxidation and reduction are :

Oxidation : Fe(s)\rightarrow Fe^{3+}+3e^-

Reduction : \frac{1}{2}O_2+2e^-\rightarrow O^{2-}

From this reaction we conclude that the electrons are getting transferred from iron to oxygen.  That means in this reaction, oxygen is an oxidizing agent and iron is a reducing agent.

Hence, the correct option is, Oxygen was reduced over the course of this reaction.

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Determine the number of grams in each of the quantities<br><br> 1.39.0 x 1024 molecules Cl2
STatiana [176]

Mass of Cl₂ : 164.01 g

<h3>Further explanation</h3>

A mole is a number of particles(atoms, molecules, ions)  in a substance

This refers to the atomic total of the 12 gr C-12  which is equal to 6.02.10²³, so 1 mole = 6.02.10²³ particles  

Can be formulated :

N = n x No

N = number of particles

n = mol

No = 6.02.10²³ = Avogadro's number

mol Cl₂ :

\tt n=\dfrac{N}{No}\\\\n=\dfrac{1.39.10^{24}}{6.02.10^{23}}\\\\n=2.31

mass Cl₂(MW=71 g/mol) :

\tt mass=mol\times MW\\\\mass=2.31\times 71=164.01

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Number of protons for oxygen 16
DanielleElmas [232]
8 protons in oxygen bc it has an atomic number of 8
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They saw that the rock was deep purple. A quantitative B qualitative
Anika [276]
I think B qualitative is the answer
7 0
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Are any substance made up of matter and can be natural or man-made.
Mila [183]

Answer:

Yes.

Explanation:

Substances are made up of matter or matter are made up of tiny molecules or atoms that occur naturally or some are synthetic or man made.

All matter are made up of substances called elements and each elements have its own physical and chemical properties and cannot be broken easily by ordinary chemical reactions.

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3 0
3 years ago
0.60 atm of SO3 and 0.30 atm of SO2are placed in a container and the system is allowed to reach equilibrium. Calculate the press
frozen [14]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The pressure is [O_2] =  4.8 *10^{-5} \ atm

Explanation:

From the question we are told that

     The  pressure of  SO_3 is  [SO_3 ] =  0.63 \ atm

     The  pressure of  SO_2 is  [SO_ 2]  =  0.30 \ atm

      The equilibrium constant is K_p  =  1.2 *10^{-5}

     The  reaction is

           2SO_3 _{(g)} ⇔ 2SO_2_{(g)} +  O_2 _{(g)}

Generally the equilibrium constant is mathematically represented as

           K_p  =  \frac{(SO_2)^2 *  (O_2)}{(SO_3)^2}

=>         [O_2] =  \frac{k_p * [SO_3] ^2 }{[SO_2]^2}

substituting values

            [O_2] =  \frac{1.2 *10^{-5} * 0.60 ^2 }{0.30^2}

             [O_2] =  4.8 *10^{-5} \ atm

     

8 0
3 years ago
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