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Brrunno [24]
3 years ago
15

One of the most recognizable corrosion reactions is the rusting of iron. rust is caused by iron reacting with oxygen gas in the

presence of water to create an oxide layer. iron can form several different oxides, each having its own unique color. red rust is caused by the formation of iron(iii) oxide trihydrate. in the space provided, write the balanced reaction for the formation of fe2o3•3h2o(s). phases are optional.
Chemistry
2 answers:
lesya692 [45]3 years ago
8 0

Answer : The balanced chemical reaction for rusting of irons is :

4Fe(s)+3O_2(g)+6H_2O(l)\rightarrow 2[Fe_2O_3(s).3H_2O(s)]

Explanation :

Iron rusting : It is a type of chemical process where an iron nail react with the water in the presence of moisture (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 : It is the reaction in which a substance looses its electrons. In this oxidation state increases.

Reduction reaction : It is the reaction in which a substance gains electrons. In this oxidation state decreases.

The balanced chemical reaction for rusting of irons is :

4Fe(s)+3O_2(g)+6H_2O(l)\rightarrow 2[Fe_2O_3(s).3H_2O(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-}

zloy xaker [14]3 years ago
3 0

Answer:  Fe(s) + 6H_2O(l) + 3O_2(g)  -> 4Fe(OH)_3(s)

Explanation:

The Chemical equation for the formation of rust is:

Iron + Water + Oxygen ----> Rust

4 Fe(s) + 6 H2O(l) + 3 O2(g) → 4 Fe(OH)3(s)

The Iron Hydroxide However dehydrates to produce Fe2O3 * nH2O

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Acetylene gas (C2H2) burns completely in the presence of oxygen gas (O2) to yield carbon dioxide
Luda [366]

A. The balanced equation for the reaction is

2C₂H₂ + 5O₂ —> 4CO₂ + 2H₂O

B. The volume of the oxygen gas required to burn 0.700 L of acetylene gas is 1.75 L

C. The volume of carbon dioxide gas produced is 1.4 L

D. The volume of water vapor produced is 0.7 L

<h3>A. Balanced equation </h3>

The balanced equation for the reaction between acetylene gas (C₂H₂) and oxygen gas (O₂) is given below

2C₂H₂ + 5O₂ —> 4CO₂ + 2H₂O

<h3>B. How to determine the volume of oxygen </h3>

2C₂H₂ + 5O₂ —> 4CO₂ + 2H₂O

Since the reaction occurred at standard temperature and pressure, we can thus say that:

From the balanced equation above,

2 L of C₂H₂ reacted with 5 L O₂.

Therefore,

0.7 L of C₂H₂ will react with = (0.7 × 5) / 2 = 1.75 L of O₂

Thus, 1.75 L of O₂ is needed for the reaction

<h3>C. How to determine the volume of carbon dioxide </h3>

2C₂H₂ + 5O₂ —> 4CO₂ + 2H₂O

From the balanced equation above,

2 L of C₂H₂ reacted to produce 4 L of CO₂

Therefore,

0.7 L of C₂H₂ will react to produce = (0.7 × 4) / 2 = 1.4 L of CO₂

Thus, 1.4 L of CO₂ were produced

<h3>D. How to determine the volume of water. </h3>

2C₂H₂ + 5O₂ —> 4CO₂ + 2H₂O

From the balanced equation above,

2 L of C₂H₂ reacted to produce 2 L of H₂O

Therefore,

0.7 L of C₂H₂ will also react to produce 0.7 L of H₂O

Thus, 0.7 L of H₂O was produced

Learn more about stoichiometry:

brainly.com/question/14735801

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When modeling a scientific process, it is more important to organize the parts in a way that makes sense to you than to list the
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The answer would actually be false. I just took the test.

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6 0
3 years ago
if 100. mL of 0.800 M Na2SO4 is added to 200. mL of 1.20 M NaCl, what is the concentration of Na+ ions in the final solution? As
Black_prince [1.1K]
Compounds Na₂SO₄ and NaCl are mixed together are we are asked to find the concentration of Na⁺ in the mixture 
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1 mol of Na₂SO₄ gives out 2 mol of Na⁺ ions 

the number of Na₂SO₄ moles added - 0.800 M/1000 * 100 ml
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therefore number of Na⁺ ions from Na₂SO₄ = 0.08 * 2 = 0.16 mol

NaCl ----> Na⁺ + Cl⁻ 
1 mol of NaCl gives 1 mol of Na⁺ ions
number of NaCl moles added = 1.20 M/1000 * 200 ml
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number of Na⁺ ions from NaCl = 0.24 mol

total number of Na⁺ ions in the mixture = 0.16 mol + 0.24 mol = 0.4 mol
as stated the volumes are additive, 
therefore total volume  = 100 ml + 200 ml = 300 ml 
the concentration of Na⁺ ions = number of moles / volume 
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