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3 years ago

Very quick oxidation reactions result in _____.

2 answers:

8 0

Very quick oxidation reactions result in combustion. Combustion happens when oxidation is fast. An example of combustion is the light batteries. We know that it is really dangerous if we will burn it because it has a fast oxidation process that can cause fire and its gas will explode.

astraxan [27]3 years ago

4 0

Very quick oxidation reactions result in combustion.

Reason

Oxidation reaction involved reaction with oxygen. Consider following example:
C6H12O6 + 6O2 → 6CO2 + 6H2O
(glucose) (carbon dioxide)

From the above reaction, it can be seen that, oxidation state of C in glucose in zero, while that in CO2 is +4. Thus, oxidation state of C is increased from 0 to +4. This process is called oxidation. Further, complete oxidation of reactant molecule is referred as combustion.

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Sulfur dioxide and oxygen react to form sulfur trioxide during one of the key steps in sulfuric acid synthesis. An industrial ch

guajiro [1.7K]

Answer:

Kp = 0.049

Explanation:

The equilibrium in question is;

2 SO₂ (g) + O₂ (g) ⇄ 2 SO₃ (g)

Kp = p SO₃² / ( p SO₂² x p O₂ )

The initial pressures are given, so lets set up the ICE table for the equilibrium:

atm SO₂ O₂ SO₃

I 3.3 0.79 0

C -2x -x 2x

E 3.3 - 2x 0.79 - x 2x

We are told 2x = partial pressure of SO₃ is 0.47 atm at equilibrium, so we can determine the partial pressures of SO₂ and O₂ as follows:

p SO₂ = 3.3 -0.47 atm = 2.83 atm

p O₂ = 0.79 - (0.47/2) atm = .56 atm

Now we can calculate Kp:

Kp = 0.47² /[ ( 2.83 )² x 0.56 ] = 0.049 ( rounded to 2 significant figures )

Note that we have extra data in this problem we did not need since once we setup the ICE table for the equilibrium we realize we have all the information needed to solve the question.

7 0

3 years ago

Samuel needs to dissolve a solid compound in water. what are some things samuel could do to cause this dissolution to occur at a

ziro4ka [17]

In order to make the dissolution of the solid compound in water to occur at a faster rate, Samuel could do the following:
1. Break down the solid into tiny particles: breaking down the solid into tiny particles increases the surface area of the solid and thus increase the quantity of the substance that comes in contact with the solvent per time, this leads to a faster dissolution of the solid.
2. Stir the liquid with iron rod: Samuel can increase the dissolution rate of the substance by stirring it continuously with iron rod.
3. Increasing the temperature:Samuel could also increase the rate of dissolution of the substance by increasing the temperature of the water.

5 0

3 years ago

Read 2 more answers

Completely reacting 150.0 g of a substance with oxygen releases 395.1 J of energy. How much energy would be released if 450.0 g

MrMuchimi

To solve this problem we just need to use the rule of three:
150g..................395.1J
450g................xJ

x = 450*395.1/150 = 1185,3J

450.0 g of the substance completely reacted with oxygen will produce 1.1853 kJ(kiloJoule)

4 0

3 years ago

Read 2 more answers

When molten sulfur reacts with chlorine gas, a vile-smelling orange liquid forms that has an empirical formula of SCl. The struc

egoroff_w [7]

Answer:

The structure is shown below.

Explanation:

The formal charge (FC) is the charge that is more close to the actual charge in the real molecules and ions. It can be calculated based on the number of valence electrons (V), the shared electrons (S) and the electrons in the lone pairs (L) by the equation:

FC = V - (L + S/2)

Sulfur is in group 16 of the periodic table, so it has 6 valence electrons, and chlorine is from group 17 of the periodic table, and so it has 7 valence electrons. Chlorine can share only one electron, so it is stable. Sulfur can expand its octet (because it's from the third period) and can have more than 8 electrons when stable.

The possible formulas, from the empiric one, are:

SCl, S₂Cl₂, and S₃Cl₃.

To have FC = 0, chlorine must done only one bond, because S = 2, and L = 6, so:

FC = 7 - (6 + 2/2) = 0

So, it can not be the central atom of a structure. In the SCl, it will hav only a simple bond, so for sulfur, S = 2, and L = 4 (only the lone pairs are counted)

FC = 6 - (4+ 2/2) = +1

For S₂Cl₂, the two sulfurs must be bonded to a simple bond, and each one to one chlorine, thus, for both od them S = 4, and L = 4. so

FC = 6 - (4 + 4/2) = 0

So, it is the correct structure. The lewis structure represents the bonds by lines and the lone pairs of electrons by dots, and it is shown below.

3 0

3 years ago

What mass of precipitate forms when 185.5 ml of 0.533 m naoh is added to 627 ml of a solution that contains 15.8 g of aluminum s

Law Incorporation [45]

Answer:

2,57 g of precipitate.

Explanation:

For the reaction:

6 NaOH + Al₂(SO₄)₃ → 2 Al(OH)₃ + 3 Na₂SO₄

The precipitate is Al(OH)₃.

185,5mL of 0,533M NaOH are:

0,1855L × 0,533M = 0,0989 moles NaOH

Moles of Al₂(SO₄)₃ are:

15,8g × $\frac{1mol}{342,15g}$ = 0,0462 moles Al₂(SO₄)₃

For the total reaction of 0,0989 moles NaOH with Al₂(SO₄)₃ you need:

0,0989moles NaOH × $\frac{1molAl_{2}(SO_{4})_{3}}{6 moles NaOH}$ = 0,0165 moles Al₂(SO₄)₃

As you have 0,0462 moles Al₂(SO₄)₃ the limiting reactant is NaOH.

0,0989 moles of NaOH produce:

0,0989moles NaOH × $\frac{2molAl(OH)_{3}}{6 moles NaOH}$ = 0,0330 moles of Al(OH)₃

These moles are:

0,0330 moles of Al(OH)₃ × (78 g/mol) = 2,57 g of Al(OH)₃ ≡ mass of precipitate

I hope it helps!

3 0

3 years ago