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Yuri [45]
2 years ago
15

Pls helps asap plsssssss

Chemistry
2 answers:
nata0808 [166]2 years ago
6 0

Answer:

D. An ion that is not directly involved in the reaction.

Explanation:

Spectator ions are ions that are not directly involved in a chemical reaction. They are commonly seen in complete ionic equations and are often taken out to make a net ionic equation because the ions aren't involved in forming the precipitate.

  • Complete ionic equation: Gives all the ions participated in the chemical reaction, even spectator ions.
  • Net ionic equation: Leaves out spectator ions so the equation only shows the ions that formed the precipitate.

For example, this is very useful when dealing with redox reactions because are complicated enough without spectator ions being in the way, so they help focus on those ions that actually form the precipitate.

Lisa [10]2 years ago
4 0

Answer:

A spectator ion refers to a charged atom in a chemical reaction that does not undergo a chemical change.

Hope this helps!!!

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What are two ways condensation differs from boiling?
NARA [144]
Condensation is when vapor liquidizes and gathers on top of a surface, while boiling is turning a liquid into a gas or vapor. They are opposites. In the first, the molecules of the substance split and separate and increase the volume, while in the second they come together and reduce the volume.
6 0
2 years ago
PLSSS HELP ANYONE ASAP!
algol [13]
B I hope it’s right I don’t really help a lot but yeah lol
7 0
2 years ago
Consider the following equilibrium: 2SO^2(g) + O2(9) = 2 SO3^(g)
saul85 [17]

Answer:

At equilibrium, the forward and backward reaction rates are equal.

The forward reaction rate would decrease if \rm O_2 is removed from the mixture. The reason is that collisions between \rm SO_2 molecules and \rm O_2\! molecules would become less frequent.

The reaction would not be at equilibrium for a while after \rm O_2 was taken out of the mixture.

Explanation:

<h3>Equilibrium</h3>

Neither the forward reaction nor the backward reaction would stop when this reversible reaction is at an equilibrium. Rather, the rate of these two reactions would become equal.

Whenever the forward reaction adds one mole of \rm SO_3\, (g) to the system, the backward reaction would have broken down the same amount of \rm SO_3\, (g)\!. So is the case for \rm SO_2\, (g) and \rm O_2\, (g).

Therefore, the concentration of each species would stay the same. There would be no macroscopic change to the mixture when it is at an an equilibrium.

<h3>Collision Theory</h3>

In the collision theory, an elementary reaction between two reactants particles takes place whenever two reactant particles collide with the correct orientation and a sufficient amount of energy.

Assume that \rm SO_2\, (g) and \rm O_2\, (g) molecules are the two particles that collide in the forward reaction. Because the collision has to be sufficiently energetic to yield \rm SO_3\, (g), only a fraction of the reactions will be fruitful.

Assume that \rm O_2\, (g) molecules were taken out while keeping the temperature of the mixture stays unchanged. The likelihood that a collision would be fruitful should stay mostly the same.

Because fewer \!\rm O_2\, (g) molecules would be present in the mixture, there would be fewer collisions (fruitful or not) between \rm SO_2\, (g) and \rm O_2\, (g)\! molecules in unit time. Even if the percentage of fruitful collisions stays the same, there would fewer fruitful collisions in unit time. It would thus appear that the forward reaction has become slower.

<h3>Equilibrium after Change</h3>

The backward reaction rate is likely going to stay the same right after \rm O_2\, (g) was taken out of the mixture without changing the temperature or pressure.

The forward and backward reaction rates used to be the same. However, right after the change, the forward reaction would become slower while the backward reaction would proceed at the same rate. Thus, the forward reaction would become slower than the backward reaction in response to the change.

Therefore, this reaction would not be at equilibrium immediately after the change.

As more and more \rm SO_3\, (g) gets converted to \rm SO_2\, (g) and \rm O_2\, (g), the backward reaction would slow down while the forward reaction would pick up speed. The mixture would once again achieve equilibrium when the two reaction rates become equal again.

5 0
2 years ago
Engineers are starting on a new project to develop technology. Which of these
KATRIN_1 [288]

Answer:

D.

Explanation:

Deciding whether the best product has been designed,should be the last step.

3 0
2 years ago
What is the theoretical yield of aluminum oxide if 1.40 mol of aluminum metal is exposed to 1.35 mol of oxygen?
jasenka [17]

Answer:

71.372 g or 0.7 moles

Explanation:

We are given;

  • Moles of Aluminium is 1.40 mol
  • Moles of Oxygen 1.35 mol

We are required to determine the theoretical yield of Aluminium oxide

The equation for the reaction between Aluminium and Oxygen is given by;

4Al(s) + 3O₂(g) → 2Al₂O₃(s)

From the equation 4 moles Al reacts with 3 moles of oxygen to yield 2 moles of Aluminium oxide.

Therefore;

1.4 moles of Al will require 1.05 moles (1.4 × 3/4) of oxygen

1.35 moles of Oxygen will require 1.8 moles (1.35 × 4/3) of Aluminium

Therefore, Aluminium is the rate limiting reagent in the reaction while Oxygen is the excess reactant.

4 moles of aluminium reacts to generate 2 moles aluminium oxide.

Therefore;

Mole ratio Al : Al₂O₃ is 4 : 2

Thus;

Moles of Al₂O₃ = Moles of Al × 0.5

                         = 1.4 moles × 0.5

                         = 0.7 moles

But; 1 mole of Al₂O₃ = 101.96 g/mol

Thus;

Theoretical mass of Al₂O₃ = 0.7 moles × 101.96 g/mol

                                            = 71.372 g

3 0
3 years ago
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