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3 years ago

For the reaction CH3COOH→CH3COO– + H+, which statement is true?

Chemistry

1 answer:

kow [346]3 years ago

7 0

Answer:

For the reaction CH3COOH→CH3COO– + H+

Among the given statements which is the correct statement?

CH3COOH is a Brønsted-Lowry base.

CH3COO– is an Arrhenius base.

CH3COOH is a Brønsted-Lowry acid.

CH3COO– is a Lewis base.

Explanation:

Bronsted acid is the proton donor.

Bronsted base is the proton acceptor.

Arrhenius base is the one that releases OH- ions when dissolved in water.

Lewis base is the one that is an electron-pair donor.

In the given reaction,

CH3COOH→CH3COO– + H+

Acetic acid releases a proton.

So, it is a proton donor.

Hence, It is a Bronsted-Lowry acid.

Third option is the correct answer.

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In science, we like to develop explanations that we can use to predict the outcome of events and phenomena. Try to develop an ex

Kay [80]

The question is incomplete. The complete question is :

In science, we like to develop explanations that we can use to predict the outcome of events and phenomena. Try to develop an explanation that tells how much NaOH needs to be added to a beaker of HCl to cause the color to change. Your explanation can be something like: The color change will occur when [some amount] of NaOH is added because the color change occurs when [some condition]. The goal for your explanation is that it describes the outcome of this example, but can also be used to predict the outcome of other examples of this phenomenon. Here's an example explanation: The color of the solution will change when 40 ml of NaOH is added to a beaker of HCl because the color always changes when 40ml of base is added. Although this explanation works for this example, it probably won't work in examples where the flask contains a different amount of HCl, such as 30ml. Try to make an explanation that accurately predicts the outcome of other versions of this phenomenon.

Solution :

Consider the equation of the reaction between NaOH and $HCl$

NaOH (aq) + HCl (aq) → NaCl(aq) + $H_2O (l)$

The above equation tells us that $1 \text{mole}$ of $NaOH$ reacts with $1 \text{mole}$ of $HCl$ .

So at the equivalence point, the moles of NaOH added = moles of $HCl$ present.

If the volume of the $HCl$ taken = $V_1$ mL and the conc. of $HCl$ = $M_1$ mole/L

The volume of NaOH added up to the color change = $V_2 \text{ and conc of NaOH = M}_2$ mole/L

Moles of $HCl$ taken = $V_1 \ mL \times M_1 \ mol/100 \ mL = V_2M_2 \times 10^{-3}$ moles.

The color change will occur when the moles of NaOH added is equal to the moles of $HCl$ taken.

Thus when $V_1 M_1 \times 10^{-3} = V_2M_2 \times 10^{-3}$

or when $V_1M_1 = V_2M_2$

or $V_2=\frac{V_1M_1}{M_2}$ mL of NaOH added, we observe the color change.

Where $V_1, M_1$ are the volume and molarity of the $HCl$ taken.

$M_2$ is the molarity of NaOH added.

When both the NaOH and $HCl$ are of the same concentrations, i.e. if $M_1=M_2$ , then $V_2=V_1$

Or the 40 mL of $HCl$ will need 40 mL of NaOH for a color change and

30 mL of $HCl$ would need 30 mL of NaOH for the color change (provided the concentration $M_1=M_2$ )

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3 years ago

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vova2212 [387]

Answer:

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The Na+ ions and NO3- ions remain separate in the sodium nitrate solution and do not form a precipitate. Ions that remain essentially unchanged during a reaction are called spectator ions.This means these can be ignored when writing the ionic equation. Only how the solid silver chloride forms is needed to be shown:

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In a balanced ionic equation:

the number of positive and negative charges is the same

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Displacement reactions take place when a reactive element displaces a less reactive element from one of its compounds.

A common type of displacement reaction takes place when a reactive metal reacts with the salt of a less reactive metal. For example, copper reacts with silver nitrate solution to produce silver and copper(II) nitrate solution:

2AgNO3(aq) + Cu(s) → 2Ag(s) + Cu(NO3)2(aq)

In this reaction, the NO3- ions remain in the solution and do not react - they are the spectator ions in this reaction. So, they can be removed from the ionic equation:

2Ag+(aq) + Cu(s) → 2Ag(s) + Cu2+(aq)

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There are the same numbers of atoms of each element on both sides of the equation. The total charge on both sides is also the same (zero).

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Explanation:

this will help, I used this for my work x

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3 years ago

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almond37 [142]

Answer:

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Explanation:

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