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Veronika [31]
3 years ago
14

Which of the following is not an example of a molecule?

Chemistry
2 answers:
satela [25.4K]3 years ago
6 0

Mn is the answer i just took the test

I am Lyosha [343]3 years ago
3 0

Answer : The option is, (D) Mn

Explanation :

Pure substance : It is defined as a substance that is made by the combination of only one type of atom or only one type of molecule.

Compound : It is a pure substance which is made from atoms of different elements combined together in a fixed ratio by mass.

Element : It is a pure substance which is composed of atoms of similar elements.

Molecule : It is the smallest particle in the chemical compound or chemical element that has the same chemical properties of that compound or element.

Option A :

O_3 is molecule of element.

Option B :

KOH is molecule of compound.

Option C :

H_2S is molecule of compound.

Option D :

Mn is and element.

Hence, the element Mn is not an example of a molecule.

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25 mL of 0.10 M aqueous acetic acid is titrated with 0.10 M NaOH(aq). What is the pH after 30 mL of NaOH have been added?
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Answer:

pH = 11.95≈12

Explanation:

Remember  the reaction among aqueous acetic acid (CH_3COOH) and aqueous sodium hydroxide (NaOH)

CH_3COOH + NaOH ->  CH_3COONa + H_2O

First step. Need to know how much moles of the substances are present

0.1 \frac{mol NaOH}{L} * 0.030 L = 0.003 mol NaOH[tex]0.1 \frac{mol CH_3COOH}{L} * 0.025 L= 0.0025 mol CH_3COOHSecond step. Know wich substance is in excess.0.0025 mol CH_3COOH * [tex]1 mol NaOH/ 1 mol CH_3COOH = 0.0025 mol NaOH

0.003 mol NaOH * 1 mol CH_3COOH/ 1 mol NaOH = 0.003 mol CH_3COOH[/tex]

NaOH is in excess. Now, how much?

0.003 mole NaOH - 0.0025 mole NaOH = 0.0005 mole NaOH

Then, that amount in excess would be responsable for the pH.

Third step. Know the pH

Remember that pH= -log[H+]

According to the dissociation of water equilibrium

Kw=[H+]*[OH-]= 10^(-14)

The dissociation of NaOH is

NaOH -> Na^{+} + OH^{-}

Now, concentration of OH^{-}[/tex] would be given for the excess of NaOH.

[OH-]= 0.0005 mole / 0.055 L = 0.00909 M

Careful: we have to use the total volumen

Les us to calculate pH

pH= -log [H+]\\pH= -log \frac{K_w}{[OH-]} \\pH= 11.95

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