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bonufazy [111]
2 years ago
9

Indicate the oxidation number of oxygen for each compound in the following reaction: 2H2O2(aq)→ 2H2O(l) O2(g) Select the choice

that gives the oxidation numbers of H2O2, H2O(l), and O2(g), respectively.
Select the choice that gives the oxidation numbers of , , and , respectively. −1, −2, 0 −1, 0, −2 0, −2, −1 −2, −1, 0
Chemistry
1 answer:
Aleksandr [31]2 years ago
3 0

Answer:

2H2O2(aq)→ 2H2O(l) O2(g) : The oxidation number of oxygen for each compound is -1, -2, 0

Explanation:

In peroxides the oxidation state of oxygen is -1, since one oxygen bonds to the other oxygen and a hydrogen and the bound oxygen captures the electron of the remaining hydrogen. Through a scheme would be

H --- O --- O --- H

We remember that oxygen needs two electrons to get to have the configuration of the nearest noble gas (Lewis octet rule). In Peroxides, the oxygen is linked by covalent bonds. If we take it strictly, peroxide is a grouping of two oxygen, having the whole valence -2. which is why it is usually said that it is when oxygen has a valence -1

As we said the oxidation state is -2, the one that appears in the water molecule, since Hydrogen acts with valence +1 and it is 2 atoms that give up electrons to compensate for oxygen.

In the O2 it acts with valence 0 since we talk about gas in its elementary state. All diatomic molecules in their elemental state, generally gases or metals in solid state, act with a valence of 0.

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We want to calculate the concentrations of all species in a 0.58 M Na 2 SO 3 (sodium sulfite) solution. The ionization constants
NeX [460]

Explanation:

Reaction equation is as follows.

      Na_{2}SO_{3}(s) \rightarrow 2Na^{+}(aq) + SO^{2-}_{3}(aq)

Here, 1 mole of Na_{2}SO_{3} produces 2 moles of cations.

[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58

                                  = 1.16 M

[SO^{2-}_{3}] = [Na_{2}SO_{3}] = 0.58 M

The sulphite anion will act as a base and react with H_{2}O to form HSO^{-}_{3} and OH^{-}.

As,     K_{b} = \frac{K_{w}}{K_{a_{2}}}

                       = \frac{10^{-14}}{6.3 \times 10^{-8}}

                       = 1.59 \times 10^{-7}

According to the ICE table for the given reaction,

          SO^{2-}_{3} + H_{2}O \rightleftharpoons HSO^{-}_{3} + OH^{-}

Initial:        0.58             0              0

Change:     -x               +x             +x

Equilibrium: 0.58 - x     x               x

So,

        K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}

 1.59 \times 10^{-7} = \frac{x^{2}}{0.58 - x}

        x^{2} = 1.59 \times 10^{-7} \times (0.58 - x)

                x = 0.0003 M

So,   x = [HSO^{-}_{3}] = [OH^{-}] = 0.0003 M

[SO^{2-}_{3}] = 0.58 - 0.0003

                     = 0.579 M

Now, we will use [HSO^{-}_{3}] = 0.0003 M

The reaction will be as follows.

              HSO^{2-}_{3} + H_{2}O \rightleftharpoons H_{2}SO_{3} + OH^{-}

Initial:   0.0003

Equilibrium:  0.0003 - x        x             x

              K_{b} = \frac{x^{2}}{0.0003 - x}

        K_{b} = \frac{K_{w}}{K_{a_{1}}}

                      = \frac{10^{-14}}{1.4 \times 10^{-2}}

                      = 7.14 \times 10^{-13}

Therefore,  7.14 \times 10^{-13} = \frac{x^{2}}{0.0003 - x}

As,  x <<<< 0.0003. So, we can neglect x.

Therefore,  x^{2} = 7.14 \times 10^{-13} \times 0.0003

                              = 0.00214 \times 10^{-13}

                     x = 0.0146 \times 10^{-6}

x = [OH^{-}] = [H_{2}SO_{3}] = 1.46 \times 10^{-8}

    [H^{+}] = \frac{10^{-14}}{[OH^{-}]}

                = \frac{10^{-14}}{0.0003}

                = 3.33 \times 10^{-11} M

Thus, we can conclude that the concentration of spectator ion is 3.33 \times 10^{-11} M.

5 0
3 years ago
Describe the properties of alkali metals. Based on their electronic arrangement, explain whether they exist alone in nature.
bearhunter [10]

Answer:

- They are highly reactive metals

- They have low electro negativity

- They have low ionization energy

- They don't exist alone in nature

- They have low densities

Explanation:

Alkali metals are the elements in group 1 of the periodic table. They include Sodium, Lithium, Potassium e.t.c.

Due to the fact they have one atom in their outermost shell, they are very unstable because they easily react with other elements and are therefore don't exist alone in nature but combined with other elements for this same reason.

Since alkali metals don't easily attract other elements due to it's lone pair in the outer most shell, it can be said to have low electro negativity.

Also, they don't need energy to discharge their electrons since they are highly reactive due to their lone pair in the outermost shell and so we say they have low ionization energy.

Due to this reason, they also have low densities.

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galina1969 [7]

Answer:

Robert Boyle

Explanation:

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Answer:

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Explanation:

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