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Klio2033 [76]
2 years ago
9

Given bh3, o3, co2, and nh3, which of these compound is an exception to the octet rules?.

Chemistry
1 answer:
Ghella [55]2 years ago
5 0

Answer:

a. BH₃

Explanation:

According to the octet rules, atoms reach stability when are surrounded by eight electrons in their valence shell when they combine to form a chemical compound.

From the options, the only compound in which the central atom does not meet the octet rules is BH₃. The central atom is boron (B), which has 3 electrons in its valence shell. When B is combined with hydrogen (H), 3 electrons from the 3 atoms of H are added. The total amount of electrons is 6, fewer than 8 electrons needed to meet the rule.

hope this helps

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The following is what type of reaction: 2Al(s) + 3CuCl2(aq) ---> 2AlCl3(aq) + 3Cu(s)
Volgvan

Answer:

single replacment

Explanation:

7 0
3 years ago
The equation below represents a double replacement/displacement reaction;
Naddik [55]
AB+XY->AY+XB
We know that the answer would be KCl because of the switching that takes place during a double displacement reaction. Just like Zn and MnO4 switched and combined, the remaining elements, K and Cl, will combine.
We know that the answer is simply KCl because both K and Cl have an ion of only +/-1, meaning when they cross, no suffixes are made, since their ions are only 1.
For example, if you combined Mg with Cl, you would get MgCl2, because Mg has an ion of +2.
I hope this helps!
4 0
3 years ago
Which of the following statements related to S N1 reactions is not true? - The heterolysis of a bond between atoms which do not
Varvara68 [4.7K]

Answer:

The charged carbon atom of a carbocation has a complete octet of valence shell electrons

Explanation:

A charged carbon atom of a carbocation has a valence shell that is not filled, <u>that's why it acts as an electrophile (or a Lewis base)</u>. This unfilled valence shell is also the reason of the nucleophilic attack that takes place during the second step of a SN1 reaction.

6 0
2 years ago
Each mineral group has a distinctive chemical feature such as
disa [49]

Answer:

A

Explanation:

6 0
2 years ago
Consider the following reaction:
adell [148]

Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

    2H₂O₂     ⟶      2H₂O     +     O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt  

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

 d[H₂O]/dt =  2d[O₂]/dt =  2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ =  6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂  

(a) Initial moles of H₂O₂

\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}

8 0
3 years ago
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