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bulgar [2K]
3 years ago
8

Os subníveis mais energéticos de um dado átomo são: ...4s2 3d10 4p5 a) indique o seu número atomico b) quantos electrões de valê

ncia apresenta esse átomo c) a que família pertence?
Chemistry
1 answer:
givi [52]3 years ago
7 0

Answer:

A. 35

B. 7

C. halogênios

Explanation:

Aqui, para responder a essa pergunta, precisaremos conhecer o elemento particular em questão.

..... 4s ^ 2 3d ^ 10 4p ^ 5 significa que está a cinco elétrons da configuração eletrônica do último elemento na primeira camada dos metais pesados.

O último elemento da 1ª série do elemento de transição é o zinco, portanto, como está a apenas 5 elementos de distância, o átomo de que estamos falando é o átomo de Bromo de Bromo.

A. O zinco tem um número atômico 30 e como o bromo está a 5 elementos de distância, seu número atômico é 35

B. Uma vez que pertence ao grupo halogênio, tem 7 elétrons de valência como o resto da família

C. Pertence à família dos halogênios

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In a voltaic cell, where does the reduction take place?
algol13

the cathode is where reduction occurs!

5 0
3 years ago
The carbon-14 content of a wooden harpoon handle found in an Inuit archaeological site was found to be 61.9% of the carbon-14 co
astraxan [27]

Answer:

3,964 years.

Explanation:

  • It is known that the decay of a radioactive isotope isotope obeys first order kinetics.
  • Half-life time is the time needed for the reactants to be in its half concentration.
  • If reactant has initial concentration [A₀], after half-life time its concentration will be ([A₀]/2).
  • Also, it is clear that in first order decay the half-life time is independent of the initial concentration.

  • The half-life of the element is 5,730 years.

  • For, first order reactions:

<em>k = ln(2)/(t1/2) = 0.693/(t1/2).</em>

Where, k is the rate constant of the reaction.

t1/2 is the half-life of the reaction.

∴ k =0.693/(t1/2) = 0.693/(5,730 years) = 1.21 x 10⁻⁴ year⁻¹.

  • Also, we have the integral law of first order reaction:

<em>kt = ln([A₀]/[A]),</em>

where, k is the rate constant of the reaction (k = 1.21 x 10⁻⁴ year⁻¹).

t is the time of the reaction (t = ??? year).

[A₀] is the initial concentration of the sample ([A₀] = 100%).

[A] is the remaining concentration of the sample ([A] = 61.9%).

∴ t = (1/k) ln([A₀]/[A]) = (1/1.21 x 10⁻⁴ year⁻¹) ln(100%/61.9%) = 3,964 years.

7 0
4 years ago
In which of these diatomic molecules would you NOT find an octet of electrons
Semmy [17]

Answer:

The hydrogen molecule is the only one in which can not find an octet of electrons around each atom.

Explanation:

Let's evaluate each case.  

1. Nitrogen (N₂):

With Z = 7, nitrogen has the following electronic configuration

1s²

2s² 2p³  → valence electrons

Since its valence electrons are 5, in the molecule one nitrogen atom shares 3 electrons with the other one, and each remains with an electron pair, so <u>each atom has an octet of electrons.</u>

2. Hydrogen (H₂):

With Z = 1, its electronic configuration is:

1s¹  → valence electron

In the molecule, the hydrogen atoms share the only electron they have, so they will have only 2 electrons around. In this diatomic molecule, <em><u>we can not find an octet.</u></em>

3. Oxygen (O₂):

Z = 8. Electronic configuration:

1s²

2s² 2p⁴  → valence electrons

In the diatomic molecule, each oxygen atom shares 2 electrons with the other one and remains with 2 pairs of electrons, therefore, <u>each oxygen atom has an octet</u>.      

4. Fluorine (F₂)

Z = 9. Electronic configuration:

1s²

2s² 2p⁵  → valence electrons

In this molecule, each fluorine atom shares 1 electron with the other and remains with 3 pairs of electrons, hence, <u>each fluorine atom has an octet of electrons around</u>.

Finally, we can say that the hydrogen molecule is the only one in which can not find an octet of electrons around each atom.

I hope it helps you!  

8 0
3 years ago
What should you do during a titration when you notice the indicator start to indicate the approach of the equilibrium point? Add
Leni [432]

B. Add the second reactant slower.

6 0
3 years ago
Read 2 more answers
Balance the following skeleton reactions and identify the oxidizing and reducing agents:(b) P₄(s) → HPO₃²⁻(aq) + PH₃(g) [acidic]
m_a_m_a [10]

The balanced chemical equation is :

5P₄ + 36OH → 12HPO₃⁻² (aq) + 8PH₃ (acidic)

Here the oxidation number of P changed from 0 to -3 in PH₃ and increases  from 0 to +3 in HPO₃⁻². When P₄ changes to PH₃ reduction reaction is taking place as there is addition of hydrogen and when P₄ changes to HPO₃⁻² oxidation takes place as there is addition of oxygen.

Thus clearly both reduction and oxidation are taking place.

Thus, we can infer that here P₄ is both oxidizing as well as reducing agent.

To know more about oxidation number here:

brainly.com/question/13182308

#SPJ4

5 0
2 years ago
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