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lutik1710 [3]
2 years ago
12

Somone PLZZZ BELP ME asappp

Chemistry
1 answer:
Elenna [48]2 years ago
5 0

Answer:

I believe it's the last answer choice.

Explanation:

You might be interested in
When carbon undergoes sp2 hybridization it forms
Artist 52 [7]
When carbon undergoes sp2 hybridization it forms methane ? I believe.

6 0
3 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
2 years ago
Propiedades químicas del óxido​
tangare [24]

Óxidos básicos: Son formados por metales. El metal presente en su fórmula puede presentar carga eléctrica +1 y +2, o sea, poseer carácter iónico. Ejemplos: Na2O (óxido de sodio), BaO (óxido de bario).

Óxidos neutros: Son compuestos por no metales. No reaccionan con agua, ácido o base, en razón del enlace covalente que une sus componentes; de ahí el por qué de ser llamados óxidos inertes. Ejemplos: monóxido de dinitrógeno (N2O) y monóxido de carbono (CO).

Óxidos ácidos: También conocidos como anhídridos de ácidos, son formados por no metales y presentan carácter covalente. En la presencia de agua, producen ácidos y en la presencia de bases, origina sal y agua. Ejemplo: CO2 (dióxido de carbono o gas carbono) y el SO2 (dióxido de azufre)

Óxidos dobles o mixtos: La combinación de dos óxidos de un mismo elemento, da origen a este tipo de óxidos. Ejemplo: magnetita (Fe2O4), unión de los óxidos de hierro (Fe) y oxígeno (O).

Óxidos anfóteros: Presentan ambigüedad, en la presencia de un ácido se comportan como óxidos básicos y en la presencia de una base, como óxidos ácidos. Ejemplos: óxido de aluminio (Al2O3 ) y el óxido de zinc (ZnO).

Peróxidos: Compuestos que poseen en su fórmula el grupo (O2)2- . Los peróxidos más comunes son formados por hidrógeno, metales alcalinos y metales alcalinos térreos. Ejemplos: agua oxigenada (H2O) y peróxido de sodio (Na2O2).

8 0
3 years ago
. A bright violet line occurs at 435.8 nm in the emission spectrum of mercury vapor. What amount of energy, in joules, must be r
Tom [10]

Answer :  The energy released by an electron in a mercury atom to produce a photon of this light must be, 4.56\times 10^{-19}J

Explanation : Given,

Wavelength = 435.8nm=435.8\times 10^{-9}m

conversion used : 1nm=10^{-9}m

Formula used :

E=h\times \nu

As, \nu=\frac{c}{\lambda}

So, E=h\times \frac{c}{\lambda}

where,

\nu = frequency

h = Planck's constant = 6.626\times 10^{-34}Js

\lambda = wavelength = 435.8\times 10^{-9}m

c = speed of light = 3\times 10^8m/s

Now put all the given values in the above formula, we get:

E=(6.626\times 10^{-34}Js)\times \frac{(3\times 10^{8}m/s)}{(435.8\times 10^{-9}m)}

E=4.56\times 10^{-19}J

Therefore, the energy released by an electron in a mercury atom to produce a photon of this light must be, 4.56\times 10^{-19}J

3 0
3 years ago
Activation energy is (blank)
timurjin [86]

B. The energy barrier between reactants and products

hope this helps!

(:

7 0
2 years ago
Read 2 more answers
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