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Semmy [17]
3 years ago
11

Formula semidesarrollada de los hidrocarburos aromáticos de 1 a 10 átomos de carbono

Chemistry
1 answer:
Tju [1.3M]3 years ago
3 0

Answer:

El principal componente del gas natural es también el hidrocarburo más simple: el metano. Este compuesto está formado por un átomo de carbono y cuatro átomos de hidrógeno y se representa de dos formas:

El hidrocarburo que le sigue en simplicidad es aquel que está constituido por dos átomos de carbono. Su fórmula condensada es C2H6 y se le conoce como etano.

Si se continúan colocando átomos de carbono con enlaces sencillos entre ellos e hidrógenos en los enlaces libres, se crean largas cadenas de compuestos. Al etano le sigue el propano (C2H8) y a éste, el butano (C4H10). Todos estos compuestos forman parte de la familia de los alcanos, y sus nombres terminan con el sufijo –ano para indicar que pertenecen a la misma familia.

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Vinegar is acidic => acetic acid (HC₂H₃O₂)

Explanation:

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Which of the following explains the difference between genes and chromosomes
Irina-Kira [14]

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B Genes determine specific traits while the chromosomes contain these genes

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The shielding of electrons gives rise to an effective nuclear charge, Zeff, which explains why boron is larger than oxygen. Esti
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Zeff = Z - S

Here, Z is the number of protons in the nucleus, that is, atomic number, and S is the number of nonvalence electrons.

For boron, the electronic configuration is 1s₂ 2s₂ 2p₄

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Zeff = 5-2 = +3

For O, electronic configuration is 1s₂ 2s₂ 2p₄

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3 0
3 years ago
A lithium flame has a characteristic red color due to emissions of wavelength 671 nm. What is the mass equivalence of 1 mol of p
e-lub [12.9K]

Answer:

1.98x10⁻¹² kg

Explanation:

The <em>energy of a photon</em> is given by:

  • E= hc/λ

h is Planck's constant, 6.626x10⁻³⁴ J·s

c is the speed of light, 3x10⁸ m/s

and λ is the wavelenght, 671 nm (or 6.71x10⁻⁷m)

  • E = 6.626x10⁻³⁴ J·s * 3x10⁸ m/s ÷ 6.71x10⁻⁷m = 2.96x10⁻¹⁹ J

Now we multiply that value by <em>Avogadro's number</em>, to <u>calculate the energy of 1 mol of such protons</u>:

  • 1 mol =  6.023x10²³ photons
  • 2.96x10⁻¹⁹ J *  6.023x10²³ = 1.78x10⁵ J

Finally we <u>calculate the mass equivalence</u> using the equation:

  • E=m*c²
  • m=E/c²
  • m =  1.78x10⁵ J / (3x10⁸ m/s)² = 1.98x10⁻¹² kg

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4 years ago
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