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lutik1710 [3]
2 years ago
7

Cuál es la importancia de una fórmula química y explique las clases de fórmulas con sus ejemplos respectivos.

Chemistry
1 answer:
Ksenya-84 [330]2 years ago
8 0

Answer and Explanation:

Las fórmulas químicas sirven para conocer qué elementos químicos y en qué cantidad están presentes en un compuesto químico. Por ejemplo: la fórmula química del dióxido de carbono es CO₂. Esta fórmula nos indica que la molécula de dióxido de carbono está formada por 1 átomo del elemento químico carbono y 2 átomos del elemento oxígeno.

Los diferentes tipos de fórmulas químicas son: fórmula empírica, fórmula molecular, fórmula semi-desarrollada y fórmula desarrollada. Tomando como ejemplo al compuesto <em>butano</em>:

- Fórmula empírica: C₂H₅ (indica que hay 2 átomos de C por cada 5 átomos de H; es una fórmula mínima)

- Fórmula molecular: C₄H₁₀ (indica que en realidad cada molécula de butano tiene 4 átomos de C y 10 átomos de H).

- Fórmula semi-desarrollada: CH₃CH₂CH₂CH₃ (indica el órden en que se unen los átomos en la molécula).

-Fórmula desarrollada: ver dibujo adjunto (indica cómo están unidos los átomos dentro de la molécula. Ahora vemos los enlaces entre los átomos y que los 2 átomos de C de los extremos están unidos a 3 átomos de H y a otro C, y los dos átomos de C del medio están unidos cada uno a 2 C y 2 H).

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This is an incomplete question, the image for the given question is attached below.

Answer : The wavelength of photon would be absorbed, 3.06\times 10^{-7}m

Explanation :

From the given diagram of energy we conclude that,

Energy at ground state, A = 400 zJ

Energy of 2nd excited state, C = 1050 zJ

Now we have to calculate the energy of the photon.

E=E_c-E_A

E=(1050-400)zJ= 650zJ=650\times 10^{-21}J

Now we have to calculate the wavelength of the photon.

Formula used :

E=h\times \nu

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

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

where,

E = energy of photon = 650\times 10^{-21}J

\nu = frequency of photon

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

\lambda = wavelength of photon  = ?

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

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

650\times 10^{-21}J=(6.626\times 10^{-34}Js)\times \frac{(3\times 10^{8}m/s)}{\lambda}

\lambda=3.06\times 10^{-7}m

Therefore, the wavelength of photon would be absorbed, 3.06\times 10^{-7}m

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