Chemistry, 14.07.2020 19:01 brisamauro27

Sabiendo que a 20 °C, 0,275 moles de una sustancia desconocida cuya densidad es 0,924 g/cm3 ocupan un volumen de 45,9 cm3
, calcular:
a) la masa molar de la sustancia;
b) el número de moléculas presentes en 120 cm3
;
c) la masa, expresada en gramos, de una molécula de dicha sustancia.

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Answer from: biennhan9796

a) M=154.224g/mol

b) $moleculas=4.33x10^{23}moleculas$

c) $m=2.56x10^{-22}g$

Explanation:

Hola.

Para resolver los numerales a), b) y c), debemos seguir los siguientes procedimientos:

a) En este caso, primeo calculamos la masa que 45.9 cm³ ocupan con una densidad 0.924 g/cm³ como se muestra a continuación:

$\rho=\frac{m}{V}\\ \\m=\rho V =0.924\frac{g}{cm^3}*45.9cm^3=42.4g$

Seguidamente, sabiendo que estos gramos ocupan 0.275 moles, podemos facilmente calcular la masa molar como la relación entre ellas:

$M=\frac{m}{n}=\frac{42.4g}{0.275mol}\\ \\M=154.224g/mol$

b) En este caso, como se tienen 120 cm³ calculamos la nueva masa:

$m=0.924\frac{g}{cm^3} *120cm^3=110.88g$

Luego, con la masa molar, las moles:

$n=110.88g*\frac{1mol}{154.224g} =0.719mol$

Finalmente, con el número de Avogadro, las moléculas presentes:

$moleculas=0.719mol*\frac{6.022x10^{23}moleculas}{1mol} \\\\moleculas=4.33x10^{23}moleculas$

c) Aquí, consideramos que hay una molécula de la sustancia, por lo que usando el número de Avogadro, podemos calcular las moles que una molécula tiene:

$n=1molecula*\frac{1mol}{6.022x10^{23}moleculas}=1.66x10^{-24}mol$

Luego, usando la masa molar, calculamos la masa de una molécula:

$m=1.66x10^{-24}mol*\frac{154.224g}{1mol} \\\\m=2.56x10^{-22}g$

¡Saludos!

Answer from: Quest

assuming that your periodic table has eighteen columns, you should highlight the four elements in the third column from the left of the table.

explanation

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the $3d$ have energy higher than those in the $4 s$ orbital while lower than those of the $4 p$ orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the $4s$ orbital is therefore filled right after $3p$ orbitals and before $3 d$ orbitals.

potassium has electron configuration $[\text{ar}] 4 s^{1}$ and calcium $[\text{ar}] 4 s^{2}$ . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration $[\text{ar}] 4 s^{2} \; 3d^{1}$ and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the $ns^{2} \; (n-1) p^{1}$ and should be highlighted as well.

Answer from: Quest

Particle exchange energy though elastic collision , hope this .

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