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3 years ago

When 2.0 mol of methanol is dissolved in 45 grams of water, what is the mole fraction of methanol

Chemistry

1 answer:

kaheart [24]3 years ago

7 0

The mole fraction of methanol in the mixture is 0.444

We'll begin by calculating the number of mole of water.

Mass of water = 45 g

Molar mass of water = 18 g/mol

Mole of water =?

Mole = mass / molar mass

Mole of water = 45 / 18

Mole of water = 2.5 moles

Finally, we shall determine the mole fraction of methanol.

Mole of water = 2.5 moles

Mole of methanol = 2 moles

Total mole = 2 + 2.5 = 4.5 moles

Mole fraction of methanol =?

Mole fraction = mole / total mole

Mole fraction of methanol = 2 / 4.5

Mole fraction of methanol = 0.444

Thus, the mole fraction of methanol is 0.444

Learn more about mole fraction:

brainly.com/question/15444997

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Which response includes all the following that are properties of most metals, and no other properties?

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Answer:

The correct options are: Z1) They tend to form cations, and Z4) They tend to form ionic compounds when they combine with the elements of Group VIIA.

Explanation:

The chemical elements that have a tendency of losing their valence electrons and are generally present on the left-side of the periodic table are known as metals.

The general characteristics of metals are malleability, ductility, lustre, high thermal and electrical conductivity.

The metals readily lose their valence electrons present in their valence electron shell to form positively charged ions called cations and thus have low ionization energy.

They also tend to form ionic compounds or salts with the reactive non-metals belonging to the Group VIIA of the periodic table.

Therefore, Z1 and Z4 are the properties of most metals.

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3 years ago

La densidad del óxido de magnesio. MgO, es de 3.581 g/cm3 El MgO, es de 3.581 g/cm3 El MgO cristaliza con ordenamiento cúbico co

Jobisdone [24]

Answer:

$a=4.213cm$

$r=1.490x10^{-8}cm$

Explanation:

Hola.

En este caso, para calcular la longitud (a) de una cara de celda unitaria, consideramos la siguiente ecuación:

$\rho =\frac{#at*M}{a^3N_A}$

En la que consideramos el número de átomos por celda (4 para FCC), la masa molar (40.3 g/mol para MgO) y el número de avogadro para obtener:

$3.581g/mol = \frac{4atom/celda*40.3g/mol}{a^3*6.02x10^{23}atom/mol}$

Despejando para a, obtenemos:

$a^3 = \frac{4atom/celda*40.3g/mol}{3.581g/cm^3*6.02x10^{23}atom/mol}\\\\a=\sqrt[3]{7.478cm^3} \\\\a=4.213cm$

Finalmente, el radio lo calculamos como:

$r=\frac{\sqrt{2}*a}{4}=\frac{\sqrt{2}*4.213x10^{-8}cm}{4}\\\\r=1.490x10^{-8}cm$

¡Saludos!

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