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3 years ago

How many moles of magnesium chloride (mgcl2) are in 1.31 l of 0.38 m mgcl2?

1 answer:

5 0

Molarity can be defined as the number of moles dissolved in 1 L of solution.
The molarity of MgCl₂ is 0.38 mol/L
In 1 L of MgCl₂ solution, amount go 0.38 moles is dissolved.
Therefore in 1.31 L of MgCl₂ - 0.38 mol /1 L x 1.31 L = 0.4978 mol rounded off is 0.50 mol
In 1.31 L of solution, amount of 0.50 mol of MgCl₂ is present.

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A sample compound contains 5.723g Ag, 0.852g S and 1.695g O. Determine its empirical formula.

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Answer:

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Explanation:

Formula for the calculation of no. of Mol is as follows:

$mol=\frac{mass\ (g)}{molecular\ mass}$

Molecular mass of Ag = 107.87 g/mol

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$mol\ of\ Ag=\frac{5.723\ g}{107.87\ g/mol} =0.05305\ mol$

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In order to get integer value, divide mol by smallest no.

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$O, \frac{0.10594}{0.02657} \approx 4$

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Explanation:

¡Hola!

En este caso, considerando el escenario dado, se hace necesario para nosotros saber que la posible reacción de disociación la experimenta el cloruro de plomo (II) como se muestra a continuación:

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Y que en términos de la solubilidad molar, s, se resuelve como:

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