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boyakko [2]
3 years ago
13

Calcula la concentración porcentual de una disolución que se preparó disolviendo 30 g de Hidróxido de Litio en suficiente cantid

ad de agua para obtener 100 ml
Chemistry
1 answer:
Eduardwww [97]3 years ago
3 0

Answer:

30 % m/V o 23 % m/m

Explanation:

Al leer el problema y entender el enunciado como que se usó suficiente cantidad de agua para obtener 100 mL, con la masa de LiOH podemos determinar la concentración en m/V.

m/V es una concentración que establece la relación de gramos de soluto en 100 mL de solución.

Como los gramos de soluto son 30g, la concentración en m/V es

30/100 = 0.3 % m/V

Podemos también establecer concentración m/m la cual establece la relación como masa de soluto en 100 g de solución. Como el volumen de solución coincide con el de solvente, la masa de solución la podemos obtener sumando masa de solvente + masa de soluto

30 g + 100 g = 130 g

Si tomamos el agua como solvente puro, y por densidad (1 g/mL) la masa de agua es 100g

Así el % m/m lo calculamos como (masa de soluto/masa solucion) . 100

(30g  / 130g) . 100 = 23 % m/m

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The correct answer is 9.7 grams because mg are 1,000 difference to grams.

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3 years ago
The most unusual characteristic of Uranus is _____.
fiasKO [112]

Answer:

the axis is tilted at 98 degrees

Explanation:

Uranus is blue, not green; it has an atmosphere of hydrogen and helium, not methane.  While it does have an icy cold temperature, that's not a very unusual characteristic for an outer planet.  However, the tilt of the axis is unusual; no other planet is tilted almost completely on its side!

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3 years ago
13. A mixture of MgCO3 and MgCO3.3H2O has a mass of 3.883 g. After heating to drive off all the water the mass is 2.927 g. What
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Answer:

63.05% of MgCO3.3H2O by mass

Explanation:

<em>of MgCO3.3H2O in the mixture?</em>

The difference in masses after heating the mixture = Mass of water. With the mass of water we can find its moles and the moles and mass of MgCO3.3H2O to find the mass percent as follows:

<em>Mass water:</em>

3.883g - 2.927g = 0.956g water

<em>Moles water -18.01g/mol-</em>

0.956g water * (1mol/18.01g) = 0.05308 moles H2O.

<em>Moles MgCO3.3H2O:</em>

0.05308 moles H2O * (1mol MgCO3.3H2O / 3mol H2O) =

0.01769 moles MgCO3.3H2O

<em>Mass MgCO3.3H2O -Molar mass: 138.3597g/mol-</em>

0.01769 moles MgCO3.3H2O * (138.3597g/mol) = 2.448g MgCO3.3H2O

<em>Mass percent:</em>

2.448g MgCO3.3H2O / 3.883g Mixture * 100 =

<h3>63.05% of MgCO3.3H2O by mass</h3>
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