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Describe the whole process of Aluminum extraction from bauxite.

AlekseyPX

Answer:

Extraction of aluminium

Aluminium ore is called bauxite . The bauxite is purified to produce aluminium oxide, a white powder from which aluminium can be extracted. The extraction is done by electrolysis. The ions in the aluminium oxide must be free to move so that electricity can pass through it.

4 0

2 years ago

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1.0 gallon to mm<br>N.B 1 gallon=3.8L

I am Lyosha [343]

Answer:

Gallon (US) to Milliliter Conversion Table

Gallon (US) [gal (US)] Milliliter [mL]

1 gal (US) 3785.411784 mL

2 gal (US) 7570.823568 mL

3 gal (US) 11356.235352 mL

5 gal (US) 18927.05892 mL

4 0

2 years ago

How can you increase the momentum of an object?

Nataliya [291]

Answer:

(c) by increasing it's mass

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2 years ago

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When 61.6 g of alanine (C3H7NO2) are dissolved in 1150. g of a certain mystery liquid X, the freezing point of the solution is 2

MrRissso [65]

Answer:

Explanation:

From the given information:

TO start with the molarity of the solution:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ C_3H_7 NO_3}{89.1 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

= 0.601 mol/kg

= 0.601 m

At the freezing point, the depression of the solution is $\Delta \ T_f = T_{solvent}- T_{solution}$

$\Delta \ T_f = 2.9 ^0 \ C$

Using the depression in freezing point, the molar depression constant of the solvent $K_f = \dfrac{\Delta T_f}{m}$

$K_f = \dfrac{2.9 ^0 \ C}{0.601 \ m}$

$K_f = 4.82 ^0 C / m}$

The freezing point of the solution $\Delta T_f = T_{solvent} - T_{solution}$

$\Delta T_f = 7.3^ 0 \ C$

The molality of the solution is:

$= \dfrac{61.6 \ g \times \dfrac{1 \ mol \ NH_4Cl}{53.5 \ g} }{1150 \ g \times \dfrac{1 \ kg}{1000 \g }}$

Molar depression constant of solvent X, $K_f = 4.82 ^0 \ C/m$

Hence, using the elevation in boiling point;

the Vant'Hoff factor $i = \dfrac{\Delta T_f}{k_f \times m}$

$i = \dfrac{7.3 \ ^0 \ C}{4.82 ^0 \ C/m \times 1.00 \ m}$

$\mathbf {i = 1.51 }$

3 0

2 years ago

A penny has a mass of 2.50g and the Moon has a mass of ×7.351022kg, How many moles of pennies have a mass equal to the mass of t

baherus [9]

Answer: 122 moles

Procedure:

1) Convert all the units to the same unit

2) mass of a penny = 2.50 g

3) mass of the Moon = 7.35 * 10^22 kg (I had to arrage your numbers because it was wrong).

=> 7.35 * 10^22 kg * 1000 g / kg = 7.35 * 10^ 25 g.

4) find how many times the mass of a penny is contained in the mass of the Moon.

You have to divide the mass of the Moon by the mass of a penny

7.35 * 10^ 25 g / 2.50 g = 2.94 * 10^25 pennies

That means that 2.94 * 10^ 25 pennies have the mass of the Moon, which you can check by mulitiplying the mass of one penny times the number ob pennies: 2.50 g * 2.94 * 10^25 = 7.35 * 10^25.

5) Convert the number of pennies into mole unit. That is using Avogadros's number: 6.022 * 10^ 23

7.35 * 10^ 25 penny * 1 mol / (6.022 * 10^ 23 penny) = 1.22* 10^ 2 mole = 122 mol.

Answer: 122 mol

8 0

2 years ago