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3 years ago

Aluminium extraction from its ore

1 answer:

8 0

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. Aluminium oxide has a very high melting point (over 2000°C) so it would be expensive to melt it. Aluminium oxide does not dissolve in water, but it does dissolve in molten cryolite. This is an aluminium compound with a lower melting point than aluminium oxide. The use of cryolite reduces some of the energy costs involved in extracting aluminium.

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Answer:

Divergent boundaries – where two plates are moving apart. ...

Transform boundaries – where plates slide passed each other

Explanation:

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A neutral atom posses an atomic number of 13 and an atomic mass of 27 three electrons are lost. From which regio of an atom arr

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3 years ago

Which of the following is a halogen?

evablogger [386]

Answer:

$\boxed{\mathrm{Iodine}}$

Explanation:

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3 years ago

Read 2 more answers

How many mL of water would be displaced by 408 g of lead?

Galina-37 [17]

Answer:

36 mL of water can be displace by the 408 g lead

Explanation:

Given data:

Mass of lead = 408 g

Volume of water displaced by lead = ?

Solution:

The volume of water displace by the lead would be equal to the given volume of lead.

So we will calculate the volume from given mass of lead from density formula.

Density:

Density is equal to the mass of substance divided by its volume.

Units:

SI unit of density is Kg/m3.

Other units are given below,

g/cm3, g/mL , kg/L

Formula:

D=m/v

D= density

m=mass

V=volume

Density of lead is 11.36 g/cm³.

d = m/v

11.36 g/cm³ = 408 g/ v

v = 408 g/ 11.36 g/cm³

v =36 cm³

1 cm³ = 1 mL

volume = 36 mL

36 mL of water can be displace by the 408 g lead.

4 0

3 years ago

The osmotic pressure of an aqueoussolution of urea at 300 K is

den301095 [7]

<u>Answer:</u> The freezing point of solution is -0.09°C

<u>Explanation:</u>

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

$\pi=imRT$

where,

$\pi$ = osmotic pressure of the solution = 120 kPa

i = Van't hoff factor = 1 (for non-electrolytes)

m = concentration of solute in terms of molality = ?

R = Gas constant = $8.31\text{L kPa }mol^{-1}K^{-1}$

T = temperature of the solution = 300 K

Putting values in above equation, we get:

$120kPa=1\times m\times 8.31\text{ L kPa }mol^{-1}K^{-1}\times 300K\\\\m=0.05m$

To calculate the depression in freezing point, we use the equation:

$\Delta T=i\times K_f\times m$

where,

i = Vant hoff factor = 1 (for non-electrolytes)

$K_f$ = molal freezing point depression constant = 1.86°C/m

m = molality of solution = 0.05 m

Putting values in above equation, we get:

$\Delta T=1\times 1.86^oC/m\times 0.05m\\\\\Delta T=0.09^oC$

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

The equation used to calculate freezing point of solution is:

$\Delta T=\text{freezing point of water}-\text{freezing point of solution}$

where,

$\Delta T$ = Depression in freezing point = 0.09 °C

Freezing point of water = 0°C

Freezing point of solution = ?

Putting values in above equation, we get:

$0.09^oC=0^oC-\text{Freezing point of solution}\\\\\text{Freezing point of solution}=-0.09^oC$

Hence, the freezing point of solution is -0.09°C

4 0

3 years ago

Aluminium extraction from its ore​

Aluminium extraction from its ore