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Komok [63]
3 years ago
13

I need a simplified and summarized version of the extraction of Aluminum using electrolysis.

Chemistry
2 answers:
vladimir1956 [14]3 years ago
8 0

Answer:

Alcoa Process:

Al Reaction 2.74: 23+2+32→23++2

Explanation:

2O3 from the Bayer process is chlorinated under reducing condition in the presence of C at (700-900)0 C to produce AlCl3, CO, CO2. AlCl3 vapour from the reaction is condensed at 700C in a fluid bed containing AlCl3 particles. The solid AlCl3 particles containing feed into an electrolytic cell containing a fused chloride electrolyte (AlCl3+NaCl+LiCl) at 7000 C. On electrolysis, Al (liquid) formed at the cathode, gaseous Cl2 liberated at the anode.

kobusy [5.1K]3 years ago
6 0

Instead, it is extracted by electrolysis. The ore is first converted into pure aluminium oxide by the Bayer Process, and this is then electrolysed in solution in molten cryolite - another aluminium compound. The aluminium oxide has too high a melting point to electrolyse on its own. The usual aluminium ore is bauxite.

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The spontaneous reaction that occurs when the cell in the picture operates is as follows: 2Ag+ + Cd(s) ???? 2 Ag(s) + Cd2+ (A) V
Snowcat [4.5K]

Answer:

14. B    15. D    16. C     17. B

Explanation:

The spontaneous reaction that occurs when the cell operates is shown below:

2Ag^{+} + Cd_{(s)} ⇒2Ag_{(s)} + Cd^{2+}

We need to select the correct option from the list below for the following questions.

(A) Voltage increases. (B) Voltage decreases but remains > zero. (C) Voltage becomes zero and remains at zero. (D) No change in voltage occurs. (E) Direction of voltage change cannot be predicted without additional information.

14. A 50-milliliter sample of a 2-molar Cd(NO_{3})_{2} solution is added to the left beaker.

If a 50-milliliter sample of a 2-molar Cd(NO_{3})_{2}  solution is added to the left beaker, the voltage decreases but its value remains greater than zero. The correct option is B

15. The silver electrode is made larger.

If the silver electrode is made larger, no change in the value of the voltage since we don't have the idea of the initial value. The correct option is D.

16. The salt bridge is replaced by a platinum wire.

If the salt bridge is replaced by a platinum wire, there will be no passage of electrons because electrons can't pass through a platinum wire. Therefore, the voltage will be zero and remains at zero. The correct option is C.

17. Current is allowed to flow for 5 minutes.

If current is allowed to flow for 5 minutes, the voltage decreases but its value remains greater than zero. The correct option is B.

7 0
4 years ago
A solution is made by mixing 33.0 ml of ethanol, C2H6O and 67.0 ml of water. Assuming ideal behavior, what is the vapor pressure
ivann1987 [24]
Grams ethanol = 33 ml times .789 gms/ml = 26.037 gms 

<span>Moles ethanol = 26.037 gms / 46 gms/mole = .57 moles </span>

<span>Moles water = 67 ml or 67 grams/18 gms/mole = 3.22 moles </span>

<span>total moles = .57 + 3.72 = 4.29 moles </span>

<span>Mole fraction ethanol = .57 moles ethanol / 4.29 moles total = 0.13</span>

<span>Moles fraction water = 3.72 moles water / 4.29 moles total = 0.87</span>

<span>Partial pressure of ethanol = mole fraction ethanol (.13) _ times VP ethanol 43.9 torr) = 5.707 torr </span>

<span>partial pressure water = mole fraction water .87) times VP water (l7.5 torr) = 15.23 torr </span>

<span>Total vapor pressure over solution = 5.71 torr + 15.23 torr = 20.94 torr</span>
7 0
4 years ago
What is the empirical formula for a compound that contains 10.89% magnesium 31.77% chloride and 57.34% oxygen
Blizzard [7]

Answer: Mg_{1}Cl_{2}O_{8}

Explanation:

If percentage are given then we are taking total mass is 100 grams.So, the mass of each element is equal to the percentage given.

Mass of Mg = 10.89 g

Mass of Cl = 31.77 g

Mass of O = 57.34 g

Step 1 : convert given masses into moles.

Moles of Mg=\frac {\text{ given mass of Mg}}{\text{ molar mass of Mg}}= \frac {10.89g}{24g/mole}=0.45moles

Moles of Cl = \frac {\text{ given mass of Cl}}{\text{ molar mass of Cl}}= \frac {31.77g}{35.5g/mole}=0.89moles

Moles of O =\frac {\text{ given mass of O}}{\text{ molar mass of O}}=\frac {57.34g}{16g/mole}=3.58moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Mg = \frac {0.45}{0.45}=1

For Cl = \frac {0.89}{0.45}=2

For O= \frac {3.58}{0.45}=8

The ratio of Mg :Cl : O= 1 : 2 : 8

Hence the empirical formula is Mg_{1}Cl_{2}O_{8}

4 0
3 years ago
What element is NaCl?
ale4655 [162]
Sorry but NaCl is a ionic compound not an element though it's made up of two elements sodium and chlorine
4 0
3 years ago
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k0ka [10]

Answer:

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4 years ago
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