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1 year ago

What material is the source for commercial production of each of the following elements:(a) aluminum;

Chemistry

1 answer:

AleksandrR [38]1 year ago

3 0

The material which is used as source for commercial production aluminum is bauxite.

The aluminum can be extracted from bauxite ore by the process of Bayer process.

In the Bayer process, bauxite ore is heated in the pressure vessel along with a caustic soda solution (sodium hydroxide) at a temperature between 150 to 200 °C. At this temperatures, the aluminium is dissolved in the solution as sodium aluminate in the extraction process. After separation of the residue by filtering, when the liquid is cooled gibbsite is precipitated and then it is seeded with fine-grained aluminum hydroxide crystals from previous extractions. The precipitation take 7-19 days without the addition of seed crystals.

This extraction process converts the aluminium oxide to soluble sodium aluminate, NaAlO2, which afterward converted into aluminum hydroxide and then into aluminum oxide.

Thus, we concluded that the material which is used as source for commercial production aluminum is bauxite ore.

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Question 2 of 10

Temka [501]

Scientists tell people which issues are the most important describes an important role scientists have in society.

In addition to advising policymakers and other stakeholders about the best and wisest steps to take toward a human-centered society, scientists have a crucial role to play in preventing unwise and dangerous decisions. This promotes scientific knowledge and strengthens cross-cultural relationships and collaborative research. Additionally, they shouldn't overlook the objective, perpetually unfinished nature of science. With the help of a transdisciplinary cross-talks method, we emphasize the significance of information transfer across all scientific disciplines.

Due to the specific ways in which they aid in understanding how the world functions, scientists are crucial to society. It is a mode of thinking that helps us organize our knowledge in order to comprehend how things operate on a deeper level.

Learn more about Scientists here

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8 0

1 year ago

For the reaction KClO2⟶KCl+O2 KClO2⟶KCl+O2 assign oxidation numbers to each element on each side of the equation. K in KClO2:K i

frosja888 [35]

Answer:

Explanation:

The formula of the reaction:

KClO₂ → KCl + O₂

To assign oxidation numbers, we have to obey some rules:

Elements in an uncombined state or one whose atoms combine with one another to form molecules have an oxidation number of zero.
The charge on simple ions signifies their oxidation number.
The algebraic sum of all the oxidation number of all atoms in a neutral compound is zero. For radicals with charges, their oxidation number is the charge.

The oxidation number of K in KClO₂:

K + (-1) + 2(-2) = 0

K-5 = 0

K = +5

The oxidation number of K in KCl:

K + (-1) = 0

K = +1

The oxidation number Cl in KClO₂ is -1

For Cl in KCl, the oxidation number is -1

For O in KClO₂, the oxidation number is (2 x -2) = -4

For O in O₂, the oxidation number is 0

K moves from an oxidation state of +5 to +1. This is a gain of electrons and K has undergone reduction. We then say K is reduced.

O moves from an oxidation state of -4 to 0. This is a loss of electrons and O has undergone oxidation. We say O is oxidized.

7 0

2 years ago

Read 2 more answers

In Keynes's liquidity preference framework, as the expected return on bonds increases (holding everything else unchanged), the e

Julli [10]

Answer:

the correct words to fill up the blank spaces are falls and money

Explanation:

.1. the expected return on money falls

.2. causing the demand for money to fall.

3 0

2 years ago

The sun radiates different type of Electromagnetic energy (TRUE or FALSE)

Paha777 [63]

That statment is true

7 0

3 years ago

A 2.912 g sample of a compounds containing only C, H, and O was completely oxidized in a reaction that yielded 3.123 g of water

Taya2010 [7]

Answer:

Explanation:

18 gram of water contains 2 g of hydrogen

3.123 gram of water will contain 2 x 3.123 / 18 = .347 g of hydrogen .

44 gram of carbon dioxide contains 12 g of carbon

7.691 gram of carbon dioxide will contain 12 x 7.691 / 44 = 2.1 g of carbon .

So the sample will contain 2.912 - ( .347 + 2.1 ) g of oxygen .

= .465 g of oxygen .

moles of Carbon = 2.1 / 12 = .175

moles of hydrogen = .347 / 1 = .347

moles of oxygen = .465 / 16 = .029

Ratio of moles of carbon , oxygen and hydrogen ( C,O,H )

= 0.175 : 0.029 : 0.347

= .175/ .029 : 1 : .347 / .029

= 6 : 1 : 12

So empirical formula = C₆H₁₂O

Let the molecular formula be $(C_6H_{12}O)_n$

molecular weight = n ( 6 x 12 + 12x 1 + 16)

= 100 n

Given 100 n = 100.1

n = 1

Molecular formula = C₆H₁₂O.

3 0

2 years ago