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BaLLatris [955]
3 years ago
14

what are we/what has someone did to prevent/lessen the negative side effects of aluminium extraction?

Chemistry
1 answer:
kogti [31]3 years ago
7 0

Aluminum is one of the main factors that reduce plant growth in acid soils. Although it is generally harmful to plants in soils with a neutral medium, the concentration of positive aluminum ions in acid soils increases and malfunctions in root and function growth.

Most acid soils are saturated with aluminum rather than hydrogen ions. Soil acidity is the result of hydrolysis of aluminum compounds. This principle (lime correction) to determine the degree of base saturation in the soil has become the basis of the methods used in soil testing laboratories to determine the lime requirements for soil. Application of lime to soil reduces the toxicity of aluminum to plants. Note This connector loads slowly.

Adaptation of wheat to allow aluminum to be carried out is due to the fact that aluminum releases organic compounds that in turn combine with harmful aluminum cations. It is believed that sorghum has the same endurance. The first genes found to withstand aluminum were found in wheat. Aluminum sulphide bearing has been found to be governed by an individual gene, such as in wheat. This is not the case in all plants.

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Identify the oxidized substance, the reduced substance, the oxidizing agent, and the reducing agent in the redox reaction. Cu(s)
spin [16.1K]

<u>Answer:</u> Copper is getting oxidized and is a reducing agent. Silver is getting reduced and is oxidizing agent.

<u>Explanation:</u>

Oxidation reaction is defined as the reaction in which an atom looses its electrons. Here, oxidation state of the atom increases.

X\rightarrow X^{n+}+ne^-

Reduction reaction is defined as the reaction in which an atom gains electrons. Here, the oxidation state of the atom decreases.

X^{n+}+ne^-\rightarrow X

Oxidizing agents are defined as the agents which oxidize other substance and itself gets reduced. These agents undergoes reduction reactions.

Reducing agents are defined as the agents which reduces the other substance and itself gets oxidized. These agents undergoes reduction reactions.

For the given chemical reaction:

Cu(s)+2AgNO_3(aq.)\rightarrow 2Ag(s)+Cu(NO_3)_2(aq.)

The half reactions for the above reaction are:

<u>Oxidation half reaction:</u>  Cu(s)\rightarrow Cu^{2+}(aq.)+2e^-

<u>Reduction half reaction:</u>  2Ag^+(aq.)+2e^-\rightarrow 2Ag(s)

From the above reactions, copper is loosing its electrons. Thus, it is getting oxidized and is considered as a reducing agent.

Silver is gaining electrons and thus is getting reduced and is considered as an oxidizing agent.

4 0
3 years ago
Air, Sea water, alloy afe the examples of
ratelena [41]

Answer:

B. Mixture

because air,sea water and alloy are example of Mixture

8 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Chuge%7B%5Cboxed%7B%5Cmathfrak%7BQuestion%3A%7D%7D%7D%5C%20%20%5Ctextless%20%5C%20br%20%2F%5
Kisachek [45]

Explanation:

when hot copper metal reacts with chlorine gas it forms CuCl2 which is yellow in colour.

non metal carbon burns in Oxygen gas they form non metal oxide.

5 0
3 years ago
Read 2 more answers
Which scenario is most similar to the type of collision that gas particles have according to kinetic molecular theory?
Ugo [173]
I’m not for sure but I think it’s A.
6 0
3 years ago
Read 2 more answers
What is the maximum mass of P2I4 that can be prepared from 8.80g of P4O6 and 12.37g of iodine according to the reaction:
Dvinal [7]
The balanced chemical reaction would be as follows:

<span>5P4O6 +8I2 ---> 4P2I4 +3P4O10

We are given the amount of reactants used for the reaction. We first need to determine the limiting reactant from the given amounts. We do as follows:

8.80 g P4O6 (1 mol / </span><span>219.88 g) = 0.04 mol P4O6
12.37 g I2 ( 1 mol / </span><span>253.809 g ) = 0.05 mol I2

Therefore, the limiting reactant is iodine since less it is being consumed completely in the reaction. We calculate the amount of P2I4 prepared as follows:

0.05 mol I2 ( 4 mol P2I4 / 8 mol I2 ) (</span><span>569.57 g / 1 mol) = 14.24 g P2I4</span>
8 0
3 years ago
Read 2 more answers
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