We can store the copper sulphate solution in alumiun container, if cover on alumiun is present.
<h3>Can you store cuso4 in an aluminum container?</h3>
Aluminium is more reactive than copper so the Aluminium will displace copper sulphate from its solution by reacting with it but if there is cover on the aluminium then the alumium can't react with copper.
So we can store the copper sulphate solution in alumiun container.
Learn more about container here: brainly.com/question/11459708
In the periodic table the lanthanoid and the actinides are place separately at the bottom because of their electronic configuration and their properties compared to the other elements.
The the lanthanoid and the actinides are place separately at the bottom in the periodic table due to their electronic configuration and the properties. and to make the periodic table more convenient . if we place the f block elements that is he lanthanoid and the actinides then the size of the periodic table will increase. the f block elements are called as the inner transition element.
Thus , to make the periodic table more convenient and to group the elements in the block the the lanthanoid and the actinides are place separately at the bottom.
To learn more about lanthanoid and actinide here
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Answer:
and 
Explanation:
The equation for the reaction is AgNO3(aq) + KCl(aq) ==> AgCl(s) + KNO3(aq)
With all the ions, it is
(aq) +
(aq) +
(aq) +
(aq) ==> AgCl(s) +
and
do not change, so they are the spectator ions and are removed
The ionic equation is:
(aq) +
(aq) ==> AgCl(s)
There should be mass balance and the charge balance between the reactants and the products
Mass balance : total no of individual atoms of each type should be balanced before and after the reaction
Charge balance : Overall charge of the reactants should be balanced with the overall charge of the products
You can balance,
1)by just looking at it
2)by Algebraic method given above or
3)by the redox method
You need to know how to get the oxidation numbers in order to use the oxidation method