When one mole of Na3PO4.3H2O is heated extensively, three moles of water are released.
The water molecules in Na3PO4.3H2O are called molecules of water of crystallization. These molecules are not covalently bonded to the Na3PO4 molecule. They are only loosely attached to the substance.
Strong heating will drive away these molecules of water of crystallization to give three moles of water in the product.
Hence, when one mole of Na3PO4.3H2O is heated extensively, three moles of water are released.
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Answer:
It Is Considered The "negative" Electrode
Explanation:
An electrochemical cell is an electrolytic cell that drives a non-spontaneous redox reaction through the application of electrical energy. This cell is used to decompose chemical compounds, in a process called electrolysis. An electrode at which reduction take place is called the cathode. In reduction, electrons travel toward the site of reduction such that the negative charge is on the cathode.
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.
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