What is the oxidation number of cl− in the hypochlorite ion clo−?
1 answer:
Answer : The oxidation number of chlorine (Cl) is, (+1)
Explanation :
Oxidation number : It represent the number of electrons lost or gained by the atoms of an element in a compound.
Oxidation numbers are generally written with the sign (+) and (-) first and then the magnitude.
When the atoms are present in their elemental state then the oxidation number will be zero.
Rules for Oxidation Numbers :
The oxidation number of a free element is always zero.
The oxidation number of a monatomic ion equals the charge of the ion.
The oxidation number of Hydrogen (H) is +1, but it is -1 in when combined with less electronegative elements.
The oxidation number of oxygen (O) in compounds is usually -2, but it is -1 in peroxides.
The oxidation number of a Group 1 element in a compound is +1.
The oxidation number of a Group 2 element in a compound is +2.
The oxidation number of a Group 17 element in a binary compound is -1.
The sum of the oxidation numbers of all of the atoms in a neutral compound is zero.
The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.
The given compound is,
Let the oxidation state of 'Cl' be, 'x'
Therefore, the oxidation number of chlorine (Cl) is, (+1)
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Explanation:
The mass of a pot is 300g and contains 90% aluminum. Find the number of moles of aluminum in the pot. P.A. (Al = 27)
The mass of aluminum present in the pot is:
Hence, in the given pot 270g Al is present.
The gram atomic mass of Al -27 g/mol
Given the mass of Al is 270 g
Substitute these values in the above formula:
Answer is 10.0 mol of Al is present.
42.4 ml is the volume in milliliters of the lead ball if a lead ball is added to a graduated cylinder containing 50.6 ml of water.
<h3>What is a graduated cylinder?</h3>A tall narrow container with a volume scale is used especially for measuring liquids.
The graduated cylinder contains water
mL is a volume unit.
Water volume = 50.6 ml
The lead ball caused an increase in volume from 50.6 ml to 93.0 mL.
The new volume is the lead ball volume plus the original water volume :
Final volume = Vlead ball+ Water original volume
Hence, 42.4 ml is the volume in milliliters of the lead ball.
Learn more about the graduated cylinder here:
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