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earnstyle [38]
3 years ago
5

What is the oxidation state of an element that is not involved in bonding

Chemistry
1 answer:
masha68 [24]3 years ago
6 0

Answer:

Zero

Explanation:

The oxidation state of an element that is not involved in bonding is zero. An atom in its free and uncombined state has zero oxidation number.

The extent of oxidation of each atom of the elements in a molecular formula or formula unit or an ionic radical is given by the oxidation number.

  • The formal charge assigned to an atom present in a molecule or a formula unit or ion based on some arbitrary rules is given by the oxidation number.
  • Elements in their uncombined state or one whose atom combine with themselves to form molecules  have zero oxidation number.
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How many molecules of CaCl2 are equivalent to 75.9 g CaCl2
sergij07 [2.7K]
First, you need to find:
One mole of CaCl_{2} is equivalent to how many grams?

Well, for this you have to look up the periodic table. According to the periodic table:
The atomic mass of Calcium Ca = 40.078 g (See in group 2)
The atomic mass of <span>Chlorine Cl = 35.45 g (See in group 17)
</span>
As there are two atoms of Chlorine present in CaCl_{2}, therefore, the atomic mass of CaCl_{2} would be:

Atomic mass of CaCl_{2}  = Atomic mass of Ca + 2 * Atomic mass of Cl

Atomic mass of CaCl_{2} = 40.078 + 2 * 35.45 = 110.978 g

Now,

110.978 g of CaCl_{2} = 1 mole.
75.9 g of CaCl_{2} = \frac{75.9}{110.978} moles = 0.6839 moles.

Hence,
The total number of moles in 75.9g of CaCl_{2} = 0.6839 moles

According to <span>Avogadro's number,
1 mole = 1 * </span>6.022 * 10^{23} molecules
0.6839 moles = 0.6839 * 6.022 * 10^{23} molecules = 4.118*10^{23} molecules

Ans: Number of molecules in 75.9g of  CaCl_{2} =  4.118*10^{23} molecules

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3 0
3 years ago
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A) NOPE!
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