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Tanzania [10]
4 years ago
6

How many atoms are in 1.45 mol of pure aluminum

Chemistry
1 answer:
Serhud [2]4 years ago
7 0
Using Avogadro's number:

(1.45 mol)(6.022140857 x 10^23 atoms/mol) = 8.73 x 10^23 atoms

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Calculate %ic of the interatomic bonds for the intermetallic compound al6mn. on the basis of this result, what type of interatom
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Explanation: For the calculation of % ionic character, we use the formula

\% \text{ ionic character}= [1-e^{\frac{-(X_A-X_B)^2)}{4}}]\times(100\%)

where X_A & X_B are the Pauling's electronegativities.

The table attached has the values of electronegativities, by taking the values of Al and Mn from there,

X_{Al}=1.61

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Putting the values in the electronegativity formula, we get

\% \text{ ionic character}= [1-e^{\frac{-(1.61-1.55)^2)}{4}}]\times(100\%)

                                               = 0.20%

Now, there are 3 types of inter atomic bonding

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2) Covalent Bonding: It refers to the chemical bond involving the sharing of electron pairs between 2 atoms.

3) Metallic Bonding: It refers to the chemical bond in which there is an electrostatic force between the positively charged metal ions and delocalised electrons.

In Al_6Mn compound, the % ionic character is minimal that is \sim0.20% and there are two metal ions present, therefore this compound will have metallic bonding.

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