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muminat
3 years ago
11

How can electronegativity be used to explain the reaction we see in terms of the numbers and types of bonded being formed in lit

hium?

Chemistry
1 answer:
kolezko [41]3 years ago
5 0

Explanation:

Lithium is an electropositive element that readily loses electrons.

Oxygen is electronegative and it will readily accept electrons.

 Due to this significant electronegativity differences between the two species they form electrovalent or ionic bonds between them.

  2atoms of Li lose two electrons:

      Li  →  Li²⁺ + e⁻

             Lithium isoelectronic with helium

 For oxygen;

      O + 2e⁻   →  O²⁻

                          Oxygen is isoelectronic with Neon

Two ions of the lithium combines with the oxygen to form the bond;

                   4Li + O₂  →  2Li₂O

The electrostatic attraction between the two ions forms the ionic bond

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In the reaction below how would adding more of product C affect the equilibrium of the system? A+B arrows both ways C+D
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Answer:

1. The reaction will proceed backward, shifting the equilibrium position to the left.

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Answer:

a) Ba(OH)₂.8H₂O(s) + <em>2 </em>NH₄SCN(s) → Ba(SCN)₂(s) +<em>10</em> H₂O(l) + <em>2</em> NH₃(g)

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a) For the reaction:

Ba(OH)₂.8H₂O(s) + NH₄SCN(s) → Ba(SCN)₂(s) + H₂O(l) + NH₃(g)

As you see, there are 8 moles of water in reactants and 2 moles of oxygen in octahydrate, thus, water moles must be 10:

Ba(OH)₂.8H₂O(s) + NH₄SCN(s) → Ba(SCN)₂(s) +<em>10</em> H₂O(l) + NH₃(g)

To balance hydrogens, the other coefficients are:

Ba(OH)₂.8H₂O(s) + <em>2 </em>NH₄SCN(s) → Ba(SCN)₂(s) +<em>10</em> H₂O(l) + <em>2</em> NH₃(g)

b) As you see in the balanced reaction, 1 mole of barium hydroxide octahydrate reacts with 2 moles of NH₄SCN. 6.5g of Ba(OH)₂.8H₂O are:

6.5 g × (1mol / 315.48g) =<em> 0.0206moles of  Ba(OH)₂.8H₂O</em>. Thus, moles of NH₄SCN that must be used for a complete reaction are:

0.0206moles of  Ba(OH)₂.8H₂O × ( 2 mol NH₄SCN / 1 mol Ba(OH)₂.8H₂O) = <em>0.0412moles of NH₄SCN</em>. In grams:

0.0412moles of NH₄SCN × ( 76.12g / 1mol) = <em>3.14g must be added</em>

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