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aliya0001 [1]
3 years ago
5

What kind of bond is formed when lithium and fluorine combine to form lithium fluoride?

Chemistry
2 answers:
Aleksandr [31]3 years ago
6 0

Answer:

The answer to your question is Ionic bonding.

Explanation:

There are three main types of bonding:

- Ionic bonding is when a metal attaches to a nonmetal and the electronegativity is higher than 1.7.

- Covalent bonding is when a non metal attaches to another non metal. and the electronegativity is between 0 and 1.7.

- Metallic bonding is a bond between metals.

In this question, we have      Lithium plus fluorine

Formula                                               LiF    

In this compound there a metal and a nonmetal

Electronegativity    Fluorine = 3.98     Lithium = 0.98

Difference of electronegativity =  3.98 - 0.98 = 3.0

                                             3.0 > 1.7

Then we conclude that the bond is Ionic.                                                

cupoosta [38]3 years ago
4 0

Answer:

ionic bond

Explanation:

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Which particles may be gained lost or shared by an atom when it forms a chemical bond?
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Calculate the change in pH when 71.0 mL of a 0.760 M solution of NaOH is added to 1.00 L of a solution that is 1.0O M in sodium
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Explanation:

It is known that pK_{a} value of acetic acid is 4.74. And, relation between pH and pK_{a} is as follows.

                    pH = pK_{a} + log \frac{[CH_{3}COOH]}{[CH_{3}COONa]}

                          = 4.74 + log \frac{1.00}{1.00}

So, number of moles of NaOH = Volume × Molarity

                                                   = 71.0 ml × 0.760 M

                                                    = 0.05396 mol

Also, moles of  CH_{3}COOH = moles of CH_{3}COONa

                                          = Molarity × Volume

                                          = 1.00 M × 1.00 L

                                          = 1.00 mol

Hence, addition of sodium acetate in NaOH will lead to the formation of acetic acid as follows.

            CH_{3}COONa + NaOH \rightarrow CH_{3}COOH

Initial :    1.00 mol                                  1.00 mol

NaoH addition:               0.05396 mol

Equilibrium : (1 - 0.05396 mol)    0           (1.00 + 0.05396 mol)

                    = 0.94604 mol                       = 1.05396 mol

As, pH = pK_{a} + log \frac{[CH_{3}COONa]}{[CH_{3}COOH]}

               = 4.74 +  log \frac{0.94604}{1.05396}

               = 4.69

Therefore, change in pH will be calculated as follows.

                         pH = 4.74 - 4.69

                               = 0.05

Thus, we can conclude that change in pH of the given solution is 0.05.

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