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garik1379 [7]
3 years ago
5

Ionic bonds generally form when the bonded elements have a difference in electronegativity greater than 1.5 to 1.7. Subtract the

electronegativities for the following pairs of elements and predict whether they form ionic bonds.
Electronegativity difference of Na and F:


yes


no
Chemistry
2 answers:
aksik [14]3 years ago
8 0

Electronegativity difference of Na and F: 3.1

yes.


Electronegativity difference of Mg and O: 2.3

yes.

krek1111 [17]3 years ago
6 0

Answer: Yes, sodium and fluorine form ionic bonds.

Explanation:

Electronegativity of sodium (Na) is 0.93.

Electronegativity of fluorine (F) is 3.98.

Electronegativity difference of Na and F will be calculated as follows.

   Electronegativity difference = Electronegativity of F - Electronegativity of Na

                                                   = 3.98 - 0.93

                                                   = 3.05

Thus, difference in electronegativity greater than 1.5 to 1.7, hence, bond between sodium and fluorine is ionic.

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<u>Answer:</u> The time required will be 19.18 years

<u>Explanation:</u>

All the radioactive reactions follows first order kinetics.

The equation used to calculate half life for first order kinetics:

k=\frac{0.693}{t_{1/2}}

We are given:

t_{1/2}=4.7\times 10^1yrs

Putting values in above equation, we get:

k=\frac{0.693}{4.7\times 10^1yr}=0.015yr^{-1}

Rate law expression for first order kinetics is given by the equation:

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}

where,  

k = rate constant  = 0.015yr^{-1}

t = time taken for decay process = ?

[A_o] = initial amount of the reactant = 2 g

[A] = amount left after decay process =  (2 - 0.5) = 1.5 g

Putting values in above equation, we get:

0.015yr^{-1}=\frac{2.303}{t}\log\frac{2}{1.5}\\\\t=19.18yrs

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