Which property is best to use when determining the strength of an ionic bond in a solid?

1 answer:

6 0

I think the best property to use is electronegativity since that is how much it attracts/wants electrons. The larger the difference in electronegativity is between two atoms, the stronger the ionic bond will be. This is also why two atoms need to have a minimum difference between the electronegativity to be considered ionic and not covalent.

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Step-by-step explanation:

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<h3>Answer: 5 cakes</h3>

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Explanation:

Let's start off converting the mixed number 12 & 1/4 to an improper fraction.

$a \frac{b}{c} = \frac{a*c+b}{c}\\\\12 \frac{1}{4} = \frac{12*4+1}{4}\\\\12 \frac{1}{4} = \frac{49}{4}\\\\$

Do the same for the other mixed number 2 & 1/3.

$a \frac{b}{c} = \frac{a*c+b}{c}\\\\2 \frac{1}{3} = \frac{2*3+1}{3}\\\\2 \frac{1}{3} = \frac{7}{3}\\\\$

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From here, we divide the two fractions. I converted them to improper fractions to make the division process easier.

$\frac{49}{4} \div \frac{7}{3} = \frac{49}{4} \times \frac{3}{7}\\\\\frac{49}{4} \div \frac{7}{3} = \frac{49\times 3}{4\times 7}\\\\\frac{49}{4} \div \frac{7}{3} = \frac{7\times 7\times 3}{4\times 7}\\\\\frac{49}{4} \div \frac{7}{3} = \frac{7\times 3}{4}\\\\\frac{49}{4} \div \frac{7}{3} = \frac{21}{4}\\\\$