Chemistry, 26.09.2019 05:50 rhettperkins

Which of the following has the bonds correctly arranged in order of increasing polarity? a) be - f, mg - f, n - f, o - f b) o - f, n - f, be - f, mg - f c) o - f, be - f, mg - f, n - f d) n - f, be - f, mg - f, o - f e) mg-f, be - f, n - f, o - f

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Answer from: beenbad510

The right order of increasing polarity of bonds is: O−F, N−F, Be−F, Mg−F

The polarity of bond is known commonly to be directly proportional to the electronegativity difference.

Therefore the more greater the electronegativity difference, greater the polarity of bond.

Electronegativities can make one to know if a given bond will be non-polar, polar covalent, or ionic.

The greater the difference in electronegativity, the more polar the bond.

When looking for the element with high polarity, look at the electronegativities of each atom, find the difference in electronegativity for each atom in the bond. due observe and the Larger difference, the more polar and vice versa.

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Answer from: ximenareyna07

option b) O - F < N - F < Be - F < Mg - F

Explanation:

To calculate polarity, you need a table of electronegativities, to find the difference of electronegativities between each pair of atoms. The higher the electronegativity difference between the bonded atoms the higher the polarity of the bond.

Nevertheless, you can calculate the relative order for the given pairs by using periodic table trends, following this reasoning:

1) Be, N, O, and F are in the same period (row), in that same order from left to right.

2) In a period the electronegativity increases from left to right, so Be is the least electronegative, followed by N, followed by O, and F is the most electronegative.

So, you can predict this order for the polarities: O - F < N - F < Be - F

3) Now you have to deal with Mg - F.

You must use the fact that the electronegativity decreases in the periodic table in a group (column) from top to bottom. So, since Mg is directly below Be, you conclude that Mg is less electronegative and Mg - F is more polar than Be - F, and now you complete the order:

O - F < N - F < Be - F < Mg - F, which is the option b).

You can verify the conclusion using a table of electronegativities as follow:

1) Start by writing the electronegativities of the elements stated in the question:

Be: 1.57F: 3.98Mg: 1.31N: 3.04O: 3.44

2) Find the electronegative differences between each pair:

Be - F: | 1.57 - 3.98 | = 2.41Mg - F: | 1.31 - 3.98 | = 2.67N - F: | 3.04 - 3.98 | = 0.94O - F: | 3.44 - 3.98 | = 0.54

3) Compare and order:

0.54 < 0.94 < 2.41 < 2.67 ⇒ O-F < N - F < Be - F < Mg - F, which is the same conclusion from above, option b).

Answer from: Quest

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Answer from: Quest

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = $25^oc$ = 298 k $)0^oc=273kelvins$

$o_3+3no\rightarrow 3no_2,\delta h_{reaction}=? [/)</p><p>[tex]\delta h_{no}=90.4kj/mol$

$\delta h_{no_2}=33.85kj/mol$

$\delta h_{o_3}=142.2kj/mol$

$\delta h_{reaction}=\sum \delta h_{products}-\sum\delta h_{reactants}=3\times\delta h_{no_2}-(3\times \delta h_{no}+\delta h_{o_3})$

$\delta h_{reaction}=33.85-(90.4+142.2)=-311.85kj/mol$

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

$v_{no}=12.00 l,v_{o_3}=8.50 l$

$pv_{no}=n_{no}rt=1atm \times 12.00 l=n_{no}\times 0.0821 atml/k mol \times 298 k$

$n_{no}=0.49 moles$

$pv_{o_3}=n_{o_3}rt=1atm \times 8.50 l=n_{o_3}\times 0.0821 atml/k mol \times 298 k$

$n_{o_3}=0.34 moles$

according to reaction 3 moles no are reacting with 1 mol of $o_3$ ,then 0.49 mol of no will react with $\frac{1}{3}\times 0.49$ moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and $o_3$ is an excessive reagent.