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garri49 [273]
3 years ago
14

Choose the selection which correctly characterizes all three of the following substances in terms of whether they are polar or n

onpolar: SiH4 and BBr3 and SiF4 a) SiH4 is nonpolar and BBr3 is polar and SiF4 is nonpolar. b) SiH4 is nonpolar and BBr3 is polar and SiF4 is polar. c) SiH4 is nonpolar and BBr3 is nonpolar and SiF4 is polar. d) SiH4 is polar and BBr3 is nonpolar and SiF4 is polar. e) SiH4 is nonpolar and BBr3 is nonpolar and SiF4 is nonpolar.
Chemistry
1 answer:
ohaa [14]3 years ago
4 0

Answer:

SiH4 is nonpolar and BBr3 is nonpolar and SiF4 is nonpolar.

Explanation:

SiH4 is a non-polar compound. Though the Si–H bonds are polar, as a result of different electronegativities of Si and H. However, as there are 4 electron repulsions around the central Si atom, the polar bonds are arranged symmetrically around the central atom having a tetrahedral shape hence they cancel out making the compound nonpolar.

SiF4 is a nonpolar molecule because the fluorine atoms are arranged symetrically around the central silicon atom in a tetrahedral molecule with all of the regions of negative charge cancelling each other out just like in SiH4.

The 3 bromine atoms all lie in the same plane thus the geometry of the compound will be trigonal planar. The BBr3 will be non polar because the three B-Br bonds will cancel out each others' dipole moment given that they are in the same plane.

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Answer : The H_3O^+ ion concentration is, 1.12\times 10^{-3}M and the pH of a buffer is, 2.95

Explanation : Given,

K_a=7.1\times 10^{-4}

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Concentration of KNO_2 (conjugate base or salt)= 0.89 M

First we have to calculate the value of pK_a.

The expression used for the calculation of pK_a is,

pK_a=-\log (K_a)

Now put the value of K_a in this expression, we get:

pK_a=-\log (7.1\times 10^{-4})

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pK_a=3.15

Now we have to calculate the pH of the solution.

Using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}

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Now put all the given values in this expression, we get:

pH=3.15+\log (\frac{0.89}{0.26})

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pH=-\log [H_3O^+]

2.95=-\log [H_3O^+]

[H_3O^+]=1.12\times 10^{-3}M

The H_3O^+ ion concentration is, 1.12\times 10^{-3}M

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