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IrinaK [193]
1 year ago
10

Which statement is correct abound chemical bonding in BF3 molecule O The molecule is not stable and hardly exist since the numbe

r of valence electrons around a central atom in a molecule are less than 8 O The molecule is stable and can exist even though the number of valence electrons around central atom in the molecule are less than 8 The molecule is not stable and hardly exist since the number of valence electrons around an atom in a molecule are more than 8 and covalent bonding is not possible None of the above is correct​
Chemistry
1 answer:
IceJOKER [234]1 year ago
5 0

The molecule is stable and can exist even though the number of valence electrons around central atom in the molecule are less than 8.

<h3>Is BF3 molecule stable or not?</h3>

BF3 molecule is a stable molecule because all the electrons present in the outermost shell of boron are covalently bonded with fluorine. Boron in BF3, three bonds is the maximum possible because boron only has 3 electrons to share.

So we can conclude that the molecule is stable and can exist even though the number of valence electrons around central atom in the molecule are less than 8.

Learn more about molecule here: brainly.com/question/26044300

#SPJ1

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Answer:

m_{PbI_2}=18.2gPbI_2

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Hello,

In this case, we write the reaction again:

Pb(NO_3)_2(aq) + 2 KI(aq)\rightarrow PbI_2(s) + 2 KNO_3(aq)

In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:

n_{Pb(NO_3)_2}=\frac{0.14gPb(NO_3)_2}{1g\ sln}*\frac{1molPb(NO_3)_2}{331.2gPb(NO_3)_2}  *\frac{1.134g\ sln}{1mL\ sln} *96.7mL\ sln\\\\n_{Pb(NO_3)_2}=0.04635molPb(NO_3)_2\\\\n_{KI}=\frac{0.12gKI}{1g\ sln}*\frac{1molKI}{166.0gKI}  *\frac{1.093g\ sln}{1mL\ sln} *99.8mL\ sln\\\\n_{KI}=0.07885molKI

Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:

0.04635molPb(NO_3)_2*\frac{2molKI}{1molPb(NO_3)_2} =0.0927molKI

But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:

m_{PbI_2}=0.07885molKI*\frac{1molPbI_2}{2molKI} *\frac{461.01gPbI_2}{1molPbI_2} \\\\m_{PbI_2}=18.2gPbI_2

Best regards.

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Explanation:

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