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2 years ago

Identify the dominant intermolecular attraction in bh3.

1 answer:

6 0

The molecule BH3 is trigonal planar, with B in the center and H in the three vertices. Ther are no free electrons. All the valence electrons are paired in and forming bonds.

There are four kind of intermolecular attractions: ionic, hydrogen bonds, polar and dispersion forces.

B and H have very similar electronegativities, Boron's electronegativity is 2.0 and Hydrogen's electronegativity is 2.0.

The basis of ionic compounds are ions and the basis of polar compounds are dipoles.

The very similar electronegativities means that B and H will not form either ions or dipoles. So, that discards the possibility of finding ionic or polar interactions.

Regarding, hydrogen bonds, that only happens when hydrogen bonds to O, N or F atoms. This is not the case, so you are sure that there are not hydrogen bonds.

When this is the case, the only intermolecular force is dispersion interaction, which present in all molecules.

Then, the answer is dispersion interaction.

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3 years ago

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If 30.0 mL of 0.150 M CaCl₂ is added to 38.5 mL of 0.100 M AgNO3, what is the mass of the AgCl precipitate?

Ksenya-84 [330]

If 30.0 mL of 0.150 M CaCl₂ is added to 38.5 mL of 0.100 M AgNO3. The mass of the AgCl precipitate is 0.552 g.

<h3>What is Stoichiometry ?</h3>

Stoichiometry helps us use the balanced chemical equation to measure quantitative relationships and it is to calculate the amounts of products and reactants that are given in a reaction.

<h3>What is Balanced Chemical Equation ?</h3>

The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.

Now we have to write the balanced equation

2AgNO₃ + CaCl₂ → 2AgCl + Ca(NO₃)₂

Molecular Weight of CaCl₂ = 110.98 g/mol

Molecular Weight of AgNO₃ = 170.01 g/mol

Molecular Weight of AgCl = 143.45 g/mol

Here,

Volume of CaCl₂ = 30.0 mL = 0.03L

Volume of AgNO₃ = 38.5 mL = 0.0385 L

Now find the number of moles

Number of moles = Volume × Molarity

Number of moles of CaCl₂ = 0.03 L × 0.150

= 0.00456 mol

Number of moles of AgNO₃ = 0.0385 L × 0.100

= 0.00385 mol

The stoichiometric ratio of AgNO₃ to CaCl₂ is 2:1.

= $\frac{0.00385}{2}$

= 0.001925 mol

According to Stoichiometry Mass of AgCl

$= 0.0385 \times \frac{0.1}{1\ \text{mol}} \times \frac{2\ \text{mol} AgCl}{2\ \text{mol} AgNO_3} \times \frac{143.4\ g}{1\ \text{mol}}$

= 0.552 g AgCl

Thus from the above conclusion we can say that If 30.0 mL of 0.150 M CaCl₂ is added to 38.5 mL of 0.100 M AgNO3. The mass of the AgCl precipitate is 0.552 g.

Learn more about the Stoichiometry here: brainly.com/question/16060223

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1 year ago

A conductivity apparatus was inserted into a container of distilled water. The light bulb did not light up. What is the BEST exp

aliya0001 [1]

Distilled water is neutral, so there are no ions to allow the flow of electricity

4 0

2 years ago

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As you move from left to right across a period, what happens to the atomic radil?

egoroff_w [7]

Answer:

Moving Across a Period

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3 years ago

Consider the combustion of octane (C8H18)

sesenic [268]

The balanced equation for the above reaction is as follows;
2C₈H₁₈ + 25O₂ ---> 16CO₂ + 18H₂O
stoichiometry of octane to CO₂ is 2:16
number of C₈H₁₈ moles reacted - 191.6 g / 114 g/mol = 1.68 mol
when 2 mol of octane reacts it forms 16 mol of CO₂
therefore when 1.68 mol of octane reacts - it forms 16/2 x 1.68 = 13.45 mol of CO₂
number of CO₂ moles formed - 13.45 mol
therefore mass of CO₂ formed - 13.45 mol x 44 g/mol = 591.8 g
mass of CO₂ formed is 591.8 g

3 0

3 years ago