I need help on this.

Identify the dominant intermolecular attraction in bh3.

k0ka [10]

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.

6 0

3 years ago

The activation energy for a reaction is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a catalyst. If t

11111nata11111 [884]

Answer:

The catalyzed reaction will take 2.85 seconds to occur.

Explanation:

The activation energy of a reaction is given by:

$k = Ae^{-\frac{E_{a}}{RT}}$

For the reaction without catalyst we have:

$k_{1} = Ae^{-\frac{E_{a_{1}}}{RT}}$ (1)

And for the reaction with the catalyst:

$k_{2} = Ae^{-\frac{E_{a_{2}}}{RT}}$ (2)

Assuming that frequency factor (A) and the temperature (T) are constant, by dividing equation (1) with equation (2) we have:

$\frac{k_{1}}{k_{2}} = \frac{Ae^{-\frac{E_{a_{1}}}{RT}}}{Ae^{-\frac{E_{a_{2}}}{RT}}}$

$\frac{k_{1}}{k_{2}} = e^{\frac{E_{a_{2}} - E_{a_{1}}}{RT}$

$\frac{k_{1}}{k_{2}} = e^{\frac{59.0 \cdot 10^{3}J/mol - 184 \cdot 10^{3} J/mol}{8.314 J/Kmol*600 K} = 1.31 \cdot 10^{-11}$

Since the reaction rate is related to the time as follow:

$k = \frac{\Delta [R]}{t}$

And assuming that the initial concentrations ([R]) are the same, we have:

$\frac{k_{1}}{k_{2}} = \frac{\Delta [R]/t_{1}}{\Delta [R]/t_{2}}$

$\frac{k_{1}}{k_{2}} = \frac{t_{2}}{t_{1}}$

$t_{2} = t_{1}\frac{k_{1}}{k_{2}} = 6900 y*1.31 \cdot 10^{-11} = 9.04 \cdot 10^{-8} y*\frac{365 d}{1 y}*\frac{24 h}{1 d}*\frac{3600 s}{1 h} = 2.85 s$

Therefore, the catalyzed reaction will take 2.85 seconds to occur.

I hope it helps you!

4 0

3 years ago

If a young girl learned that it would be 0° C, or 32° F, the next

Ad libitum [116K]

Answer:

Cold

Explanation:

Because 0°C And below or 32°F Refers To Cold Weather

7 0

3 years ago

What is the difference between mass number and average atomic mass?

myrzilka [38]

Answer:

the answer is c. I think

Explanation:

4 0

3 years ago

What does density have to do with the cycling of matter in earths interior

Anastaziya [24]

Answer:

The rocks of higher density inter the mantle. Due to density they sink and melt. When they melt the density decreases and rises to the top. They then are recycled through volcanoes.

Explanation:

8 0

3 years ago