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.
Answer:
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Answer:
The equation that gives the overall equilibrium in terms of the equilibrium constants K and Ky is K1 = K^6 * Ky
Explanation:
we have the following balanced reaction:
CaC2 + 2H2O = C2H2 + Ca(OH)2
the value of K for this reaction will be equal to:
K = ([C2H2] * [Ca(OH)2])/([CaC2] * [H2O]^2)
if we multiply the reaction by the value of 6, we have:
6CaC2 + 12H2O = 6C2H2 + 6Ca(OH)2
Again, the value of K for this reaction will be equal to:
K,´ = ([C2H2] ^6 * [Ca(OH)2]^6)/([CaC2]^6 * [H2O]^12) = K^6
For the second reaction:
6C2H2 + 3CO2 + 4H2O = 5CH2CHCO2H
The value of K for this reaction:
K2 = ([CH2CHCO2H]^5)/([C2H2]^6 * [CO2]^3 * [H2O]^4)
we also have:
K1 = ([CH2CHCO2H]^5)/([C2H2]^6 * [CO2]^3 * [H2O]^16)
Thus:
K1 = K^6 * Ky
Answer: The total pressure of air in lungs of an individual is 760.28 mm Hg
Explanation:
According to Dalton's law, the total pressure is the sum of individual pressures.
Given : 
=total pressure of gases = ?
= partial pressure of oxygen = 100 mm Hg
= partial pressure of nitrogen = 573 mm Hg
= partial pressure of Carbon dioxide = 0.053 atm = 40.28 mm Hg(1 atm = 760 mmHg)
= partial pressure of water vapor = 47 torr = 47 mm Hg (1torr=1 mm Hg)
putting in the values we get:
Thus the total pressure of air in lungs of an individual is 760.28 mm Hg
Mass, if you know what element you are working with.
