The triple bond is in the first place, not the 3rd because we have to choose the smallest possible number. That means the name will be: ‘but-1-yne’ or ‘1-butyne’
As you move across a period, the atomic radii decreases. ... As you move across a period, electrons are added to the same energy level while protons are also being added. The concentration of more protons creates a higher effective nuclear charge.
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:
382.5J
Explanation:
<em>Use the formula:</em>
E = mcΔθ or Q = mcΔT
m = 100g
c = 0.45 J/g°C
ΔT or Δθ = 110 - 25 = 85°
<em>Sub in the values:</em>
E = 100 × 0.45 × 85
= 382.5J
