It's NF3 (one Nitrogen and 3 Flouride)
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 is: not enough <span>colorless syrupy liquid.
</span>n(H₂SO₄) = 1,2 mol.
M(H₂SO₄) = 2Ar(H) + Ar(S) + 4Ar(O) · g/mol.
M(H₂SO₄) = 2·1 + 32 + 4·16 · g/mol.
M(H₂SO₄) = 98 g/mol.
m(H₂SO₄) = n(H₂SO₄) · M(H₂SO₄).
m(H₂SO₄) = 1,2 mol · 98 g/mol.
m(H₂SO₄) = 117,6 g needed.
100 g is less that 117,6 g.
Okay I did the math and I'm guessing around 18*C
Balanced there are 2 Al, 6 oxygens on both sides.
