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:
B.
Explanation:
One mole is the amount of substance that contain the Avogadro number which is equal to 6.022×10^23 atom, molecules or ions.
The answer is c. number of protons and d. atomic number. The proton number can identify an element. And also the atomic number is equal to the number of protons.
Answer:
ΔH = - 272 kJ
Explanation:
We are going to use the fact that Hess law allows us to calculate the enthalpy change of a reaction no matter if the reaction takes place in one step or in several steps. To do this problem we wll add two times the first step to second step as follows:
N2(g) + 3H2(g) → 2NH3(g) ΔH=−92.kJ Multiplying by 2:
2N2(g) + 6H2(g) → 4NH3(g) ΔH=− 184 kK
plus
4NH3(g) + 5O2(g) → 4NO(g) +6H2O(g) ΔH=−905.kJ
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2N2(g) + 6H2(g) + 5O2(g)→ 4NO(g) + 6H2O(g) ΔH = (-184 +(-905 )) kJ
ΔH = -1089 kJ
Notice how the intermediate NH3 cancels out.
As we can see this equation is for the formation of 4 mol NO, and we are asked to calculate the ΔH for the formation of one mol NO:
-1089 kJ/4 mol NO x 1 mol NO = -272 kJ (rounded to nearest kJ)
Answer:
Sr 2+(aq) + SO42-(aq) → SrSO4(s)
Explanation:
<u>Step 1</u>: Write a properly balanced equation with states:
K2SO4(aq) + Srl2(aq) → 2KI(aq) + SrSO4(s)
<u>Step 2</u>: write the full ionic equation with states. Remember to keep molecules intact. Only states (aq) will dissociate, (s) will not dissociate
. This means SrSO4 won't dissociate.
2K+(aq) + SO42-(aq) + Sr 2+(aq) + 2I-(aq) → 2K+(aq) + 2I-(aq) + SrSO4(s)
<u>Step 3</u>: Balanced net ionic equation
Sr 2+(aq) + SO42-(aq) → SrSO4(s)
