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:
We need 226 grams of FeS
Explanation:
Step 1: Data given
Mass of FeCl2 = 326 grams
Molar mass FeCl2 = 126.75 g/mol
Step 2: The balanced equation
FeS + 2 HCl → H2S + FeCl2
Step 3: Calculate moles FeCl2
Moles FeCl2 = 326 grams / 126.75 grams
Moles FeCl2 = 2.57 moles
Step 4: Calculate moles FeS needed
For 1 mol H2S and 1 mol FeCl2 produced, we need 1 mol FeS and 2 moles HCl
For 2.57 moles FeCl2 we need 2.57 moles FeS
Step 5: Calculate mass FeS
Mass FeS = 2.57 moles * 87.92 g/mol
Mass FeS = 226 grams FeS
We need 226 grams of FeS
The stoichiometric ratio of CuCl2 to NaCl is 1 is to 2. The stoichiometric ratio of 31.0 g CuCl2 is 26.95 grams of NaCl by converting the amount of CuCl2 to mole and multiplying by 0.5 and molar mass of NaCl.This amount is equal to 78.65% yield.
Answer:
The concentration of I at equilibrium = 3.3166×10⁻² M
Explanation:
For the equilibrium reaction,
I₂ (g) ⇄ 2I (g)
The expression for Kc for the reaction is:
![K_c=\frac {\left[I_{Equilibrium} \right]^2}{\left[I_2_{Equilibrium} \right]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%20%7B%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%7D%7B%5Cleft%5BI_2_%7BEquilibrium%7D%20%5Cright%5D%7D)
Given:
![\left[I_2_{Equilibrium} \right]](https://tex.z-dn.net/?f=%5Cleft%5BI_2_%7BEquilibrium%7D%20%5Cright%5D)
= 0.10 M
Kc = 0.011
Applying in the above formula to find the equilibrium concentration of I as:
![0.011=\frac {\left[I_{Equilibrium} \right]^2}{0.10}](https://tex.z-dn.net/?f=0.011%3D%5Cfrac%20%7B%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%7D%7B0.10%7D)
So,
![\left[I_{Equilibrium} \right]^2=0.011\times 0.10](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%3D0.011%5Ctimes%200.10)
![\left[I_{Equilibrium} \right]^2=0.0011](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%5E2%3D0.0011)
![\left[I_{Equilibrium} \right]=3.3166\times 10^{-2}\ M](https://tex.z-dn.net/?f=%5Cleft%5BI_%7BEquilibrium%7D%20%5Cright%5D%3D3.3166%5Ctimes%2010%5E%7B-2%7D%5C%20M)
<u>Thus, The concentration of I at equilibrium = 3.3166×10⁻² M</u>
