The balanced chemical equation is written as:
<span>CsF(s) + XeF6(s) ------> CsXeF7(s)
We are given the amount of </span>cesium fluoride and <span>xenon hexafluoride used for the reaction. We need to determine first the limiting reactant to proceed with the calculation. From the equation and the amounts, we can say that the limiting reactant would be cesium fluoride. We calculate as follows:
11.0 mol CsF ( 1 mol </span>CsXeF7 / 1 mol CsF ) = 11.0 mol <span>CsXeF7</span>
Answer:
PBr3 - Molecule , Polar
N2H2 - Molecule , (Polar in E- form and Non- polar in Z form)
C2H2 - Molecule , Non- Polar
N2 - Molecule , Polar
NCl3 - Molecule , Polar
SiF4 - Molecule , Non- Polar
NH3 - Molecule , Polar
F - Not- Molecule (atom)
H2 - Molecule and Non- Polar
Explanation:
Molecule : these are group of two or more atoms joined by strong force of attraction.
H2 is non- polar because it is homoatomic molecule.(made up of same element)
N2 is non- polar because it is homoatomic molecule.
<span>The nitartion of methyl benzoate is expected to proceed as given in the equation below:
</span>
In methyl benzoate there are 3 types of 1 H proton. The two ortho to the C=O group is a doublet at 8 ppm the 2 metal to the C=O is a multiple at 7.5 ppm and one para to the C=O is a multiplet at 7.5 ppm.
On nitration the ortho will probably show two signal one being a single with 3 proton integration and one a doublet with 1 H integration
The meta will show a highly down field singlet (coresponding to 1 proton), two unequal doublets (corresponding to 1 H each) and one multiplets (corresponding to 1H). This is the major product as seen from the 1H NMR.
The para isomer will come as two doublets which will be very close to each other there is a small signal for this set between 8.2 and 8.3 ppm.
Hello!
The correct answer is 1. KCI.
I really hope this helped you out! c: