<span>The nitartion of methyl benzoate is expected to proceed as given in the equation below:
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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.
The Ti 2+ ions is represented by electron configuration (Ar)3d2. Titanium is in atomic number 22 and its electronic configuration is (Ar)3d2 4s2. Titanium loss two electron that is 4s2 electrons hence the electronic configuration ( Ar)3d2. 4s2 is the valence electron hence it the one which is lost to form Ti2+
Answer:
≅ 16.81 kJ
Explanation:
Given that;
mass of acetone = 31.5 g
molar mass of acetone = 58.08 g/mol
heat of vaporization for acetone = 31.0 kJ/molkJ/mol.
Number of moles = 
Number of moles of acetone = 
Number of moles of acetone = 0.5424 mole
The heat required to vaporize 31.5 g of acetone can be determined by multiplying the number of moles of acetone with the heat of vaporization of acetone;
Hence;
The heat required to vaporize 31.5 g of acetone = 0.5424 mole × 31.0 kJ/mol
The heat required to vaporize 31.5 g of acetone = 16.8144 kJ
≅ 16.81 kJ
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Hydrogen ion reacts with zinc to produce
gas
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