A mixture of reactants and products for the reaction shown below is at
1 answer:
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Answer:
0.0177 L of nitrogen will be produced
Explanation:
The decomposition reaction of sodium azide will be:
As per the balanced equation two moles of sodium azide will give three moles of nitrogen gas
The molecular weight of sodium azide = 65 g/mol
The mass of sodium azide used = 100 g
The moles of sodium azide used =
so 1.54 moles of sodium azide will give = mol
the volume will be calculated using ideal gas equation
PV=nRT
Where
P = Pressure = 1.00 atm
V = ?
n = moles = 2.31 mol
R = 0.0821 L atm / mol K
T = 25 °C = 298.15 K
Volume =
this is the formula and answer of this question
<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.