Nitrogen in this compound has 2 atoms. 2 multiplied by its mass, 14.007, equals 28.014. Divide 28.014 by the molar mass of calcium nitrate: 28.014/164 = 0.17081. Multiply this by 100 to achieve its percentage: 0.17081 x 100 = 17.08%
The following reaction was monitored as a function of time: AB --> A + B A plot of 1/[AB] versus time is a straight line with slope, K = 5.5×10^−2 M * s.
Now,
<span>Now, Since at 70 s, [AB] = 0.116 M,
then amount of AB lost: </span> <span>0.210 M - 0.116 M = 0.094 M </span> Now, according to the stoichiometry of the reaction, <span>AB : A : B = 1 : 1 : 1, </span> <span>so both [A] and [B] gained the same number of moles and thus have same concentration, as [AB] lost.