Answer : All of the above are valid expressions of the reaction rate.
Explanation :
The given rate of reaction is,

The expression for rate of reaction for the reactant :
![\text{Rate of disappearance of }NH_3=-\frac{1}{4}\times \frac{d[NH_3]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20disappearance%20of%20%7DNH_3%3D-%5Cfrac%7B1%7D%7B4%7D%5Ctimes%20%5Cfrac%7Bd%5BNH_3%5D%7D%7Bdt%7D)
![\text{Rate of disappearance of }O_2=-\frac{1}{7}\times \frac{d[O_2]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20disappearance%20of%20%7DO_2%3D-%5Cfrac%7B1%7D%7B7%7D%5Ctimes%20%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D)
The expression for rate of reaction for the product :
![\text{Rate of formation of }NO_2=+\frac{1}{4}\times \frac{d[NO_2]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20formation%20of%20%7DNO_2%3D%2B%5Cfrac%7B1%7D%7B4%7D%5Ctimes%20%5Cfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D)
![\text{Rate of formation of }H_2O=+\frac{1}{6}\times \frac{d[H_2O]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20formation%20of%20%7DH_2O%3D%2B%5Cfrac%7B1%7D%7B6%7D%5Ctimes%20%5Cfrac%7Bd%5BH_2O%5D%7D%7Bdt%7D)
From this we conclude that, all the options are correct.
= 24.3
The average atomic mass of X is the <em>weighted average</em> of the atomic masses of its isotopes.
We multiply the atomic mass of each isotope by a number representing its <em>relative importance</em> (i.e., its % abundance).
Thus,
0.790 × 24 u = 18.96 u
0.100 × 25 u = 2.50 u
0.110 × 26 u = <u>2.86 u</u>
TOTAL = 24.3 u
∴ The relative atomic mass of X is 24.3.
Answer:
4244.48 g to the nearest hundredth.
Explanation:
The molar mass of Glucose = 6*12.011 + 12*1.008+ 6*15.999
= 180.156.
So 23.56 moles = 180.156 * 23.56 = 4244.48 g