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.
A weak bond between two molecules resulting from an electrostatic attraction between a proton in one molecule and an electronegative atom in another
According to florida wildlife group who experimentally tape magnets to crocodile heads to disrupt their homing ability so they don't wander into residential areas
A. The Average Ecological footprint will need to decrease . 6-11-18
Answer:
185.05 g.
Explanation
Firstly, It is considered as a stichiometry problem.
From the balanced equation: 2LiCl → 2Li + Cl₂
It is clear that the stichiometry shows that 2.0 moles of LiCl is decomposed to give 2.0 moles of Li metal and 1.0 moles of Cl₂, which means that the molar ratio of LiCl : Li is (1.0 : 1.0) ratio.
We must convert the grams of Li metal (30.3 g) to moles (n = mass/atomic mass), atomic mass of Li = 6.941 g/mole.
n = (30.3 g) / (6.941 g/mole) = 4.365 moles.
Now, we can get the number of moles of LiCl that is needed to produce 4.365 moles of Li metal.
Using cross multiplication:
2.0 moles of LiCl → 2.0 moles of Li, from the stichiometry of the balanced equation.
??? moles of LiCl → 4.365 moles of Li.
The number of moles of LiCl that will produce 4.365 moles of Li (30.3 g) is (2.0 x 4.365 / 2.0) = 4.365 moles.
Finally, we should convert the number of moles of LiCl into grams (n = mass/molar mass).
Molar mass of LiCl = 42.394 g/mole.
mass = n x molar mass = (4.365 x 42.394) = 185.05 g.
