NH3 + H2O -> NH4+ + OH- is the answer I think I just did this in class today
Explanation:
(a) The given data is as follows.
Pressure on top (
) = 140 bar =
(as 1 bar =
)
Temperature =
= (15 + 273) K = 288 K
Density of gas = 


= 0.4548

=
= 
Hence, pressure at the natural gas-oil interface is
.
(b) At the bottom of the tank,

= 2.206 \times 10^{7} Pa + 700 \times 9.81 \times (6000 - 4700)[/tex]
= 
= 309.8 bar
Hence, at the bottom of the well at
pressure is 309.8 bar.
These are dissolved in water to form colourless solutions, and then mixed together. This mixing leads to a double displacement reaction, essentially resulting in the metals 'swapping' their places in the two compounds, producing lead (II) iodide, and potassium nitrate.
Answer:
See explanation
Explanation:
The reaction that we are considering here is quite a knotty reaction. It is difficult to decide if the mechanism is actually E1 or E2 since both are equally probable based on the mass of scientific evidence regarding this reaction. However, we can easily assume that the methylenecyclohexane was formed by an E1 mechanism.
Looking at the products, one could convincingly assert that the reaction leading to the formation of the two main products proceeds via an E1 mechanism with the formation of a carbocation intermediate as has been shown in mechanism attached to this answer. Possible rearrangement of the carbocation yields the 3-methylcyclohexene product.
2 mol KClO₃ - 3 mol O₂
x mol KClO₃ - 9 mol O₂
x=9*2/3= 6 mol