The ideal gas under STP is 22.4 L/mol. While the gas has a rule of P1V1/T1=P2V2/T2. So the volume under 101 kPa and 273 K is 0.2*22.4=4.48 L.
Because i think melting is kinda the biproduct of evaporation.
Answer:
Less
Explanation:
Since [Cu(NH3)4]2+ and [Cu(H2O)6]2+ are Octahedral Complexes the transitions between d-levels explain the majority of the absorbances seen in those chemical compounds. The difference in energy between d-levels is known as ΔOh (ligand-field splitting parameter) and it depends on several factors:
- The nature of the ligand: A spectrochemical series is a list of ligands ordered on ligand strength. With a higher strength the ΔOh will be higher and thus it requires a higher energy light to make the transition.
- The oxidation state of the metal: Higher oxidation states will strength the ΔOh because of the higher electrostatic attraction between the metal and the ligand
A partial spectrochemical series listing of ligands from small Δ to large Δ:
I− < Br− < S2− < Cl− < N3− < F−< NCO− < OH− < C2O42− < H2O < CH3CN < NH3 < NO2− < PPh3 < CN− < CO
Then NH3 makes the ΔOh higher and it requires a higher energy light to make the transition, which means a shorter wavelength.
Answer:
The right solution according to the question is provided below.
Explanation:
According to the question,
(a)
The initial conditions will be:
DO = 
= 
= 
The initial oxygen defict will be:
Do = 
= 
The initial BOD will be:
Lo = 
= 
= 
(b)
The time reach minimum DO:
tc = ![\frac{1}{(kr-kd)} ln{(\frac{0.76}{0.61} )[1-\frac{1.674(0.76-0.61)}{0.61\times 6.453} ]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%28kr-kd%29%7D%20ln%7B%28%5Cfrac%7B0.76%7D%7B0.61%7D%20%29%5B1-%5Cfrac%7B1.674%280.76-0.61%29%7D%7B0.61%5Ctimes%206.453%7D%20%5D%7D)
= 
By putting the values of log, we get
= 
The distance to reach minimum DO will be:
Xc = 
= 
= 
