The first step of the scientific method would be Observations.
Answer:
Elemental S reacts with O2 to form SO3 according to the reaction 2S+3O2→2SO3 Part B: What is the theoretical yield of SO3 produced by the quantities described in Part A? Express your answer numerically in grams.
Part A: 1.88x10^23 O2 molecules are needed to react with 6.67 g of S.
We address the equation...
S
(
s
)
+
3
2
O
2
(
g
)
→
S
O
3
(
g
)
Explanation:
The question specifies that we got
1.88
×
10
23
dioxygen molecules
...i.e. a molar quantity of...
1.88
×
10
23
⋅
molecules
6.022
×
10
23
⋅
molecules
⋅
m
o
l
−
1
=
0.312
⋅
m
o
l
...
But we gots with respect to sulfur,
6.67
⋅
g
32.06
⋅
g
⋅
m
o
l
−
1
=
0.208
⋅
m
o
l
...
And a bit of arithmetic later, we establish that we got stoichiometric quantities of dioxygen, and sulfur….in the reaction we produce a mass of ………..
0.208
⋅
m
o
l
×
80.07
⋅
g
⋅
m
o
l
−
1
=
16.65
⋅
g
.
Note that when
sulfur trioxide
is made industrially (and this a very important commodity chemical), sulfur is oxidized to
S
O
2
, and this is then oxidized up to
S
O
3
with some catalysis...
S
O
2
(
g
)
+
1
2
O
2
(
g
)
V
2
O
5
−−→
S
O
3
(
g
)
S
O
3
(
g
)
+
H
2
O
(
l
)
→
H
2
S
O
4
(
a
q
)
sulfuric acid
The industrial sulfur cycle must be a dirty, smelly, unfriendly process. The process is undoubtedly necessary to support our civilization....
<u>Answer:</u> The frequency of the light is ![3.088\times 10^{15}s^{-1}](https://tex.z-dn.net/?f=3.088%5Ctimes%2010%5E%7B15%7Ds%5E%7B-1%7D)
<u>Explanation:</u>
The equation used to calculate the energy for a transition, we use the equation:
![E_n=-2.18\times 10^{-18}J(\frac{1}{n^2})](https://tex.z-dn.net/?f=E_n%3D-2.18%5Ctimes%2010%5E%7B-18%7DJ%28%5Cfrac%7B1%7D%7Bn%5E2%7D%29)
where,
n = principal energy level
Calculating the energy difference:
![\Delta E=E_4-E_1](https://tex.z-dn.net/?f=%5CDelta%20E%3DE_4-E_1)
![\Delta E=-2.18\times 10^{-18}\left (\frac{1}{4^2}-\frac{1}{1^1}\right)\\\\\Delta E=2.044\times 10^{-18}J](https://tex.z-dn.net/?f=%5CDelta%20E%3D-2.18%5Ctimes%2010%5E%7B-18%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B4%5E2%7D-%5Cfrac%7B1%7D%7B1%5E1%7D%5Cright%29%5C%5C%5C%5C%5CDelta%20E%3D2.044%5Ctimes%2010%5E%7B-18%7DJ)
To calculate the energy of the light for a given frequency, we use the equation given by Planck, which is:
![E=h\nu](https://tex.z-dn.net/?f=E%3Dh%5Cnu)
where,
E = energy of the light = ![2.044\times 10^{-18}J](https://tex.z-dn.net/?f=2.044%5Ctimes%2010%5E%7B-18%7DJ)
h = Planck's constant = ![6.62\times 10^{-34}Js](https://tex.z-dn.net/?f=6.62%5Ctimes%2010%5E%7B-34%7DJs)
= frequency of the light = ?
Putting values in above equation, we get:
![2.044\times 10^{-18}J=6.62\times 10^{-34}Js\times \nu\\\\\nu=\frac{2.044\times 10^{-18}J}{6.62\times 10^{-34}Js}=3.088\times 10^{15}s^{-1}](https://tex.z-dn.net/?f=2.044%5Ctimes%2010%5E%7B-18%7DJ%3D6.62%5Ctimes%2010%5E%7B-34%7DJs%5Ctimes%20%5Cnu%5C%5C%5C%5C%5Cnu%3D%5Cfrac%7B2.044%5Ctimes%2010%5E%7B-18%7DJ%7D%7B6.62%5Ctimes%2010%5E%7B-34%7DJs%7D%3D3.088%5Ctimes%2010%5E%7B15%7Ds%5E%7B-1%7D)
Hence, the frequency of the light is ![3.088\times 10^{15}s^{-1}](https://tex.z-dn.net/?f=3.088%5Ctimes%2010%5E%7B15%7Ds%5E%7B-1%7D)
Answer:
a) 21.133 minutes for series
b) 84.54 minutes when split into 4
Explanation:
Data provided:
Total flow, V₀ = 45 MGD
total volume of each basin, V = 2500 m³
Now,
1 MGD = 3785.4118 m³/day
also,
1 day = 1440 minutes
thus,
45 MGD = 45 × 3785.4118 m³/day
or
45 MGD = 170343.5305 m³/day
and,
170343.5305 m³/day in 'm³/min'
= 170343.5305 / 1440
= 118.2941 m³/min.
Therefore,
The retention time, t = ![\frac{\textup{Volume of tank}}{\textup{Volumetric Flowrate }}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctextup%7BVolume%20of%20tank%7D%7D%7B%5Ctextup%7BVolumetric%20Flowrate%0A%7D%7D)
or
The retention time, t = ![\frac{\textup{2500}}{\textup{118.2941 }}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctextup%7B2500%7D%7D%7B%5Ctextup%7B118.2941%0A%7D%7D)
or
The retention time, t = 21.13 min
Hence,
for series of tanks the retention time is 21.133 min.
Now,
on splitting the tanks in 4
the volumetric flow rate will be
= ![\frac{\textup{118.2941}}{\textup{4}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctextup%7B118.2941%7D%7D%7B%5Ctextup%7B4%7D%7D)
= 29.57 m³/min.
Therefore,
The retention time = ![\frac{\textup{Total Volume of tank}}{\textup{Volumetric Flowrate }}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctextup%7BTotal%20Volume%20of%20tank%7D%7D%7B%5Ctextup%7BVolumetric%20Flowrate%0A%7D%7D)
or
The retention time, t = ![\frac{\textup{2500}}{\textup{29.57 }}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctextup%7B2500%7D%7D%7B%5Ctextup%7B29.57%0A%7D%7D)
or
The retention time, t = 84.54 min