The period is the end of the sentence!!!
Answer:
<em>Green</em><em> </em><em>vitriol</em><em>-</em><em> </em><em>FeSO4·7H2O</em>
<em>white</em><em> </em><em>vitriol</em><em>-</em><em> </em><em>ZnSO₄</em>
The inner diameter for a steel stack that exhausts 1,200 m3/min of gases at 1 atm and 400 k is 1.45 m
<h3>What is Stack Height ?</h3>
Stack height means the distance from the ground-level elevation at the base of the stack to the crown of the stack.
If a stack arises from a building or other structure, the ground-level elevation of that building or structure will be used as the base elevation of the stack.
Given is a steel stack that exhausts 1,200 cu.m/min of gases
P= 1 atm and
T= 400 K
maximum expected wind speed at stack height of 12 m/s
The formula for the diameter of chimney
![\rm d=\sqrt{\dfrac{4Q}{\pi v} }](https://tex.z-dn.net/?f=%5Crm%20d%3D%5Csqrt%7B%5Cdfrac%7B4Q%7D%7B%5Cpi%20v%7D%20%7D)
Q =1200 cu.m/min
= 1200 * 0.0166 = 19.92 cu.m/sec
Velocity = 12m/s
![\rm d=\sqrt{\dfrac{4\times 19.92}{3.14*12} }\\](https://tex.z-dn.net/?f=%5Crm%20d%3D%5Csqrt%7B%5Cdfrac%7B4%5Ctimes%2019.92%7D%7B3.14%2A12%7D%20%7D%5C%5C)
d= 1.45 m
Therefore The inner diameter for a steel stack that exhausts 1,200 m3/min of gases at 1 atm and 400 k is 1.45 m.
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0.040 mol / dm³. (2 sig. fig.)
<h3>Explanation</h3>
in this question acts as a weak base. As seen in the equation in the question,
produces
rather than
when it dissolves in water. The concentration of
will likely be more useful than that of
for the calculations here.
Finding the value of
from pH:
Assume that
,
.
.
Solve for
:
![\dfrac{[\text{OH}^{-}]_\text{equilibrium}\cdot[(\text{CH}_3)_3\text{NH}^{+}]_\text{equilibrium}}{[(\text{CH}_3)_3\text{N}]_\text{equilibrium}} = \text{K}_b = 1.58\times 10^{-3}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D_%5Ctext%7Bequilibrium%7D%5Ccdot%5B%28%5Ctext%7BCH%7D_3%29_3%5Ctext%7BNH%7D%5E%7B%2B%7D%5D_%5Ctext%7Bequilibrium%7D%7D%7B%5B%28%5Ctext%7BCH%7D_3%29_3%5Ctext%7BN%7D%5D_%5Ctext%7Bequilibrium%7D%7D%20%3D%20%5Ctext%7BK%7D_b%20%3D%201.58%5Ctimes%2010%5E%7B-3%7D)
Note that water isn't part of this expression.
The value of Kb is quite small. The change in
is nearly negligible once it dissolves. In other words,
.
Also, for each mole of
produced, one mole of
was also produced. The solution started with a small amount of either species. As a result,
.
,
,
.
Answer:
The mass of C2H2 in the mixture is 0.56gram using the ratio of carbon in the products contributed by the C2H2.
Explanation:
The balanced equation for the reaction is: C3H8 + 2C2H2 + 10O2 >> 7CO2 + 6H2O.
From the reaction, we know that the oxygen was in excess, this will make the Carbon sources the limiting agents in the reaction. The details of the reaction showed that the ratio of water to the carbon dioxide is 1.6:1. This also means that the expected mole of carbon dioxide will be 7/1.6, which is 3.75moles.
The individual balanced equation of reaction is:
C3H3 +5O2 >> 3CO2 + 4H2O
and 2C2H2 + 5O2 >>4CO2 + 2H2O. From this one can quickly tell that the propane is in sufficient supply as it produces 3 moles of CO2 out of the expected 3.75 moles obtained above. Leaving 0.75moles of CO2 to the ethyne.
The mass of ethyne in the mixture will therefore be: 0.75/3.75 X 2.8 = 0.56g.