The effluent flow in concentration and particulate mass flow will be 0.198m³/sec.
<h3>How to calculate the effluent flow?</h3>
It should be noted that the total inflow will be equal to the total outflow. Therefore,
0.2 + 0.048 = 0.05 + We
Collect like terms
Qe = 0.2 + 0.048 - 0.05
Qe = 0.198m³/sec
The concentration will be:
= (360 × 1000)/0.05
= 7200mg/L.
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Answer:
1) C2H4(OH)2
Explanation:
A 1,2-ethanediol has an ethane structure consisting of two Carbon atoms with a hydrogen from each carbon substituted by a hydroxyl group. This makes it a 1,2-diol.
The pH value of the solution is mathematically given as
pH=2.35
<h3>What
pH value of the
solution?</h3>
Question Parameters:
pH during the titration of 20.00 mL of 0.1000 M dimethylamine,
with 0.1000 M HCl(aq) after 21.23 mL of the acid
Generally, the equation for the Chemical Reaction is mathematically given as
(CH3)2NH(aq), +Hcl ---> <---- (CH3)2NH2Cl(aq)
Therefore
HCL=0.00444M
WHere
HClaq--->H+(aq)+Cl-(aq)
Hence
H+=0.00444M
pH= -log{H+}
pH=log(0.00444)
pH=2.35
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Answer:
P and V: inversely proportional
P and T: directly proportional
V and T: inversely proportional
Explanation:
For pressure and volume, as the volume goes up, meaning the container gets bigger, the pressure would go down. There would be more room in the container, so there would be less collisions between the molecules themselves and between the molecules and the container. This makes them inversely proportional.
For pressure and temperature, as the pressure goes up, there are more collisions, so the particles move faster. Temperature is the speed of the particles, so, since both pressure and temperature would go up at the same time, they are directly proportional.
For volume and temperature, this is similar to the PV relationship. As volume increases, there are less collisions between the particles. This means that the particles are going to move slower. Therefore, as volume goes up, temperature goes down, so they are inversely proportional.
Sorry this is super long, but I hope it fully explains the question for you! ☺
The 18o-labeled methanol (CH3O*H) will appear in the products side at position b.
<h3>
Position of 18o-labeled methanol in the products</h3>
The 18O label will appear at position b in the product as indicated in the image.
This methoxy group in the product formed in position b comes from the 18O-labeled methanol (CH3OH).
While the oxygens at positions a and c in the product come from the unlabeled hemiacetal.
Thus, the 18o-labeled methanol (CH3O*H) will appear in the products side at position b.
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