Answer:
11.8.4 Distillation Columns
Distillation columns present a hazard in that they contain large inventories of flammable boiling liquid, usually under pressure. There are a number of situations which may lead to loss of containment of this liquid.
The conditions of operation of the equipment associated with the distillation column, particularly the reboiler and bottoms pump, are severe, so that failure is more probable.
The reduction of hazard in distillation columns by the limitation of inventory has been discussed above. A distillation column has a large input of heat at the reboiler and a large output at the condenser. If cooling at the condenser is lost, the column may suffer overpressure. It is necessary to protect against this by higher pressure design, relief valves, or HIPS. On the other hand, loss of steam at the reboiler can cause underpressure in the column. On columns operating at or near atmospheric pressure, full vacuum design, vacuum breakers, or inert gas injection is needed for protection. Deposition of flammable materials on packing surfaces has led to many fires on opening of distillation column for maintenance.
Another hazard is overpressure due to heat radiation from fire. Again pressure relief devices are required to provide protection.
The protection of distillation columns is one of the topics treated in detail in codes for pressure relief such as APIRP 521. Likewise, it is one of the principal applications of trip systems.
Another quite different hazard in a distillation column is the ingress of water. The rapid expansion of the water as it flashes to steam can create very damaging overpressures.
Answer:
34.3 g
Explanation:
Step 1: Write the balanced equation
2 CH₃CH₂OH ⇒ CH₃CH₂OCH₂CH₃ + H₂O
Step 2: Calculate the moles corresponding to 50.0 g of CH₃CH₂OH
The molar mass of CH₃CH₂OH is 46.07 g/mol.
50.0 g × 1 mol/46.07 g = 1.09 mol
Step 3: Calculate the theoretical moles of CH₃CH₂OCH₂CH₃ produced
The molar ratio of CH₃CH₂OH to CH₃CH₂OCH₂CH₃ is 2:1. The moles of CH₃CH₂OCH₂CH₃ theoretically produced are 1/2 × 1.09 mol = 0.545 mol.
Step 4: Calculate the real moles of CH₃CH₂OCH₂CH₃ produced
The percent yield of the reaction is 85%.
0.545 mol × 85% = 0.463 mol
Step 5: Calculate the mass corresponding to 0.463 moles of CH₃CH₂OCH₂CH₃
The molar mass of CH₃CH₂OCH₂CH₃ is 74.12 g/mol.
0.463 mol × 74.12 g/mol = 34.3 g
The number of hours required : 37.2 hours
<h3>Further explanation</h3>
Given
⁴²K (potassium -42)
Required
The number of hours
Solution
The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.
Usually, radioactive elements have an unstable atomic nucleus.
Based on Table N(attached), the half-life for ⁴²K is 12.4 hours, which means half of a sample of ⁴²K will decay in 12.4 hours
For three half-life periods :
Answer:
pH= 11.49
Explanation:
Ethanolamine is an organic chemical compound of the formula; HOCH2CH2NH2. Ethanolamine, HOCH2CH2NH2 is a weak base.
From the question, the parameters given are; the concentration of ethanolamine which is = 0.30M, pH value= ??, pOH value= ??, kb=3.2 ×10^-5
Using the formula below;
[OH^-]=√(kb×molarity)----------------------------------------------------------------------------------------------------------(1)
[OH^-] =√(3.2×10^-5 × 0.30M)
[OH^-]= √(9.6×10^-6)
[OH^-]=3.0984×10^-3
pOH= -log[OH^-]
pOH= -log 3.1×10^-3
pOH= 3-log 3.1
pH= 14-pOH
pH= 14-(3-log3.1)
pH= 11+log 3.1
pH= 11+ 0.4914
pH= 11.49
Answer:
Higher than 59 °C because dipole-dipole interactions in iodine monochloride are stronger than dispersion forces in bromine.
Explanation:
I just took the test and i got it right
