Answer: V2 = 35.54L
Explanation:
Applying
P1= 67.4, V1= 85, T1= 245, P2= 179.6, V2= ?,. T2=273
P1V1/ T1= P2V2/T2
Substitute and simplify
(67.4*85)/245 = (179.6*V2)/273
V2= 35.54L
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:
HNO3 + H2S = H2O + NO + S - Chemical Equation Balancer. Balanced Chemical Equation. 2 HNO 3 + 3 H 2 S → 4 H 2 O + 2 NO + 3 S. Reaction Information. Nitric Acid + Hydrogen Sulfide = Water + Nitric Oxide (radical) + Sulfur . Reactants.
Explanation:
A . O2- because it is a negatively charged ion
The mass in grams of butane at standard room temperature is 53.21 grams.
<h3>How can we determine the mass of an organic substance at room temperature?</h3>
The gram of an organic substance at room temperature can be determined by using the ideal gas equation which can be expressed as:
PV = nRT
- Pressure = 1.00 atm
- Volume = 22.4 L
- Rate = 0.0821 atm*L/mol*K
- Temperature = 25° C = 298 k
1 × 22.4 L = n × (0.0821 atm*L/mol*K× 298 K)
n = 22.4/24.4658 moles
n = 0.91556 moles
Recall that:
- number of moles = mass(in grams)/molar mass
mass of butane = 0.91556 moles × 58.12 g/mole
mass of butane = 53.21 grams
Learn more about calculating the mass of an organic substance here:
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