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gogolik [260]
3 years ago
14

Why is oxygen gas usually stored in metal cylinders?

Physics
2 answers:
olga2289 [7]3 years ago
4 0
The reason oxygen and other gasses are stored in metal cylinders is that of the simple fact it's a gas.  The reasoning behind it is to safely store the gas with no leaking etc.

"The purpose of this safety policy and procedure is to establish guidelines for the protection and safety of ([COMPANY]) employees who handle and use compressed gases. Compressed gases are typically stored under pressure in metal cylinders. These cylinders are designed and constructed to withstand high pressures. Improper handling and use of compressed gases can result in devastating consequences.

This safety policy and procedure provide guidelines for the safe handling and use of compressed gases. It includes provisions for training and presents safe handling guidelines. It also presents the types, uses, inspection, and marking requirements of compressed gas cylinders. Additionally, this safety policy and procedure present transportation and storage requirements for compressed gas cylinders."


quoted answers belong to: 


<span>https://www.safetyinfo.com/written-safety-programs-compressed-gas-cylinder-safety-program-free-index...
</span>


I take NO CREDIT in this answer.  All quoted answers are copied from the link above,  I am only here to help, not to get into trouble.  

worty [1.4K]3 years ago
4 0

Federal and state guidelines offer regulations related to the storage of oxygen cylinders, which state that the cylinders should be racked and kept away from combustible agents. Oxygen cylinders are used in industrial, educational, medical and consumer areas and should always be handled with care.

<span> Oxygen cylinders are hazardous because of the pressure inside them. They should always be stored in an upright and secure position since a fall could damage the tank, could cause a rupture, or could cause the tank to become an airborne hazard itself. The oxygen inside the cylinder may be very cold, particularly if the gas is to be used for cryogenic or freezing purposes. Such gases cause skin and other materials to become very brittle and can result in injuries that are similar to thermal burns.Oxygen cylinders must be stored away from flammable sources as their contents could ignite. Oxygen should also not be stored in the same space with other tanks of flammable gases or liquids. Additionally, federal and state regulations provide clear guidelines for how many tanks may be stored in a single area. Individual tanks for medicinal purposes present similar hazards and should be handled in according to the same precautions.Oxygen tanks should be stored in rooms designated for the purpose. The room should have sufficient area to move the cylinders and should be secured with locking doors. The room must meet appropriate fire ratings, ventilation requirements and construction codes. The racks used to secure both filled and empty tanks must meet similar code requirements.</span>
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What change in entropy occurs when a 0.15 kg ice cube at -18 °C is transformed into steam at 120 °c 4.
Studentka2010 [4]

<u>Answer:</u> The change in entropy of the given process is 1324.8 J/K

<u>Explanation:</u>

The processes involved in the given problem are:

1.)H_2O(s)(-18^oC,255K)\rightarrow H_2O(s)(0^oC,273K)\\2.)H_2O(s)(0^oC,273K)\rightarrow H_2O(l)(0^oC,273K)\\3.)H_2O(l)(0^oC,273K)\rightarrow H_2O(l)(100^oC,373K)\\4.)H_2O(l)(100^oC,373K)\rightarrow H_2O(g)(100^oC,373K)\\5.)H_2O(g)(100^oC,373K)\rightarrow H_2O(g)(120^oC,393K)

Pressure is taken as constant.

To calculate the entropy change for same phase at different temperature, we use the equation:

\Delta S=m\times C_{p,m}\times \ln (\frac{T_2}{T_1})      .......(1)

where,

\Delta S = Entropy change

C_{p,m} = specific heat capacity of medium

m = mass of ice = 0.15 kg = 150 g    (Conversion factor: 1 kg = 1000 g)

T_2 = final temperature

T_1 = initial temperature

To calculate the entropy change for different phase at same temperature, we use the equation:

\Delta S=m\times \frac{\Delta H_{f,v}}{T}      .......(2)

where,

\Delta S = Entropy change

m = mass of ice

\Delta H_{f,v} = enthalpy of fusion of vaporization

T = temperature of the system

Calculating the entropy change for each process:

  • <u>For process 1:</u>

We are given:

m=150g\\C_{p,s}=2.06J/gK\\T_1=255K\\T_2=273K

Putting values in equation 1, we get:

\Delta S_1=150g\times 2.06J/g.K\times \ln(\frac{273K}{255K})\\\\\Delta S_1=21.1J/K

  • <u>For process 2:</u>

We are given:

m=150g\\\Delta H_{fusion}=334.16J/g\\T=273K

Putting values in equation 2, we get:

\Delta S_2=\frac{150g\times 334.16J/g}{273K}\\\\\Delta S_2=183.6J/K

  • <u>For process 3:</u>

We are given:

m=150g\\C_{p,l}=4.184J/gK\\T_1=273K\\T_2=373K

Putting values in equation 1, we get:

\Delta S_3=150g\times 4.184J/g.K\times \ln(\frac{373K}{273K})\\\\\Delta S_3=195.9J/K

  • <u>For process 4:</u>

We are given:

m=150g\\\Delta H_{vaporization}=2259J/g\\T=373K

Putting values in equation 2, we get:

\Delta S_2=\frac{150g\times 2259J/g}{373K}\\\\\Delta S_2=908.4J/K

  • <u>For process 5:</u>

We are given:

m=150g\\C_{p,g}=2.02J/gK\\T_1=373K\\T_2=393K

Putting values in equation 1, we get:

\Delta S_5=150g\times 2.02J/g.K\times \ln(\frac{393K}{373K})\\\\\Delta S_5=15.8J/K

Total entropy change for the process = \Delta S_1+\Delta S_2+\Delta S_3+\Delta S_4+\Delta S_5

Total entropy change for the process = [21.1+183.6+195.9+908.4+15.8]J/K=1324.8J/K

Hence, the change in entropy of the given process is 1324.8 J/K

4 0
3 years ago
An object has a mass of 50.0 g and a volume of 10.5 cm3. What is the object's density?
seraphim [82]
Volume=mass/density

volume=455.6/19.3

volume=23.6 mL

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3 years ago
1. What happens to an atom when it gains electrons? *
Deffense [45]

Answer: An atom that gains or loses an electron becomes an ion. If it gains a negative electron, it becomes a negative ion. If it loses an electron it becomes a positive ion

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3 years ago
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Is there a change in speed from when a tennis ball first starts bouncing to when it is near the end of the bounce? How can you t
spin [16.1K]

Explanation:

The potential energy is stronger at the start of the bounce and when it is near the end bounce the balls accelerated rate will be lower because the force has decreased . Also I love your profile picture miraculous rocks !

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3 years ago
On a horizontal frictionless floor, a worker of weight 0.900 kN pushes horizontally with a force of 0.200 kN on a box weighing 1
Ad libitum [116K]

Answer:

D) The worker will accelerate at 2.17  m/s²  and the box will accelerate at 1.08  m/s² , but in opposite directions.

Explanation:

Newton's third law

Newton's third law or principle of action and reaction states that when two interaction bodies appear equal forces and opposite directions. in each of them.

F₁₂= -F₂₁

F₁₂: Force of the box on the worker

F₂₁: Force of the worker on the box

Newton's second law

∑F = m*a

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Formula to calculate the mass (m)

m =  W/g

Where:

W : Weight (N)

g : acceleration due to gravity  (m/s²)

Data

W₁ =1.8 kN   : box weight

W₂ = 0.900 kN : worker weight

F₂₁ = 0.200 kN

F₁₂ = - 0.200 kN

g = 9.8 m/s²

Newton's second law for the box

∑F = m*a

F₂₁ = m₁*a₁    m₁=W₁/g

0.2 kN = (1.8kN)/(9.8 m/s² ) *a₁

a_{1} =\frac{(0.2kN)*9.8\frac{m}{s^{2} } }{1.8 kN}

a₁= 1.08 m/s² : acceleration of the box

Newton's second law for the worker

∑F = m*a

F₁₂ = m₂*a₂ , m₂=W₂/g

- 0.2 kN =( (0.9 kN) /(9.8 m/s² ) )*a₂

a_{1} =\frac{(0.2kN)*9.8\frac{m}{s^{2} } }{0.9 kN}

a₂=  -2.17 m/s² : acceleration of the worker

5 0
3 years ago
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