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3 years ago

Explain why the flow of a compressed gas must be controlled for practical and safe use? use

1 answer:

8 0

Compressed gas is pressurized, which means it has very high pressure and can flow out really quickly in huge amounts which makes it very dangerous if it is not controlled correctly.

That's why the flow of a compressed gas must be controlled for practical and safe use.

You might be interested in

Changes in appearance...

Serjik [45]

I'm not quite sure but I think if you field the corner or cut it it would physically change it while not creating something new (chemical change).

3 0

4 years ago

Student exploitation: electron configuration

adell [148]

E:2

M ummmmm

try simplifying

4 0

4 years ago

2. A balloon full of air has a volume of 2.75 L at a temperature of 18°C. What is the balloon's volume

Sphinxa [80]

Answer:

3.01 L

Explanation:

: 2.75L

∘

: ?

∘

If you know your gas laws, you have to utilize a certain gas law called Charles' Law:

is the initial volume,

is initial temperature,

is final volume,

is final temperature.

Remember to convert Celsius values to Kelvin whenever you are dealing with gas problems. This can be done by adding 273 to whatever value in Celsius you have.

Normally in these types of problems (gas law problems), you are given all the variables but one to solve. In this case, the full setup would look like this:

2.75

291

318

By cross multiplying, we have...

291

= 874.5

Dividing both sides by 291 to isolate

, we get...

= 3.005...

In my school, we learnt that we use the Kelvin value in temperature to count significant figures, so in this case, the answer should have 3 sigfigs.

Therefore,

= 3.01 L

5 0

4 years ago

How do you find the limiting reactant if 2.50 mol copper and 5.50 mol silver nitrate react by single displacement?

Crank

The limiting reagent is Copper (Cu).

Explanation:

To identify the limiting reactant., first balance the reaction.

The balanced equation is:

Cu + 2AgNO3 -----> Cu(NO3)2 + 2Ag

To find the limiting reactant, take the amount of initial substance and find the number of moles of one of the products. The reactant that gives a product with the least number of moles, is the limiting reactant

$5.50 mol Cu \times (\frac{1.00 mol Cu}{2.00 mol AgNO3} )$ = 2.75 mol Cu.

Since there are only 2.50 mol Cu, copper is the limiting reactant, because, with that quantity of copper, only 2.50 mol AgNO3 will be reacted.

6 0

3 years ago

An object at 20∘C absorbs 25.0 J of heat. What is the change in entropy ΔS of the object? Express your answer numerically in jou

nevsk [136]

Question:

Part D) An object at 20∘C absorbs 25.0 J of heat. What is the change in entropy ΔS of the object? Express your answer numerically in joules per kelvin.

Part E) An object at 500 K dissipates 25.0 kJ of heat into the surroundings. What is the change in entropy ΔS of the object? Assume that the temperature of the object does not change appreciably in the process. Express your answer numerically in joules per kelvin.

Part F) An object at 400 K absorbs 25.0 kJ of heat from the surroundings. What is the change in entropy ΔS of the object? Assume that the temperature of the object does not change appreciably in the process. Express your answer numerically in joules per kelvin.

Part G) Two objects form a closed system. One object, which is at 400 K, absorbs 25.0 kJ of heat from the other object,which is at 500 K. What is the net change in entropy ΔSsys of the system? Assume that the temperatures of the objects do not change appreciably in the process. Express your answer numerically in joules per kelvin.

Answer:

D) 85 J/K

E) - 50 J/K

F) 62.5 J/K

G) 12.5 J/K

Explanation:

Let's make use of the entropy equation: ΔS = $\frac{Q}{T}$

Part D)

Given:

T = 20°C = 20 +273 = 293K

Q = 25.0 kJ

Entropy change will be:

ΔS = $\frac{25*1000}{293}$

= 85 J/K

Part E)

Given:

T = 500K

Q = -25.0 kJ

Entropy change will be:

ΔS = $\frac{-25*1000}{500}$

= - 50 J/K

Part F)

Given:

T = 400K

Q = 25.0 kJ

Entropy change will be:

ΔS = $\frac{25*1000}{400}$

= 62.5 J/K

Part G:

Given:

T1 = 400K

T2 = 500K

Q = 25.0 kJ

The net entropy change will be:

ΔS = $(\frac{25*1000}{400}) + (\frac{-25*1000}{500}$

= 12.5 J/K

7 0

3 years ago