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OverLord2011 [107]
3 years ago
13

Why should the oil be removed from a low pressure system at 130F

Chemistry
1 answer:
grin007 [14]3 years ago
6 0
When removing oil from a low pressure system, the temperature should be 130°F because, less refrigerant will be contained in the oil at higher temperature. Liquid charge into a deep vacuum will boil and may lower temperatures enough to freeze water in the tubes.
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Plz can anyone help me I'm a bit stuck in this one ​
trasher [3.6K]

Answer:

just see if i am not wrong

learning balancing in chemistry it take time

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8 0
2 years ago
Picture of gas laws question below:
Vera_Pavlovna [14]

Volume of the tank is 5.5 litres.

Explanation:

mass of the CO2 is given 8.6 grams

Pressure of the gas is 89 Kilopascal which is 0.8762 atm

Temperature of the gas is 29 degrees ( 0 degrees +273.5= K) so (29+273)

R = gas constant 0.0821 liter atmosphere per kelvin)

FROM THE IDEAL GAS LAW

PV=nRT ( P Pressure, V Volume, n is number of moles of gas, R gas constant, Temperature in Kelvin)

no of moles = mass/atomic mass

                    =  8.6/44

                    = 0.195 moles

now putting the values in equation

V=nRT/P

  = 0.195*0.0821*302/ 0.8762

  = 5.5 litres.

As the carbon dioxide gas occupies the volume os the tank hence volume of tank is 5.5 litres.

4 0
3 years ago
Which part of the atom forms chemical bonds?
Leokris [45]
Out of the choices given, the part of the atom that forms chemical bonds are the outermost electrons. The correct answer is B. 
8 0
3 years ago
Read 2 more answers
How many electrons in an atom could have these sets of quantum numbers n=7 l=3 ml=-1?
zysi [14]

Solution:

Since we have ml=-1

it shows that it has two 2e- i;e it fond in 2nd subshell in f orbital. And each subshell can hold 2 e-.

Thus the required answer is 2 electrons hold by an atom.

4 0
3 years ago
A student needs to prepare 50.0 mL of a 1.20 M aqueous H2O2 solution. Calculate the volume of 4.9 M H2O2 stock solution that sho
Nonamiya [84]

Answer : The volume of 4.9 M H_2O_2 stock solution used to prepare the solution is, 12.24 ml

Solution : Given,

Molarity of aqueous H_2O_2 solution = 1.20 M = 1.20 mole/L

Volume of aqueous H_2O_2 solution = 50.0 ml = 0.05 L

(1 L = 1000 ml)

Molarity of H_2O_2 stock solution = 4.9 M = 4.9 mole/L

Formula used :

M_1V_1=M_2V_2

where,

M_1 = Molarity of aqueous H_2O_2 solution

M_2 = Molarity of H_2O_2 stock solution

V_1 = Volume of aqueous H_2O_2 solution

V_2 = Volume of H_2O_2 stock solution

Now put all the given values in this formula, we get the volume of H_2O_2 stock solution.

(1.20mole/L)\times (0.05L)=(4.9mole/L)\times V_2

By rearranging the term, we get

V_2=0.01224L=12.24ml

Therefore, the volume of 4.9 M H_2O_2 stock solution used to prepare the solution is, 12.24 ml

3 0
3 years ago
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