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Anna [14]
2 years ago
12

If the power to a condensing unit has been turned off for an extended period of time, the _________________________ should be en

ergized first for the recommended time to protect the compressor from liquid floodback. A. blower motor B. thermostat C. crankcase heater D. condenser fan motor
Engineering
1 answer:
Mama L [17]2 years ago
7 0

Answer:

C. Crankcase heater

Explanation:

The crankcase heater of the compressor was developed to mitigate the problematic variation of temperature on the in and out sides of the crankcase that has the capacity to change the physical states of the refrigerant and oil, cause them migrate and possibly mix together. It does so by allowing the regulation of the temperature on the inside of the crankcase

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A 13.7g sample of a compound exerts a pressure of 2.01atm in a 0.750L flask at 399K. What is the molar mass of the compound?a. 3
Katarina [22]

Answer: Option D) 298 g/mol  is the correct answer

Explanation:

Given that;

Mass of sample m = 13.7 g

pressure P = 2.01 atm

Volume V = 0.750 L

Temperature T = 399 K

Now taking a look at the ideal gas equation

PV = nRT

we solve for n

n = PV/RT

now we substitute

n = (2.01 atm x 0.750 L) / (0.0821 L-atm/mol-K x 399 K )

= 1.5075 / 32.7579

= 0.04601 mol

we know that

molar mass of the compound = mass / moles

so

Molar Mass = 13.7 g / 0.04601 mol

= 297.7 g/mol  ≈ 298 g/mol

Therefore Option D) 298 g/mol  is the correct answer

4 0
3 years ago
(Practice work, not graded)
luda_lava [24]

Answer: ALL CAREFULLY ANSWERED CORRECTLY.

Explanation:

1) A loaf of Bread PHYSICAL SYSTEM

✓ How can the environment affect the edibility of the bread

✓ What are the constituents that makes up the bread

✓ What process is involved in these constituents mixing to form the loaf.

2) The law of thermodynamics makes us to understand that when heat/energy passes through a system, the systems internal energy changes with respect to the conservation of energy law. That is energy lost = energy gained. Typically, ice would melt in a cup of hot tea because of the thermal energy in the molecules of the hot tea. When you heat a material, you are adding thermal kinetic energy to its molecules and usually raising its temperature. The temperature of the ice raises due to the kinetic energy added to it and it melts to water.

3) The theory of systems view the world as a complex system of interconnected parts. If we consider the society; (financial systems, political systems, etc) we will agree that they individually have their own components and it's the summation of this components that makes the system, this implies that system thinking could be applicable in this kinda of systems as long as they are made up of components.

4) Technology has boosted every sector of our lives and it has the capacity to do more. Restricting it's importance to entertainment alone would be an underusing of its potentials. Engineering students infact should not need any drive to be encouraged about maximizing all it can do in shaping our world.

5) ~ Nature shows its splendid soul

~Never ceases to leave us in amazement

~And we are in love

7 0
3 years ago
For methyl chloride at 100°C the second and third virial coefficients are: B = −242.5 cm 3 ·mol −1 C = 25,200 cm 6 ·mol −2 Calcu
bogdanovich [222]

Answer:

a)W=12.62 kJ/mol

b)W=12.59 kJ/mol

Explanation:

At T = 100 °C the second and third virial coefficients are

B = -242.5 cm^3 mol^-1

C = 25200 cm^6  mo1^-2

Now according isothermal work of one mole methyl gas is

W=-\int\limits^a_b {P} \, dV

a=v_2\\

b=v_1

from virial equation  

\frac{PV}{RT}=z=1+\frac{B}{V}+\frac{C}{V^2}\\   \\P=RT(1+\frac{B}{V} +\frac{C}{V^2})\frac{1}{V}\\

And  

W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V}  } \, dV

a=v_2\\

b=v_1

Now calculate V1 and V2 at given condition

\frac{P1V1}{RT} = 1+\frac{B}{v_1} +\frac{C}{v_1^2}

Substitute given values P_1\\ = 1 x 10^5 , T = 373.15 and given values of coefficients we get  

10^5(v_1)/8.314*373.15=1-242.5/v_1+25200/v_1^2

Solve for V1 by iterative or alternative cubic equation solver we get

v_1=30780 cm^3/mol

Similarly solve for state 2 at P2 = 50 bar we get  

v_1=241.33 cm^3/mol

Now  

W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V}  } \, dV

a=241.33

b=30780

After performing integration we get work done on the system is  

W=12.62 kJ/mol

(b) for Z = 1 + B' P +C' P^2 = PV/RT by performing differential we get  

         dV=RT(-1/p^2+0+C')dP

Hence work done on the system is  

W=-\int\limits^a_b {P(RT(-1/p^2+0+C')} \, dP

a=v_2\\

b=v_1

by substituting given limit and P = 1 bar , P2 = 50 bar and T = 373 K we get work  

W=12.59 kJ/mol

The work by differ between a and b because the conversion of constant of virial coefficients are valid only for infinite series  

8 0
3 years ago
How would you describe what would happen to methane if the primary bonds were to break?
erastova [34]

Answer:

All the bonds in methane (CH4CH4) are equivalent, and all have the same dissociation energy.

The product of the dissociation is methyl radical (CH3CH3). All the bonds in methyl radical are equivalent, and all have the same dissociation energy.

The product of that dissociation is methylene (CH2CH2). All the bonds in methylene are equivalent, and all have the same dissociation energy.

The product of that dissociation is methyne (CHCH) .

The C-H bonds in methane do not have the same dissociation energy as C-H bonds in methyl radical, which in turn do not have the same dissociation energy as the C-H bonds in methylene, which are again different from the C-H bond in methyne.

If (by some miracle) you were able to get all four bonds in methane to dissociate absolutely simultaneously, they would all show the same dissociation energy… but that energy, per bond broken, would be different than the energy required to break just one C-H bond in methane, because the products are different.

(In this case, it’s CH4→C+4HCH4→C+4H versus CH4→CH3+HCH4→CH3+H.)

To alter hydrocarbons you add enough energy to break a C-H bond. Why does only one bond break? What concentrates the energy on one C-H bond?

the weakest CH bond is the one that breaks. in plain alkanes it has to do with the molecular orbital interactions between neighboring carbon atoms. look at propane for example. the middle carbon has two C-C bonds, and each of those C-C bonds is strengthened by slight electron delocalization from the C-H bonds overlapping with the antibonding orbitals of the adjacent carbons.

since the C-H bonds on the middle carbon donate electron density to both of its neighbors, those two are weakest.

one of them will break preferentially.

which one actually breaks depends on the reaction conditions (kinetics). frankly it's whichever one ramdomly approaches a nucleophile first. when the nucleophile pulls of one of the H's, the other C-H bonds start to share (delocalize) the negative charge across the whole molecule. so while the middle C feels the majority of the negative charge character, the other two C's take on a fair amount as well...

by the way, alkanes don't really like to break and form anions like that.

a better example would be something like isopropyl iodide, where the C-I bond breaks and the I carries away the electron pair, forming a carbocation (also not particularly stable, but more so than the carbanion).

7 0
2 years ago
Can anyone help me ?
MrRa [10]
That’s too hard for me lol oof
3 0
3 years ago
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