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Irina18 [472]
3 years ago
8

Compare how entropy changes for the following two systems:

Chemistry
1 answer:
postnew [5]3 years ago
7 0

System A undergoes an increase in entropy while system B undergoes a decrease in entropy.

Entropy is the degree of disorderliness of a system. The entropy of a system depends on the number of particles present in the system as well as the state of matter.

Entropy is increased when solid particles dissolve in water because more particles are produced thereby increasing the level of disorderliness in the system.

On the other hand, when vapor is condensed, the degree of disorderliness decreases as gases are converted to liquids.

Learn more: brainly.com/question/13146879

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Pressure gauge at the top of a vertical oil well registers 140 bars. The oil well is 6000 m deep and filled with natural gas dow
andreyandreev [35.5K]

Explanation:

(a)  The given data is as follows.

              Pressure on top (P_{o}) = 140 bar = 1.4 \times 10^{7} Pa       (as 1 bar = 10^{5})

              Temperature = 15^{o}C = (15 + 273) K = 288 K

         Density of gas = \frac{PM}{ZRT}

                \frac{dP}{dZ} = \rho \times g

               \frac{dP}{dZ} = \int \frac{PM}{ZRT}

                \int_{P_{o}}^{P_{1}} \frac{dP}{dZ} = \frac{Mg}{ZRT} \int_{0}^{4700} dZ

           ln (\frac{P_{1}}{P_{o}}) = \frac{18.9 \times 10^{-3} \times 9.81 \times 4700 m}{0.80 \times 8.314 J/mol K \times 288 K}

                              = 0.4548

                     P_{1} = P_{o} \times e^{0.4548}

                                 = 1.4 \times 10^{7} Pa \times 1.5797

                                 = 2.206 \times 10^{7} Pa

Hence, pressure at the natural gas-oil interface is 2.206 \times 10^{7} Pa.

(b)   At the bottom of the tank,

                 P_{2} = P_{1}  + \rho \times g \times h

                             = 2.206 \times 10^{7} Pa + 700 \times 9.81 \times (6000 - 4700)[/tex]

                             = 309.8 \times 10^{5} Pa

                             = 309.8 bar

Hence, at the bottom of the well at 15^{o}C pressure is 309.8 bar.

6 0
4 years ago
a particular application calls for N2 g with a density of 1.80 g/L at 32 degrees C what must be the pressure of the n2 g in mill
baherus [9]

Answer:

1223.38 mmHg

Explanation:

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 62.3637\text{ L.mmHg }mol^{-1}K^{-1}

Also,  

Moles = mass (m) / Molar mass (M)

Density (d)  = Mass (m) / Volume (V)

So, the ideal gas equation can be written as:

PM=dRT

Given that:-

d = 1.80 g/L

Temperature = 32 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (32 + 273.15) K = 305.15 K

Molar mass of nitrogen gas = 28 g/mol

Applying the equation as:

P × 28 g/mol  = 1.80 g/L × 62.3637 L.mmHg/K.mol × 305.15 K

⇒P = 1223.38 mmHg

<u>1223.38 mmHg must be the pressure of the nitrogen gas.</u>

5 0
3 years ago
The most negative electron affinity is most likely associated with which type of atoms?
Llana [10]

Explanation:

elements are based on electrical conductivity

7 0
4 years ago
Read 2 more answers
The addition of hydrochloric acid to a silver nitrate solution precipitates silver chloride according to the reaction:
Nimfa-mama [501]

Answer:

Enthalpy change for the reaction is -67716 J/mol.

Explanation:

Number of moles of AgNO_{3} in 50.0 mL of 0.100 M of AgNO_{3}

= Number of moles of HCl in 50.0 mL of 0.100 M of HCl

= \frac{0.100}{1000}\times 50.0 moles

= 0.00500 moles

According to balanced equation, 1 mol of AgNO_{3} reacts with 1 mol of HCl to form 1 mol of AgCl.

So, 0.00500 moles of AgNO_{3} react with 0.00500 moles of HCl to form 0.00500 moles of AgCl

Total volume of solution = (50.0+50.0) mL = 100.0 mL

So, mass of solution = (100.0\times 1.00) g = 100 g

Enthalpy change for the reaction = -(heat released during reaction)/(number of moles of AgCl formed)

= \frac{-m_{solution}\times C_{solution}\times \Delta T_{solution}}{0.00500mol}

= \frac{-100g\times 4.18\frac{J}{g.^{0}\textrm{C}}\times [24.21-23.40]^{0}\textrm{C}}{0.00500mol}

= -67716 J/mol

[m = mass, c = specific heat capacity, \Delta T = change in temperature and negative sign is included as it is an exothermic reaction]

4 0
3 years ago
What is the charge associated with each side of the HBr molecule?
nevsk [136]

Answer:

Partial positive on hydrogen and partial negative charge on oxygen atom

Explanation:

5 0
3 years ago
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