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creativ13 [48]
2 years ago
8

If one benzene molecule is placed in water, the total entropy of the water + benzene system:___________

Chemistry
1 answer:
valkas [14]2 years ago
3 0

Answer:

d) increases

Explanation:

Benzene is an aromatic hydrocarbon which is obtained from the destructive distillation of coal. It is a colourless volatile liquid with a sweet smell. It boils ar 80° C (353 K) and freezes at 5°C (2278 K). It is insoluble in water but mixes in all proportions with ethanol, ethoxyethane and methylbenzene. The reason which benzene is insoluble in water is that benzene is a non-polar compound and water is polar, meanwhile only "like dissolves like". So, when the benzene molecule is placed in water. There will be distortion and disturbance between the benzene molecule and the water. Thus, the particle of each molecule will be distant from each other. This state results to change in the entropy of the system as the entropy of the system increases.

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Calculate the standard potential, e∘, for this reaction from its equilibrium constant at 298 k. x(s)+y4+(aq)↽−−⇀x4+(aq)+y(s)k=4.
laila [671]
First, we need to get the number of moles:

from the reaction equation when Y4+ takes 4 electrons and became Y, X loses 4 electrons and became X4+  

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㏑4.98 x 10^-5 = 4*96500*E / 8.314*298

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A 200.0mL closed flask contains 2.000mol of carbon monoxide gas and 2.000mol of oxygen gas at the temperature of 300.0K. How man
max2010maxim [7]

Answer:

There will react 0.400 moles of oxygen.

Explanation:

<u>Step 1:</u> Data given

Volume of the closed flask = 200.00 mL = 0.2 L

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Number of moles of O2 = 2.000 mol

Temperature = 300.0 K

Pressure decreases with 10%

<u>Step 2</u>: The balanced equation

2CO(g)+O2(g)⟶2CO2(g)

<u>Step 3</u>: Calculate the initial pressure of the flask before the reaction

P = nRT/V

⇒ with n = the number of moles (2.000 moles CO + 2.000 moles O2 = 4.000 moles)

⇒ R is gas constant (0.08206 atm*L/mol*K)

⇒T = the  temperature = 300.0K

⇒ V = the volume = 200.0 mL = 0.2 L

P = (4 * 0.08206*300)/0.2

P = 492.36 atm

<u>Step 4:</u> When the pressure is 10 % decreased:

The final pressure = 492.36 - 49.236 = 443.124 atm

<u>Step 5:</u> Calculate the number of moles

n = PV/RT

⇒ with n = the number of moles

⇒ with P = the pressure = 443.124 atm

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⇒ R is gas constant (0.08206 atm*L/mol*K)

⇒T = the  temperature = 300.0K

n =(443.124*0.2)/(0.08206*300)

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<u>Step 6:</u> Calculate number of moles

For the reaction :2CO(g) + O₂(g) ⟶ 2CO₂(g)

For each mole of O2 we have 2 moles of CO, to produce 2 moles of CO2

Moles CO = (2 -2X) moles

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Moles CO2 = 2X

The total number of moles (4 -X)= 3.6 moles

Where X are moles that react

X = 0.400 moles

There will react 0.400 moles of oxygen.

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sattari [20]
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PV = k
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