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SCORPION-xisa [38]
3 years ago
10

Ethanol and water appear the saame to the naked eye at 25C. Which properties can help a chemist distinguish between? denisty, ma

lting point, specific heat capacity, mass.
Chemistry
1 answer:
Oxana [17]3 years ago
7 0

Explanation:

As it is known that there are two types of properties. These are extensive and intensive.

Extensive properties : Properties that depend on the size or amount of system. For example, mass, volume etc.

Intensive properties : Properties that do not depend on the size or amount of system. For example, density, melting point, specific heat capacity etc.

On the basis of these properties water and ethanol are distinguished as follows.

  • Density of water is 997 kg/m^{3} whereas density of ethanol is 789 kg/m^{3}. Both these liquids can be separated by intensive properties.
  • Melting point of water is zero degree celsius whereas melting point of ethanol is -114.1 degree celsius.
  • Specific heat capacity of water is 4.184 J/g ^{o}C whereas specific heat capacity of ethanol is 2.46 J/g ^{o}C.
  • Mass of the given liquids cannot be differentiated because they will keep on changing depending on the quantity required. As mass is an extensive property, therefore, it is difficult to differentiate between the two liquids.

Thus, we can conclude that properties like density, melting point, specific heat capacity can help a chemist distinguish between ethanol and water.

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When ΔG° is the change in Gibbs free energy

So according to ΔG° formula:

ΔG° =  - R*T*(㏑K)

here when K = [NH3]^2/[N2][H2]^3 = Kc 

and Kc = 9 

and when T is the temperature in Kelvin = 350 + 273 = 623 K

and R is the universal gas constant = 8.314 1/mol.K

So by substitution in ΔG° formula:

∴ ΔG° = - 8.314 1/ mol.K * 623 K *㏑(9)

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Answer:

219.95 °C

Explanation:

Given data:

Volume of gas = 9.71 L

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Initial temperature = 10.1 °C (10.1 +273 = 283.1 K)

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Final pressure = 364 torr (364/760 =0.479 atm)

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According to Gay-Lussac Law,

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Now we will put the values in formula:

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T₂ = 0.479 atm × 283.1 K/ 0.275 atm

T₂ = 135.6 atm. K /0.275 atm

T₂ = 493.1 K

Kelvin to °C:

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