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kow [346]
2 years ago
7

Difference between saturated unsaturted and supersaturatted

Chemistry
1 answer:
Yakvenalex [24]2 years ago
3 0

Explanation:

Unsaturated solution: A solution with less solute that completely dissolves without leaving any remaining substance

Saturated solution: A solution with solute that dissolves until it can't dissolve anymore, leaving some of the remaining substance left.

Supersaturated solution: A solution that has more remaining substance left than a saturated solution and tends to crystallize more than a saturated solution.

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What is the hottest object on earth? And why is it the hottest object on earth?
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Answer:

The hottest thing on earth is the man-made quark-gluon plasma that is generated at the LHC at CERN by colliding two lead nuclei together at 7 GeV /c2.

Explanation:

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3 years ago
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What is the vapor pressure of CS2CS2 in mmHgmmHg at 26.5 ∘C∘C? Carbon disulfide, CS2CS2, has PvapPvap = 100 mmHgmmHg at −−5.1 ∘C
Digiron [165]

Answer: 26.5 mm Hg

Explanation:

The vapor pressure is determined by Clausius Clapeyron equation:

ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1= initial pressure at 26.5^oC = ?

P_2 = final pressure at -5.1^oC = 100 mm Hg

= enthalpy of vaporisation = 28.0 kJ/mol =28000 J/mol

R = gas constant = 8.314 J/mole.K

T_1= initial temperature = 26.5^oC=273+26.5=299.5K

T_2 = final temperature =-5.1^oC=273+(-5.1)=267.9K

Now put all the given values in this formula, we get

\log (\frac{P_1}{100})=\frac{28000}{2.303\times 8.314J/mole.K}[\frac{1}{299.5}-\frac{1}{267.9}]

\log  (\frac{P_1}{100})=-0.576

\frac{P_1}{100}=0.265

P_1=26.5mmHg

Thus the vapor pressure of CS_2CS_2 in mmHg at 26.5 ∘C is 26.5

7 0
3 years ago
On Darwin's tree of life, organisms at the base of
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B - Evolve into the closest species on the tree’s trunk
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Electrical energy can be transformed into other types of energy. We often experience this transformation of energy in our everyd
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3 years ago
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What is δg for the formation of solid uranium hexafluoride from uranium and fluorine at 25∘c when the partial pressure of f2 is
Sladkaya [172]

The reaction is

U(s)+3F_{2} (g)-->UF_{6} (s)

the Q of reaction will be

Q = \frac{1}{(pF_{2})^{3}}

Q = \frac{1}{(0.054)^{3}  } = 6350.66

ΔG=ΔG^{0}+RTlnQ

Putting values

ΔG = -2068000 + (8.314)(298)lnQ

ΔG = -2068000 + (8.314)(298)ln(6350.66) = -2046305 J /mol = -2046.3 kJ / mol



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3 years ago
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