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Ierofanga [76]
3 years ago
15

Ethanol is a common laboratory solvent and has a density of 0.789 g/ml. What is the mass, in grams, of 125 ml of ethanol?

Chemistry
1 answer:
olga55 [171]3 years ago
3 0

The density of a material is its unique property by which a unknown material can be identified and also the impurity (if present) in a material can be concluded. Mathematically the density can be expressed as-\frac{mass}{volume}. Thus from the mathematical expression we can say that the density is the mass per unit volume of a material. Here the density of ethanol is given 0.789 g/mL. Thus the weight of the 1 mL ethanol is 0.789g. Thus the weight of the 125 mL of ethanol will be (125×0.789) = 98.625 g.

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Organic: sugar
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if a solution is saturated, how does the rate of dissolution of a solute compare with it's rate of crystallization?
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The same

Explanation:

In a saturated solution, the rate of dissolution is equal and the same to the rate of crystallization.

  • A saturated solution of as substance (solute) at a particular temperature is one which contains the maximum amount of the substance that can dissolve at that temperature  in the presence of the crystals of the substance.
  • It is an equilibrium system in which a solid substance is in equilibrium with its own ions in solution.
  • Therefore the rate of dissolution will the same with that of crystallization.
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The movement of one or more from one reactant to another is called
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Answer: Redox Reaction

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2C2H6 + 8O2 = 4CO2 + 6H2O
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4 0
2 years ago
Three 1.0-l flasks, maintained at 308 k, are connected to each other with stopcocks. initially the stopcocks are closed. one of
Licemer1 [7]

Answer:

0.6103 atm.

Explanation:

  • We need to calculate the vapor pressure of each component after the stopcocks are opened.
  • Volume after the stopcocks are opened = 3.0 L.

<u><em>1) For N₂:</em></u>

P₁V₁ = P₂V₂

P₁ = 1.5 atm & V₁ = 1.0 L & V₂ = 3.0 L.

P₂ of N₂ = P₁V₁ / V₂ = (1.5 atm) (1.0 L) / (3.0 L) = 0.5 atm.

<u><em>2) For H₂O:</em></u>

Pressure of water at 308 K is 42.0 mmHg.

we need to convert from mmHg to atm: <em>(1.0 atm = 760.0 mmHg)</em>.

P of H₂O = (1.0 atm x 42.0 mmHg) / (760.0 mmHg) = 0.0553 atm.

We must check if more 2.2 g of water is evaporated,

n = PV/RT = (0.0553 atm) (3.0 L) / (0.082 L.atm/mol.K) (308 K) = 0.00656 mole.

m = n x cmolar mass = (0.00656 mole) (18.0 g/mole) = 0.118 g.

It is lower than the mass of water in the flask (2.2 g).

<em><u>3) For C₂H₅OH:</u></em>

Pressure of C₂H₅OH at 308 K is 102.0 mmHg.

we need to convert from mmHg to atm: (1.0 atm = 760.0 mmHg).

P of C₂H₅OH = (1.0 atm x 102.0 mmHg) / (760.0 mmHg) = 0.13421 atm.

We must check if more 0.3 g of C₂H₅OH is evaporated,

n = PV/RT = (0.13421 atm) (3.0 L) / (0.082 L.atm/mol.K) (308 K) = 0.01594 mole.

m = n x molar mass = (0.01594 mole) (46.07 g/mole) = 0.7344 g.

<em>It is more than the amount in the flask (0.3 g), so the pressure should be less than 0.13421 atm.</em>

We have n = mass / molar mass = (0.30 g) / (46.07 g/mole) = 0.00651 mole.

So, P of C₂H₅OH = nRT / V = (0.00651 mole) (0.082 L.atm/mole.K) (308.0 K) / (3.0 L) = 0.055 atm.

  • <em>So, </em><em>total pressure</em><em> = </em><em>P of N₂ + P of H₂O + P of C₂H₅OH</em><em> = 0.5 atm + 0.0553 atm + 0.055 atm = </em><em>0.6103 atm</em><em>.</em>
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2 years ago
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