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3 years ago

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?

1 answer:

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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Give an example of organic and inorganic compound

Olenka [21]

Organic: sugar
inorganic: salt

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3 years ago

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

LenKa [72]

Answer:

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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3 years ago

The movement of one or more from one reactant to another is called

mrs_skeptik [129]

Answer: Redox Reaction

Explanation:

Redox reaction is the key chemical events in an oxidation-reduction also called Redox. It is the net movement of electrons from one reactants to another

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3 years ago

2C2H6 + 8O2 = 4CO2 + 6H2O

Neporo4naja [7]

Answer:

4 moles of carbon

6 moles of water

Explanation:

I think as there no data given u have to is the numbers infront of the equation e.g 4CO2 so 4.

hope this helps :)

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.

1) For N₂:

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.

2) For H₂O:

Pressure of water at 308 K is 42.0 mmHg.

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

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).

3) For C₂H₅OH:

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.

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

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.

So, total pressure = P of N₂ + P of H₂O + P of C₂H₅OH = 0.5 atm + 0.0553 atm + 0.055 atm = 0.6103 atm.

3 0

2 years ago