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Olenka [21]
2 years ago
14

Identify each described physical separation technique. a liquid is carefully poured out, leaving the solid behind in the contain

er. choose... a solid is caught on paper or a membrane while the liquid passes through. choose... a solid mixture is heated. one component transitions directly between solid and gas. choose... a solvent is added to dissolve only one mixture component and then the liquids are separated.
Chemistry
2 answers:
Salsk061 [2.6K]2 years ago
6 0

The physical separation techniques used are :

  • Decanting
  • Filtration
  • sublimation
  • distillation

<h3>What are physical separation techniques ?</h3>

Physical Separation techniques are procedures applied for the separation of a mixture of two different states pf matter based on their physical properties. usually solid and liquid mixtures. When a mixture of liquid and solid is separated by carefully pouring out the liquid while leaving the solid behind that separation technique is known as decanting.

Learn more about separation techniques : brainly.com/question/10862519

#SPJ1

Vikki [24]2 years ago
5 0

Decanting, filtration, sublimation and distillation are the separation techniques.

<h3>Identify the physical separation technique?</h3>

Decanting is a process in which a liquid is carefully poured out, leaving the solid behind in the container.

A solid is caught on paper or a membrane while the liquid passes through, this process is called Filtration.

Sublimation is a process in which a solid mixture is heated in which one component transitions directly between solid and gas.

A solvent is added to dissolve only one mixture component and then the liquids are separated, this process is known as distillation.

Learn more about filtration here: brainly.com/question/23945157

#SPJ1

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Answer:

9.9652g of water

Explanation:

The establishment of the liquid-vapor equilibrium occurs when the vapour of water is equal to vapour pressurem 26.7 mmHg. Using gas law it is possible to know how many moles exert that pressure, thus:

n = PV / RT

Where P is pressure 26,7 mmHg (0.0351atm), V is volume (1.350L), R is gas constant (0.082 atmL/molK) and T is temperature (27°C + 273,15 = 300.15K)

Replacing:

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n = 1.93x10⁻³ moles of water are in gaseous phase. In grams:

1.93x10⁻³ moles × (18.01g / 1mol) = <u><em>0.0348g of water</em></u>

<u><em /></u>

As the initial mass of water was 10g, the mass of water that remains in liquid phase is:

10g - 0.0348g = <em>9.9652g of water</em>

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I hope it helps!

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