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quester [9]
3 years ago
7

Explain why the solutions of SbCl3(aq) in our lab also contain HCl. Why would adding a few drops of this solution to 300 mL of w

ater result in a foggy or cloudy appearance?
Chemistry
1 answer:
IRISSAK [1]3 years ago
7 0

Answer:

Here's what I get  

Explanation:

SbCl₃ reacts with water to form slightly soluble antimony oxychloride.

SbCl₃(aq) +H₂O(ℓ) ⇌ SbOCl(s) + 2HCl(aq)

Your observation is an example of Le Châtelier's Principle in action,

The SbCl₃(aq) in your lab has enough HCl added to push the position of equilibrium to the left and keep the SbOCl in solution.

If a few drops of the SbCl₃(aq) were added to 300 mL of water, the solution would turn cloudy. The HCl would be so dilute that the position of equilibrium would lie to the right, and a cloudy precipitate of antimony oxychloride would form.

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Explain why the freezing of water into ice causes the cracks to widen
horrorfan [7]

Answer:

The freezing of water into ice causes the cracks to widen because it occupy more space.

Explanation:

When the water change its state of liquid to solid state due to decrease in temperature, it contracts means occupy less space but when the temperature reaches to 4 degree Celsius, it starts to expand. When the temperature goes below 0 degree Celsius, it will cause more expansion and occupy more space that causes the cracks to widen.

4 0
3 years ago
An aqueous solution containing 17.5 g of an unknown molecular (nonelectrolyte) compound in 100.0 g of water has a freezing point
Sonbull [250]

Answer:

molar mass = 180.833 g/mol

Explanation:

  • mass sln = mass solute + mass solvent

∴  solute: unknown molecular (nonelectrolyte)

∴ solvent: water

∴ mass solute = 17.5 g

∴ mass solvent =  100.0 g = 0.1 Kg

⇒ mass sln = 117.5 g

freezing point:

  • ΔTc = - Kc×m

∴ ΔTc = -1.8 °C

∴ Kc H2O = 1.86 °C.Kg/mol

∴ m: molality (mol solute/Kg solvent)

⇒ m = ( - 1.8 °C)/( - 1.86 °C.Kg/mol)

⇒ m = 0.9677 mol solute/Kg solvent

  • molar mass (Mw) [=] g/mol

∴ mol solute = ( m )×(Kg solvent)

⇒ mol solute = ( 0.9677 mol/Kg) × ( 0.100 Kg H2O )

⇒ mol solute = 0.09677 mol

⇒ Mw solute = ( 17.5 g ) / ( 0.09677 mol )

⇒ Mw solute = 180.833 g/mol

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