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alina1380 [7]
4 years ago
9

During science class, while studying mixtures, you mix together table salt and sand. Your teacher challenges you to separate the

salt from the sand. What property could you use and how would you use it?
Chemistry
2 answers:
-Dominant- [34]4 years ago
5 0
Solubility,,,,,add water to the mixture,salt will dissolve to form a solution while sand is insoluble,,,,,decant the salt solution as filtrate and remain with sand as residue,,,,,if salt crystals are needed,,,,evaporate the filtrate to dryness.
Ivanshal [37]4 years ago
4 0

Answer:

Use solubility

Explanation:

Salt will dissolve in water while sand will not.

Dissolve the mixture of sand and salt in water, make sure all salt is dissolved. Filtrate the solution a couple of times and use vacuum filtration at the end to obtain a clear solution. At this point you will have wet sand and a salty solution. Take the salty solution and evaporate water to obtain the salt.

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If a small amount of the unknown solute fails to dissolve in the lauric acid, will the molar mass that you calculate for unknown
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To avoid that this happens, ensure that all of the solute dissolves well in solvent by mechanical means (stirring, shaking, etc) or introducing heat. 
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Which of these statements about the Sun is accurate?
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C.

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A reaction is in equilibrium as shown: A + B C + D. Calculate the equilibrium constant in the final concentrations stabilized at
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Answer:

K = 0.167

Explanation:

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For the reactions:

A + B ⇄ C + D

For the definition, K is:

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K = [4.0M] [4.0M] / [9.6M] [10.0M]

<h3>K = 0.167</h3>

8 0
3 years ago
Using the appropriate Ksp values, find the concentration of K+ ions in the solution at equilibrium after 600 mL of 0.45 M aqueou
Alekssandra [29.7K]

Answer:

[K⁺] = 0.107 M

[OH⁻] = 1.13 ×  10⁻⁹ M

Explanation:

600 mL of 0.45 M Cu(NO3)2 gives equal mole of Cu²⁺ and (NO₃)²⁻

⇒ 0.45 × 600 × 10⁻³

= 0.27 moles of Cu²⁺ and (NO₃)²⁻

450 mL of 0.25 M KOH gives equal moles of K⁺ and OH⁻

⇒ 0.25 × 450 × 10⁻³

= 0.1125 moles of K⁺ and OH⁻

Now after mixing 0.1125 moles of OH⁻ precipitates 0.05625 moles of Cu²⁺  (because 1 Cu²⁺  needs 2 OH⁻)

Therefore , moles of remaining Cu²⁺  = 0.27 - 0.05625

=0.21375 moles which is equal to :

⇒ 0.21375/(( 600+450))× 10⁻³

= 0.21375/1050 × 10⁻³

= 0.20357 M

Given that :

(Ksp for Cu(OH)2 is 2.6 ×  10⁻¹⁹)

We know that , Ksp = [Cu²⁺][OH⁻]²

2.6 ×  10⁻¹⁹ = 0.20357 × [OH⁻]²

[OH⁻]² = 2.6 ×  10⁻¹⁹/0.20357

[OH⁻] = 1.13 ×  10⁻⁹ M

[K⁺] = moles of K⁺ /total volume

[K⁺] = 0.1125 / 1050 × 10⁻³

[K⁺] = 0.107 M

6 0
3 years ago
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