Physical and chemical means cannot separate which into smaller parts?

Chemistry

2 answers:

Hunter-Best [27]3 years ago

6 0

Your answer would be A-mixtures

o-na [289]3 years ago

4 0

A:mixtures i hope that helps (dont mind this i have to have more then 20 words so i can send you this answer)

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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

Describe the formation of an aqueous KI solution when KI dissolves in water

Sidana [21]

Potassium Iodide or KI is an ionic compound and dissolves to water since the two are polar substances. Like dissolves like. As KI is dissolved into water, the KI molecules dissociates into ions, the K+ and I- ions. Water molecules can stabilize these ions. The K+ ions attracts the negative side of H2O molecules which is the oxygen side while the I- ions attracts the positive side of H2O molecules or the hydrogen side.

8 0

3 years ago

You have 363 mL of a 1.25M potassium chloride solution, but you need to make a 0.50M potassium chloride solution. How many milli

maxonik [38]

Answer:- 544.5 mL of water need to be added.

Solution:- It is a dilution problem. The equation used for solving this type of problems is:

$M_1V_1=M_2V_2$

where, $M_1$ is initial molarity and $M_2$ is the molarity after dilution. Similarly, $V_1$ is the volume before dilution and $V_2$ is the volume after dilution.