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expeople1 [14]
3 years ago
14

Which of the following techniques can be used to separate a heterogeneous mixture into its component parts?

Chemistry
2 answers:
fredd [130]3 years ago
8 0

Answer:

                  Separation by density

Explanation:

                    Mixtures are made up of two or more pure substances which tends to keep their individual identities. These components can be separated from each other by different physical techniques.

                    Mixtures are further classified as;

  (i)  Homogenous Mixture:

                                            In this type of the mixtures the components are uniformly mixed and their properties as well as composition as uniform throughout. Such mixtures are also called as solutions.

The physical methods used to separate these components from each other are distillation (taking heat and pressure into account), Solvent extraction, Magnetic separation, Chromatography e.t.c.

  (ii)  Homogenous Mixture:

                                            In this type of the mixtures the components are not uniformly mixed and their physical properties and composition are also not uniform.

The physical methods used to separate these components from each other are Filtration, Magnetic Separation, Centrifugation, Flotation e.t.c.

So, in given options the density can play role by settling the massive components of heterogenous mixture to sit at the bottom and separated

Mkey [24]3 years ago
8 0

Answer:

B)

Explanation:

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Answer:

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Explanation:

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8 0
3 years ago
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A gravel can be broken down into different kinds of substances by physical processes. No chemical reactions are needed to separa
butalik [34]

answer:

heterogenous mixture

Explanation:

6 0
3 years ago
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You notice that the water in your friend's swimming pool is cloudy and that the pool walls are discolored at the water line. A q
Flauer [41]

Answer:

6,78 mL of 12,0 wt% H₂SO₄

Explanation:

The equilibrium in water is:

H₂O (l) ⇄ H⁺ (aq) + OH⁻ (aq)

The initial concentration of [H⁺] is 10⁻⁸ M and final desired concentration is [H⁺] = 10^{-7,20}, thus,

Thus, you need to add:

[H⁺] = 10^{-7,2} -10^{-8,0} = 5,31x10⁻⁸ M

The total volume of the pool is:

9,00 m × 15,0 m ×2,50 m = 337,5 m³ ≡ 337500 L

Thus, moles of H⁺ you need to add are:

5,31x10⁻⁸ M × 337500 L = 1,792x10⁻² moles of H⁺

These moles comes from

H₂SO₄ → 2H⁺ +SO₄²⁻

Thus:

1,792x10⁻² moles of H⁺ × \frac{1H_{2}SO_4 mol}{2H^{+} mol} = 8,96x10⁻³ moles of H₂SO₄

These moles comes from:

8,96x10⁻³ moles of H₂SO₄ × \frac{98,1 g}{1 mol} × \frac{100 gSolution}{12 gH_{2}SO_4 } × \frac{1 mL}{1,080 g}  =

6,78 mL of 12,0wt% H₂SO₄

I hope it helps!

8 0
3 years ago
Label the particles, charges, & mass for Elements: 6, 8 & 10?
pashok25 [27]

Answer:

charges, & mass

Explanation:

3 0
2 years ago
If 450.0 mL of a 0.500 M solution is mixed with 200.0 mL of water, what is the molarity of the new
Oksana_A [137]

Answer:

Answer: A) .346 M

Explanation:

Given:

- 450 mL

- .5 M soln

-200 mL water

1) Convert mL to L

450 mL = .45 L

200 mL = .2 L

2) Find mols in solution

.5 M = x/.45 L

x = .225 mol

3) Find total volume of solution

.45 L + .2 L =.65 L

4) Find new molarity

molarity (M) = mols solute/ L solution

y = .225 mol (from step 2)/ .65 L (from step 3)

y = .346 M

Answer: A) .346 M

7 0
3 years ago
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