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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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For water to change from a liquid to solid, thermal energy needs to be <br><br> .
telo118 [61]

answer: dispersed from the liquid so cold air can take its place

6 0
2 years ago
Most metals allow electricity to flow freely through them. So, most metals are.....
algol [13]

Answer:

Conductors

Explanation:

Metals that are conductors let electric currents flow freely. Insulators have a resistance of a charge to flow through them.

5 0
2 years ago
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The ksp of pbbr2 is 6.60× 10–6. what is the molar solubility of pbbr2 in 0.500 m kbr solution
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The solubility of PbBr₂(s) with the presence of 0.500 M of KBr is 2.64 x 10⁻⁵ M.

4 0
3 years ago
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A sample of gas occupies 9.0 mL at a pressure of 500.0 mm Hg. A new volume of the same sample is at a pressure of 750.0 mm Hg.
Anuta_ua [19.1K]
<h3>Answer:</h3>

                The New pressure (750 mmHg) is greater than the original pressure (500 mmHg) hence, the new volume (6.0 mL) is smaller than the original volume (9.0 mL).

<h3>Solution:</h3>

              According to Boyle's Law, " <em>The Volume of a given mass of gas at constant temperature is inversely proportional to the applied Pressure</em>". Mathematically, the initial and final states of gas are given as,

                                     P₁ V₁  =  P₂ V₂    ----------- (1)

Data Given;

                  P₁  =  500 mmHg

                  V₁  =  9.0 mL

                  P₂  =  750 mmHg

                  V₂  =  ??

Solving equation 1 for V₂,

                   V₂  =  P₁ V₁ / P₂

Putting values,

                   V₂  =  (500 mmHg × 9.0 mL) ÷ 750 mmHg

                   V₂  =  6.0 mL

<h3>Result:</h3>

            The New pressure (750 mmHg) is greater than the original pressure (500 mmHg) hence, the new volume (6.0 mL) is smaller than the original volume (9.0 mL).

4 0
3 years ago
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A mixture of methane and carbon dioxide gases contains methane at a partial pressure of 431 mm Hg and carbon dioxide at a
KatRina [158]

Answer:

XCH₄ = 0.461

XCO₂ = 0.539

Explanation:

Step 1: Given data

  • Partial pressure of methane (pCH₄): 431 mmHg
  • Partial pressure of carbon dioxide (pCO₂): 504 mmHg

Step 2: Calculate the total pressure in the container

We will sum both partial pressures.

P = pCH₄ + pCO₂

P = 431 mmHg + 504 mmHg = 935 mmHg

Step 3: Calculate the mole fraction of each gas

We will use the following expression.

Xi = pi / P

XCH₄ = pCH₄/P = 431 mmHg/935 mmHg = 0.461

XCO₂ = pCO₂/P = 504 mmHg/935 mmHg = 0.539

3 0
2 years ago
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