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liraira [26]
3 years ago
14

A mixture that separates into different layers when you stop stirring it is Select one: a. a solution. b. a suspension. c. a col

loid. d. an emulsion.
Chemistry
1 answer:
leva [86]3 years ago
8 0

Answer: b suspension

a suspension is a heterogeneous mixture that contains solid particles sufficiently large for sedimentation . The particles may be

visible to the naked eye, usually must be larger than one micrometer , and will eventually settle, although the mixture is only classified as a suspension when and while the particles have not settled out. A suspension is a heterogeneous mixture in which the solute particles do not dissolve , but get suspended throughout the bulk of the solvent , left floating around freely in the medium. [1] The internal phase (solid) is dispersed throughout the external phase (fluid) through mechanical agitation , with the use of certain excipients or suspending agents.

An example of a suspension would be sand in water. The suspended particles are visible under a

microscope and will settle over time if left undisturbed. This distinguishes a suspension from a colloid , in which the suspended particles are smaller and do not settle.

Colloids and suspensions are different from

solution , in which the dissolved substance (solute) does not exist as a solid, and solvent and solute are homogeneously mixed.

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675 g of carbon tetrabromide is equivalent to how many
VARVARA [1.3K]
<h3>Answer:</h3>

2.04 mol CBr₄

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Organic</u>

  • Writing Organic Compounds
  • Writing Covalent Compounds
  • Organic Prefixes

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

675 g CBr₄

<u>Step 2: Identify Conversions</u>

Molar Mass of C - 12.01 g/mol

Molar Mass of Br - 79.90 g/mol

Molar Mass of CBr₄ - 12.01 + 4(79.90) = 331.61 g/mol

<u>Step 3: Convert</u>

<u />\displaystyle 675 \ g \ CBr_4(\frac{1 \ mol \ CBr_4}{331.61 \ g \ CBr_4}) = 2.03552 \ mol \ CBr_4<u />

<u />

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

2.03552 mol CBr₄ ≈ 2.04 mol CBr₄

7 0
3 years ago
Read 2 more answers
*multiple choice*
OleMash [197]
I think the answer is A but I could be wrong
4 0
3 years ago
Percent Composition and Determining Empirical And Molecular Formulas
raketka [301]

Answer:

% comp. H = 2.8%, % comp. Cl = 97.2%

Explanation:

HCl mass = mass of H + mass of Cl

HCL mass = 1.00794 + 35.453 = 36.46094

% comp. of H = 1.007694 / 36.46094 x 100 = around 2.8% (2.76376308455)

% comp. of Cl = 35.453 / 36.46094 x 100 = around 97.2% (97.2355622208)

3 0
2 years ago
PLEASE HELP 20 POINTS What is a key event found in CO2 and glucose?
irina [24]

the answer is carbon

5 0
3 years ago
13. A mixture of MgCO3 and MgCO3.3H2O has a mass of 3.883 g. After heating to drive off all the water the mass is 2.927 g. What
rjkz [21]

Answer:

63.05% of MgCO3.3H2O by mass

Explanation:

<em>of MgCO3.3H2O in the mixture?</em>

The difference in masses after heating the mixture = Mass of water. With the mass of water we can find its moles and the moles and mass of MgCO3.3H2O to find the mass percent as follows:

<em>Mass water:</em>

3.883g - 2.927g = 0.956g water

<em>Moles water -18.01g/mol-</em>

0.956g water * (1mol/18.01g) = 0.05308 moles H2O.

<em>Moles MgCO3.3H2O:</em>

0.05308 moles H2O * (1mol MgCO3.3H2O / 3mol H2O) =

0.01769 moles MgCO3.3H2O

<em>Mass MgCO3.3H2O -Molar mass: 138.3597g/mol-</em>

0.01769 moles MgCO3.3H2O * (138.3597g/mol) = 2.448g MgCO3.3H2O

<em>Mass percent:</em>

2.448g MgCO3.3H2O / 3.883g Mixture * 100 =

<h3>63.05% of MgCO3.3H2O by mass</h3>
6 0
3 years ago
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