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Sati [7]
4 years ago
11

Which criteria determine whether a heterogeneous mixture is a colloid or a suspension?

Chemistry
2 answers:
lianna [129]4 years ago
6 0
The correct answer is B. whether the particles do not settle for an extended period of time

Uniformity and transparency are not affected by these, which also applies to ratio fixedness.
vivado [14]4 years ago
5 0
Correct Answer: Option B: <span>whether the particles do not settle for an extended period of time

Reason:
Particle size in colloidal solution ranges from 1-100 nm, while that is suspension is more than 100 nm. Due to large particle size, particles in suspension settle down, upon standing for long. Thus, the best </span>criteria determine whether a heterogeneous mixture is a colloid or a suspension is to check <span>whether the particles do not settle for an extended period of time</span>
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Margarita [4]

the are 18.01528 grams

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3 years ago
• Average molarity of acetic acid
kipiarov [429]

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6 0
4 years ago
if one mole of calcium carbonate (the limiting reactant) is used, how much calcium chloride should the reaction produce? look at
salantis [7]

Answer:

The answer to your question is <u>111 g of CaCl₂</u>

Explanation:

Reaction

                 2HCl  +  CaCO₃   ⇒    CaCl₂  +  CO₂  +  H₂O

Process

1.- Calculate the molecular mass of Calcium carbonate and calcium chloride

CaCO₃ = (1 x 40) + (1 x 12) + ((16 x 3) = 100 g

CaCl₂ = (1 x 40) + (35.5 x 2) = 111 g

2.- Calculate the amount of calcium chloride produced using proportions.

The proportion CaCO₃ to CaCl₂ is   1 : 1.

                      100 g of CaCO₃  ------------- 111 g of CaCl₂

Then 111g of CaCl₂ will be produced.

5 0
4 years ago
How do you set up a proper experiment
vova2212 [387]
First off, With a hypothesis
5 0
4 years ago
If 00810 mol neon gas at the particular temperature and pressure occupies a volume of 214 mL ehat volume would 0.00684 mol neon
Grace [21]

Answer:

V_2=181mL

Explanation:

Hello,

In this case, Avogadro's relationship allows us to relate the moles and the volume of a gas at an initial condition and an a final one as shown below:

\frac{n_1}{V_1}=\frac{n_2}{V_2}

Thus, the volume that 0.00684 mol of neon the same pressure and temperature is computed below as required:

V_2=\frac{V_1*n_2}{n_1}=\frac{214mL*0.00684mol}{0.00810mol}  \\\\V_2=181mL

Best regards.

6 0
3 years ago
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