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Nat2105 [25]
2 years ago
7

What two categories are used in classifying particulate matter?

Chemistry
1 answer:
Allushta [10]2 years ago
3 0
<span>The two categories for classifying particulate matter are through analysis of the intensive and extensive properties. Intensive properties are independent properties that can be measured independent of the amount of matter while extensive properties are measured dependent on the amount.</span>
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100 points to the person who can answer these 4 questions
Vinil7 [7]

Answer:

....,................................

Explanation:

1= A

2=D

3=C

4=C

8 0
2 years ago
A sample of sugar (C12H22O11) contains
Oduvanchick [21]

Answer: 0.25 mol

Explanation:

Use the formula n=N/NA

n= number of mols

N =  number of particles

Nᵃ = Avogadros constant = 6.02 x

So, n=

The 10 to the power of 23 cancels out and you are left with 1.505/6.02, which is approximately 1/4. This is the same as 0.25 mol.

Hope this helped :)

5 0
3 years ago
(a) ions in a certain volume of .20M NaCl (aq) are represented in the box above on the left. in the box above on the right, draw
mamaluj [8]

Answer:

  • The first picture attached is the diagram that accompanies the question.

  • The<u> second picture attached</u> is the diagram with the answer.

Explanation:

In the box on the left there are 8  Cl⁻ ions and 8 Na⁺ ions.

The dissociaton equation for NaCl(aq) is:

  • NaCl (aq) → Na⁺ (aq) + Cl⁻(aq)

The dissociation equation for CaCl₂ (aq)  is:

  • CaCl₂ (aq) → Ca²⁺ (aq) + 2Cl⁻(aq)

A 0.10MCaCl₂ (aq) solution will have half the number of CaCl₂ units as the number of NaCl units in a 0.20M NaCl (aq) solution.

Thus, while the 0.20M NaCl (aq) solution yields 8 ions of Na⁺ and 8 ions of Cl⁻, the 0.10MCaCl₂ (aq) solution will yield 4 ions of Ca²⁺ (half because the concentration if half)  and 8 ions of Cl⁻ (first take half and then multiply by 2 because the dissociation reaction).

Thus, your drawing must show 4 dots representing Ca²⁺ ions and 8 dots representing Cl⁻ ions in the box on the right.

4 0
2 years ago
A standard 10.00 g mass is weighed on an analytical balance 100 times. The average and standard deviation obtained gives 10.12 ±
Mekhanik [1.2K]

Answer:

There was an improvement in accuracy. There was no change in precision.

Explanation:

<em>The average mass after recalibration is closer to the mass of the standard, </em>so the recalibration improved the accuracy<em> </em>(the measurement is closer to an accepted 'true' value).

The standard deviation did not change, so the precision (or how disperse the measurements are) was not affected.

5 0
3 years ago
PLEASE HELP!!!! WILL GIVE BRAINLIEST!!!!
Marina CMI [18]

Answer:

C

Explanation:

It afffects changes in pressure and temperature not melting and boiling points

5 0
2 years ago
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