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Katen [24]
3 years ago
8

How do properties, such as Density, help us to classify and identify matter?

Chemistry
1 answer:
kumpel [21]3 years ago
4 0

Density can be defined in terms of the mass of a substance present in a certain volume.

i.e. Density = Mass/ volume

It helps classify and identify matter due to the following reasons:

1) Density is an intensive property. For a given substance the density will remain the same irrespective of the amount in which it is present.

For example: The density of Aluminum (Al) = 2.7 g/cm3. This value is a constant at a certain temperature and pressure. Therefore, the density will remain the same for 1 kilogram or 1 mg of Al

2) Different elements/compounds have different values of densities which is characteristic of that substance

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Una sustancia tiene una masa de 81632 g. Se logró desplazar aplicando un trabajo de 2800 J. Averigüe la distancia que logró move
Rudik [331]

Answer:

Supongo que su suma 81632 + 2800 = 84432, por lo que la distancia que logró mover el objeto fue 2800.

Explanation:

7 0
3 years ago
What are some common examples of solids, liquids and gas?
Sonja [21]
Solids- wood, sand, brick, rock

Liquids- water, milk, blood, coffee

Gases- air, helium, nitrogen, hydrogen
3 0
3 years ago
A buffer can be prepared by mixing two solutions. Determine if each of the following mixtures will result in a buffer solution o
GenaCL600 [577]

Answer:

The answers are in the explanation

Explanation:

A buffer is the mixture of a weak acid with its conjugate base or vice versa. Thus:

<em>1)</em> Mixing 100.0 mL of 0.1 M HF with 100.0 mL of 0.05 M mol KF. <em>Will </em>result in a buffer because HF is a weak acid and KF is its conjugate base.

<em>2)</em> Mixing 100.0 mL of 0.1 M NH₃ with 100.0 mL of 0.1 M NH₄Br. <em>Will not </em>result in a buffer because NH₃ is a strong base.

<em>3) </em>Mixing 100.0 mL of 0.1 M HCN with 100.0 mL of 0.05 M KOH. <em>Will </em>result in a buffer because HCN is a weak acid and its reaction with KOH will produce CN⁻ that is its conjugate base.

<em>4)</em> Mixing 100.0 mL of 0.1 M HCl with 100.0 mL of 0.1 M KCl <em>Will not </em>result in a buffer because HCl is a strong acid.

<em>5)</em> Mixing 100.0 mL of 0.1 M HCN with 100.0 mL of 0.1 M KOH <em>Will not </em>result in a buffer because each HCN will react with KOH producing CN⁻, that means that you will have just CN⁻ (Conjugate base) without HCN (Weak acid).

I hope it helps!

6 0
3 years ago
Investigations were carried out in a science lab to explore the topic of chemical and physical changes. Investigation A Step 1.
Angelina_Jolie [31]

Answer : B) In step 2, there was a chemical change which was observed in sugar.

Explanation : In A step 2 there was a physical change that was seen. By just boiling the dissolved salt solution salt was obtained. Therefore, it is a physical change. In B step 2 there was a chemical change seen as sugar solution was thickened and turned brown. It was not obtained in its original form; there was a chemical reaction that took place during sugar evaporation. As chemical change is the one where the reaction is irreversible.

Therefore only in B step 2 there was a chemical change that was observed.

4 0
3 years ago
Read 2 more answers
A solution made by dissolving 9.81 g of a nonvolatile, nonelectrolyte in 90.0 g of water boiled at 100.37 °C at 760 mm Hg. What
NISA [10]

Answer:

151 g/mol

Explanation:

In order to solve this problem we need to keep in mind the formula for the <em>boiling point elevation</em>:

  • ΔT = Kb * <em>m</em> * i

Where:

  • ΔT is the temperature difference between the boiling point of the solution and that of pure water. 100.37 °C - 100.00 °C = 0.37 °C.
  • <em>m</em> is the molarity of the solution
  • i is the van't Hoff factor. As the solute is a nonelectrolyte, the factor is 1.

Input the data and <u>calculate </u><em><u>m</u></em>:

  • 0.37 °C = 0.51 °C/m * <em>m</em> * 1
  • <em>m</em> =0.72 m

We now can <u>calculate the number of moles of the substance</u>, using the <em>definition of molarity</em>:

  • molarity = moles of solute / kg of solvent

In this case kg of solvent = 90.0 g / 1000 = 0.090 kg

  • 0.72 m = moles / 0.090 kg
  • moles = 0.065 mol

Finally we <u>calculate the molar mass</u>, using the <em>number of moles and the mass</em>:

  • 9.81 g / 0.065 mol = 151 g/mol
8 0
3 years ago
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