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ohaa [14]
3 years ago
5

Using your knowledge of properties of matter, choose the correct description: Question 3 options: physical prop. are luster, col

or & reactivity Chemical prop. are conductivity, flammability & reactivity Chemical prop. are tarnishing, flammability & reactivity physcial prop are malleability, conductivity & rusting
Chemistry
1 answer:
MrRissso [65]3 years ago
3 0

Based on the knowledge of properties of matter, Chemical prop. are tarnishing, flammability & reactivity.

In Chemistry, matter can be defined as anything that has mass and occupies space. Any physical object that is found on earth is typically composed of matter.

Basically, matter are known to be made up of atoms and as a result has the property of existing in states;

  • Solid.
  • Liquid.
  • Gas.

A chemical change gives rise to the chemical properties of matter and it is typically characterized by a change in chemical composition and the formation of a new substance.

In this context, we can deduce that the chemical properties of matter are;

  • Tarnishing.
  • Flammability.
  • Reactivity.

Find more information: brainly.com/question/23529643

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450 nanometers to meters
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Answer: 4.5 x 10e-7

Explanation: 450 x 1e+9 = correct answer

Multiply amount of nanometers by 1e+9 to get the approximate result in meters.

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Determine the freezing point and boiling point of a solution that has 68.4 g of sucrose
Ymorist [56]

Answer:

Freezing T° of solution = - 3.72°C

Boiling T° of solution =  101.02°C

Explanation:

To solve this we apply colligative properties. Firstly, freezing point depression:

ΔT = Kf . m . i

ΔT = Freezing T° of pure solvent - Freezing T° of solution

Kf = Cryoscopic constant, for water is 1.86 °C/m

m = molality (moles of solute in 1kg of solvent)

i = Ions dissolved in solution

Our solute is sucrose, an organic compound so no ions are defined. i = 1.

Let's determine the moles: 68.4 g . 1mol/ 342g = 0.2 moles

molality = 0.2 mol / 0.1kg of water = 2 m

We replace data: ΔT = 1.86°C/m . 2m . 1

Freezing T° of solution = - 3.72°C

Now, we apply elevation of boiling point: ΔT = Kb . m . i

ΔT = Boiling T° of solution - Boiling T° of  pure solvent

Kf = Ebulloscopic constant, for water is 0.512 °C/m

We replace:

Boiling T° of solution - Boiling T° of pure solvent = 0.512 °C/m . 2 . 1

Boiling T° of solution = 0.512 °C/m . 2 . 1 + 100°C → 101.02°C

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3 years ago
What is the pH of a Koh solution that has [H+]=1.87×10^-13M
xenn [34]
PH=-log[H⁺]
pH=-log(1.87×10⁻¹³)
pH=12.72

I hope this helps.  Let me know if anything is unclear.
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