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skelet666 [1.2K]
3 years ago
5

Compare and contrast the movements of particles that make up solid, liquid and gaseous samples of matter.

Chemistry
2 answers:
slavikrds [6]3 years ago
7 0

Explanation:

  • In a solid, molecules are held together by strong intermolecular forces of attraction. As a result, they are unable to move from their initial place but they can vibrate at their mean position.  

Hence, in solid substances the particles have low kinetic energy.

  • Whereas in liquids, the molecules are held by less strong intermolecular forces of attraction as compared to solids. Due to which they are able to slide past each other. Hence, particles of a liquid have medium kinetic energy.
  • In gases, the molecules are held by weak Vander waal forces. Hence, they have high kinetic energy due to which they move rapidly from one place to another leading to more number of collisions.  

Hence, particles of a gas are able to expand more rapidly as compared to particles of a liquid or solid.

Marta_Voda [28]3 years ago
4 0
Matter made up of gaseous would exert highest movement, due to high randomness; and solid will be least in that due to low randomness
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4 years ago
2. Write the chemical equations for the neutralization reactions that occurred when HCL and NaOH were added to the buffer soluti
lutik1710 [3]

Answer:

HCI(aq)+CH3COONa(s) ----> CH3COOH(aq)+NaCl(s)

NaOH(aq)+CH3COOH(aq) ----> CH3COONa(s)+H2O(l)

Explanation:

A buffer is a solution that resists changes in acidity or alkalinity. A buffer is able to neutralize a little amount of acid or base thereby maintaining the pH of the system at a steady value.

A buffer may be an aqueous solution of a weak acid and its conjugate base or a weak base and its conjugate acid.

The equations for the neutralizations that occurred upon addition of HCl or NaOH are;

HCI(aq)+CH3COONa(s) ----> CH3COOH(aq)+NaCl(s)

NaOH(aq)+CH3COOH(aq) ----> CH3COONa(s)+H2O(l)

5 0
3 years ago
calculate the number of oxygen atoms in a 90.0g sample of vanadium(v) oxide v2o5. be sure your answer has a unit symbol if neces
Alborosie

90.0 g of vanadium (V) oxygen sample has 1.50 x 10 24 oxygen atoms.

To determine the number of oxygen atoms in a vanadium oxide sample, the following steps must be made.

Step 1: Convert the mass of vanadium(V) oxide to mol of vanadium(V) oxide

mol vanadium(V) oxide = mass / molar mass

mol vanadium (V) oxide = 90.0 g / 181.88 g/mol

mol vanadium (V) oxide = 0.49483 mol

Step 2: Convert mol of vanadium (V) oxide to moles of oxygen using mole ratio

mol oxygen = mol vanadium (V) oxide * mole ratio O / mol V2O5

mol oxygen = 0.49483 mol * 5 mol O / mol V2O5

mol oxygen = 2.47416 mol O

Step 3: Convert to oxygen atoms using Avogadro's number

atoms oxygen = mol oxygen * 6.02214 x 10 23 atoms O /mol O

atoms oxygen = 2.47416 mol O * 6.02214 x 10 23 atoms O/mol O

atoms oxygen = 1.48997 x 10 24 atoms or 1.50 x 10 24 atoms O

To learn more about Avogadro's number, please refer to the link brainly.com/question/859564.

#SPJ4

4 0
2 years ago
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