THE KINETIC MOLECULAR THEORY STATES THAT ALL PARTICLES OF AN IDEAL GAS ARE IN CONSTANT MOTION AND EXHIBITS PERFECT ELASTIC COLLISIONS.
Explanation:
An ideal gas is an imaginary gas whose behavior perfectly fits all the assumptions of the kinetic-molecular theory. In reality, gases are not ideal, but are very close to being so under most everyday conditions.
The kinetic-molecular theory as it applies to gases has five basic assumptions.
- Gases consist of very large numbers of tiny spherical particles that are far apart from one another compared to their size.
- Gas particles are in constant rapid motion in random directions.
- Collisions between gas particles and between particles and the container walls are elastic collisions.
- The average kinetic energy of gas particles is dependent upon the temperature of the gas.
- There are no forces of attraction or repulsion between gas particles.
<h2><em>Solid. Because molecules densely packed it has a high resistance to flow thereby lower kinetic energy.
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Answer:
7.1
Explanation:
equation to calculate pH is
Answer:
A = 2A + 3B → 5C
Explanation:
The two molecule of A and three molecules of B will react to form the five molecules of C.
2A + 3B → 5C
Other options are incorrect because,
B = A₂ + B₃ → C₅
in this reaction one molecule of A₂ and one molecule of B₃ combine to form one molecule of C₅.
C = 2A + 5B → 3C
in this reaction two molecules of A and five molecules of B combine to form three molecule of C.
D = A₂ + B₃ → C₃
in this reaction one molecule of A₂ and one molecule of B₃ combine to from one molecule of C₃.
The balanced chemical equation for the above reaction is as follows;
2LiOH + H₂SO₄ ---> Li₂SO₄ + 2H₂O
stoichiometry of base to acid is 2:1
Number of OH⁻ moles reacted = number of H⁺ moles reacted at neutralisation
Number of LiOH moles reacted = 0.400 M / 1000 mL/L x 20.0 mL = 0.008 mol
number of H₂SO₄ moles reacted - 0.008 mol /2 = 0.004 mol
Number of H₂SO₄ moles in 1 L - 0.500 M
This means that 0.500 mol in 1 L solution
Therefore 0.004 mol in - 1/0.500 x 0.004 = 0.008 L
therefore volume of acid required = 8 mL
