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GaryK [48]
4 years ago
7

How did democtitis define atom?

Chemistry
1 answer:
klemol [59]4 years ago
8 0

Democritus, theorized that atoms were specific to the material which they composed. In addition, Democritus believed that the atoms differed in size and shape, were in constant motion in a void, collided with each other; and during these collisions, could rebound or stick together.

<u>Explanation:</u>

  • One of the main atomic theorists was Democritus, a Greek philosopher who lived in the fifth century BC. Democritus realized that if a stone was partitioned fifty-fifty, the two parts would have indistinguishable properties from the whole.
  • Therefore, he contemplated that if the stone were to be constantly cut into littler and littler pieces at that point; sooner or later, there would be a piece that would be so little as to be inseparable. He called these small pieces of matter as "atomos", the Greek word for inseparable.  
  • Democritus estimated that atoms were explicit to the material which they made. Also, Democritus accepted that the particles varied in size, were an inconsistent shape, crashed into one another; and during these impacts, could bounce back or stay together. Hence, changes in the matter were a consequence of separations or mixes of the atoms as they moved all through the void.
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5. Calculate the mass percent of carbon in C3H8
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3 years ago
Materials expand when heated. Consider a metal rod of length L0 at temperature T0. If the temperature is changed by an amount ΔT
marysya [2.9K]

Answer:

(a) The length at temperature 180°C is 40.070 cm

(b) The length at temperature 90°C is 64.976 inches

(c) L(T, α) = 60·α·T - 9000·α + 60

Explanation:

(a) The given parameters are

The thermal expansion coefficient, α for steel = 1.24 × 10⁻⁵/°C

The initial length of the steel L₀ = 40 cm

The initial temperature, t₀ = 40°C

The length at temperature 180°C = L

Therefore, from the given relation, for change in length, ΔL, we have;

ΔL = α × L₀ × ΔT

The amount the temperature changed ΔT = 180°C - 40°C = 140°C

Therefore, the change in length, ΔL, is found as follows;

ΔL = α × L₀ × ΔT = 1.24 × 10⁻⁵/°C × 40 × 140°C = 0.07 cm

Therefore, L =  L₀ + ΔL = 40 + 0.07 = 40.07 cm

The length at temperature 180°C = 40.07 cm

(b) Given that the length at T = 120°C is 65 in., we have;

The temperature at which the new length is sought = 90°C

The amount the temperature changed ΔT = 90°C - 120°C = -30°C

ΔL = α × L₀ × ΔT = 1.24 × 10⁻⁵/°C × 65 × -30°C = -0.024375 inches

The length, L at 90°C is therefore, L = L₀ + ΔL = 65 - 0.024375 = 64.976 in.

The length at temperature 90°C = 64.976 inches

(c) L = L₀ + ΔL  = L₀ +  α × L₀ × ΔT = L₀ +  α × L₀ × (T - T₀)

Therefore;

L = 60 +  α × 60 × (T - 150°C)

L = 60 + α × 60 × T - 9000 × α

L(T, α) = 60·α·T - 9000·α + 60

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