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Dmitry [639]
3 years ago
11

How is energy related to the change of state represented by

Chemistry
1 answer:
pychu [463]3 years ago
7 0

Answer:

Energy is related to the change of state represented by the model in this way - C. Atoms lose energy as a gas changes to a solid. As a certain gas is changing its state into that of a solid material, its atoms are going to lose some energy.

Explanation: hope this helped

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How many significant digits are in this number: 0.0030670?
yarga [219]

Answer:

Answer:

Number of Significant Figures: 5

The Significant Figures are 3 0 6 7 0

Explanation:

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3 years ago
If eggs are so fragile, how do they actually protect the growing bird?​
Step2247 [10]

Although birds' eggs appear to be fragile, they are in fact extremely robust. The oval shape applies the same rules of engineering as an arched bridge; the convex surface can withstand considerable pressure without breaking.

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3 years ago
Number of
Karolina [17]
Mass plus atomic and number

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3 years ago
How do you balance this chemical equation?
garik1379 [7]

Answer:

i will help hold up

Explanation:

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8 0
3 years ago
Naphthalene, C10H8, melts at 80.2°C. If the vapour pressure of the liquid is 1.3 kPa at 85.8°C and 5.3 kPa at 119.3°C, use th
sweet-ann [11.9K]

(a) One form of the Clausius-Clapeyron equation is

ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂); where in this case:

  • P₁ = 1.3 kPa
  • P₂ = 5.3 kPa
  • T₁ = 85.8°C = 358.96 K
  • T₂ = 119.3°C = 392.46 K

Solving for ΔHv:

  • ΔHv = R * ln(P₂/P₁) / (1/T₁ - 1/T₂)
  • ΔHv = 8.31 J/molK * ln(5.3/1.3) / (1/358.96 - 1/392.46)
  • ΔHv = 49111.12 J/molK

(b) <em>Normal boiling point means</em> that P = 1 atm = 101.325 kPa. We use the same formula, using the same values for P₁ and T₁, and replacing P₂ with atmosferic pressure, <u>solving for T₂</u>:

  • ln(P₂/P₁) = (ΔHv/R) * (1/T₁ - 1/T₂)
  • 1/T₂ = 1/T₁ - [ ln(P₂/P₁) / (ΔHv/R) ]
  • 1/T₂ = 1/358.96 K - [ ln(101.325/1.3) / (49111.12/8.31) ]
  • 1/T₂ = 2.049 * 10⁻³ K⁻¹
  • T₂ = 488.1 K = 214.94 °C

(c)<em> The enthalpy of vaporization</em> was calculated in part (a), and it does not vary depending on temperature, meaning <u>that at the boiling point the enthalpy of vaporization ΔHv is still 49111.12 J/molK</u>.

3 0
3 years ago
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