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GrogVix [38]
3 years ago
15

Scientists have changed the model of the atom as they have gathered new evidence. One of the atomic models is shown below. A pur

ple center outlined in black with two concentric black circles around the center, the inner circle having 2 small green balls on it and the outer circle having 8 small green balls on it. What experimental evidence led scientists to change from the previous model to this one? A few of the positive particles aimed at a gold foil seemed to bounce back. The colors of light emitted from heated atoms had very specific energies. Experiments with water vapor showed that elements combine in specific proportions. Cathode rays were bent in the same way whenever a magnet was brought near them.
Chemistry
1 answer:
Julli [10]3 years ago
6 0

Answer:

dude that doesnt maek sense

Explanation:

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How much energy is required to vaporize 155 g of butane at its boiling point? the heat of vaporization for butane is 23.1 kj/mol
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The energy required to vaporize 155 g of butane at its boiling point: 61,723 kJ

<h3>Further explanation</h3>

Enthalpy is the amount of system heat at constant pressure.

The enthalpy is symbolized by H, while the change in enthalpy is the difference between the final enthalpy and the initial enthalpy symbolized by ΔH.

\large{\boxed{\boxed{\bold{\Delta H=H_{End}-H_{First}}}}

Delta H reaction (ΔH) is the amount of heat change between the system and its environment

(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)

The standard unit is kilojoules (kJ)

The enthalpy change symbol (ΔH) is usually written behind the reaction equation.

Change in Standard Evaporation Enthalpy (ΔH vap) is a change in enthalpy at the evaporation of 1 mol liquid phase to the gas phase at its boiling point and standard pressure.

Examples of water evaporation:

 H₂O (l) ---> H₂O (g); ΔH vap = + 44kJ

The enthalpy of evaporation is positive because its energy is needed to break the attraction between molecules in a liquid

  • 155 g of butane

relative molecular mass of butane (C₄H₁₀) = 4.12 + 10.1 = 58 gram / mol

tex]\large{\boxed{mole\:=\:\frac{grams}{relative\:molecular\:mass}}}[/tex]

\large mole\:=\:\large \frac{155}{58}

mole = 2,672

Since the heat of vaporization for butane is 23.1 kj / mol, the energy needed to evaporate 2,672 moles of butane is:

23.1 kJ / mol x 2,672 mol = 61,723 kJ

<h3>Learn more</h3>

the heat of vaporization

brainly.com/question/11475740

The latent heat of vaporization

brainly.com/question/10555500

brainly.com/question/4176497

Keywords: the heat of vaporization, butane, mole, gram, exothermic, endothermic

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