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Gwar [14]
3 years ago
6

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

?

Chemistry
2 answers:
klemol [59]3 years ago
6 0

\boxed{{\text{61}}{\text{.65 kJ}}} of energy will be required to vaporize 155 g of butane.

Further explanation:

Heat of vaporization is the amount of energy that is needed to convert a substance from its liquid state to a gaseous state. It is represented by \Delta {H_{{\text{vap}}}}.

The relation between the amount of energy and heat of vaporization is as follows:

{\text{q}} = {\text{n}}\Delta {H_{{\text{vap}}}}  …… (1)                                                                                        

Here,

q is the amount of energy.

n is the number of moles of the substance.

\Delta {H_{{\text{vap}}}} is the heat of vaporization.

The moles of butane can be calculated by the following expression:

{\text{Moles of butane}} = \dfrac{{{\text{Given mass of butane}}}}{{{\text{Molar mass of butane}}}}                                                  ...... (2)

The given mass of butane is 155 g.

The molar mass of butane is 58.12 g/mol.

Substitute these values in equation (2).

\begin{aligned}{\text{n}}&= \left( {155{\text{ g}}} \right)\left( {\frac{{1{\text{ mol}}}}{{58.12{\text{ g}}}}} \right)\\&= 2.669{\text{ mol}}\\\end{aligned}  

The value of n is 2.669 mol.

The value of \Delta {H_{{\text{vap}}}} is 23.1 kJ/mol.

Substitute these values in equation (1) to calculate the energy required for butane.

 \begin{aligned}{\text{q}}&= \left( {{\text{2}}{\text{.669 mol}}} \right)\left( {\frac{{23.1{\text{ kJ}}}}{{1{\text{ mol}}}}} \right)\\&= 61.65{\text{kJ}}\\\end{aligned}

Learn more:

  1. What is the enthalpy of the given reaction? brainly.com/question/10412973
  2. Find the enthalpy of decomposition of 1 mole of MgO: brainly.com/question/2416245

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Thermodynamics

Keywords: enthalpy of vaporization, q, n, butane, 58.12 g/mol, 23.1 kJ/mol, 61.65 kJ, 155 g, 2.669 g, amount of energy, moles, molar mass, energy, substance, liquid state, gaseous state.

netineya [11]3 years ago
4 0

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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KCl and KBr are both ionic solids. A mixture of KCl and KBr has a mass of 3.595 g. When this mixture is heated in the presence o
monitta

Answer:

The percentage (by mass) of KBr in the original mixture was 33.1%.

Explanation:

The mixture of KCl and KBr has a mass of 3.595g, thus the sum of the moles of KCl (<em>x</em>) multiplied by it molar mass (74.5g/mol) and the moles of KBr (<em>y</em>) multiplied by it molar mass (119g/mol) is the total mass of the mixture:

x.74.5g/mol + y.119g/mol = 3.595g

Also, after the conversion of KBr into KCl, the total mass of 3.129 g is only from KCl moles, hence

\frac{3.129g}{74.5g/mol} = 0.042 moles

But the 0.042 moles came from the originals KCl and KBr moles, thus

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Now it is possible to propose a system of equations:

x.74.5g/mol + y.119g/mol = 3.595g

x + y = 0.042moles

Solving the system of equations,

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0.010 moles of KBr multiplied it molar mass is

0.010molesx119g/mol = 1.19g

Therefore, the percentage (by mass) of KBr in the original mixture was:

\frac{1.19g}{3.595g}x100% = 33.1%%

4 0
3 years ago
1. Something must have mass in order to create change.
nignag [31]

These are three questions each with its complete answer.

Question 1. Something must have mass in order to create change.

A. True

B. False

Answer:

  • <u><em>False</em></u>

Explanation:

Two counter examples will prove that the statement is false.

  • light and fire

Neither light nor fire have mass, so they are not considered matter, but yet both can create changes.

Light is an electromagnetic wave, so it is energy, and as such it, of course, can cause some changes. One example are the chemical reactions, like photosynthesis, which require light to happen.

Ligth may also be described either as quanta named photons. Although photons behave like particles they do not have mass.

So, that example contradicts the statement that something must have mass in order to create change.

Fire also contradicts the statement. Fire is a manifestation of energy, it does not have mass, and sure you know how fire create changes in living and no living things.

Question 2. Powdered medicine fills a pill capsule. Which of the following is the best description of the state of the powder?

A. It is a solid because all powders are solid.

B. It is a gas because it takes up the entire volume of the pill capsule.

C. It is a liquid because it does not hold a definite shape.

Answer: the best description is the choice A:

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Explanation:

Indeed, although sometimes the powder may be very fine (dust), yet it is not liquid or gas. Powders are always solid.

There are four main states of matter: solid, liquid, gas, and plasma.

Liquids are able to take the form of the container but do not occupy the entire container. Gases do occupy the entire container.

When powder is very fine it could resemble some properties of liquid, but the structure of powders is not crystaline, this is it is not an ordered arrangement. The structure of powders is an ordered arrangement, so powders are solids and not liquids.

It is easy to discard that a powdered medicine could be a gas, because it definetly do not occupy the entire volume of the pill capsule.

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Answer:

Two examples come to my mind:

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Explanation:

I will use the example of Nutella to complete the answer.

Nutella is not a gas because when you open the jar it will not escape and fill the room around it.

When you put Nutella in a plate it will not flow over the whole plate, is it a solid ? But yet you can spread it easyly over a cookie. Is it a liquid?

So, Nutella behaves like a solid until you apply some force over it and then it seems a liquid.

I would guess that pastes are solids because they should have an inner structure that keep the particles together.

If you do some research you will find that pastes are neither liquid nor solids, they are named gels. And gels may be seen as a dispersion of molecules of a liquid within a solid, i.e. in liquid particles dispersed in the solid medium

This sure meets the requirements of this question.

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Explanation:

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