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2 years ago

Which substance absorbs 58.16 kJ of energy when 3.11 mol vaporizes? a)CH4 b)H2S c)CO2 d)NaCl

Chemistry

1 answer:

schepotkina [342]2 years ago

8 0

Answer:

$H_2S$

Explanation:

Given the amount of heat absorbed and the amount of substance in moles, we may calculate the heat of vaporization. Heat of vaporization is defined as the amount of heat per 1 mole of substance required to evaporate that specific substance.

Based on the value of heat of vaporization, we will identify the substance. Firstly, let's calculate the heat of vaporization:

$\Delta H^o = \frac{58.16 kJ}{3.11 mol} = 18.7 kJ/mol$

Secondly, let's use any table for heat of vaporization values for substances. We identify that the heat of vaporization of $H_2S$ is 18.7 kJ/mol

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An air sample consists of oxygen and nitrogen gas as major components. It also contains carbon dioxide and traces of some rare g

jekas [21]

Explanation:

A mixture is defined as the substance that contains two or more different number of substances that are physically mixed together.

For example, a mixture of air which contains oxygen, nitrogen and other gases.

A mixture in which solute particles are unevenly distributed into the solvent then it is known as a heterogeneous mixture.

For example, sand in water is a heterogeneous mixture.

A homogeneous mixture is defined as the mixture in which solute particles are evenly distributed in a solvent.

A homogeneous mixture is a clear solution.

For example, salt dissolved in water is a homogeneous mixture.

A solution is defined as the substance in which two or more substances are mixed together.

A compound is defined as the substance that contains two or more different elements that chemically combined together in a fixed ratio by mass.

A element is defined as the substance that contains only one type of atoms.

For example, a piece of sodium element will contain only atoms of sodium.

Whereas a pure substance is defined as the substance which contains only one type of molecule or one type of atom.

For example, $O_{2}$ , $N_{2}$ etc are pure substances.

Thus, we can conclude that the terms which could be used to describe the given sample of air is as follows.

pure chemical substance.

heterogenous mixture.

mixture.

4 0

2 years ago

A sample of an unknown compound with a mass of 0.847 g has the following composition: 50.51 % fluorine and 49.49 % iron. When th

sergejj [24]

Answer: 0,4278g of F and 0,4191g of Fe

Explanation: it's possible to calculate the mass of each element by multiplying the percentage (decimal) of the element by the mass of the compound.

For Fluorine (F)

0,847g * 0,5051 = 0,4278g of F

For iron (Fe)

0,847 * 0,4949 = 0,4191g of Fe

This is determined because even when the compound is decomposed, due to conservative law of mass, the decomposition process do not affect the amount of matter, so the mass of the elements remain even if they are separated from the original molecule.

At the end, the sum of the elements masses should be the total mass of the compound.

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2 years ago

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A gas is contained in a thick-walled balloon. When the pressure changes from 319 mm Hg to 215 mm Hg, th volume changes from 0.55

Dmitry_Shevchenko [17]

In this question we need to find the new volume of the gas. Since we have been given the pressure and temperature change, we can used to combined gas law equation.
$\frac{P1V1}{T1} = \frac{P2V2}{T2}$
the parameters for 1st instance are given on the left side and parameters for the second instance are given on the right side of the equation
(319 mmHg x 0.558 L)/ 115 K = (215 mmHg x V)/387 K
V = 2.79 L

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2 years ago

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5. Before vehicle emissions were well-regulated CO emissions were 66 g/mile. Assume this emission rate applies for an airshed. T

Aleks [24]

Answer:CALINE4 Model and Geographical Information System were used for

the present study to predict the CO concentrations and prepare thematic

maps for the study area.

CALINE4 is a latest model that predicts the concentration of carbon

monoxide impacts near the roadways. The California Department of

Transformation (CALTRANS) developed the model and its purpose is to help

planners protect public health from the adverse effects of carbon monoxide.

The model predictions along with the aid of GIS based model help to arrive at

air shed levels of carbon monoxide. CALINE4 is a simple line source

Gaussian plume dispersion model. The user defines the proposed roadway

geometry, worst-case meteorological parameters, anticipated traffic volumes

and receptor positions to predict the concentrations of pollutant.

CALINE4 is graphical with windows based user interface, designed to

ease data entry and increase the online help capabilities. The model was

developed for predicting the concentrations of relatively inert pollutants such

as carbon monoxide and it is now used for several other pollutants like NO2

and SPM. The model is based on fine tuned Gaussian diffusion equation and

it employs mixing zone concept to characterize the pollutant dispersion over

the roadways. Given the exhaust emission concentrations, meteorology and

site geometry, the model can predict the concentration of pollutants for any experimentation

Explanation:

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2 years ago

A 0.72-mol sample of PCl5 is put into a 1.00-L vessel and heated. At equilibrium, the vessel contains 0.40 mol of PCl3(g) and 0.

Sonja [21]

Answer:

Equilibrium constant for $PCl_5$ is 0.5

Equilibrium constant for decomposition of $NO_2$ is $1.79 \times 10^{-14}$

Explanation:

$PCl_5$ dissociates as follows:

$PCl_5 \rightleftharpoons PCl_3+Cl_2$

initial 0.72 mol 0 0

at eq. 0.72 - 0.40 0.40 0.40

Expression for the equilibrium constant is as follows:

$k=\frac{[PCl_3][Cl_2]}{[PCl_5]}$

Substitute the values in the above formula to calculate equilibrium constant as follows:

$k=\frac{[0.40/1][0.40/1]}{0.32/1} \\=\frac{0.40 \times 0.40}{0.32} \\=0.5$

Therefore, equilibrium constant for $PCl_5$ is 0.5

Now calculate the equilibrium constant for decomposition of $NO_2$

It is given that $3.3 \times 10^{-3} \%$ is decomposed.

$NO_2$ decomposes as follows:

$2NO_2 \rightleftharpoons 2NO + O_2$

initial 1.0 M 0 0

at eq. concentration of $NO_2$ is:

$[NO_2]_{eq}=1-(0.000066) = 0.999934\ M$

$[NO]_{eq}=6.6 \times 10^{-5}\ M$

$[O_2]_{eq}=3.3\times 10^{-5} = 3.3\times 10^{-5}\ M$

Expression for equilibrium constant is as follows:

$K=\frac{[NO]^2[O_2]}{[NO_2]^2}$

Substitute the values in the above expression

$K=\frac{[6.6\times 10^{-5}]^2[3.3 \times 10^{-5}]}{[0.999934]^2} \\=1.79\times 10^{-14}$

Equilibrium constant for decomposition of $NO_2$ is $1.79 \times 10^{-14}$

8 0

2 years ago