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3 years ago

Why was it important to establish the Clean Air Act?

Chemistry

2 answers:

frosja888 [35]3 years ago

8 0

Answer:

It was important to establish the clean air act to create a regulation that makes any activities pollute the air illegal

Explanation:

romanna [79]3 years ago

7 0

Answer:

C. to create a regulation that makes any activities that pollute the air illegal

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Be sure to answer all parts. The standard enthalpy of formation and the standard entropy of gaseous benzene are 82.93 kJ/mol and

skelet666 [1.2K]

Answer : The values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

Explanation :

The given balanced chemical reaction is,

$C_6H_6(l)\rightarrow C_6H_6(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{C_6H_6(g)}\times \Delta H_f^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta H_f^0_{(C_6H_6(l))}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0_{(C_6H_6(g))}$ = standard enthalpy of formation of gaseous benzene = 82.93 kJ/mol

$\Delta H_f^0_{(C_6H_6(l))}$ = standard enthalpy of formation of liquid benzene = 49.04 kJ/mol

Now put all the given values in this expression, we get:

$\Delta H^o=[1mole\times (82.93kJ/mol)]-[1mole\times (49.04J/mol)]$

$\Delta H^o=33.89kJ/mol=33890J/mol$

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{C_6H_6(g)}\times \Delta S^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta S^0_{(C_6H_6(l))}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S^0_{(C_6H_6(g))}$ = standard entropy of formation of gaseous benzene = 269.2 J/K.mol

$\Delta S^0_{(C_6H_6(l))}$ = standard entropy of formation of liquid benzene = 173.26 J/K.mol

Now put all the given values in this expression, we get:

$\Delta S^o=[1mole\times (269.2J/K.mol)]-[1mole\times (173.26J/K.mol)]$

$\Delta S^o=95.94J/K.mol$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is $25^oC\text{ or }298K$ .

$\Delta G^o=(33890J)-(298K\times 95.94J/K)$

$\Delta G^o=5299.88J/mol=5.299kJ/mol$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

6 0

3 years ago

Read 2 more answers

Why does salt dissolve in water?

tigry1 [53]

The water dissolves the salt because the water molecules are able to interact with the salt-forming particles, called ions. When the water Interacts with the trainers of the salt ions, the solid salt crystal structure Suffers breakdown, until the trainers of the salt ions are completely surrounded by water molecules. At this time the salt is fully dissolved by water.

hope this helps!

4 0

3 years ago

Someone please help me with this!!

pantera1 [17]

I would think that it’s C, but try to get a second opinion as well!!

4 0

3 years ago

Why was 1,2 dichlorobenzene used as the solvent for the diels alder reaction we performed in the lab?

pav-90 [236]

Answer:

1,2 dichlorobenzene was used as the solvent for the diels alder reaction: <em>because the co elimination part of the reaction needs high temp and a high boiling point solvent such as 1,2 dichlorobenzene</em>

Explanation:

Diels-Alder Reaction is a useful synthetic tool to prepare cyclohexane rings. It is a process, which occurs in a single step that consists of a cyclic redistribution of its electrons. The two reagents are bond together through a cyclic transition state in which the two new C-C bonds are formed at the same time. For this to occur, most of the time, it is necessary a high temperature and high-pressure conditions. Since 1,2 dichlorobenzene has a boiling point of 180ºC is a good solvent for this type of reactions.

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3 years ago

Propiedades quimicas de la plata

timofeeve [1]

Elemento químico, símbolo Ag, número atómico 47 y masa atómica 107.870. Es un metal lustroso de color blanco-grisáceo. La plata, que posee las más altas conductividades térmica y eléctrica de todos los metales, se utiliza en puntos de contacto eléctricos y electrónicos.

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3 years ago