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svetlana [45]
3 years ago
9

You have 0.5 L of air at 203 k in an expandable container at constant pressure. You heat the container to 273 k. What is the vol

ume of the air?
A. 0.25 L
B. 1.5 L
C. 0.67 L
0.37 L
Chemistry
1 answer:
Jobisdone [24]3 years ago
4 0
You can use P1V1/T1 = P2V2/T2 but since pressure is constant is becomes V1/T1=V2/T2

V1=0.5 L
T1=203 K
T2=273 K
V2=unknown

0.5L/203 = V2/273
V2= 0.67 L so C

Hope this helps :)
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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
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The reaction takes place in water. What happens to the equilibrium when the pressure is increased? a)It favors formation of prod
OLga [1]

Answer:

I don't really get the options but it favoures the reactant side.

Explanation:

Increasing pressure favours the side with fewer moles of gas while decreasing pressure favours the side with the more moles of gas. E.g

If there is 0 moles of gas particles in the reactant side and 1 mole of gas particle in the product side, increasing pressure favours the reactants while decreasing pressure favours the product side.

With the explanations I have made, I hope the question is now clear to you.

6 0
4 years ago
Lithium reacts with chlorine to form a new substance.What other element will also with chlorine to form a new substance with sim
quester [9]

Answer:

Flourine

Explanation:

Chlorine is a member of the halogen family. Halogens are the elements that make up Group 17 (VIIA) of the periodic table, a chart that shows how elements are related to one another. They include fluorine, bromine, iodine, and astatine.

4 0
3 years ago
Yeast and other organisms can convert glucose (C6H12O6) to Ethanol (CH3CH2OH) through a process called Alcoholic Fermentation. T
Murljashka [212]

Answer:

The answer to your question is 8.28 g of glucose

Explanation:

Data

Glucose  (C₆H₁₂O₆) = ?

Ethanol (CH₃CH₂OH)

Carbon dioxide (CO₂) = 2.25 l

Pressure = 1 atm

T = 295°K

Reaction

                             C₆H₁₂O₆    ⇒    2C₂H₅OH(l) +2CO₂(g)

- Calculate the number of moles

                             PV = nRT

Solve for n

                             n = \frac{PV}{RT}

Substitution

                             n = \frac{(1)(2.25)}{(0.082)(295)}

Simplification

                            n = 0.092

- Calculate the mass of glucose

                         1 mol of glucose --------------- 2 moles of carbon dioxide

                          x                         --------------- 0.092 moles

                          x = (0.092 x 1) / 2

                          x = 0.046 moles of glucose

Molecular weight of glucose = 180 g

                       180 g of glucose ---------------  1 mol

                         x  g                    ---------------0.046 moles

                         x = (0.046 x 180) / 1

                         x = 8.28 g of glucose                                    

5 0
3 years ago
What are the coefficients of this chemical equation?
777dan777 [17]

Answer:

i think 2 ,1, 2 i may be wrong sorry if i am

8 0
3 years ago
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