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kolbaska11 [484]
3 years ago
14

Which option is an example of a chemical property?

Chemistry
2 answers:
saveliy_v [14]3 years ago
5 0
<h3>Answer </h3>

Option B= flammability

<h3>Explanation </h3>

Flammability is a chemical property because combustion is a chemical change. A chemical change is a process in which one or more substances are converted into one or more new substances with different properties.

Gasoline is highly flammable. When it burns, it is converted into carbon dioxide and water. These have different properties from gasoline, so combustion is a chemical process. Flammability is a measure of how easily something will burn.

<h3>Examples of chemical properties are:  </h3>

Heat of combustion, reactivity with water, PH, and electromotive force.

While density color and melting point are examples of physical properties.


Delvig [45]3 years ago
4 0
Yes, the correct answer is b
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Gay-Lussac's Law: This law states that pressure is directly proportional to the temperature of the gas at constant volume and number of moles.

P\propto T     (At constant volume and number of moles)

\frac{P_1}{T_1}=\frac{P_2}{T_2}

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P_1 = initial pressure of gas  = 2.79 atm

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T_1 = initial temperature of gas  = 273K

T_2 = final temperature of gas = 315 K

\frac{2.79}{273}=\frac{P_2}{315}

P_2=3.22atm

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Use the given data at 500 K to calculate ΔG°for the reaction
Anton [14]

Answer : The  value of \Delta G^o for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(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_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]

where,

\Delta H^o = enthalpy of reaction = ?

n = number of moles

\Delta H_f^0 = standard enthalpy of formation

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

\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]

\Delta H^o=-1035.6kJ=-1035600J

conversion used : (1 kJ = 1000 J)

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_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S_f^0 = standard entropy of formation

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

\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]

\Delta S^o=-153J/K

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 500 K.

\Delta G^o=(-1035600J)-(500K\times -153J/K)

\Delta G^o=-959100J=-959.1kJ

Therefore, the value of \Delta G^o for the reaction is -959.1 kJ

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