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3 years ago

Fe+CuCI2> FeCI2+Cu balancing equations

Chemistry

1 answer:

Volgvan3 years ago

3 0

Explanation:

To balance a chemical equation implies that we are conserving the number and mass of the reacting species.

Chemical equations must obey the law of conservation of mass "in a chemical reaction, matter is neither created nor destroyed".

This suggests that mass is conserved.

The number of atoms on the reactant side must be equal to that on the product end.

Before proceeding, check the species to see if they have equal number of atoms on both sides:

Fe + CuCI₂ → FeCI₂ + Cu

Elements reactants products

Fe 1 1

Cu 1 1

Cl 2 2

We can see that the equation is already balanced

learn more:

balanced equation brainly.com/question/2612756

#learnwithbrainly

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Liquid Boiling point (K) Enthalpy of vaporization (kJ/mol) benzene 353.1 30.72 toluene 383.6 33.18 nonane 423.8 37.18 chloroform

nata0808 [166]

Answer :

(a) The compound benzene follow the Trouton's rule.

(b) The compound toluene follow the Trouton's rule.

(c) The compound nonane follow the Trouton's rule.

(d) The compound chloroform follow the Trouton's rule.

(e) The compound ethanol does not follow the Trouton's rule.

(f) The compound water does not follow the Trouton's rule.

Explanation :

Trouton's rule : According to this rule, the molar entropy of vaporization should be constant that means,

$\Delta S_{vap}=85-88J/K.mole$

Formula used : $\Delta S=\frac{\Delta H_{vap}}{T_b}$

Now we have to determine the molar entropy of vaporization for the following compounds.

(a) For Benzene:

Boiling point = 353.1 K

Enthalpy of vaporization = 30.72 kJ/mole = 30720 J/mole

$\Delta S=\frac{30720J/mole}{353.1K}=87.00J/K.mole$

Thus, this compound is follow the Trouton's rule.

(b) For Toluene:

Boiling point = 383.6 K

Enthalpy of vaporization = 33.18 kJ/mole = 33180 J/mole

$\Delta S=\frac{33180J/mole}{383.6K}=86.49J/K.mole$

Thus, this compound is follow the Trouton's rule.

(c) For Nonane:

Boiling point = 423.8 K

Enthalpy of vaporization = 37.18 kJ/mole = 37180 J/mole

$\Delta S=\frac{37180J/mole}{423.8K}=87.73J/K.mole$

Thus, this compound is follow the Trouton's rule.

(d) For Chloroform :

Boiling point = 334.2 K

Enthalpy of vaporization = 29.24 kJ/mole = 29240 J/mole

$\Delta S=\frac{29240J/mole}{334.2K}=87.49J/K.mole$

Thus, this compound is follow the Trouton's rule.

(e) For Ethanol:

Boiling point = 351.3 K

Enthalpy of vaporization = 38.56 kJ/mole = 38560 J/mole

$\Delta S=\frac{38560J/mole}{351.3K}=109.76J/K.mole$

Thus, this compound is not follow the Trouton's rule.

(f) For Water:

Boiling point = 373 K

Enthalpy of vaporization = 40.65 kJ/mole = 40650 J/mole

$\Delta S=\frac{40650J/mole}{373K}=108.98J/K.mole$

Thus, this compound is not follow the Trouton's rule.

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