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devlian [24]
4 years ago
3

In the simulation, were you able to use noninteger numbers (like ½ or 0.43) for the coefficients in a balanced equation? Why do

you think this is?
Chemistry
2 answers:
BARSIC [14]4 years ago
4 0
Because half the time the number are either fractions or decimals.
GalinKa [24]4 years ago
4 0

Answer:

By standard convention, decimals and fractions are avoided or rather do away with when balancing chemical equations.

We do not have decimals when balancing equations, because we are balancing atoms and molecules. You can't cut an atom or a molecule in half after all, so you must use whole numbers.

By doing away with decimals and fractions, mole relationships between reactants and products are simply presented.

It is much more understandable and presentable to say; "2 moles of Hydrogen gas reacts with 1 mole of oxygen gas to form2moles of water molecule"

Rather than; "1 moles of Hydrogen gas reacts with 1/2 mole of oxygen gas to form 1 moles of water molecule"

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Use the thermochemical equations shown below to determine the enthalpy for the reaction: COCl2(g) + H2O(l) --> CH2Cl2(l) + O2
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Answer:

ΔH° (Enthalpy change) for the reaction is equal to (-151.5 KJ)

Explanation:

Determine the enthalpy for the reaction

           COCl₂ (g) + H₂O(l) --> CH₂Cl₂(l) + O₂(g)

It becomes easier to find the unknown value of enthalpy of a particular reaction by applying the Hess's law on the given reaction with the known value of standard enthalpy change.

\frac{1}{2} H₂(g) + \frac{1}{2}Cl₂(g) --> HCl(g) ΔH=-46 kJ -----------------------------(1)

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CH₂Cl₂(l) + H₂(g) + \frac{3}{2}O₂(g) --> COCl₂(g) + 2 H₂O(l) ΔH = 80.5 kJ ----------(3)

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Revers the equation (2) and (3) we get

2 HCl(g) + \frac{1}{2}O₂ (g) --> H₂O(g) + Cl₂(g)  ΔH=+21 KJ ----------------(5)

COCl₂(g) + 2 H₂O(l) --> CH₂Cl₂(l) + H₂(g) + \frac{3}{2}O₂(g) ΔH = - 80.5 kJ  -----------(6)

Now add these three equations i.e., (4) + (5) + (6)

we can get these equation

                COCl₂ (g) + H₂O(l) --> CH₂Cl₂(l) + O₂(g)

The enthalpy change during the reaction = -92 + 21 -80.5 = - 151.5 KJ

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