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3 years ago

How can balancing equations support the law of conservation

Chemistry

1 answer:

amm18123 years ago

4 0

Answer:

detail is given below.

Explanation:

This law was given by French chemist Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.

Law of conservation of mass:

According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.

For example:

In given photosynthesis reaction:

6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂

The given equation is balanced chemical equation of photosynthesis. There are six carbon atoms, eighteen oxygen atoms and twelve hydrogen atoms on the both side of equation so this reaction followed the law of conservation of mass.

If equation is not balanced,

CO₂ + H₂O → C₆H₁₂O₆ + O₂

It can not follow the law of conservation of mass because mass is not equal on both side of equation.

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In a lab experiment 80.0 g of ammonia [NH3] and 120 g of oxygen are placed in a reaction vessel. At the end of the reaction 72.2

valentinak56 [21]

The percent yield of the reaction : 89.14%

<h3>Further explanation</h3>

Reaction of Ammonia and Oxygen in a lab :

mass NH₃ = 80 g

mol NH₃ (MW=17 g/mol):

$\dfrac{80}{17}=4.706$

mass O₂ = 120 g

mol O₂(MW=32 g/mol) :

$\tt \dfrac{120}{32}=3.75$

Mol ratio of reactants(to find limiting reatants) :

$\tt \dfrac{4.706}{4}\div \dfrac{3.75}{5}=1.1765\div 0.75\rightarrow O_2~limiting~reactant(smaller~ratio)$

mol of H₂O based on O₂ as limiting reactants :

mol H₂O :

$\tt \dfrac{6}{5}\times 3.75=4.5$

mass H₂O :

4.5 x 18 g/mol = 81 g

The percent yield :

$\tt \%yield=\dfrac{actual}{theoretical}\times 100\%\\\\\%yield=\dfrac{72.2}{81}\times 100\%=89.14\%$

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Answer:

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Explanation:

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Mrac [35]

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3 years ago

What is the maximum number of orbitalks for each set of quantum numbers? If it is not possible to determine the maximum number o

kolbaska11 [484]

Answer:

See Below for answer!

Explanation:

In order to find the number of orbitals given the principal quantum number, n, we can use $2n^{2}$ . For example n=2 would have 8 orbitals.

A) 32

B) 98

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D) 18

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