Calculate the free energy of formation of NaBr(s) given the following information: NaBr(s) → Na(s) + 1/2Br2(l), ∆G° = 349 kJ/mol
1 answer:
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Answer:
<h2>8.92 g/cm³</h2>Explanation:
The density of a substance can be found by using the formula
From the question
mass = 4.75 g
volume = 0.5325 cm³
We have
We have the final answer as
<h3>8.92 g/cm³</h3>Hope this helps you
Answer : The one part nitrogen gas combines with one part oxygen gas to form one part of dinitrogen monoxide.
Explanation :
Balanced chemical reaction : It is defined as the reaction in which the number of atoms of individual elements present on reactant side must be equal to the product side.
When nitrogen gas combines with oxygen gas then it react to give dinitrogen monoxide or nitrous oxide.
The balance chemical reaction will be:
By the stoichiometry we can say that, 2 parts of nitrogen gas combines with 1 part of oxygen gas to give 2 parts of dinitrogen monoxide or nitrous oxide.
First we have to determine the limiting reagent.
From the reaction we conclude that,
As, 2 moles of nitrogen gas combine with 1 mole of oxygen gas
So, 1 moles of nitrogen gas combine with 0.5 mole of oxygen gas
It means that, oxygen gas is an excess reagent because the given moles are greater than the required moles and nitrogen gas is a limiting reagent and it limits the formation of product.
Now we have to determine the moles of dinitrogen monoxide.
As, 2 moles of nitrogen gas combine to give 2 mole of dinitrogen monoxide
So, 1 mole of nitrogen gas combine to give 1 mole of dinitrogen monoxide
Thus, the one part nitrogen gas combines with one part oxygen gas to form one part of dinitrogen monoxide.