The mole ratio of the reaction shows that equal volumes of hydrogen gas will be produced by the two reactions.
<h3>What is the mole ratio of a reaction?</h3>
The mole ratio of a reaction is the ratio in which the reactants and products of a given reaction occur for the reaction to proceed to completion.
The mole ratio of a reaction is also known as the stoichiometry of the reaction.
The equation of the two reactions are given below:


From the equation of the reaction reaction, an equal volume of hydrogen gas will be produced by the two reactions.
Therefore, the mole ratio of the reaction shows that equal volumes of hydrogen gas will be produced by the two reactions.
Learn more about mole ratio at: brainly.com/question/19099163
#SPJ1
Answer:
Each nitrogen molecule consists of two atoms of nitrogen that are bonded by a triple covalent bond. This is a direct consequence of the fact that each nitrogen atom has 5 valence electrons. Each atom can thus complete its octet by sharing three electrons.
Explanation:
Sorry i had to look it up i didn't know this answer
Please mark brainliest:)
The combined gas law combines the three gas laws: Boyle's Law, Charles'Law, and Gay-Lussac's Law. It states that the ratio of the product of pressure and volume and the absolute temperature of a gas is equal to a constant. When Avogadro's law is added to the combined gas law, the ideal gas law res
Answer is volume and pressure only
Answer : The amount of formaldehyde permissible are, 
Explanation : Given,
Density of air =

First we have to calculate the mass of air.



Now we have to calculate the amount of formaldehyde.
Permissible exposure level of formaldehyde = 0.75 ppm = 
Amount of formaldehyde in 7.2 g of formaldehyde = 
Amount of formaldehyde in 7.2 g of formaldehyde = 
Thus, the amount of formaldehyde permissible are, 