Taking into account the rule of three, the volume of air that you inhale (and exhale) in 12.0 hours is 4,384.8 liters.
<h3>Rule of three</h3>In first place, the rule of three is a way of solving problems of proportionality between three known values and an unknown value, establishing a relationship of proportionality between all of them.
That is, what is intended with it is to find the fourth term of a proportion knowing the other three.
If the relationship between the magnitudes is direct, that is, when one magnitude increases, so does the other (or when one magnitude decreases, so does the other) , the direct rule of three must be applied.
To solve a direct rule of three, the following formula must be followed, being a, b and c known data and x the variable to be calculated:
a ⇒ b
c ⇒ x
So:
To calculated the volume of air inhaled and exhaled in 12 hours, you know that each breath has a volume of 0.435 liters, and that you are breathing 14 times a minute.
The you can apply the following rule of three: If each breath has a volume of 0.435 liters, 14 breaths have how much volume?
<u><em>volume= 6.09 liters</em></u>
So, 14 breaths a minute have a volume of 6.09 liters, this means that in 1 minute you inhale and exhale 6.09 liters.
Now, knowing that 1 hour has 60 minutes, 12 hours has 720 minutes. So you can apply the following rule of three: in 1 minute you inhale and exhale 6.09 liters, in 720 minutes you inhale and exhale how much volume?
<u><em>volume= 4,384.8 liters</em></u>
In summary, the volume of air that you inhale (and exhale) in 12.0 hours is 4,384.8 liters.
Learn more with this similar problems:
Answer:
The acceleration is:
Explanation:
Given
--- The initial velocity
--- time
-- The final velocity
Required
Determine the acceleration
To do this, we make use of the first equation of motion
We used negative because the car was coming to stop.
This gives:
Collect like terms
Solve for a
Answer:
Explanation:
In this case we want to know the structures of A (C6H12), B (C6H13Br) and C (C6H14).
A and C reacts with two differents reagents and conditions, however both of them gives the same product.
Let's analyze each reaction.
First, C6H12 has the general formula of an alkene or cycloalkane. However, when we look at the reagents, which are HBr in ROOR, and the final product, we can see that this is an adition reaction where the H and Br were added to a molecule, therefore we can conclude that the initial reactant is an alkene. Now, what happens next? A is reacting with HBr. In general terms when we have an adition of a molecule to a reactant like HBr (Adding electrophyle and nucleophyle) this kind of reactions follows the markonikov's rule that states that the hydrogen will go to the carbon with more hydrogens, and the nucleophyle will go to the carbon with less hydrogen (Atom that can be stabilized with charge). But in this case, we have something else and is the use of the ROOR, this is a peroxide so, instead of follow the markonikov rule, it will do the opposite, the hydrogen to the more substituted carbon and the bromine to the carbon with more hydrogens. This is called the antimarkonikov rule. Picture attached show the possible structure for A. The alkene would have to be the 1-hexene.
Now in the second case we have C, reacting with bromine in light to give also B. C has the formula C6H14 which is the formula for an alkane and once again we are having an adition reaction. In this case, conditions are given to do an adition reaction in an alkane. bromine in presence of light promoves the adition of the bromine to the molecule of alkane. In this case it can go to the carbon with more hydrogen or less hydrogens, but it will prefer the carbon with more hydrogens. In this case would be the terminal hydrogens of the molecules. In this case, it will form product B again. the alkane here would be the hexane. See picture for structures.
