<span>C7H8
First, lookup the atomic weight of all involved elements
Atomic weight of carbon = 12.0107
Atomic weight of hydrogen = 1.00794
Atomic weight of oxygen = 15.999
Then calculate the molar masses of CO2 and H2O
Molar mass CO2 = 12.0107 + 2 * 15.999 = 44.0087 g/mol
Molar mass H2O = 2 * 1.00794 + 15.999 = 18.01488 g/mol
Now calculate the number of moles of each product obtained
Note: Not interested in the absolute number of moles, just the relative ratios. So not going to get pedantic about the masses involved being mg and converting them to grams. As long as I'm using the same magnitude units in the same places for the calculations, I'm OK.
moles CO2 = 3.52 / 44.0087 = 0.079984
moles H2O = 0.822 / 18.01488 = 0.045629
Since each CO2 molecule has 1 carbon atom, I can use the same number for the relative moles of carbon. However, since each H2O molecule has 2 hydrogen atoms, I need to double that number to get the relative number of moles for hydrogen.
moles C = 0.079984
moles H = 0.045629 * 2 = 0.091258
So we have a ratio of 0.079984 : 0.091258 for carbon and hydrogen. We need to convert that to a ratio of small integers. First divide both numbers by 0.079984 (selected since it's the smallest), getting
1: 1.140953
The 1 for carbon looks good. But the 1.140953 for hydrogen isn't close to an integer. So let's multiply the ratio by 1, 2, 3, 4, ..., etc and see what each new ratio looks like (Effectively seeing what 1, 2, 3, 4, etc carbons look like)
1 ( 1 : 1.140953) = 1 : 1.140953
2 ( 1 : 1.140953) = 2 : 2.281906
3 ( 1 : 1.140953) = 3 : 3.422859
4 ( 1 : 1.140953) = 4 : 4.563812
5 ( 1 : 1.140953) = 5 : 5.704765
6 ( 1 : 1.140953) = 6 : 6.845718
7 ( 1 : 1.140953) = 7 : 7.986671
8 ( 1 : 1.140953) = 8 : 9.127624
That 7.986671 in row 7 looks extremely close to 8. I doubt I'd get much closer unless I go to extremely high integers. So it looks like the empirical formula for toluene is C7H8</span>
Answer : The pOH of pure water is, 6.68
Explanation :
As we are given that:
![K_w=4.38=\times 10^{-14}](https://tex.z-dn.net/?f=K_w%3D4.38%3D%5Ctimes%2010%5E%7B-14%7D)
First we have to calculate the concentration of hydroxide ion.
As, ![K_w=[H^+]\times [OH^-]](https://tex.z-dn.net/?f=K_w%3D%5BH%5E%2B%5D%5Ctimes%20%5BOH%5E-%5D)
As we know that in pure water the hydrogen ion and hydroxide ion concentration are equal. That means,
![[H^+]=[OH^-]](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%5BOH%5E-%5D)
So, ![K_w=[OH^-]\times [OH^-]](https://tex.z-dn.net/?f=K_w%3D%5BOH%5E-%5D%5Ctimes%20%5BOH%5E-%5D)
![K_w=[OH^-]^2](https://tex.z-dn.net/?f=K_w%3D%5BOH%5E-%5D%5E2)
![4.38\times 10^{-14}=[OH^-]^2](https://tex.z-dn.net/?f=4.38%5Ctimes%2010%5E%7B-14%7D%3D%5BOH%5E-%5D%5E2)
![[OH^-]=2.09\times 10^{-7}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D2.09%5Ctimes%2010%5E%7B-7%7DM)
Now we have to calculate the pOH.
![pOH=6.68](https://tex.z-dn.net/?f=pOH%3D6.68)
Therefore, the pOH of pure water is, 6.68
Potassium is an element with a symbol of K. When it forms an oxide, it means that it has undergone combustion. The complete reaction for this is
K + O₂ ⇒ K₂O
This is a combination reaction which involves two reactants and one single product. It is important to note that the Law of Conservation of Mass mass be obeyed at all times. Technically, the law involved in balancing reactions is the Law of Definite Proportions. This is obeyed by making sure that the number of a specific element is the same before and after the reaction. So, the complete balanced reaction for this is
2 K + 1/2 O₂ ⇒ K₂O
The coefficient of O₂ in the reaction is 1/2.
Explanation:
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