B.5 as the electronic configuration is 2.5. 2 on the first shell abd 5 on the valency or most outer sgell.
Answer:
At standard room temperature, ![[{\rm OH^{-}] \approx 2.5 \times 10^{-7}\; \rm M](https://tex.z-dn.net/?f=%5B%7B%5Crm%20OH%5E%7B-%7D%5D%20%5Capprox%202.5%20%5Ctimes%2010%5E%7B-7%7D%5C%3B%20%5Crm%20M)
when ![[{\rm H^{+}] = 4.0 \times 10^{-8}\; \rm M](https://tex.z-dn.net/?f=%5B%7B%5Crm%20H%5E%7B%2B%7D%5D%20%3D%204.0%20%5Ctimes%2010%5E%7B-8%7D%5C%3B%20%5Crm%20M)
.
Explanation:
The following equilibrium goes on in water:
.
The forward reaction is known as the self-ionization of water. The ionization constant of water, 
, gives the equilibrium position of this reaction:
![K_{\rm w} = [{\rm H^{+}] \cdot [{\rm OH^{-}}]](https://tex.z-dn.net/?f=K_%7B%5Crm%20w%7D%20%3D%20%5B%7B%5Crm%20H%5E%7B%2B%7D%5D%20%5Ccdot%20%5B%7B%5Crm%20OH%5E%7B-%7D%7D%5D)
.
At standard room temperature (
), 
. Also, ![[{\rm H^{+}}] = 4.0 \times 10^{-8}\; \rm mol \cdot L^{-1}](https://tex.z-dn.net/?f=%5B%7B%5Crm%20H%5E%7B%2B%7D%7D%5D%20%3D%204.0%20%5Ctimes%2010%5E%7B-8%7D%5C%3B%20%5Crm%20mol%20%5Ccdot%20L%5E%7B-1%7D)
. Substitute both values into the equation and solve for ![[{\rm OH^{-}}]](https://tex.z-dn.net/?f=%5B%7B%5Crm%20OH%5E%7B-%7D%7D%5D)
.
![\begin{aligned} {[}{\rm OH^{-}}{]} &= \frac{K_{\rm w}}{[{\rm H^{+}}]} \\ &\approx \frac{10^{-14}}{4.0 \times 10^{-8}} = 2.5 \times 10^{-7}\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%20%7B%5B%7D%7B%5Crm%20OH%5E%7B-%7D%7D%7B%5D%7D%20%26%3D%20%5Cfrac%7BK_%7B%5Crm%20w%7D%7D%7B%5B%7B%5Crm%20H%5E%7B%2B%7D%7D%5D%7D%20%5C%5C%20%26%5Capprox%20%5Cfrac%7B10%5E%7B-14%7D%7D%7B4.0%20%5Ctimes%2010%5E%7B-8%7D%7D%20%3D%202.5%20%5Ctimes%2010%5E%7B-7%7D%5Cend%7Baligned%7D)
.
In other words, in an aqueous solution at standard room temperature, when .
The amine here is the easiest to spot since there’s only one structure that has a nitrogen atom, which would be the first (the first structure is a primary amine).
The distinguishing functional group of an alcohol is the hydroxy group (—OH). Both the second and third structures have an —OH group, but the —OH in the third structure is part of a carboxyl group (—COOH or —C(=O)OH). A carboxyl group takes priority over hydroxy group. Thus, the second structure would be an alcohol and the third structure would be a carboxylic acid.
That leaves us with the fourth structure, a hydrocarbon with a halogen substitutent, or, aptly named, a halocarbon.
Answer:
2KClO3 ==> 2 KCl + 3O2
393g O2 x 1 mole/32 g = 12.28 moles O2
moles KClO3 needed = 12.28 mol O2 x 2 mol KClO3/3mol O2 = 8.19 mol
grams KClO3 = 8.19 moles x 123g/mol = 1007g
Explanation:
