The 4:00 time because later that afternoon would be later in the day
Answer:
b i think because i think so
Answer:
the final product is called a product
Explanation:
The pH of the buffer is 6.1236.
Explanation:
The strength of any acid solution can be obtained by determining their pH. Even the buffer solution strength of the weak acid can be determined using pH. As the dissociation constant is given, we can determine the pKa value as the negative log of dissociation constant value.
![pKa=-log[H] = - log [ 5.66 * 10^{-7}]\\ \\pka = 7 - log (5.66)=7-0.753=6.247\\\\pka = 6.247](https://tex.z-dn.net/?f=pKa%3D-log%5BH%5D%20%3D%20-%20log%20%5B%205.66%20%2A%2010%5E%7B-7%7D%5D%5C%5C%20%5C%5Cpka%20%3D%207%20-%20log%20%285.66%29%3D7-0.753%3D6.247%5C%5C%5C%5Cpka%20%3D%206.247)
The pH of the buffer can be known as
![pH = pK_{a} + log[\frac{[A-]}{[HA]}}]](https://tex.z-dn.net/?f=pH%20%3D%20pK_%7Ba%7D%20%2B%20log%5B%5Cfrac%7B%5BA-%5D%7D%7B%5BHA%5D%7D%7D%5D)
The concentration of ![[A^{-}] = Moles of [A]/Total volume = 0.608/2 = 0.304 M\\](https://tex.z-dn.net/?f=%5BA%5E%7B-%7D%5D%20%3D%20Moles%20of%20%5BA%5D%2FTotal%20volume%20%3D%200.608%2F2%20%3D%200.304%20M%5C%5C)
Similarly, the concentration of [HA] = 
Then the pH of the buffer will be
pH = 6.247 + log [ 0.304/0.404]

So, the pH of the buffer is 6.1236.
Answer:

Explanation:
Hello there!
In this case, according to the given chemical reaction:
2 Al + 3 Cl2 --> 2 AlCl3
Whereas there is a 2:3 mole ratio of aluminum to chlorine; it will be possible for us to calculate the required grams of aluminum by using the equality 22.4 L = 1 mol, the aforementioned mole ratio and the atomic mass of aluminum (27.0 g/mol) to obtain:

Regards!