Answer:
The answer to your question is 33.4 ml
Explanation:
Data
volume 1 = V1 = 42 ml
temperature 1 = T1 = 20°C
temperature 2 = T2 = -60°C
Volume 2 = V2 = x
Process
1.- Convert celsius to kelvin
T1 = 20 + 273 = 293°K
T2 = -60 + 273 = 233°K
2.- Use the Charles' law to solve this problem

Solve for V2
V2 = 
3.- Substitution
V2 = 
4.- Simplification
V2 = 
5.- Result
V2 = 33.4ml
Around 36 or 37 days, because of the decimal
Answer:

Explanation:
Hello,
In this case, since the chemical reaction is:

We can see that hydrochloric acid and magnesium hydroxide are in a 2:1 mole ratio, which means that the neutralization point, we can write:

In such a way, the moles of magnesium hydroxide (molar mass 58.3 g/mol) in 500 mg are:

Next, since the pH of hydrochloric acid is 1.25, the concentration of H⁺ as well as the acid (strong acid) is:
![[H^+]=[HCl]=10^{-pH}=10^{-1.25}=0.0562M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%5BHCl%5D%3D10%5E%7B-pH%7D%3D10%5E%7B-1.25%7D%3D0.0562M)
Then, since the concentration and the volume define the moles, we can write:
![[HCl]*V_{HCl}=2*n_{Mg(OH)_2}](https://tex.z-dn.net/?f=%5BHCl%5D%2AV_%7BHCl%7D%3D2%2An_%7BMg%28OH%29_2%7D)
Therefore, the neutralized volume turns out:

Best regards.
Answer:
The benthic zone is the ecological region at the lowest level of a body of water such as an ocean, lake, or stream, including the sediment surface and some sub-surface layers.
Explanation:
In order to deprotonate an acid, we must remove protons in order to achieve a more stable conjugate base. For this example, we can use the relationship between carboxylic acid and hydroxide.
Deprotonation is the removal of a proton from a specific type of acid in reaction to its coming into contact with a strong base. The compound formed from this reaction is known as the conjugate base of that acid. The opposite process is also possible and is when a proton is added to a special kind of base. This is a process referred to as protonation, which forms the conjugate acid of that base.
For the example we have chosen to give, the conjugate base is the carboxylate salt. This would be the compound formed by the deprotonated carboxylic acid. The base in question was strong enough to deprotonate the acid due to the greater stability offered as a conjugated base.
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