Answer:
0.084 M
Explanation:
Using the Henderson-Hasselbalch equation for a buffer ( a buffer is solution contain a weak acid and it conjugate base; the solution resist change in pH)
pH = pKa + log ( base/acid)
4.9 - 4.76 =log ( base / acid)
10^0.14 = ( base / acid)
1.38 = (base / acid)
since there is 0.2 M in the buffer solution
the concentration of acid = 
× 0.2 = 0.084 M
<span> When an </span>acid and a base<span> are placed together, they </span>react<span> to neutralize the </span>acid<span> and </span>base<span> properties, producing a salt. The H(+) cation of the </span>acid<span>combines with the OH(-) anion of the </span>base<span> to form water.</span>
Al
Explanation:
The limiting reactant will be Al:
4Al + 3O₂ → 2Al₂O₃
The limiting reactant is the reactant in short supply in a chemical reaction.
Given parameters:
Mass of Al = 30g Molar mass = 27g/mol
Number of moles = 
= 
Number of moles of Al = 1.111 mole
Mass of O₂ = 30g, molar mass = 32g/mol
Number of moles = 
= 0.94mol
In the reaction:
4 moles of Al reacted with 3 moles of O₂
1.11moles of Al will require 
= 0.83mole to react
But we have been given 0.94mole of O₂. This is more than required.
Therefore O₂ is in excess and Al is the limiting reactant.
Learn more:
Limiting reagents brainly.com/question/6078553
#learnwithBrainly
From the given balanced equation we have find out the amount (in gm) of Ag formed from 5.50 gm of Ag₂O.
2Ag₂O(s) → 4Ag (s) + O₂ (g)
We know, molecular mass of Ag₂O= 231.7 g/mol, and atomic mass of Ag= 107.8 g/mol. Given, mass of Ag₂O=5.50 gm. Number moles of Ag₂O=
= 0.0237 moles.
From the balanced chemical reaction we get 2 (two) moles of Ag₂O produces 4 (four) moles of Ag. So, 0.0237 moles of Ag₂O produces 
moles=0.0474 moles of Ag= 0.0474 X 107.8 g of Ag=5.11g Ag.
Therefore, 5.50 g Ag₂O produces 5.11 g of Ag as per the given balanced chemical reaction.
2.0 L
The key to any dilution calculation is the dilution factor
The dilution factor essentially tells you how concentrated the stock solution was compared with the diluted solution.
In your case, the dilution must take you from a concentrated hydrochloric acid solution of 18.5 M to a diluted solution of 1.5 M, so the dilution factor must be equal to
DF=18.5M1.5M=12.333
So, in order to decrease the concentration of the stock solution by a factor of 12.333, you must increase its volume by a factor of 12.333by adding water.
The volume of the stock solution needed for this dilution will be
DF=VdilutedVstock⇒Vstock=VdilutedDF
Plug in your values to find
Vstock=25.0 L12.333=2.0 L−−−−−
The answer is rounded to two sig figs, the number of significant figures you have for the concentration od the diluted solution.
So, to make 25.0 L of 1.5 M hydrochloric acid solution, take 2.0 L of 18.5 M hydrochloric acid solution and dilute it to a final volume of 25.0 L.
IMPORTANT NOTE! Do not forget that you must always add concentrated acid to water and not the other way around!
In this case, you're working with very concentrated hydrochloric acid, so it would be best to keep the stock solution and the water needed for the dilution in an ice bath before the dilution.
Also, it would be best to perform the dilution in several steps using smaller doses of stock solution. Don't forget to stir as you're adding the acid!
So, to dilute your solution, take several steps to add the concentrated acid solution to enough water to ensure that the final is as close to 25.0 L as possible. If you're still a couple of milliliters short of the target volume, finish the dilution by adding water.
Always remember
Water to concentrated acid →.NO!
Concentrated acid to water →.YES!
