H2O2(I)
C6H6(O)
CO2(I)
C2H6(O)
HNO3(I)
<span><span>S is for soil,</span><span>cl (sometimes c) represents climate,</span><span>o organisms including humans,</span><span>r relief,</span><span>p parent material, or lithology, and</span><span>t time.</span></span>
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!
The balanced equation for the above reaction is;
CH₄ + 2O₂ ---> CO₂ + 2H₂O
Stoichiometry of CH₄ to O₂ is 1:2
The number of methane moles present - 1.44 g/ 16 g/mol = 0.090 mol
Number of oxygen moles present - 9.5 g/ 32 g/mol = 0.30 mol
If methane is the limiting reagent,
0.090 moles of methane react with 0.090x 2 = 0.180 mol
only 0.180 mol of O₂ is required but 0.30 mol of O₂ has been provided therefore O₂ is in excess and CH₄ is the limiting reactant.
Number of moles of water that can be produced - 0.180 mol
Therefore mass of water produced - 0.180 x 18 g/mol = 3.24 g
Therefore mass of 3.24 g of water can be produced
Answer:
Molality, Solvent, Solute, Mole fraction, Molarity.
Explanation:
The expression of concentration that provides the moles of solute per kilograms of solvent is Molality. This in the only expression referred to the solvent.
A solution is made up of 0.15 grams of sodium chloride in 1 liter of water. For this solution, the Solvent is water. When water is present, it is usually considered the solvent.
A solution is made up of 0.15 grams of sodium chloride in 1 liter of water. For this solution, the Solute is sodium chloride. There can be 1 or more solutes in a solution.
If you place 5 moles of sodium chloride and 4 moles of sucrose into 11 moles of water, the Mole fraction of sodium chloride would be 0.25. The mole fraction is equal to the moles of a substance divided by the total number of moles.
A way to express concentration that provides the moles of solute per liter of solution is Molarity.
