Answer:
219.78 mL of the stock solution are needed
Explanation:
First, we take a look at the desired Al2(SO4)3 working solution. We are told that the we need 400 mL of an aqueous aluminum sulfate solution 1.0 M. Let's see how many moles of the compound we have in the desired volume:
1000 mL Al2(SO4)3 solution ----- 1 mole of Al2(SO4)3
400 mL Al2(SO4)3 solution ----- x = 0.4 moles of Al2(SO4)3
To reach the desired concentration in the working solution we need 0.4 moles of Al2(SO4)3 in 400 mL, so we calculate the volume of the stock solution needed to prepare the working solution:
1.82 moles Al2(SO4)3/L = 1.82 M → This is the molar concentration of the stock solution.
1.82 moles of Al2(SO4)3 ----- 1000 mL
0.4 moles of Al2(SO4)3 ----- x = 219.78 mL
So, if we take 219.78 mL of the 1.82 M stock solution, we put it in a graduated cylinder and we dilute it to 400 mL, we would obtain a 1.0 M Al2(SO4)3 solution.
A chemical reaction happens if you mix together an acid and a base. The reaction is called neutralisation<span>, and a neutral solution is made if you add just the right amount of acid and base together.</span>
<span>pH = pKa + log ([base]/[acid]) = -log (3.5 x 10^-4) + log (0.040/.020) = 3.46 - 0.30 = 3.16</span>
C. If the banks of the river are eroding, then the river will be wider.
V) the amount of sugar in the solution
C) whether the sugar is stirred
V) the temperature of the solution
C) the type of solute added
C) the type of solvent used
NP.
