Hi!
To make 500 mL of a 1,500 M solution of NaCl you'll require
43,83 g
To calculate that, you will need to use a conversion factor to go from the volume of the 1,500 M solution to the required grams. For this conversion factor, you'll use the definition for Molar concentration (M=mol/L) and the molar mass of NaCl. The conversion factor is shown below:

Have a nice day!
Answer:
(a) 
(b) 
(c) 
(d) 
Explanation:
Hello,
In this case, we define the pH in terms of the concentration of hydronium ions as:
![pH=-log([H^+])](https://tex.z-dn.net/?f=pH%3D-log%28%5BH%5E%2B%5D%29)
Which is directly computed for the strong hydrochloric acid (consider a complete dissociation which means the concentration of hydronium equals the concentration of acid) in (a) and (c) as shown below:
(a)
![[H^+]=[HCl]=0.1M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%5BHCl%5D%3D0.1M)
(b)
![[H^+]=[HCl]=0.05M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D%5BHCl%5D%3D0.05M)

Nevertheless, for the strong sodium hydroxide, we don't directly compute the pH but the pOH since the concentration of base equals the concentration hydroxyl in the solution:
![[OH^-]=[NaOH]](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%5BNaOH%5D)
![pOH=-log([OH^-])](https://tex.z-dn.net/?f=pOH%3D-log%28%5BOH%5E-%5D%29)

Thus, we have:
(b)

(d)

Best regards.
The three substances add up to 105% instead of 100%. Also, <span>the amount of either one of those compounds is more than according to the Stoichiometry of the reaction, meaning one of those compounds is in excess. </span>
Answer:
Increases I had this question