Need the remaining questions answered thank you

Explanation:

14) We need to find the # of moles of NaOH in 3.5 L solution. Note: 1 M = 1 mol/Ll

(3.00 mol NaOH/1 L)(3.5 L) = 10.5 mol NaOH

Next we need find the volume of 19.4M NaOH solution that contains 10.5 mol NaOH:

(10.5 mol NaOH)×(1 L NaOH/19.4 mol NaOH)

= 0.541 L NaOH

This means that 0.541 L (541 mL) of 19.4M NaOH solution must be diluted to 3.5 L in order to get 3.00M NaOH solution.

15) 425 mL (0.425 L) of 0.105M HCl solution contains

(0.105 mol HCl/1 L)×(0.425 L) = 0.0446 mol HCl

so when enough water is added to 1L, the molarity becomes

0.0446 mol HCl/1 L = 0.0446M HCl

16) CaBr2 ---> Ca^+2 + 2Br^-1

For every Ca^+2 ion in the solution, you have 2 Br^-1 ions

If the concentration of CaBr2 is 6 mol/L (or 6M),.then the concentration of Br^-1 irons is

[Br^-1] = 2×[Ca^+2] = 2×(6 mol/L) = 12 mol Br^-1/L

17) The reason you won't need 1.0 L of water is that part of the volume of the solution is taken up by the NaOH so you will always need slightly less than 1 L.