PLEASE HELP!!
The unknown concentration of NaOH was determined by titration with HCl. The following data was collected: mL NaOH = 34.8 mL mL of HCl used = 29.6 mL concentration of HCl = 0.19 M What is the concentration of the NaOH? 0.16 M 0.22 M 4.5 M 6.3 M

0.16 M

Explanation:

We are given;

We are required to determine the concentration of the NaOH

We are going to write the equation of the reaction first;

NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)

We can then determine the number of moles HCl

Moles = Molarity × volume

          = 0.19 M × 0.0296 L

          = 0.005624 moles

We the determine the number of moles of NaOH

From the equation;

1 mole of NaOH reacts with 1 mole of HCl

Therefore;

Moles of NaOH = moles of HCl

                     = 0.005624 moles

Finally, we determine the concentration of NaOH

Concentration/Molarity = Number of moles ÷ Volume

                                      = 0.005624 moles ÷ 0.0348 L

                                      = 0.1616 M

                                      = 0.16 M

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"