3. If 45.0 mL of 0.560 M KOH are needed to neutralize 125 mL of HCl of unknown concentration, what is the molarity and the grams/Liter of HCI?

g/L=
Molarity=

Molarity = 0.202 M

7.36 g/L HCl

Explanation:

KOH                        +                  HCl > KCl  +  H2O

1 mol                                           1 mol

0.560mol/L *0.045L                 x mol/L*0.125L

0.560mol/L *0.045L    =   x mol/L*0.125L

x = 0.560mol/L *0.045L  /0.125L= 0.2016 mol/L≈ 0.202 mol/L=0.202M HCl

M(HCl) = 1.0+35.5 = 36.5 g/mol

0.2016 mol/L*36.5 g/mol ≈ 7.36 g/L HCl

r just means there's a number of carbon in teh chain. for example rcooh means that after x number of carbon, the chain ends with cooh.

explanation:

"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"