If I start with 0.0350 L of a 12.0 M hydrochloric acid solution, and dilute it with water to a new volume of 1.20 L, what is the new concentration?

0.35 M

Explanation:

Using the formula; C1V1 = C2V2

Where;

C1 = concentration of initial solution

C2 = concentration of final solution

V1 = volume of initial solution

V2 = volume of final solution

Based on this question, I start with 0.0350 L of a 12.0 M hydrochloric acid solution, and dilute it with water to a new volume of 1.20 L. This means that:

C1 = 12.0 M, V1 = 0.0350 L, V2 = 1.20 L, C2 = ?

Hence, C1V1 = C2V2

12 × 0.0350 = C2 × 1.20

0.42 = 1.2C2

C2 = 0.42/1.2

C2 = 0.35 M

the grams   of   barium that   are in   2.3 x 10^23   atoms of barium   is   52.4 g (answer c)

    calculation

mass = moles   of ba   x molar   mass of ba

  molar mass of ba from periodic table=   137.3 g/mol

  use the   avogadro's law   constant   to   calculate the moles of barium

  that   is     1 mole =   6.02   x 10^23 atoms

                    ? moles = 2.3 x 10^23   atoms

by cross   multiplication

=(1 mole   x 2.3x 10^23 atoms)   / 6.02   x10^23 atoms)=   0.382   moles

mass   is therefore = 137.3   g/mol x 0.382 mol = 52.4 ( answer c)

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"