A49.48-ml sample of an ammonia solution is analyzed by titration with hcl. it took 38.73 ml of 0.0952 m hcl to titrate the ammonia. what is the concentration of the original ammonia solution?

M₂ = 0.0745 M

Explanation:

In case of titration , the following formula can be used -

M₁V₁ = M₂V₂

where ,

M₁ = concentration of acid ,

V₁ = volume of acid ,

M₂ = concentration of base,

V₂ = volume of base .

from , the question ,

M₁ = 0.0952 M

V₁ = 38.73 mL

M₂ = ?

V₂ = 49.48 mL

Using the above formula , the molarity of ammonia , can be calculated as ,

M₁V₁ = M₂V₂  

0.0952 M * 38.73 mL = M₂* 49.48 mL

M₂ = 0.0745 M

each  column  of the  table  is called a group. elements in a group share similar properties. groups are  read  from  top  to  bottom. metal. metalloid. nonmetal. solid. liquid. gas. fe. hg. o.  2. 3. 4. 5. 6. 7. 8. 9.  2. beryllium. 9.012. 4.  sodium. 22.990.  11. 3.  magnesium. 24.305.  12.  potassium. 39.098.  19. 4.  calcium. 40.078.  20. i guess  i  

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.