What is the molarity of a solution containing 49.0 g of sulfuric acid, H2SO4, in 3.00 liters of solution? Answer with 3 sig figs.

Molarity = moles of solute/volume of solution in liters.

Our solute here is sulfuric acid, of which we have 49.0 g. The number of moles of sulfuric acid is (49.0 g)/(98.079 g/mol) = 0.4996 mol H2SO4. To obtain the molarity of this solution of sulfuric acid, we divide 0.4996 mol H2SO4 by 3.00 L, which gives us 0.167 M H2SO4 (answer given to three sig figs).

The kinetic energy of the liquid increases. hope this !

Answer : option a) 84explanation :

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.