Amixture of 10.0 g of ne and 10.0 g ar have a total pressure of 1.6 atm. what is the partial pressure of ne? actually i know the answer but i didn't know how they get 2/3

The partial pressure contributed to the overall pressure is proportional to the number of moles of gas particles. Both Neon and Argon are noble gases so there's just 1 atom per molecule. So let's determine how many moles of each we have:
 Atomic weight neon = 20.1797
 Atomic weight argon = 39.948

   Moles neon = 10.0 g / 20.1797 g/mol = 0.495547506 mol
 Moles argon = 10.0 g / 39.948 g/mol = 0.250325423 mol

   So there's a total of 0.495547506 + 0.250325423 = 0.745872929 moles of gas particles. And for neon, it contributes 0.495547506/0.745872929 = 0.664385965 = 66.4385965% of them, so it contributes 66.4385965% of the total pressure. Which is 66.4385965% * 1.6 atm = 1.063017544 atm.

Rounding to 3 significant figures gives 1.06 atm, or 66.4% of the total pressure.

The answer should be c.