If you have 1.2754x1020 molecules of Barium nitrate (Ba(NO3)2), how many moles of barium nitrate do you have? ??? moles Ba(NO3)2 (round to 5 sig figs)

0.00021186 moles of barium nitrate

Explanation:

The given parameter is as follows;

The number of molecules of Barium nitrate Ba(NO₃)₂ = 1.2754 × 10²⁰ molecules

The number of molecule present per mole of Barium nitrate, is given by Avogadro's number, 6.02 × 10²³

1 mole of a substance contains 6.02 × 10²³ molecules

The umber of moles in a substance = (The number of molecules of the substance)/(The number of molecules per mole)

Therefore the number of moles of barium nitrate in 1.2754 × 10²⁰ molecules of barium nitrate = (1.2754 × 10²⁰)/(6.02 × 10²³) ≈ 0.00021186 moles (rounding to 5 significant figures)

The number of moles of barium nitrate present in 1.2754 × 10²⁰ molecules of barium nitrate ≈ 0.00021186 moles

d.) the average kinetic energy of the gas particles is greater in container b because its particles move faster.

the answer is 0.42 is that is an option. this question provides no info