Number of O atoms in 7.8 x 10^-3 g of CaSO4 · 2 H2O

The number of O atoms in 7.3 x10 ^-3 g of CaSO4 x 2H2O  are

= 1.535 x10^20  atoms

calculation

step 1: calculate the moles of CaSO4 . 2H2O

moles = mass/molar mass

molar mass is of CaSO4.H2O=Mw of Ca + S +O +H2O

= 40+32 +( 4x16)  +2( (1x2)+16)=  172 g/mol

moles is therefore= 7.3 x10^-3 g/172 g/mol=   4.244  x10^-5 moles

find the moles of O

since there are 6 atoms of O the moles of O=   6 x ( 4.244 x10^-5)= 0.000255 moles

by use  of Avogadro's law constant find the number of O atom

that is  1 mole     =       6.02 x10 ^23 atoms

                     0.000255 moles=? atoms

by cross multiplication

= 0.000255 moles  x(6.02 x10^23 atoms)/ 1 mole=  1.535  x10^20 atoms

i had this on my unit activity, i can you if you are a plato student. are you a plato student?

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.