A 180.0 mL solution of 2.523 M strontium nitrate is mixed with 215.0 mL of a 2.959 M sodium fluoride solution. Calculate the mass of the resulting strontium fluoride precipitate. mass: g Assuming complete precipitation, calculate the final concentration of each ion. If the ion is no longer in solution, enter a 0 for the concentration. [ Na + ] = M [ NO − 3 ] = M [ Sr 2 + ] = M [ F − ] =

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.