What is the mass of potassium iodide (166.00 g/mol) that yields 0.500 g of lead(ii) iodide (461.0 g/mol) precipitate? (you must balance the equation first) __pb(no3)2(aq) + __ki(s) → __pbi2(s) + __kno3(aq) 0.0900 g 2.78 g 0.694 g 0.180 g 0.360 g?

 the  mass  of  potassium iodide  is  0.360  g

calculation

 step 1:  write the balanced molecular  equation

Pb(NO3)2  (aq)  + 2 Ki (s)→ Pbi2(s)  + 2KNO3

step 2 ; calculate the  moles of Pbi2

moles =  mass /molar mass

= 0.500 g / 461 .0  g/mol  =0.0011 moles

step  3:  use the mole ratio to  calculate  the  moles of Ki

Ki: Pbi2  is  2:1  therefore  the  moles of  ki  = 0.0011  x 2/1 =  0.0022  moles

step 4 :  find the mass of Ki

mass=  moles  x molar  mass

=0.0022  moles  x 166g/mol  =0.365  g  which is  approximate  0.360 g

if two electrons have the same n, the same ℓ, and the same ml, then they are in the same orbital. the pauli exclusion principle states that no two identical fermions (a fancy word for electrons and other subatomic particles like electrons) may occupy the same quantum state in an atom simultaneously.

> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.