ayoismeisjjjjuan
17.12.2021 •
Chemistry
VERY URGENT! IB CHEMISTRY
0.0810 g of a group 2 metal iodide, MI2, was dissolved in water and made up to a total volume of 25.00 cm3.
Excess lead(II) nitrate solution (Pb(NO3)2(aq)) was added to the MI2 solution to form a precipitate of lead(II)
iodide (PbI2). The precipitate was dried and weighed and it was found that 0.1270 g of precipitate was obtained.
a Determine the number of moles of lead iodide formed.
b Write an equation for the reaction that occurs.
c Determine the number of moles of MI2 that reacted.
d Determine the identity of the metal, M.
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Ответ:
Socratic app it will be over in the bank