You determined that the molar ratio of aluminum to alum to be 1:1If you used 2.345 g aluminum, and the percent yield was 73%, determine the mass of alum produced.
Calculate 73% of 2.345 g0.73 x 2.345 g 1.71185 g This should be rounded to 1.712 g or 1.71 g.
1Go down to row 4 (that would be period 4) and count over 4 elements SKIPPING the 10 transition elements, Sc-Zn2Assuming the atom is neutral the 70 electrons the number of protonsThe atomic number is the number of protons3None of the alkali metals (lithium, sodium, postassium, rubidium, cesium and radium) are named after a country4This one is unclearIf you mean alkali metal in period 3, count down 3 rows and look at the first elementIf you mean alkaline earth metal, count down 3 rows and then count over two elements5The transition elements are the block of elements in the middle of the periodic chartThey are 10 elements across beginning with the element Sc-Zn, then Y-Cd then Lu-Hg and Lr-118Al is not in this group.
1) Thats any period 4 element due to all elements in period 4 having 4 valence electrons 2)Ytterbium (Yb) 3) No idea (I should get back to you on that) 4) Magnesium Mg 5) Aluminium Or you could pick up a periodic table.
