What mass of zinc nitrate is needed to prepare 2563 mL of a 0.450 M solution? (molar mass = 189.40 g/mol) a 2.18 x 105 grams

b 33.3 grams

c 218 grams

d 6.09 grams

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer

they are all solids.

b) nuclear ive learned that in a pedtrodic table