how many grams of iron(ii) sulfate heptahydrate is needed to prepare 250.0 ml of a 3.25 molar aqueous solution

The mass of FeSO₄.7H₂O  needed  225.89 grams

Explanation:

From the given question

The molecular weight of iron(ii) sulfate heptahydrate FeSO₄.7H₂O = 278.02 g/mol

Molarity = 3.25

Volume = 250.0 mL

The mass of FeSO₄.7H₂O  needed =

The mass of FeSO₄.7H₂O  needed  =

The mass of FeSO₄.7H₂O  needed  = 225.89125 grams

The mass of FeSO₄.7H₂O  needed  225.89 grams

i think its 100

i hope this !

here is table 2-2:

country

percent of eligible voters that voted

rank

united states

44.9%

138

canada

60.1%

107

costa rica

84.6%

14

hungary

66.9%

82

japan

57.0%

118

egypt

27.7%

159

south africa

85.5%

10

ans is no, the data set cannot be plotted as a single circle graph