(a) How many moles of ammonium ions are in 0.339 g of ammoniumcarbonate? _____ m

Question

(a) How many moles of ammonium ions are in 0.339 g of ammoniumcarbonate?      _____ mol (b) What is the mass, in grams, of 0.0587 mol of iron(III)phosphate?      _____ g (c) What is the mass, in grams, of 2.22 x 1023molecules of aspirin,C9H8O4?      _____ g (d) What is the molar mass of diazepam (Valium) if 0.05570 molweighs 15.86 g?      _____ g/mol (a) How many moles of ammonium ions are in 0.339 g of ammoniumcarbonate?      _____ mol (b) What is the mass, in grams, of 0.0587 mol of iron(III)phosphate?      _____ g (c) What is the mass, in grams, of 2.22 x 1023molecules of aspirin,C9H8O4?      _____ g (d) What is the molar mass of diazepam (Valium) if 0.05570 molweighs 15.86 g?      _____ g/mol

Explanation / Answer

This is an issue of significant figures. Sorry about that. a) 0.339 g (NH4)2CO3 * (1mol /96.1 g) *(2 NH+ / 1 (NH4)2CO3) = =0.00705588511 ~ 0.00706 (don't use scientific notation it seems) d) (15.86/0.05570) = 284.739677 ~ 284.7 g/mol (4 sig figs)

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