How many atoms of nitrogen are present in 3.93 g of each of the following?
1.)calcium nitrate
2.) dinitrogen tetroxide
2007-01-21 17:41:26 · 2 answers · asked by Moe 2 in Science & Mathematics ➔ Chemistry
1. Ca(NO3)2 has 2 atoms of nitrogen per formula unit, and a formula mass of (40.1 + 2(14.0 + 3*16.0)) = 164.1. So in 3.93 g there are 3.93 / 164.1 = 0.0239 mol of Ca(NO3)2 and hence 2*(0.0239) = 0.0479 mol of N atoms. The number of atoms is this multiplied by Avogadro's number, i.e. 0.0479 * 6.02 * 10^23 = 2.88 * 10^22 atoms.
2. N2O4 has two nitrogen atoms per molecule, and formula mass 92.0, so in 3.93 g there are 3.93/92.0 = 0.0427 mol of N2O4 and hence 0.0854 mol of N atoms, or 5.14 * 10^22 atoms.
2007-01-21 17:57:47 · answer #1 · answered by Scarlet Manuka 7 · 1⤊ 0⤋
a million. Calculate the molar weight of each and every molecule utilising the periodic table. 2) Calculate how many moles are in 6.37 grams utilising trouble-free branch. 3) Multiply the respond by skill of Avagadro's quantity. 4) completed.
2016-11-26 01:31:23 · answer #2 · answered by riddle 4 · 0⤊ 0⤋