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How many particles are in a 91.0 gram sample of potassium carbonate?

2007-12-15 12:27:41 · 1 answers · asked by Kataya B 1 in Science & Mathematics Chemistry

1 answers

=91g K2CO3 X (1molK2CO3/138.2058g K2CO3) X (6.022*10^23 particles/mol)
=3.97X10^23 particles

*6.022X10^23 = avrogadro's constant

2007-12-15 12:32:09 · answer #1 · answered by ¿ /\/ 馬 ? 7 · 0 0

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