Question #106627
How many moles of Al2(CO3)3 contain 2.63x1025 atoms of oxygen (O)?
1
Expert's answer
2020-03-26T03:02:50-0400

n=2.63×10256.022×1023=0.44×102n=\frac {2.63\times10^25}{6.022\times10^23}=0.44\times10^2


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