Chapter 6, Problem 6.40EP

(a)

Interpretation Introduction

Interpretation:

Among $2.00mole$ of $Ne$ and $6.02x{10}^{23}$ atoms of $Ar$, the one that contains greater number of atoms has to be selected using periodic table and not by performing calculations.

(b)

Interpretation Introduction

Interpretation:

Among $10.00g$ of $O_2$ and $10.00g$ of $N_2$, the one that contains greater number of atoms has to be selected using periodic table and not by performing calculations.

(c)

Interpretation Introduction

Interpretation:

Among $1.00mole$ of $Cu$ and $1.00mole$ of $Ni$, the one that contains greater number of atoms has to be selected using periodic table and not by performing calculations.

(d)

Interpretation Introduction

Interpretation:

Among $26.98g$ of $Al$ and $6.02x{10}^{22}$ atoms of $Al$, the one that contains greater number of atoms has to be selected using periodic table and not by performing calculations.