Chemistry

Chemistry

Problem

The number of atoms present in 4.25 grams of NH3NH_3 is approximately 1) 1×10231 \times 10^{23} 2) 1.5×10231.5 \times 10^{23} 3) 2×10232 \times 10^{23} 4) 6×10236 \times 10^{23}

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