Solution:
ν(H)=4×ν(C4H4S2)=4×4.15×10−23 mol=16.6×10−23 mol\nu(H)=4\times\nu(C_4H_4S_2)=4\times4.15\times10^{-23}\,mol=16.6\times10^{-23}\,molν(H)=4×ν(C4H4S2)=4×4.15×10−23mol=16.6×10−23mol
N(H)=ν(H)×NA=16.6×10−23 mol ×6.02×1023mol−1≈100N(H)=\nu(H)\times N_A=16.6\times10^{-23}\,mol\,\,\times6.02\times10^{23}{mol^{-1}}\approx100N(H)=ν(H)×NA=16.6×10−23mol×6.02×1023mol−1≈100
Answer:
100 atoms
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