a) r= 0,135 nm= 0,135 .10^-7 cm

\( V_{NT}= 4/3 \pi r^3= 4/3 \pi (0,135.10^{-7})^3=1,03.10^{-23}\)

\(m_{Zn}= 65.u=65.1,6605 .10^{-24}=1,08.10^{-22}\)

=> \(D_{NT}=\dfrac{{1,08.10^{-22}}}{{1,03.10^{-23}}}=10,47g/cm^3\)

b)

\( V_{NT}= 4/3 \pi r^3= 4/3 \pi (2.10^{-13})^3=3,35.10^{-38}\)

\(m_{Zn}= 65.u=65.1,6605 .10^{-24}=1,08.10^{-22}\)

=> \(D_{NT}=\dfrac{{1,08.10^{-22}}}{{3,35.10^{-38}}}=3,22.10^{15}g/cm^3\)

Đáp án:

Giải thích các bước giải: a, ta có m nguyên tử Zn=65 ×1,6605 ×10^-24 ×6,022 ×10^23=64,997(g) r=1,35.10^-1nm=1,35.10^-8cm V1mol nguyên tử= $\frac{4}{3}$ × $\pi$ ×(1,35.10^-8) ³ ×6,022 ×10^23=6,2cm ³ ⇒d= $\frac{m}{V}$= $\frac{64,997}{6,2}$ =10,48(g/cm ³) b. r=2.10^-6nm=2.10^-13cm m nguyên tử Zn=65 ×1,6605 ×10^-24=1,079 ×10^-22 V hạt nhân nguyên tửZn= $\frac{4}{3}$ × $\pi$ ×(2.10^-13) ³=3,349 ×10^-38 cm ³ ⇒d= $\frac{m}{V}$ = $\frac{1,079.10^-22}{3,349.10^-38}$ =3,22.10^15 (g/cm ³)