Đáp án:
\({d_{nt}} = 1384259,723g/{m^3}\)
Giải thích các bước giải:
Ta có:
\({M_{Zn}} = 65{\text{ gam}} \)
\(\to {{\text{m}}_{1{\text{ nguyên tử Zn}}}} = \frac{{65}}{{6,{{023.10}^{23}}}} = 1,{079.10^{ - 23}}{\text{ gam}}\)
\({r_{nt}} = 1,{23.10^{ - 1}}nm = 1,{23.10^{ - 10}}m \to {V_{nt}} = \frac{4}{3}\pi {r^3} = 7,{79478.10^{ - 30}}{m^3} \)
\(\to {d_{nt}} = \frac{{1,{{079.10}^{ - 23}}}}{{7,{{79478.10}^{ - 30}}{m^3}}} = 1384259,723g/{m^3}\)
\({r_{hn}} = {2.10^{ - 6}}nm = {2.10^{ - 15}}m \to {V_{hn}} = \frac{4}{3}\pi {r_{hn}}^3 = 3,{351.10^{ - 44}}{m^3} \)
\(\to {d_{hn}} = \frac{{1,{{079.10}^{ - 23}}}}{{3,{{351.10}^{ - 44}}}} = 3,{22.10^{20}}gam/{m^3}\)