Bài 1:
a,
$n_{Fe}=\dfrac{1000}{56}=\dfrac{125}{7}(mol)$
$\to n_e=26n_{Fe}=\dfrac{3250}{7}(mol)$
Số hạt electron: $N_e=\dfrac{3250}{7}.6,022.10^{23}=2,7959.10^{26}$
$\to m_e=9,1.10^{-28}N_e=0,254(kg)$
b,
$m_e=1000g$
$\to$ số electron $=\dfrac{1000}{9,1.10^{-31}}=1,0989.10^{33}$
$\to n_e=\dfrac{1,0989.10^{33}}{6,022.10^{23}}=\text{ 1 824 809 304} (mol)=1824809,304(kmol)$
$\to n_{Fe}=\dfrac{n_e}{26}=70185(kmol)$
$\to m_{Fe}=3930360(kg)$
Bài 2:
a,
$n_{Na}=\dfrac{69}{23}=3(mol)$
$\to n_p=3.11=33(mol)$
Số hạt proton: $33.6,022.10^{23}=1,98726.10^{25}$
$\to m_p=1,98726.10^{25}.1,67.10^{-24}=33,187242g$
b,
$m_e=1000g$
$\to$ số electron $=\dfrac{1000}{9,1.10^{-31}}=1,0989.10^{33}$
$\to n_e=\dfrac{1,0989.10^{33}}{6,022.10^{23}}=\text{ 1 824 809 304}(mol)=1824809,304(kmol)$
$\to n_{Na}=\dfrac{n_e}{11}=165891,73(kmol)$
$\to m_{Na}=3815509,8kg$