Câu 5a): $n_{CuSO_{4}}$= $\frac{m}{M}$= $\frac{16}{160}$=0,1(mol)
200ml=0,2L
$C_{MCuSO_{4}}$= $\frac{n}{V}$= $\frac{0,1}{0,2}$= 0,5(M)
b)
C%=$\frac{m_{H_{2}SO_{4}}}{m_{ddH_{2}SO_{4}}}$.100%= $\frac{m_{H_{2}SO_{4}}}{150}$. 100%= 14%
⇔$m_{H_{2}SO_{4}}$= $\frac{150.14}{100}$=21(g)
Câu 6:
Áp dụng ĐLBTKL ta có:
$m_{R}$+ $m_{O_{2}}$= $m_{RO}$
⇒ 7,2+$m_{O_{2}}$= 12
⇒ $m_{O_{2}}$= 12-7,2=4,8(g)
$n_{O_{2}}$= $\frac{m}{M}$ =$\frac{4,8}{32}$=0,15(mol)
Ta có PTPU:
2$R_{}$ + $O_{2}$ → 2$RO_{}$
$n_{RO}$= $n_{O_{2}}$.2= 0,15.2=0,3(mol)
$m_{RO}$= $n_{RO}$ . $M_{RO}$= 0,3.($M_{R}$+16 )=12
⇒$M_{R}$= $\frac{12-0,3.16}{0,3}$=24(g/mol)
Vậy R là Magiê(Mg)
