$a,$

$Cu(OH)_2:$ Đồng (II) hidroxit

$NaOH:$ Natri hidroxit

$NaHCO_3:$ Natri hidrocacbonat

$HCl:$ Axit clohidric

$Fe(NO_3)_2:$ Sắt (II) nitrat

$b,$

 Đổi $200ml=0,2l;\ 32g=0,032kg$

$→CM_{KCl}=\dfrac{n}{V}=\dfrac{n}{V}=\dfrac{1}{0,2}5M$

$→C\%_{NaNO_3}=\dfrac{m_{ct}}{m_{dd}}×100\%=\dfrac{0,032}{2}×100\%=1,6\%$

Đáp án:

a) $Cu(OH)_{2}$: Đồng (II) hiđroxit

$NaOH$: Natri hiđroxit

$NaHCO_{3}$: Natri hiđrocacbonat

$HCl$: Axit clohiđric

$Fe(NO_{3})_{2}$: Sắt (II) nitrat

b) $C_{M_{KCl}}=\dfrac{n}{V}=\dfrac{1}{0,2}= 5 (M)$

Đổi: $2kg = 2000g$

$\text{C%$NaNO_{3}=\dfrac{m_{ct}}{m_{dd}}$ × 100%=$\dfrac{32}{2000}$ × 100%= 1,6%}$