Đáp án:

\(a)\ A_{O_3}=1,2\times 10^{23}\)(phân tử)

\(A_O=3,6\times 10^{23}\)(nguyên tử)

\(b) \ m_{C_{12}H_{22}O_{11}}=513\ \text{gam}\\ m_{CuSO_4}=240\ \text{gam}\\ m_{AlPO_4}=183\ \text{gam} \\ m_{Ca{(HCO_3)}_2}=243\ \text{gam}\\ c) \ n_{H_2O}≈44,44\ \text{mol}\)

Giải thích các bước giải:

\(a)\ n_{O_3}=\dfrac{9,6}{48}=0,2\ \text{mol}\)

\(\to A_{O_3}=0,2\times 6\times 10^{23}=1,2\times 10^{23}\)(phân tử)

\(A_O=0,2\times 3\times 6\times 10^{23}=3,6\times 10^{23}\)(nguyên tử)

\(b)\ n=\dfrac{9\times 10^{23}}{6\times 10^{23}}=1,5\ \text {mol}\)

\(m_{C_{12}H_{22}O_{11}}=1,5\times 342=513\ \text{gam}\)

\(m_{CuSO_4}=160\times 1,5=240\ \text{gam}\)

\(m_{AlPO_4}=1,5\times 122=183\ \text{gam}\\ m_{Ca{(HCO_3)}_2}=1,5\times 162=243\ \text{gam}\)

\(c)\ n_{N_2}=\dfrac{28}{28}=1\ \text{mol}\\ n_{NO}=\dfrac {15}{30}=0,5\ \text{mol}\to n_{hh}=1+0,5=1,5\ \text{mol}\to V_{hh}=1,5\times 22,4=33,6\ \text{lít}\)

\(d)\ n_{H_2O}=\dfrac{800}{18} ≈44,44\ \text{mol}\)