a) \(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\)

b) \(n_{CuO}=\dfrac{1,5.10^{23}}{6.10^{23}}=0,25\left(mol\right)\)

c) \(n_{CH_4}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

\(a.n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\)

\(b.n_{CuO}=\dfrac{1,5.10^{23}}{6.10^{23}}=0,25\left(mol\right)\)

\(c.n_{CH_4}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

\(a.n_{CO_2}=\dfrac{V}{22,4}=\dfrac{5.6}{22,4}=0,25\left(mol\right)\\ \Rightarrow m_{CO_2}=n.M=0,25.44\)

b. Số phân tử là: \(0,25.6.10^{23}=15.10^{22}\) (phân tử)

a,khối lượng mol của CO₂ là 11 gam 

b,số phân tử của CO₂ là 1,5.10²³

a,

\(nCO_2=5,6:22,4=0,25mol\\ mCO_2=0,25.44=11g\)

b,

số phtu CO2 là:

\(0,25.6.10^{23}=1,5.10^{23}\)

- \(n_{Fe}=\dfrac{m}{M}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

- \(n_{H_2}=\dfrac{V_{\left(\text{đ}ktc\right)}}{22,4}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

- \(n_{Cu}=\dfrac{3,0115.10^{23}}{6.10^{23}}=0,5\left(mol\right)\)

- K₂CO₃ \(\left\{{}\begin{matrix}\%K=\dfrac{78}{138}.100\%=56,52\%\\\%C=\dfrac{12}{138}.100\%=8,7\%\\\%O=100\%-56,52\%-8,7\%=34,78\%\end{matrix}\right.\)

a) \(n_{O_2}=\dfrac{9,6}{32}=0,3\left(mol\right)\)

PTHH: 2KClO₃ --t^o--> 2KCl + 3O₂

__________________0,2<---0,3

=> m_KCl = 0,2.74,5 = 14,9 (g)

b) 2KClO₃ --t^o--> 2KCl + 3O₂

Tỉ lệ: 2:2:3

Số phân tử MgSO₄ = Số phân tử H₂S

=> \(n_{MgSO_4}=n_{H_2S}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ m_{MgSO_4}=0,25.120=30\left(g\right)\)

Ta có 

N = 1500 = 2A + 2G

=> A+G = 750

A = 2G

=> 3G = 750

=> G = 250 nu = X

=> A = 2 x 250 = 500 nu = T

b, 

L = \(\dfrac{1500}{2}\) x 3,4 = 2550

\(n_{Al_2O_3}=\dfrac{0,9.10^{23}}{6.10^{23}}=0,15\left(mol\right)\\ m_{Al_2O_3}=102.0,15=15,3\left(g\right)\\ n_{O_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\\ m_{O_2}=0,25.32=8\left(g\right)\)

\(a,n_{CaO}=\dfrac{14}{40}=0,35(mol)\\ b,n_{C}=\dfrac{3.10^{-23}}{6.10^{-23}}=0,5(mol)\\ c,n_{H_2O}=\dfrac{9.10^{-23}}{6.10^{-23}}=1,5(mol)\\ d,n_{O_2}=\dfrac{16}{32}=0,5(mol)\)