\(m_{hh}=m_{O_2}+m_{N_2}=0.05\cdot32+0.15\cdot28=5.8\left(g\right)\)

\(\overline{M}_X=\dfrac{5.8}{0.05+0.15}=29\left(\dfrac{g}{mol}\right)\)

\(d_{X\text{/}H_2}=\dfrac{29}{2}=14.5\)

\(a,M_R=\dfrac{6}{0,15}=40\left(g/mol\right)\\ b,M_A=\dfrac{m_A}{n_A}=\dfrac{7}{\dfrac{5,6}{22,4}}=\dfrac{7}{0,25}=28\left(g/mol\right)\\ c,\overline{M_{hh}}=\dfrac{4\cdot28+1\cdot32}{4+1}=\dfrac{144}{5}=28,8\left(g/mol\right)\)

\(m_{O2}=n_{O2}.M_{O2}=0,1.32=3,2\left(g\right)\)

\(m_{H2}=n_{H2}.M_{H2}=0,15.2=0,3\left(g\right)\)

\(m_{SO2}=n_{SO2}.M_{SO2}=0,25.64=16\left(g\right)\)

\(m_{hh}=m_{O2}+m_{H2}+m_{SO2}=19,2\left(g\right)\)

P/s: Ko biết lm thế này có đc khum nhỉ ??

\(a.\)

\(n_{hh}=0.2+0.15+0.1=0.45\left(mol\right)\)

\(V_X=0.45\cdot22.4=10.08\left(l\right)\)

\(b.\)

\(m_X=0.2\cdot28+0.15\cdot71+0.1\cdot32=19.45\left(g\right)\)

\(c.\)

\(\overline{M}_X=\dfrac{19.45}{0.45}=43.22\left(g\text{/}mol\right)\)

\(d.\)

\(d_{X\text{/}kk}=\dfrac{43.22}{29}=1.4\)

Nặng hơn không khí 1.4 lần

a) \(n_{CO_2}=\dfrac{4,4}{44}=0,1\left(mol\right)=>V_{CO_2}=0,1.22,4=2,24\left(l\right)\)

\(n_{O_2}=\dfrac{3,2}{32}=0,1\left(mol\right)=>V_{O_2}=0,1.22,4=2,24\left(l\right)\)

b) \(n_{H_2O}=\dfrac{3.10^{23}}{6.10^{23}}=0,5\left(mol\right)\)

=> m_H2O = 0,5.18 = 9(g)

c) \(n_{Mg}=\dfrac{12}{24}=0,5\left(mol\right)\)

=> Số nguyên tử Mg = 0,5.6.10²³ = 3.10²³

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

=> Số nguyên tử Zn = 0,2.6.10²³ = 1,2.10²³

Số nguyên tử Ag = 0,15.6.10²³ = 0,9.10²³

Số nguyên tử Al = 0,45.6.10²³ = 2,7.10²³

m h h = m C O 2 + m O 2 = (0,3.44) + (0,15.32) = 18(g)

→ V C O  = 22,4. n C O 2 + n O 2 = 10,08(l)