M_A = 1,52.29 = 44(g/mol)

\(M_A=29\cdot1.52=44\left(g\text{/}mol\right)\)

BT1 : 

Ta có : \(d_{\dfrac{A}{B}}=\dfrac{M_A}{M_B}=1,8\)

\(\rightarrow M_A=1,8.30=54\) 

BT2 : 

Ta có : \(d_{\dfrac{Y}{SO2}}=\dfrac{M_Y}{M_{SO2}}=0,5\rightarrow M_Y=0,5.64=32\)

mà \(d_{\dfrac{X}{Y}}=\dfrac{M_X}{M_Y}=1,5\rightarrow M_X=1,5.32=48\)

a.

\(d_{A\text{/}O_2}=1.375\) \(\Rightarrow M_A=1.375\cdot32=44\left(g\text{/}mol\right)\)

b.

\(d_{B\text{/}kk}=0.586\) \(\Rightarrow M_B=0.586\cdot29=17\left(g\text{/}mol\right)\)

\(a,M_A=M_{H_2}.16=2.16=32(g/mol)\\ b,M_A=M_{H_2}.32=2.32=64(g/mol)\\ c,M_A=M_{H_2}.8=2.8=16(g/mol)\)

a,b,c,

Vì MH =1

\(\left\{{}\begin{matrix}MA=16\\MA=32\\MA=8\end{matrix}\right.\)

Từ đề có: \(M_A=1,518.29=44\) g/mol

A là khí \(CO_2\)

a)

$M_{khí} = M_{N_2}.2 = 28.2 = 56(g/mol)$
b)

$M_{khí\ 1} = M_{không\ khí}.2,45 = 29.2,45 = 71,05(g/mol)$
$M_{khí\ 2} = M_{không\ khí}.0,965 = 29.0,965 = 27,985(g/mol)$

Gọi CTTQ là : XO₃ 

\(a,\rightarrow M_A=80\)

\(\rightarrow M_X=80-\left(16.3\right)=32\left(\dfrac{g}{mol}\right)\)

Vậy X là S ( lưu huỳnh )

\(\%m_{S\left(SO_3\right)}=\dfrac{32}{80}.100\%=40\%\)

\(a.d_{\dfrac{X_2}{O_2}}=2,218\\ m_{O_2}=32\left(\dfrac{g}{mol}\right)\\ \Rightarrow M_{X_2}=32.2,218=71\left(\dfrac{g}{mol}\right)\\ \Rightarrow M_X=\dfrac{71}{2}=35,5\left(\dfrac{g}{mol}\right)\\ \Rightarrow X.là.Clo\)

\(b.d_{\dfrac{Cl_2}{kk}}=\dfrac{71}{29}\approx2,4>1\\ \Rightarrow thu.bằng.pp.đặt.bình.đứng\)

a) \(M_A=2\cdot32=64\)

b) \(M_B=0,5\cdot32=16\)

a) M_A = 2 . 32 = 64g/mol

b) M_B = 0,5 . 32 = 16g/mol