a.

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

b.

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

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

M_A = 1,52.29 = 44(g/mol)

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

\(M_A=1.52\cdot M_{kk}=1.52\cdot29=44\left(g\text{/}mol\right)\)

$M_A = 1,517.29 = 44(đvC)$

$M_B = 1,0625.M_{O_2} = 1,0625.32 = 349(đvC)$

bằng 2,746 nhé

\(M_A=2,207.29=64\left(\dfrac{g}{mol}\right)\)

\(d\dfrac{_A}{kk}=2,207\Rightarrow M_A=2,207.29\approx64\left(\dfrac{g}{mol}\right)\)

\(a,M_A=15.2=30\left(\dfrac{g}{mol}\right)\\ b,\text{Đ}\text{ặt}:C_xH_y\left(x,y:nguy\text{ê}n,d\text{ươ}ng\right)\\ M_A=12x+y\\ Th\text{ế}.d\text{ần},ch\text{ố}t:x=2;y=6\\ \Rightarrow A:C_2H_6\\ c,n_A=\dfrac{6}{30}=0,2\left(mol\right)\\ V_{A\left(\text{đ}ktc\right)}=0,2.22,4=4,48\left(l\right)\)

a) \(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)

b) \(V_{CO_2}=0,2.22,4=4,48\left(l\right)\)

c) \(M_A=1,172.29=34\left(g/mol\right)\)

\(n_A=\dfrac{33,6}{22,4}=1,5\left(mol\right)\)

=> m_A = 1,5.34 = 51(g)