1.

a. Trong 136g dd NaCl bão hoà ở 25 *C thì có 36 g NaCl

b. \(m_{NaCl}=300\cdot\dfrac{36}{136}=79,4g\\ m_{H_2O}=220,6g\)

2.

a. Trong 187,7 g dd CuSO₄ bão hoà ở 25 *C thì có 87,7 g CuSO₄

b. \(m_{CuSO_4}=\dfrac{87,7}{187,7}\cdot1877=877g\\ m_{H_2O}=1000g\)

3.

a. Ở 10 *C, 109 g dd Na₂SO₄ bão hoà có 9 g Na₂SO₄

b. \(m_{Na_2SO_4}=\dfrac{9}{109}\cdot1026,4=84,7g\\ m_{H_2O}=941,7g\)

4.

 \(m_{NaNO_3}=x\\ x:\left(x+110\right)=\dfrac{88}{188}\\ x=96,8g\)

a^3+b^3+c^3=3abc

=>(a+b)^3+c^3-3ab(a+b)-3bac=0

=>(a+b+c)(a^2+2ab+b^2-ac-bc+c^2)-3ab(a+b+c)=0

=>(a+b+c)(a^2+b^2+c^2-ab-ac-bc)=0

=>a^2+b^2+c^2-ab-bc-ac=0

=>2a^2+2b^2+2c^2-2ab-2bc-2ac=0

=>(a-c)^2+(a-b)^2+(b-c)^2=0

=>a=b=c

=>A=(1+b/b)(1+b/b)(1+c/c)

=2*2*2=8

\(n_{NaOH}=0,2\cdot1,2=0,24mol\\ n_{Al\left(OH\right)_3}=\dfrac{6,24}{78}=0,08mol=\dfrac{n_{NaOH}}{3}\Rightarrow V_{min}:vd.AlCl_3\\ V=\dfrac{0,08}{1}=0,08\left(L\right)=80\left(mL\right)\)

\(n_{NaOH}=0,1.3=0,3mol\\ n_{Al\left(OH\right)_3}=\dfrac{7,02}{78}=0,09mol\\ \dfrac{1}{3}n_{NaOH}>n_{Al\left(OH\right)_3}\Rightarrow NaOH\left(dư\right)\)

\(3NaOH+AlCl_3\rightarrow3NaCl+Al\left(OH\right)_3\)

0,27          0,09        0,09             0,27

\(V_{AlCl_3}=\dfrac{0,09}{2}=0,045l\)

NaOH dư thì hoà tan kết tủa tạo Na[Al(OH)4] rồi còn đâu?

\(a)Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2 \uparrow\left(d_{H_2}< d_{kk}\right)\\ b)Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\\ Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+CO_2+H_2O\left(d_{CO_2}>d_{KK}\right)\\ c)CuO+2HCl\rightarrow CuCl_2+H_2O\\ CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ d)Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\\ Fe_2O_3+3H_2SO_4\rightarrow Fe\left(SO_4\right)_3+3H_2O\\ e)Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+H_2O\\ Na_2SO_4+2HCl\rightarrow2NaCl+H_2SO_4\\ f)Ba\left(OH\right)_2+H_2SO_4\rightarrow BaSO_4\downarrow+2H_2O\)

tham khảo

PTHH: CO₂ + Ca(OH)₂ → CaCO₃ + H₂O

Mol:                  0,15             0,15 

Tỉ lệ: ⇒ CO₂ dư,Ca(OH)₂ pứ hết

\(n_{CO_2}=\dfrac{5,6}{22,4}=0,25mol\\ n_{BaCO_3}=\dfrac{20}{197}=0,1mol\\ Vì.n_{CO_2}>n_{BaCO_3}\\ pư.tạo.2.muối\\ CO_2+Ba\left(OH\right)_2\rightarrow BaCO_3+H_2O\)
0,1     0,1                 0,1           0,1

\(CO_2+Ba\left(OH\right)_2\rightarrow Ba\left(HCO_3\right)_2\)

0,15    0,15              0,15 

\(C_{M_X}=C_{M_{Ba\left(OH\right)_2}}=\dfrac{0,1+0,15}{0,15}=\dfrac{5}{3}M\)

\(BTKL:m_R+m_{O_2}=m_{R_2O}\\ \Rightarrow m_{O_2}=12-7,2=4,8g\\ n_{O_2}=\dfrac{4,8}{32}=0,15mol\\ 4R+O_2\xrightarrow[]{t^0}2R_2O\\ n_R=0,15.4=0,6mol\\ M_R=\dfrac{7,2}{0,6}=12\left(g/mol\right)\)

không có kim loại thoả mãn đề bài.

_________

sửa đề: kim loại R có hóa trị Il

\(BTKL:m_{O_2}=12-7,2=4,8g\\ n_{O_2}=\dfrac{4,8}{32}=0,15mol\\ 2R+O_2\xrightarrow[]{t^0}2RO\\ n_R=0,15.2=0,3mol \\ M_R=\dfrac{7,2}{0,3}=24\left(g/mol\right)\\ \Rightarrow R=24=Mg\left(magie\right)\)

\(m_{O_2}=12-7,2=4,8\left(g\right)\)

\(n_{O_2}=\dfrac{4,8}{32}=0,15\left(mol\right)\)

PT :

4R + O2 --> (t^o)2 R2O 

 0,6   0,15           0,3

\(M_R=\dfrac{7,2}{0,6}=12\left(\dfrac{g}{mol}\right)\)

Cacbon nhưng cacbon không có hóa trị 1 nên bạn xem lại đề 

Giả sử có 100 g dung dịch acid.

\(n_{MO}=n_{MSO_4}=n_{H_2SO_4}=\dfrac{a}{98}\left(mol\right)\\ m_{ddsau}=\dfrac{\left(M+16\right)a}{98}+100=\dfrac{\left(M+96\right)a}{98\cdot\dfrac{b}{100}}=\dfrac{a\left(M+96\right)}{0,98b} \)

\(\dfrac{M+16}{98}+100=\dfrac{M+96}{0,98b}\\ M+16+9800=\dfrac{100M+9600}{b}\\ bM+9816=100M+9600\\ M\left(100-b\right)=216\\ M=\dfrac{216}{100-b}\left(g\cdot mol^{-1}\right)\)