a) \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(b-c\right)\left(c-a\right)}+\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)

\(=\dfrac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)

b) \(\dfrac{\left(a^2-\left(b+c\right)^2\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(a^2+c^2-2ac-b^2\right)}\)

\(=\dfrac{\left(a-b-c\right)\left(a+b+c\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(\left(a-c\right)^2-b^2\right)}\)

\(=\dfrac{\left(a-c-b\right)\left(a-c+b\right)}{\left(a-c-b\right)\left(a-c+b\right)}=1\)

c) \(\dfrac{x-1}{x^3}-\dfrac{x+1}{x^3-x^2}+\dfrac{3}{x^3-2x^2+x}\)

\(=\dfrac{x-1}{x^3}-\dfrac{x+1}{x^2\left(x-1\right)}+\dfrac{3}{x\left(x-1\right)^2}\)

\(=\dfrac{\left(x-1\right)^3-x\left(x+1\right)\left(x-1\right)+3x^2}{x^3\left(x-1\right)^2}\)

\(=\dfrac{x^3-3x^2+3x-1-x^3+x+3x^2}{x^3\left(x-1\right)^2}\)

\(=\dfrac{4x-1}{x^3\left(x-1\right)^2}\)

d) \(\left(\dfrac{x^2-y^2}{xy}-\dfrac{1}{x+y}\left(\dfrac{x^2}{y}-\dfrac{y^2}{x}\right)\right):\dfrac{x-y}{x}\)

\(=\left(\dfrac{\left(x-y\right)\left(x+y\right)}{xy}-\dfrac{1}{x+y}.\dfrac{x^3-y^3}{xy}\right):\dfrac{x-y}{x}\)

\(=\left(\dfrac{\left(x-y\right)\left(x+y\right)}{xy}-\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{xy\left(x+y\right)}\right):\dfrac{x-y}{x}\)

\(=\dfrac{\left(x-y\right)\left(x^2+2xy+y^2-x^2-xy-y^2\right)}{xy\left(x+y\right)}.\dfrac{x}{x-y}\)

\(=\dfrac{x}{x+y}\)

Gọi x, y lần lượt là số mol của Mg, Al

\(Mg+\dfrac{1}{2}O_2\underrightarrow{t^o}MgO\) (1)

x ----> 0,1---------> x

\(2Al+\dfrac{3}{2}O_2\underrightarrow{t^o}Al_2O_3\) (2)

y -----> 0,3-----> 0,5y

(1)(2) \(\Rightarrow\left\{{}\begin{matrix}24x+27y=15,6\\40x+0,5y.102=28,4\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=0,2\\y=0,4\end{matrix}\right.\)

\(\%Mg=\dfrac{0,2.24.100}{15,6}=30,77\%\)

=> \(\%Al=100-30,77=69,23\%\)

\(V_{O_2}=\left(0,1+0,3\right).22,4=8,96\left(l\right)\)

\(n_{CaO}=\dfrac{6,72}{56}=0,12\left(mol\right)\)

\(n_{CO_2}=\dfrac{1,1792}{22,4}=0,05\left(mol\right)\)

\(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)

0,12 --------------> 0,12

\(Ca\left(OH\right)_2+CO_2\underrightarrow{t^o}CaCO_3+H_2O\)

0,12 0,05 ------------> 0,05

lập tỉ số: \(n_{Ca\left(OH\right)_2}:n_{CO_2}=0,12>0,05\)

=> Ca(OH)₂ dư , CO₂ hết

\(n_{Ca\left(OH\right)_2\left(dư\right)}=0,12-0,05=0,07\left(mol\right)\)

\(m_{Ca\left(OH\right)_2\left(dư\right)}=0,07.74=5,18\left(g\right)\)

\(m_{CaCO_3}=0,05.100=5\left(g\right)\)

\(M_{M_2O_x}=102\)

\(\Leftrightarrow2M+16x=102\)

loại loại nhận

M là Nhôm ( Al ), hóa trị M là III

\(n_{BaCO_3\left(5\right)}=\dfrac{19,7}{197}=0,1\left(mol\right)\)

\(n_{BaCO_3\left(7\right)}=\dfrac{14,775}{197}=0,075\left(mol\right)\)

\(C+\dfrac{1}{2}O_2\underrightarrow{t^o}CO\uparrow\) (1)

\(C+O_2\underrightarrow{t^o}CO_2\uparrow\) (2)

0,25 <------- 0,25

\(CO+Fe_3O_4\underrightarrow{t^o}3FeO+CO_2\uparrow\) (3)

\(FeO+CO\underrightarrow{t^o}Fe+CO_2\uparrow\) (4)

\(CO_2+Ba\left(OH\right)_2\underrightarrow{t^o}BaCO_3\downarrow+H_2O\) (5)

0,1 <------------------- 0,1

\(2CO_2+Ba\left(OH\right)_2\underrightarrow{t^o}Ba\left(HCO_3\right)_2\) (6)

0,15 <-------------------- 0,075

\(Ba\left(HCO_3\right)_2\underrightarrow{t^o}BaCO_3\downarrow+CO_2\uparrow+H_2O\) (7)

0,075 <--------- 0,075

\(Fe+CuSO_4\rightarrow FeSO_4+Cu\) (8)

0,03<-- 0,03

E: Cu, FeO, Fe₃O₄

(5)(6) \(\Rightarrow\Sigma_{n_{CO_2}}=0,1+0,15=0,25\left(mol\right)\)

\(m_C=0,25.12=3\left(g\right)\)

ý sau tự xử nha gọi ẩn, lắp hệ vào

\(n_{Al}=\dfrac{0,54}{27}=0,02\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{0,98}{98}=0,01\left(mol\right)\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)

0,02 0,01 ---------------------------> 0,01

Lập tỉ số: \(n_{Al}:n_{H_2SO_4}=\dfrac{1}{100}>\dfrac{1}{300}\)

=> Al dư, H₂SO₄ hết

\(V_{H_2}=0,01.22,4=0,224\left(l\right)\)

\(m_{Al\left(dư\right)}=\left(0,02-\dfrac{1}{150}\right).27=0,36\left(g\right)\)

1. Bài này câu b hình như tính C% chứ nhỉ ?

---------------

\(n_{Na_2CO_3}=\dfrac{100.10,6\%}{106}=0,1\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{200.9,8\%}{98}=0,2\left(mol\right)\)

Pt: \(Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+CO_2\uparrow+H_2O\)

0,1mol 0,2mol ---------> 0,1mol----> 0,1mol

Lập tỉ số: \(n_{Na_2CO_3}:n_{H_2SO_4}=0,1< 0,2\)

=> \(Na_2CO_3\) hết, H₂SO₄ dư

\(V_{CO_2}=0,1.22,4=2,24\left(l\right)\)

\(\Sigma_{m_{dd}\left(spu\right)}=100+200-0,1.44=295,6\left(g\right)\)

\(C\%_{Na_2SO_4}=\dfrac{0,1.82.100}{295,6}=2,77\%\)

\(C\%_{H_2SO_4\left(dư\right)}=\dfrac{\left(0,2-0,1\right).98.100}{295,6}=3,32\%\)

2. Ở câu b: Nếu đổi thành dd KOH 5% KLR là 1,045g/ml thì cần bao nhiêu ml dd KOH.

KOH chứ không phải NaOH. Lần sau ghi đề chú ý dùm.

\(n_{H_2SO_4}=1,5.0,05=0,075\left(mol\right)\)

\(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)

0,075mol --> 0,15mol

\(m_{dd}=\dfrac{0,15.40.100}{30}=20\left(g\right)\)

b) \(H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O\)

0,075mol----> 0,15mol

\(m_{dd}=\dfrac{0,15.56.100}{5}=168\left(g\right)\)

\(V_{KOH}=\dfrac{168}{1,045}=160,76\left(ml\right)\)

3. \(n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\)

\(Na_2O+H_2O\rightarrow2NaOH\)

0,25mol -----------> 0,5mol

\(C_{M_{NaOH}}=\dfrac{0,5}{0,5}=1M\)

\(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)

0,25mol<--- 0,5mol

\(m_{dd_{H_2SO_4}}=\dfrac{0,25.98.100}{20}=122,5\left(g\right)\)

\(V_{H_2SO_4}=\dfrac{122,5}{1,14}=107,46\left(ml\right)\)

\(2\sqrt{x+1}+\sqrt{4x+4}-\sqrt{9x+9}=4\) ; \(x\ge-1\)

\(\Leftrightarrow\sqrt{x+1}=4\)

\(\Leftrightarrow x+1=16\)

\(\Rightarrow x=15\) (TM)

\(sin45^o=0,7071\)

\(ch=\dfrac{500}{sin45^o}=707,11\)

\(\Rightarrow cgv\left(?\right)=\sqrt{\left(707,11\right)^2-500^2}=500\left(m\right)\)

Làm xong, cảm giác 50% sẽ sai :v

75/100. Nếu trừ trình bày ra. (?)