Dòng cuối câu (b) mình quên dấu nhân cho sửa lại:

\(x=\dfrac{2}{9}\times\dfrac{6}{7}=\dfrac{2\times3\times2}{3\times3\times7}=\dfrac{4}{21}\)

a

\(x-\dfrac{5}{14}=\dfrac{3}{7}-\dfrac{1}{14}\\ x-\dfrac{5}{14}-\dfrac{3}{7}+\dfrac{1}{14}=0\\ x+\dfrac{1}{14}-\dfrac{5}{14}-\dfrac{6}{14}=0\\ x+\dfrac{1-5-6}{14}=0\\ x-\dfrac{5}{7}=0\\ x=0+\dfrac{5}{7}\\ x=\dfrac{5}{7}\)

b

\(\dfrac{2}{9}:x=\dfrac{1}{2}+\dfrac{2}{3}\\ \dfrac{2}{9}:x=\dfrac{3}{6}+\dfrac{4}{6}\\ \dfrac{2}{9}:x=\dfrac{3+4}{6}=\dfrac{7}{6}\\ x=\dfrac{2}{9}:\dfrac{7}{6}\\ x=\dfrac{2}{9}\times\dfrac{6}{7}=\dfrac{2.3.2}{3.3.7}=\dfrac{4}{21}\)

c

\(\dfrac{6}{32:x}=\dfrac{12}{16}\\ 32:x=6:\dfrac{12}{16}\\ 32:x=6\times\dfrac{16}{12}\\ 32:x=\dfrac{3\times2\times4\times4}{3\times4}\\ 32:x=8\\ x=\dfrac{32}{8}\\ x=4\)

Gọi CTHH của 2 oxit A và B lần lượt là \(R_2O_n\) và \(R_3O_m\) (\(n,m\) nguyên và \(\le4\))

Theo đề có:

\(\%m_{R\left(A\right)}=59,66\%\\ \Leftrightarrow\dfrac{2R}{2R+16n}=\dfrac{59,66}{100}\\ \Rightarrow R=11,83n\left(g/mol\right)\)

\(\%m_{R\left(B\right)}=38,8\%\\ \Leftrightarrow\dfrac{2R}{2R+16m}=\dfrac{38,8}{100}\\ \Rightarrow R=5,07m\left(g/mol\right)\)

Có tỉ lệ: \(\dfrac{n}{m}=\dfrac{5,07}{11,83}=\dfrac{3}{7}\Rightarrow n=3,m=7\)

\(\Rightarrow\)\(\left\{{}\begin{matrix}A:R_2O_3\\B:R_2O_7\end{matrix}\right.\)

Lại có: \(R=5,07m=5,07.7=35,5\left(g/mol\right)\)

Vậy hóa trị R trong A và B lần lượt là 3 và 7.

Phi kim R là Cl (Clo)

a

\(5x^2-5xy+7y-7x\\ =5x\left(x-y\right)+7\left(y-x\right)\\ =5x\left(x-y\right)-7\left(x-y\right)\\ =\left(5x-7\right)\left(x-y\right)\)

b

\(x^2+2xy+x+2y\\ =x\left(x+2y\right)+\left(x+2y\right)\\ =\left(x+1\right)\left(x+2y\right)\)

c

\(x^2-6x-9y^2+9\\ =x^2-6x+9-\left(3y\right)^2\\ =x^2-2.x.3+3^2-\left(3y\right)^2\\ =\left(x-3\right)^2-\left(3y\right)^2\\ =\left(x-3-3y\right)\left(x-3+3y\right)\)

\(n_{NaOH}=0,2.1=0,2\left(mol\right)\\ n_{H_2SO_4}=0,3.1,5=0,45\left(mol\right)\)

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

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

Xét \(\dfrac{0,2}{2}< \dfrac{0,45}{1}\Rightarrow\) \(H_2SO_4\)dư

Trong dung dịch D có:

\(\left\{{}\begin{matrix}n_{H_2SO_4}=0,45-0,1=0,35\left(mol\right)\\n_{Na_2SO_4}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}CM_{H_2SO_4}=\dfrac{0,35}{0,5}=0,7M\\CM_{Na_2SO_4}=\dfrac{0,1}{0,5}=0,2M\end{matrix}\right.\)

b

\(Ca\left(OH\right)_2+H_2SO_4\rightarrow CaSO_4+2H_2O\)

0,35<---------0,35

\(V_{Ca\left(OH\right)_2}=\dfrac{0,35.74}{1,2}=\dfrac{259}{12}\approx21,58\left(ml\right)\\ \Rightarrow V_{dd.Ca\left(OH\right)_2}=\dfrac{\dfrac{259}{12}.100\%}{10\%}=\dfrac{1295}{6}\approx215,83\left(ml\right)\)

Vận tốc trung bình trên cả quãng đường là:

\(v_{tb}=\dfrac{v_1+v_2+v_3}{3}=\dfrac{120+80+40}{3}=80km/giờ\)

d

\(\sqrt{18}-\sqrt{6}\left(\sqrt{3}-\sqrt{24}\right)\\ =\sqrt{18}-\sqrt{6.3}+\sqrt{6.24}\\ =\sqrt{18}-\sqrt{18}+\sqrt{144}\\ =0+\sqrt{12^2}\\ =12\)

e

\(\sqrt{12}\left(\sqrt{5}-\sqrt{3}\right)-\sqrt{60}\\ =\sqrt{12.5}-\sqrt{12.3}-\sqrt{60}\\ =\sqrt{60}-\sqrt{60}-\sqrt{36}\\ =0-\sqrt{6^2}\\ =0-6\\ =-6\)

5

Với \(x>0,a\ne0\) có

\(\dfrac{x+a\sqrt{x}}{a\sqrt{x}}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}+a\right)}{a\sqrt{x}}\\ =\dfrac{\sqrt{x}+a}{a}\)

6 tối giản rồi: )

BÀI 1

1

10 oxit: 

`CO` (cacbon monooxit)

`CO_2` (cacbon dioxit)

`SO_2` (lưu huỳnh dioxit)

`Na_2O` (natri oxit)

`MgO` (magie oxit)

`CaO` (canxi oxit)

`Al_2O_3` (nhôm oxit)

`CuO` (đồng II oxit)

`BaO` (bari oxit)

`P_2O_5` (photpho bentoxit)

2

10 muối:

`Na_2CO_3` (natri cabonat)

`K_2CO_3` (kali cacbonat)

`MgCO_3` (magie cacbonat)

`NaHCO_3` (natri hidrocacbonat)

`KHCO_3` (kali hidrocacbonat)

`MgSO_3` (magie sunfit)

`BaCO_3` (bari cacbonat)

\(Ba\left(HCO_3\right)_2\) (bari hidrocacbonat)

`CaCO_3` (canxi cacbonat)

`CaSO_3` (canxit sunfit)

3

8 axit:

`H_2SO_4` (axit sunfuric)

`HCl` (axit clohidric)

`HNO_3` (axit nitric)

`H_3PO_4` (axit photphoric)

`HCN` (axit hidrocyanic)

`HF` (axit hydrofluoric)

`HBr` (axit bromhydric)

`H_2CO_3` (axit cacbonic)

10 bazo:

`NaOH` (natri hidroxit)

`KOH` (kali hidroxit)

\(Ba\left(OH\right)_2\left(bari.hidroxit\right)\\ Ca\left(OH\right)_2\left(caxi.hidroxit\right)\\ Mg\left(OH\right)_2\left(magie.hidroxit\right)\\ Al\left(OH\right)_3\left(nhôm.hidroxit\right)\)

\(CuOH\left(đồng.I.hidroxit\right)\\ Cu\left(OH\right)_2\left(đồng.II.hidroxit\right)\)

\(Fe\left(OH\right)_2\left(sắt.II.hidroxit\right)\\ Fe\left(OH\right)_3\left(sắt.III.hidroxit\right)\)

Bài 2

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ n_{HCl}=\dfrac{21,9}{36,5}=0,6\left(mol\right)\)

a \(Fe+2HCl\rightarrow FeCl_2+H_2\)

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

Xét \(\dfrac{0,1}{1}< \dfrac{0,6}{2}\Rightarrow\) HCl dư

b \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)

c

Chất tan sau phản ứng: \(\left\{{}\begin{matrix}n_{HCl.dư}=0,6-0,4=0,2\left(mol\right)\\n_{FeCl_2}=0,1\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}m_{HCl.dư}=0,2.3,65=7,3\left(g\right)\\m_{FeCl_2}=0,1.127=12,7\left(g\right)\end{matrix}\right.\)

a

Số nguyên tử X có khối lượng bằng 14,2 gam là: \(1,2.10^{23}.2=2,4.10^{23}\) (nguyên tử)

\(\Rightarrow n_X=\dfrac{2,4.10^{23}}{6.10^{23}}=0,4\left(mol\right)\)

\(M_{phân.tử.X}=\dfrac{14,2}{0,4}.2=71\left(g/mol\right)\)

b

\(M_{nguyên.tử.khối.X}=\dfrac{71}{2}=35,5\left(đvC\right)\)

X là nguyên tố Cl (Clo)