Fe = Fe.2= II.3

Fe=6:2

Fe= 3

=> Fe hóa trị III

Gọi hóa trị của Cu,Fe,S,Ba lần lượt là a,b,c,d>0

\(a,Cu_2^aO_1^{II}\Rightarrow a\cdot2=1\cdot II\Rightarrow a=1\Rightarrow Cu\left(I\right)\\ Cu_1^aO_1^{II}\Rightarrow a\cdot1=II\cdot1\Rightarrow a=2\Rightarrow Cu\left(II\right)\\ b,Fe_1^bO_1^{II}\Rightarrow b\cdot I=II\cdot1\Rightarrow b=2\Rightarrow Fe\left(II\right)\\ Fe_2^bO_3^{II}\Rightarrow2b=II\cdot3\Rightarrow b=3\Rightarrow Fe\left(III\right)\\ c,S_1^cO_2^{II}\Rightarrow c=II\cdot2=4\Rightarrow S\left(IV\right)\\ S_1^cO_3^{II}\Rightarrow c=3\cdot II=6\Rightarrow S\left(VI\right)\\ H_2^IS_1^c\Rightarrow c=I\cdot2=2\Rightarrow S\left(II\right)\\ d,Ba_1^d\left(CO_3\right)_1^{II}\Rightarrow d=II\cdot1=2\Rightarrow Ba\left(II\right)\)

gọi hóa trị của các nguyên tố cần tìm là \(x\)

a/

 \(\rightarrow Cu_2^xO^{II}_1\rightarrow x.2=II.1\rightarrow x=I\)

vậy Cu hóa trị I

\(\rightarrow Cu^x_1O^{II}_1\rightarrow x.1=II.1\rightarrow x=II\)

vậy Cu hóa trị II

b/

\(\rightarrow Fe^x_1O_1^{II}\rightarrow x.1=II.1\rightarrow x=II\)

vậy Fe hóa trị II

\(\rightarrow Fe_2^xO^{II}_3\rightarrow x.2=II.3\rightarrow x=III\)

vậy Fe hóa trị III

c/ 

\(\rightarrow S^x_1O_2^{II}\rightarrow x.1=II.2\rightarrow x=IV\)

vậy S hóa trị IV

\(\rightarrow S^x_1O_3^{II}\rightarrow x.1=II.3\rightarrow x=VI\)

vậy S hóa trị VI

\(\rightarrow H^I_2S^x_1\rightarrow I.2=x.1\rightarrow x=II\)

vậy S hóa trị II

d/ \(\rightarrow Ba^x_1\left(CO_3\right)^{II}_1\rightarrow x.1=II.1\rightarrow x=II\)

vậy Ba hóa trị II

\(a,CTTQ:Mg_x^{II}\left(OH\right)_y^I\\ \Rightarrow x\cdot II=y\cdot I\Rightarrow\dfrac{x}{y}=\dfrac{1}{2}\Rightarrow x=1;y=2\\ \Rightarrow Mg\left(OH\right)_2\\ b,CTTQ:Al_x^{III}\left(SO_4\right)_y^{II}\\ \Rightarrow x\cdot III=y\cdot II\Rightarrow\dfrac{x}{y}=\dfrac{2}{3}\Rightarrow x=2;y=3\\ \Rightarrow Al_2\left(SO_4\right)_3\\ c,CTTQ:Fe_x^{III}O_y^{II}\\ \Rightarrow x\cdot III=y\cdot II\Rightarrow\dfrac{x}{y}=\dfrac{2}{3}\Rightarrow x=2;y=3\\ \Rightarrow Fe_2O_3\\ d,CTTQ:Cu_x^{II}\left(CO_3\right)_y^{II}\\ \Rightarrow x\cdot II=y\cdot II\Rightarrow\dfrac{x}{y}=1\Rightarrow x=1;y=1\\ \Rightarrow CuCO_3\)

\(e,CTTQ:Na_x^I\left(PO_4\right)_y^{III}\\ \Rightarrow x\cdot I=y\cdot III\Rightarrow\dfrac{x}{y}=3\Rightarrow x=3;y=1\\ \Rightarrow Na_3PO_4\\ f,CTTQ:Ca_x^{II}\left(NO_3\right)_y^I\\ \Rightarrow x\cdot II=y\cdot I\Rightarrow\dfrac{x}{y}=\dfrac{1}{2}\Rightarrow x=1;y=2\\ \Rightarrow Ca\left(NO_3\right)_2\)

\(a,PTK_{Mg\left(OH\right)_2}=24+17\cdot2=61\left(đvC\right)\\ b,PTK_{Al_2\left(SO_4\right)_3}=27\cdot2+\left(32+16\cdot4\right)\cdot3=342\left(đvC\right)\\ c,PTK_{Fe_2O_3}=56\cdot2+16\cdot3=160\left(đvC\right)\\ d,PTK_{CuCO_3}=64+12+16\cdot3=124\left(đvC\right)\\ e,PTK_{Na_3PO_4}=23\cdot3+31+16\cdot4=164\left(đvC\right)\\ f,PTK_{Ca\left(NO_3\right)_2}=40+\left(14+16\cdot3\right)\cdot2=164\left(đvC\right)\)

bạn đăng tách cho mn giúp nhé 

Bài 6 : 

\(\Rightarrow30-3y=xy\Leftrightarrow xy+3y=30\Leftrightarrow y\left(x+3\right)=30\)

\(\Rightarrow x+3;y\inƯ\left(30\right)=\left\{\pm1;\pm2;\pm3;\pm5;\pm6;\pm10;\pm15;\pm30\right\}\)

ok bn, sry bn nhé 

= 1/1x2 + 1/2x3 + 1/3x4 + 1/4x5 + ...... + 1/9x10

= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4-1/5 +.......+ 1/9 -1/10

= 1/1 - 1/10

= 9/10

=1/1x2+1/2x3+1/3x4+1/4x5+......+1/9x10

Thế tích hình hộp chữ nhật là:

20x16x10=3200(cm2)

              Đáp số: 3200 cm2.

bạn sai đơn vị rồi