2,5 x 7,8 + 3,9 x 17 - 7,8 = 19,5 + 66,3 - 7,8 = 85,8 - 7,8 = 78

\(2.5\times7.8+3.9\times17-7.8\)

\(=2.5\times7.8+8.5\times7.8-7.8\)

\(=7.8\times\left(2.5+8.5-1\right)\)

\(=7.8\cdot10=78\)

 2,5 x 7,8 + 3,9 x 17 - 7,8

= 2,5 x 7,8 + 8,5 x 7,8 - 7,8

= ( 2,5 + 8,5 ) x 7,8 - 7,8

= 11 x 7,8 -7,8

= 85,8 - 7,8 

= 78

Ta có: \(\left\{{}\begin{matrix}n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\n_{SO_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\end{matrix}\right.\)

PTHH: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

                0,1____0,3_____0,1_____0,15     (mol)

           \(2Al+6H_2SO_{4\left(đ\right)}\xrightarrow[]{t^o}Al_2\left(SO_4\right)_3+3SO_2\uparrow+6H_2O\)

               0,1______0,3__________0,05____0,15_____0,3    (mol)

           \(Cu+2H_2SO_{4\left(đ\right)}\xrightarrow[]{t^o}CuSO_4+SO_2\uparrow+2H_2O\)

              0,15_____0,3________0,15___0,15_____0,3   (mol)

Ta có: \(m_{Al}+m_{Cu}=0,1\cdot27+0,15\cdot64=12,3\left(g\right)\)           

Bài 7:

PTHH: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)

Ta có: \(n_{H_2SO_4}=0,2\cdot1=0,2\left(mol\right)=n_{Mg}=n_{H_2}=n_{MgSO_4}\)

\(\Rightarrow\left\{{}\begin{matrix}m_{Ag}=15,6-0,2\cdot24=10,8\left(g\right)\\V_{H_2}=0,2\cdot22,4=2,24\left(l\right)\\C_{M_{MgSO_4}}=\dfrac{0,2}{0,2}=1\left(M\right)\end{matrix}\right.\)

7C 8D

Câu 7: B

Câu 8: C

thi ...

Không phải thi

câu 7 : đáp án c

câu 8 : đáp án c

Câu 7:

a: \(18+\dfrac{9}{3}=18+3=21\)

b: \(\dfrac{4}{9}-\dfrac{5}{18}=\dfrac{8}{18}-\dfrac{5}{18}=\dfrac{3}{18}=\dfrac{1}{6}\)

c: \(\dfrac{3}{5}:\dfrac{5}{7}=\dfrac{3}{5}\cdot\dfrac{7}{5}=\dfrac{21}{25}\)

d: \(\dfrac{4}{7}\cdot\dfrac{2}{7}=\dfrac{4\cdot2}{7\cdot7}=\dfrac{8}{49}\)

Câu 8:

a: \(\dfrac{5}{4}\cdot x=\dfrac{3}{8}+\dfrac{5}{12}\)

=>\(x\cdot\dfrac{5}{4}=\dfrac{9+10}{24}=\dfrac{19}{24}\)

=>\(x=\dfrac{19}{24}:\dfrac{5}{4}=\dfrac{19}{24}\cdot\dfrac{4}{5}=\dfrac{19}{6\cdot5}=\dfrac{19}{30}\)

b: \(3\cdot\dfrac{7}{10}+\dfrac{7}{10}\cdot5+2\cdot\dfrac{7}{10}\)

\(=\dfrac{7}{10}\left(3+5+2\right)\)

\(=\dfrac{7}{10}\cdot10=7\)

7/

4FeS2 + 11O2 → 2Fe2O3 + 8SO2

FexOy +(x-y)CO ----> xFeO +(x - y)CO2

Cu + 2H2SO4 → CuSO4 + SO2 + 2H2O

2KMnO4 + 16HCl → 2KCl + 2MnCl2 + 5Cl2 + 8H2O

2KClO3 ⟶ 2KCl + 3O2

2Fe(OH)3 ⟶ Fe2O3 + 3H2O

2Al + 2NaOH + 2H2O → 2NaAlO2 + 3H2

8. \(4M+nO_2-^{t^o}\rightarrow2M_2O_n\)

Theo PT : \(n_M=2n_{M_2O_n}\)

=> \(\dfrac{5,4}{M}=2.\dfrac{10,2}{2M+16n}\)

Chạy nghiệm n=1,2,3

n=1 => M= 9 (loại)

n=2 => M= 18 (loại)

n=3 => M= 27 (chọn)

Vậy M là Al

7,8.x+25=8x

7,8x-8x=-25

x.(7,8-8)=-25

x.(-0,2)=-25

x=-25:(-0,2)

x=125

Vậy x=125

Học tốt

8x = 7,8x+25

8x - 7,8x = 25

0,2x = 25

x = 25/0,2

x = 125

=.= hk tốt!!

8x=7,8x+25

8x - 7,8x=25

0,2x=25

x=25/0,2

x=125