Chọn A

= 56,8 x 0,1 = 5,68

Chọn a

A

56,8 x 76,9 - 56,8 x 76,8

= 56,8 x (76,9 - 76,8)

= 56,8 x 0,1

= 5,68

20,17 x 76,8 + 20,17 x 23,2  = 20,17 x (76,8 + 23,2)

= 20,17 x      100

= 2017

76,8 x 99 + 76 + 0,9

= 76,8 x 99 + 76,8 + 0,1

= 76,8 x (99+1) + 0,1

= 76,8 x 100 + 0,1

= 7680 + 0,1 = 7680,1

----

0,7 x 96 + 1,4 x 2

= 0,7 x 96 + 2,8

= 0,7 x 96 + 0,7 x 4

= 0,7 x (96+4)

 = 0,7 x 100 = 70

\(3,7\times3,8+6,2\times3,7\\ =3,7\times\left(3,8+6,2\right)\\ =3,7\times10\\ =37\\ 56,8\times76,9-56,8\times76,8\\ =56,8\times\left(76,9-76,8\right)\\ =56,8\times0,1\\ =5,68\)

a)\(3,7\times3,8+6,2\times3,7\)
\(=3,7\times\left(3,8+6,2\right)\)
\(=3,7\times10=37\)
b)\(56,8\times76,9-56,8\times76,8\)
\(=56,8\times\left(76,9-76,8\right)\)
\(=56,8\times0,1=5,68\)

a,=3,7 x (3,8+6,2)=37

b,=56,8x(76,9-76,8)=5,68

a,2048     b,45880

a .   1996.8

b .   45880

a,2048

b,45880

\(n_{SO_2}=\dfrac{76,8}{64}=1,2\left(mol\right)\)

\(V_{SO_2}=1,2.22,4=26,88\left(l\right)\)

\(n_{SO_2}=\dfrac{m_{SO_2}}{M_{SO_2}}=\dfrac{76,8}{64}=1,23125\left(mol\right)\)

\(\Rightarrow V_{SO_2,đkc}=1,23152\cdot24,79\approx30,5225\left(l\right)\)

a) 3,37 + X = 76,8 

X = 76,8 - 3,37

X = 73.43

b) x - 34,8 = 4,56 + 8,44 

 x - 34,8 = 13

x = 13 + 34,8 

x = 47.8

c) 8,2 x X = 28,7 

x = 28,7 : 8,2

x = 3.5

d) 45 ,56 - x = 23 ,8 

x  = 45,56 - 23,8

x = 21.76

đ) 25 : x = 1,6 

x = 25 : 1,6

x = 15.625

e) x : 3,2 =8,25

x = 8,25 x 3,2

x = 26.4

\(a,x=76,8-3,37\)

\(x=73,43\)

\(b,x-34,8=13\)

\(x=13+34,8\)

\(x=47,8\)

\(c,x=28,7:8,2\)

\(x=3,5\)

\(d,x=45,56-23,8\)

\(x=21,76\)

\(đ,x=25:1,6\)

\(x=15,625\)

\(e,x=8,25\times3,2\)

\(x=26,4\)

a) 3,37 + X = 76,8 

X = 76,8 - 3,37

X = 73.43

b) x - 34,8 = 4,56 + 8,44 

 x - 34,8 = 13

x = 13 + 34,8 

x = 47.8

c) 8,2 x X = 28,7 

a)

$V_{O_2} = 0,2.22,4 = 4,48(lít)$

b)

$n_{SO_2} = \dfrac{76,8}{64} = 1,2(mol)$
$V_{SO_2} = 1,2.22,4 = 26,88(lít)$

c)

$n_{N_2} = \dfrac{7,5.10^{23}}{6.10^{23}} = 1,25(mol)$
$V_{N_2} = 1,25.22,4 = 28(lít)$

d)

$V_X = (0,2 + 0,25).22,4 = 10,08(lít)$

a) V(O2,đktc)=0,2.22,4=4,48(l)

b) V(SO2,đktc)= (76,8/64).22,4=26,88(l)

c) V(N2,đktc)= [(7,5.10²³)/(6.10²³)].22,4= 28(l)

d) VhhX(đktc)= V(O2,đktc) + V(N2,đktc)= 0,2.22,4+0,25.22,4=10,08(l)

F20: Tính thể tích ở đkc của :

a. 0,2 mol O₂

=> \(V_{O_2}=0,2.22,4=4,48\left(l\right)\)

b. 76,8 gam SO₂

\(n_{SO_2}=\dfrac{76,8}{32+16.2}=1,2\left(mol\right)\)

=> \(V_{SO_2}=1,2.22,4=26,88\left(l\right)\)

c. 7,5 .10²³ phân tử N₂ 

\(n_{N_2}=\dfrac{7,5.10^{23}}{6.10^{23}}=1,25\left(mol\right)\)

=> \(V_{N_2}=1,25.22,4=28\left(l\right)\)