41,65+8,25+8,35+9,45+21,75+10,55

= ( 41,65 + 8,35 ) + ( 8,25 + 21,75 ) + ( 9,45 + 10,55 )

= 50 + 30 + 20

= 50 + 50

= 100

   41,65+8,25+8,35+9,45+21,75+10,55=

= (41,65+8,35)+(8,25+21,75)+(9,45+10,55)

=        50        +      30        +       20

=       100.

41,65+8,25+8,35+9,45+21,75+10,55

=(41,65+8,35)+(8,25+21,75)+(9,45+10,55)

=50+30+20

=50+50

=100

học tốt^^

#Trịnh hằng

a) 8,4

b) 8,2

c) 8,0

Ngu như chó

9,45 - 2,37 + 1,37 = 8,45

2,3 : 3,2 x 0,8 = 0,575

Nếu đúng thì tk nha

a 9.45-2.37-1.37=9.45-1=8.45

b 2.3*0.8chia3.2*0.8=0.575

dung thi cho k

a 9,45 + [ - 2,37 + 1, 37] = 9.45 + [ - 1] = 9.45 -1 = 8.45

b 0.575

21,75x15,36+21,75x6,64

=21,75x(15,36+6,64)

=21,75x22

=478,5

21,75 x 15,36 + 21,75 x 6,64

= 21,75 x ( 15,36 + 6,64 )

= 21,75 x  22

=   478,5

21,75x15,36+21,75×6,64

=21,75×(15,36+6,64)

=21,75×22

=478,5

73/247

( 12,375 - 10,55 ) : ( 7/12 + 94/15 )
= 1,825 : 6,85

= 0.2664

(12,375 - 10,55) : (\(\dfrac{7}{12}+\dfrac{94}{15}\))

a, Gọi \(\left\{{}\begin{matrix}n_{Al}=a\left(mol\right)\\n_{Zn}=b\left(mol\right)\end{matrix}\right.\left(đk:a,b>0\right)\)

\(n_{H_2}=\dfrac{7,28}{22,4}=0,35\left(mol\right)\)

PTHH:

2Al + 3H₂SO₄ ---> Al₂(SO₄)₃ + 3H₂

a--------------------------------------->1,5a

Zn + H₂SO₄ ---> ZnSO₄ + H₂

b------------------------------>b

Theo bài ra, ta có hệ: \(\left\{{}\begin{matrix}27a+65b=10,55\\1,5a+b=0,35\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=0,15\left(mol\right)\\b=0,1\left(mol\right)\end{matrix}\right.\left(TM\right)\)

\(\rightarrow\left\{{}\begin{matrix}m_{Al}=0,15.27=4,05\left(g\right)\\m_{Fe}=0,1.56=5,6\left(g\right)\end{matrix}\right.\)

\(\rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{4,05}{10,55}.100\%=38,4\%\\\%m_{Zn}=100\%-38,4\%=61,6\%\end{matrix}\right.\)

b, PTHH:

\(Zn+2H_2SO_{4\left(đặc,nguội\right)}\rightarrow ZnSO_4+SO_2\uparrow+2H_2O\)

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

\(2Al+3ZnSO_4\rightarrow Al_2\left(SO_4\right)_3+3Zn\downarrow\)

\(\dfrac{1}{15}\)<---0,1---------->\(\dfrac{1}{30}\)---------->0,1

\(Zn+2H_2SO_{4\left(đặc,nguội\right)}\rightarrow ZnSO_4+SO_2\uparrow+2H_2O\)

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

\(\rightarrow\left\{{}\begin{matrix}V=\left(0,1+0,1\right).22,4=4,48\left(l\right)\\x=\dfrac{1}{30}.342+0,1.161=27,5\left(g\right)\end{matrix}\right.\)

\(a.Đặt:\left\{{}\begin{matrix}n_{Al}=x\left(mol\right)\\n_{Zn}=y\left(mol\right)\end{matrix}\right.\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ \Rightarrow\left\{{}\begin{matrix}27x+65y=10,55\left(g\right)\\\dfrac{3}{2}x+y=\dfrac{7,28}{22,4}=0,325\left(mol\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,15\\y=0,1\end{matrix}\right.\\ \%m_{Al}=\dfrac{0,15.27}{10,55}.100=38,39\%;\%m_{Zn}=61,61\%\\ b.X+H_2SO_4đặc,nguội\Rightarrow ChỉcóZnphảnứng\\ Zn\rightarrow Zn^{2+}+2e\\ S^{+6}+2e\rightarrow S^{+4}\\ Bảotoàne:n_{Zn}.2=n_{SO_2}.2\\ \Rightarrow n_{SO_2}=0,1\left(mol\right)\\ \Rightarrow V_{SO_2}=0,1.22,4=2,24\left(l\right)\\ n_{ZnSO_4}=n_{Zn}=0,1\left(mol\right)\\ \Rightarrow m_{ZnSO_4}=161.0,1=16,1\left(g\right)\)

Lưu ý: Al bị thụ động với H2SO4 dặc nguội

tk đi rồi mình làm cho

Giúp mình đi

12,525

Tính hay đặt tính rồi tính em?

12,525

\(8\frac{7}{20}=8,35\)