\(a.n_{CO_2}=\dfrac{m_{CO_2}}{M_{CO_2}}=\dfrac{4,4}{44}=0,1\left(mol\right)\\ \Rightarrow V_{CO_2}=n_{CO_2}.22,4=0,1.22,4=2,24\left(l\right)\\ n_{O_2}=\dfrac{m_{O_2}}{M_{O_2}}=\dfrac{3,2}{32}=0,1\left(mol\right)\) 

\(\Rightarrow V_{O_2}=n_{O_2}.22,4=0,1.22,4=2,24\left(l\right)\) 

\(n_{H_2}\)  = 2mol

=>\(V_{H_2}\) = 2. 22,4 = 44,8l

\(n_{O_2}\) = 0,0875 mol

=>\(V_{O_2}\) = 0,0875 . 22,4 = 1,96l

\(n_{CO_2}\) = 0,5 mol

=>\(V_{CO_2}\)  = 0,5 .22,4 = \(11,2\left(l\right)\) 

\(n_{O_2}\)  = \(0,2\left(mol\right)\) 

=>\(V_{O_2}\) = 0,2 . 22,4 = \(4,48\left(l\right)\) 

b) Ta có:

\(n_{NH_3}=\dfrac{6.10^{23}}{6.10^{23}}=1\left(mol\right)\\ n_{O_2}=\dfrac{3.10^{23}}{6.10^{23}}=0,5\left(mol\right)\\ \Sigma n_{hh}=1+0,5=1,5\left(mol\right)\\ V_{hh}=1,5.22,4=33,6\left(l\right)\)

a: =(100-2)(100+2)

=100^2-4

=10000-4

=9996

b: =(69-31)(69+31)

=38*100=3800

c: =(1023-23)(1023+23)

=1046*1000

=1046000

d: =(50-3)(50+3)

=50^2-9

=2500-9

=2491

p=(5.311+4.312):(39.52-39.23)

  =((4+1).311+4.(311+1):(39.(52-23))

  =4.311+311+4.311+4):(39.29)

  =(311.(4+4+1)+4):(39.29)

tu lam tiep nhe

\(a^2+2ab+b^2\)

\(=a^2+ab+ab+b^2\)

\(=a\left(a+b\right)+b\left(a+b\right)\)

\(=\left(a+b\right)\left(a+b\right)\)

\(=\left(1023-23\right)\left(1023-23\right)\)[1023+(-23)]=1023-23,chỗ này là thế này,sợ bạn ko hiểu nên mik ghi ra thôi

\(=1000.1000\)

\(=1000000\)

ta có a²+2ab+b²=(a+b)²    (1)

thay a=1023 và b=-23 vào 1 ta lại có

[1023+(-23)]²=1000²

Không có kết quả tương thích:

Kết quả: 2,6568.10^-22

\(n_{Na}=\dfrac{1,2.10^{-23}}{6.10^{-23}}=0,2(mol)\)

1023 - 23 x 4 : 20 = 1018.4

1223-23x4:20

=(1223-23)x(4:20)

=1200x0,2

=240

1018,4

a)=25+(39+61)=25+100=125 

b)=34.(105-5)=34.100=3400

c)=(997+3)+(3+51)=1000+54=1054

d)=15.(25+75)=15.100=1500

e)=(5.2).(25.4).16=1000.16=16000

g)=32.(47+53)=32.100=3200

chúc bạn học tốt!