n H2=\(\dfrac{11,2}{22,4}\)=0,5 mol

=>m H2=0,5.2=1 g

ĐLBTKLg:

=>a=m Fe=30,2+1-9=22,2g

Ta có: \(\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}}\)

\(=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-2\cdot\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\right)}\)

\(=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-2\cdot\dfrac{x+y+z}{xyz}}\)

\(=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)

\(=\left|\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right|\)

1 Who have the custom of making....?

2 What do people in the Central Highlands believe?

3 What would they do if the dead weren't buried in beautiful graves

4 What do they make funeral houses and cutting wooden statues for?

5 How many tools do people in the central highlands use to make statues

6 What are the topics for funeral statues aken from?

a) \(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PTHH: CO₂ + CaO → CaCO₃

Mol:      0,1                      0,1

b) \(m_{CaCO_3}=0,1.100=10\left(g\right)\)

a sai vì 4 ko phải số nguyên tố

b sai lun vì 2x3 =6 trong đó 2 và 3 đều là số nguyên tố mà tích là 6 là số chẵn

a.\(C\%_{KCl}=\dfrac{20}{600}.100=3,33\%\)

b.2,5kg = 2500g

\(C\%_{Al_2\left(SO_4\right)_3}=\dfrac{34,2}{2500}.100=1,368\%\)

`(a): C%_[KCl] = 20 / 600 . 100 ~~ 3,33 %`

________________________________________

`(b):` Đổi `2,5 kg = 2500 g`

`C%_[Al_2 (SO_4)_3] = [ 34,2 ] / 2500 . 100 = 1,368%`

a. \(C_{\%_{KCl}}=\dfrac{20}{600}.100\%=3,33\%\)

b. \(2,5kg=2500g\)

\(C_{\%_{Al_2\left(SO_4\right)_3}}=\dfrac{34,2}{2500}.100\%=1,368\%\)