1,n=\(\dfrac{m}{M}\)=\(\dfrac{0,96}{64}\)=0,015(mol)
3,n=\(\dfrac{m}{M}=\dfrac{70}{56}=1,25\left(mol\right)\)
9,n\(=\dfrac{V}{22,4}=\dfrac{8,064}{22,4}=0,36\left(mol\right)\)
1) \(n_{Cu}=\dfrac{m}{M}=\dfrac{0,96}{64}=0,015\left(mol\right)\)
2) \(n_{Ba}=\dfrac{m}{M}=\dfrac{30,14}{137}=0,22\left(mol\right)\)
3) \(n_{CaO}=\dfrac{m}{M}=\dfrac{70}{40+16}=1,25\left(mol\right)\)
4) \(n_{NH_3}=\dfrac{m}{M}=\dfrac{6,12}{14+3}=0,36\left(mol\right)\)
5) \(n_{H_2SO_4}=\dfrac{m}{M}=\dfrac{37,24}{2+32+16\cdot4}=0,38\left(mol\right)\)
6) \(n_{HNO_3}=\dfrac{m}{M}=\dfrac{157,5}{1+14+16\cdot3}=2,5\left(mol\right)\)
7) \(n_{Al_2\left(SO_4\right)_3}=\dfrac{m}{M}=\dfrac{25,65}{27\cdot2+\left(32+16\cdot4\right)\cdot3}=0,075\left(mol\right)\)
8) \(n_{BaCO_3}=\dfrac{m}{M}=\dfrac{76,83}{137+12+16\cdot3}=0,39\left(mol\right)\)
9) \(n_{NH_3\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{8,064}{22,4}=0,36\left(mol\right)\)
10) \(n_{N_2O\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{41,664}{22,4}=1,86\left(mol\right)\)
