Răspuns :
Prima ecuatie:
[tex]m_{f}=m_{i}-m_{Ni}+m_{Cu}\\ 16=15-m_{Ni}+m_{Cu}\\ 1=m_{Cu}-m_{Ni}[/tex]
[tex]59......................64\\ Ni+CuCl_{2}=Cu+NiCl_{2}\\ m_{Ni}....................m_{Cu}\\[/tex]
[tex]m_{Cu}=m_{Ni}\frac{64}{59}\\ m_{Cu}=m_{Ni}*1,08[/tex]
Din cele 2 ecuatii:
[tex]m_{Cu}-m_{Ni}=1\\ m_{Ni}*1,08-m_{Ni}=1\\ m_{Ni}*0.08=1\\ m_{Ni}=12,5\\ m_{Cu}=13,5[/tex]
[tex]m_{f}=m_{i}-m_{Ni}+m_{Cu}\\ 16=15-m_{Ni}+m_{Cu}\\ 1=m_{Cu}-m_{Ni}[/tex]
[tex]59......................64\\ Ni+CuCl_{2}=Cu+NiCl_{2}\\ m_{Ni}....................m_{Cu}\\[/tex]
[tex]m_{Cu}=m_{Ni}\frac{64}{59}\\ m_{Cu}=m_{Ni}*1,08[/tex]
Din cele 2 ecuatii:
[tex]m_{Cu}-m_{Ni}=1\\ m_{Ni}*1,08-m_{Ni}=1\\ m_{Ni}*0.08=1\\ m_{Ni}=12,5\\ m_{Cu}=13,5[/tex]
A,,Ni= 59g/mol---->( niu= 15g/59g/mol=0,254mol Ni)
A,Cu= 64g/mol
Ni +CuCl2 =NiCl2 +Cu
cand reactioneaza 1 mol Ni , se pierd 59g dar se castiga 64g --> Δm= 5g
1molNi..................Δm=5g
niu=0,25mol....... Δm=1g cat s-a observat practic,
m= 0,25molx 59g/.mol Ni= 14,75g Ni.....apropiat de 15g, dat in text
1molNi..........64gCu
0,25mol..........m= 0,25x64g Cu
A,Cu= 64g/mol
Ni +CuCl2 =NiCl2 +Cu
cand reactioneaza 1 mol Ni , se pierd 59g dar se castiga 64g --> Δm= 5g
1molNi..................Δm=5g
niu=0,25mol....... Δm=1g cat s-a observat practic,
m= 0,25molx 59g/.mol Ni= 14,75g Ni.....apropiat de 15g, dat in text
1molNi..........64gCu
0,25mol..........m= 0,25x64g Cu