Cu is being reduced and Zn is being oxidized, hence Zn + CuSO4 --> ZnSO4 + Cu
CuSO4+Zn results in Cu+ZnSO4, but the reaction doesn`t take place the other way around CuSO4+Zn results in Cu+ZnSO4, but the reaction doesn`t take place the other way around CuSO4+Zn results in Cu+ZnSO4, but the reaction doesn`t take place the other way around
double replacement
Zn + CuSO4 -> ZnSO4 + Cu (molecular equation) Zn (s) + SO4 2+ + Cu 2+ --> Cu (s) + Zn 2+ + SO4 2+ (ionic equation) Cu + Zn(s) ---> Cu (s) + Zn (net ionic)
No. The % of Cu by mass in CuSO4 will be greater than the % of Cu by mass in the pentahydrate (5H2O) because in the hydrate there is added mass (5 H2O = 90 g) but no added Cu.
Zn + CuSO4 --> ZnSO4 + Cu
Zn + CuSO4 --> ZnSO4 + Cu 1.75g CuSO4 * (1moleCuSO4/159.62gCuSO4) * (1moleZn/1moleCuSO4) * (65.38gZn/1moleZn) = .7168g Zn 2.00g Zn - .7168g Zn = 1.2832g Zn in Excess
Cu is being reduced and Zn is being oxidized, hence Zn + CuSO4 --> ZnSO4 + Cu
Zn+CuSO4=Cu+ZnSO4 right?
CuSO4+Zn results in Cu+ZnSO4, but the reaction doesn`t take place the other way around CuSO4+Zn results in Cu+ZnSO4, but the reaction doesn`t take place the other way around CuSO4+Zn results in Cu+ZnSO4, but the reaction doesn`t take place the other way around
double replacement
Zn + CuSO4 -> ZnSO4 + Cu (molecular equation) Zn (s) + SO4 2+ + Cu 2+ --> Cu (s) + Zn 2+ + SO4 2+ (ionic equation) Cu + Zn(s) ---> Cu (s) + Zn (net ionic)
No. The % of Cu by mass in CuSO4 will be greater than the % of Cu by mass in the pentahydrate (5H2O) because in the hydrate there is added mass (5 H2O = 90 g) but no added Cu.
Cl H Zn -apex Fojus xD
copper (II) sulfate is CuSO4 ; Zinc sulfate is ZnSO4 Zn + CuSO4 --> ZnSO4 + Cu
CuSo4(aq) +Zn(s) ---> ZnSo4( aq) + Cu (s)
They fit the pattern: element + compound --> different element + different compound. Example: Zn + CuSO4 --> Cu + ZnSO4.