I already know how to do question #1 but I need help with question #3 and you need the answers from #1 to answer #3. I already solved question #1 so that you could help with #3 1. Balance each of the following equations:  A. __1_ Cu(s) + __4_ HNO3(aq) → _1__ Cu(NO3)2(aq) + _2__ NO2(g) + _2__ H2O(l)   B. _1__ Cu(NO3)2(aq) + __2_ NaOH(aq) → _1__ Cu(OH)2(s) + _2__ NaNO3(aq)   C. _1__ Cu(OH)2(s) → __1_ CuO(s) + __1_ H2O(l)   D. _1__ CuO(s) + _1__ H2SO4(aq) → _1__ CuSO4(aq) + __1_ H2O(l)   E. __1_ CuSO4(aq) + __1_ Zn(s) → _1__ Cu(s) + _1__ ZnSO4(aq)    3. If you measured out 100.0 mg of copper wire, calculate the exact amount of zinc that you would need in Part E of the experiment to complete the copper cycle. (Hint: use your balanced equations from Question 1.)

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Asked Oct 13, 2019
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I already know how to do question #1 but I need help with question #3 and you need the answers from #1 to answer #3. I already solved question #1 so that you could help with #3

 

1. Balance each of the following equations:
 
A. __1_ Cu(s) + __4_ HNO3(aq) → _1__ Cu(NO3)2(aq) + _2__ NO2(g) + _2__ H2O(l)
 
 
B. _1__ Cu(NO3)2(aq) + __2_ NaOH(aq) → _1__ Cu(OH)2(s) + _2__ NaNO3(aq)
 
 
C. _1__ Cu(OH)2(s) → __1_ CuO(s) + __1_ H2O(l)
 
 
D. _1__ CuO(s) + _1__ H2SO4(aq) → _1__ CuSO4(aq) + __1_ H2O(l)
 
 
E. __1_ CuSO4(aq) + __1_ Zn(s) → _1__ Cu(s) + _1__ ZnSO4(aq)
 
 
 

3. If you measured out 100.0 mg of copper wire, calculate the exact amount of zinc that you would need in Part E of the experiment to complete the copper cycle. (Hint: use your balanced equations from Question 1.)

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Expert Answer

Step 1

A)

As per the balanced chemical equation,

1 mol Cu  forms 1 mol Cu(NO3)2

Which is,

63.5 g. of Cu forming 187.5 g. of Cu(NO3)2

If,

 63.5 g. of Cu is forming 187.5 g. of Cu(NO3)2

      0.1 g of Cu forms------------ ?g of Cu(NO3)2

     = 0.1 X187.5/63.5

     = 0.295 g. of Cu(NO3)2

Step 2

(B)

As per the balanced chemical equation,

1 mol Cu(NO3)2  forms 1 mol Cu(OH)2

Which is,

187.5 g. of Cu(NO3)2 forming 97.5 g. of Cu(OH)2

If,

187.5 g. of Cu(NO3)2 forms 97.5 g. of Cu(OH)2

0.295 g. of Cu(NO3)2 forms-------? g of Cu(OH)2

= 0.295 X 97.5 / 187.5

= 0.153 g. of Cu(OH)2

Step 3

(C)

As per the balanced chemical equation,

1 mol Cu(OH)2  forms 1 mol CuO

Which is,

97.5 g. of Cu(OH)2  forming 79.5 g. of CuO

If,

97.5 g. of Cu(OH)2  is forming 79.5...

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