I think c I’m not positive about the question but I think c
Answer:
= 74.4 grams / mole. Ernest Z. The reaction will produce 15.3 g of KCl
Explanation:
Answer:
The answer to your question is 3 moles of AlCl₃
Explanation:
Process
1.- Write and balance the equation
Al(NO₃)₃ + 3NaCl ⇒ 3NaNO₃ + AlCl₃
2.- Determine the limiting reactant
Theoretical proportion 1 mol Al(NO₃)₃ : 3 moles of NaCl
Experimental proportion 4 moles Al(NO₃)₃ : 9 moles NaCl
From these values, we determine that the limiting reactant is NaCl because the number of moles increases three times and the number of moles of Al(NO₃)₃ increases four times.
3.- Determine the amount of AlCl₃ using proportions
3 moles of NaCl --------------- 1 mol of AlCl₃
9 moles of NaCl ---------------- x
x = (9 x 1) / 3
x = 9 /3
x = 3 moles
Answer : The value of
is, 0.34 V
Explanation :
Here, copper will undergo reduction reaction will get reduced. Zinc will undergo oxidation reaction and will get oxidized.
The oxidation-reduction half cell reaction will be,
Oxidation half reaction: 
Reduction half reaction: 
Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.
The overall balanced equation of the cell is,

To calculate the
of the reaction, we use the equation:


Putting values in above equation, we get:


Hence, the value of
is, 0.34 V