Answer: 37.5grams of Cu(NO3)2
Cu(1mol) + 2HNO3(2mol) —> Cu(NO3)2 + H2
<em>125 grams of Cu(1mol) reacts with 75 grams of HNO3(2mol)</em>
<em><u>HNO3 is the limiting substance, therefore, 75 grams is the limiting quantity.</u></em>
<em>Therefore, 2mol of HNO3 forms 1mol of Cu(NO3)2</em>
<em>75 grams of HNO3 forms...75grams x 1mol/2mol = 37.5 grams of Cu(NO3)2</em>
Answer:
24 is the correct anwer
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Answer: 0.52V
Explanation:
Ecell = Ecell(standard) - [(0.0592 logQ)/n]
Q = product of the quotient
n = no of electrons transferred = 2
Ecell = 0.63 - [(0.0592*Log(1 / 2.0 * 10-4) / 2]
Ecell = 0.63 - 0.0194
Ecell = 0.5205V
Answer:
the energy required to do work
