%yield = 88.5%
<h3>Further explanation</h3>
Given
Reaction
Cu(s) + 2 AgNO₃(aq) → Cu(NO₃)₂(aq) + 2Ag(s)
Required
The percent yield
Solution
mol AgNO₃(MW=169,87 g/mol) :
= mass : MW
= 127 : 169.87
= 0.748
mol Ag from equation :
= 2/2 x mol AgNO₃
= 2/2 x 0.748
= 0.748
Mass Ag (theoretical) :
= mol x Ar Ag
= 0.748 x 108
= 80.784
% yield = (actual/theoretical) x 100%
%yield = 71.5/80.784 x 100%
<em>%yield = 88.5%</em>
Answer:
The Unknown V was shiny and gray, but it was a lightweight metal. THE UNKNOWN METALS5V was aluminum, which has a density of 2.75 g/ml(Smith, 2019).
Explanation:
pls mark me as brainleast and folow me
It’s the first 1 M yea it’s the first one
Answer:
A.'C
Explanation:
Please answer my question
Answer:
I would help but the picture will not load for me :(
Explanation:
