%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 answer to your question is given below.
Explanation:
Potassium (K) has 19 electrons with electronic configuration of 2, 8, 8, 1.
Fluorine (F) has 9 electrons with electronic configuration of 2, 7.
Fluorine needs 1 electron to complete it's octet configuration.
Hence, potassium (K), will lose 1 electron to fluorine (F) to form potassium ion (K+) with electronic configuration of 2, 8, 8. The fluorine atom (F) will receive the 1 electron from potassium to form the fluoride ion (F-) with electronic configuration of 2, 8.
**** Please see attached photo for further details.
Pretty sure it's b but not an definitely
Answer:
Oppositely charged particles attract each other. This attractive force is often referred to as an electrostatic force. An ionic bond is the electrostatic force that holds ions together in an ionic compound.
Explanation:
Hope this helps :)