Answer:
Explanation:
Ba(s) + Mn²⁺ (aq,1M)  →      Ba²⁺ (aq,1M) + Mn(s)
Ba⁺²(aq) +2e →  Ba(s) ,  E° =  −2.90 V  
Mn⁺²(aq) +2e → Mn(s),  E⁰  =0.80 V
Anode reaction : 
Ba(s)  →  Ba⁺²(aq) +2e        E° =  −2.90 V  
Cathode reaction : 
Mn⁺²(aq) +2e → Mn(s)          E⁰  =0.80 V
Cell potential = Ecathode  - Eanode 
Ecell  = .80 - ( - 2.90 )
Ecell = 3.7 V .
equilibrium constant ( K ) : 
Ecell = .059 log K  / n 
n = 2 
3.7  = .059 log K  / 2 
log K = 125.42 
K = 2.63 x 10¹²⁵ . 
Free energy change : 
ΔG = - n F Ecell 
= - 2 x 96500 x 3.7 
= 714100 J 
= 7.141 x 10⁵ J . 
 
        
             
        
        
        
False, they have a balance between positive and negitive characters. 6 protons, 6 electrons
        
             
        
        
        
<span>Electrons (negatively charged) are found in almost all the matter we see. They "buzz" around positively charged atomic nuclei; you may be familiar with one not-very-correct representation of an atom.
I hope helped ^-^</span>
        
             
        
        
        
IE1 = 578 kJ/mol 
IE2 = 1820 kJ/mol 
IE3 = 2750 kJ/mol 
IE4 = 11600 kJ/mol
If the following set of successive ionization energies are
your ionization values this would likely belong to Aluminum. Since there is a
huge point between the third and fourth ionization energies, which designates
that the atom reached noble gas configuration after the third electron was
removed. The element which has 3 valence electrons in the third period is
aluminum.
 
        
                    
             
        
        
        
Answer:
Explanation: AH, AG and AS for the reaction at 873 K
delta G =-RT ln Kp 
G= -8.314 x 873 ln 0.9 
= 764.7J/mol
AG = AH-TAS 
AH = 764.7 - 8314(0.9) 
Kp =pCO/pCo2 
Substitute the values at different temp. Solve simultaneously 
Since mole fraction =kp/pCo
Mf =1.5