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
