O3 + M2+(aq) + H2O(l) => O2(g) + MO2(s) + 2 H+
Eo(cell) = Eo(O3/O2) - Eo(MO2/M2+)
0.44 = 2.07 - Eo(MO2/M2+)
Eo(MO2/M2+) = 1.59 V
C. a symmetrical molecule is always nonpolar
Formation of ammonia by nitrogen and hydrogen is habers process wher 28g N2 results in formation of 34g NH3
so 35g N2 will form 34*35/28=42.5g NH3 where it given that reaction takes place in excess of H2
N2+3H2 gives 2NH3
I'm gonna go ahead and guess, and say its B.
Answer:
The heat released by the combustion is 20,47 kJ
Explanation:
Bomb calorimeter is an instrument used to measure the heat of a reaction. The formula is:
Q = C×m×ΔT + Cc×ΔT
Where:
Q is the heat released
C is specific heat of water (4,186kJ/kg°C)
m is mass of water (1,00kg)
ΔT is temperature change (23,65°C - 20,45°C)
And Cc is heat capacity of the calorimeter (2,21kJ/°C)
Replacing these values the heat released by the combustion is:
<em>Q = 20,47 kJ</em>
