The reaction;
O(g) +O2(g)→O3(g), ΔH = sum of bond enthalpy of reactants-sum of food enthalpy of products.
ΔH = ( bond enthalpy of O(g)+bond enthalpy of O2 (g) - bond enthalpy of O3(g)
-107.2 kJ/mol = O+487.7kJ/mol =O+487.7 kJ/mol +487.7kJ/mol =594.9 kJ/mol
Bond enthalpy (BE) of O3(g) is equals to 2× bond enthalpy of O3(g) because, O3(g) has two types of bonds from its lewis structure (0-0=0).
∴2BE of O3(g) = 594.9kJ/mol
Average bond enthalpy = 594.9kJ/mol/2
=297.45kJ/mol
∴ Averange bond enthalpy of O3(g) is 297.45kJ/mol.
Had to look for the options and here is my answer.
When we say that a redox reaction is spontaneous, this would mean that there is a formation of positive voltage <span>across the electrodes of a voltaic cell. Therefore, the system that this kind of reaction produces electrical energy is in a GALVANIC CELL. Hope this helps.</span>
Answer:
The veins that carry oxygenated bloof back into the heart are the pulmonary arteries.
Explanation:
Oxygen-rich blood flows from the lungs back into the left atrium (LA), or the left upper chamber of the heart, through four pulmonary veins. Oxygen-rich blood then flows through the mitral valve (MV) into the left ventricle (LV), or the left lower chamber.
In order to emit electrons, the cesium will have to absorb photons. Each photon will knock out one electron by transferring its energy to the electron. Therefore, by the principle of energy conservation, the energy of the removed electron will be equal to the energy of the incident photon. That energy is calculated using Planck's equation:
E = hf
E = 6.63 x 10⁻³⁴ * 1 x 10¹⁵
E = 6.63 x 10⁻¹⁹ Joules
The electron will have 6.63 x 10⁻¹⁹ Joules of kinetic energy
So to put them all in the same units we have
<span>2500 mL </span>
<span>250 mL </span>
<span>25mL </span>
<span>2,500,000,000mL </span>
<span>So the third one is the smallest</span>
