DeltaH formation = deltaH of broken bonds - deltaH of formed bonds
Broken bonds: tiple bond N-N and H-H bond
Formed bonds: N-H and N-N bonds
You also have to take note of the molar coefficients
deltaH formation = <span> [(N≡N) + 2 * (H-H)] - [4 * (N-H) + (N-N)]
= (945 + 2*436) - (4*390 + 240)
= 17 kJ/mol
The answer is 17 kJ/mol.</span>
Answer: It takes 3.120 seconds for the concentration of A to decrease from 0.860 M to 0.260 M.
Explanation:
Integrated rate law for second order kinetics is given by:
k = rate constant = 
= initial concentration = 0.860 M
a= concentration left after time t = 0.260 M
Thus it takes 3.120 seconds for the concentration of A to decrease from 0.860 M to 0.260 M.
Answer:
ΔH = 57.04 Kj/mole H₂O
Explanation:
60ml(0.300M Ba(OH)₂(aq) + 60ml(0.600M HCl(aq)
=> 0.06(0.3)mole Ba(OH)₂(aq) + 0.60(0.6)mole HCl(aq)
=> 0.018mole Ba(OH)₂(aq) + 0.036mole HCl(aq)
=> 100% conversion of reactants => 0.018mole BaCl₂(aq) + 0.036mole H₂O(l) + Heat
ΔH = mcΔT/moles H₂O <==> Heat Transfer / mole H₂O
=(120g)(4.0184j/g°C)(27.74°C - 23.65°C)/(0.036mole H₂O)
ΔH = 57,042 j/mole H₂O = 57.04 Kj/mole H₂O
117.22 g are needed to react with an excess of Fe2O3 to produce 156.2 g of Fe.
Explanation:
Moles of Fe = Mass of Fe in grams / Atomic weight of Fe
= 156.2 / 55.847
Moles of Fe = 2.79.
The ratio between CO and Fe id 3 : 2.
Moles CO needed = 2.79 * (3 / 2)
= 4.185.
To calculate Atomic weight of CO,
Atomic weight of carbon = 12.011
Atomic weight of oxygen= 15.9994
Atomic weight of CO = 12.011 + 15.9994 = 28.01 g / mol.
Mass of CO = 4.185 * 28.01 = 117.22 g.
