The given reaction is:
H2(g) + CO2(g) → H2O(g) + CO(g)
i.e.
H-H + O=C=O → H-O-H + C≡O
Reaction enthalpy (ΔH) is given as:
ΔH = ∑ΔH(Bonds broken) + ∑ΔH(Bonds formed)
= [1*ΔH(H-H) + 2*ΔH(C=O)] + [2*ΔH(O-H) + 1*ΔH(C≡O)]
=[+436 +2(799)] + [2(-463) +1(-1072)] = 36 kJ/mol
The enthalpy change for the given reaction is 36 kJ/mol
<h2>Solutions:</h2>
<u>Case a:</u> Finding pH for [H⁺] = 1.75 × 10⁻⁵ mol/L :
As we know pH is given as,
pH = -log [H⁺]
Putting value,
pH = -log [1.75 x 10⁻⁵]
pH = 4.75
<u>Case b:</u> Finding pH for [H⁺] = 6.50 × 10⁻¹⁰ mol/L :
As we know pH is given as,
pH = -log [H⁺]
Putting value,
pH = -log [6.50 × 10⁻¹⁰]
pH = 9.18
<u>Case c:</u> Finding pH for [H⁺] = 1.0 × 10⁻⁴ mol/L :
As we know pH is given as,
pH = -log [H⁺]
Putting value,
pH = -log [1.0 × 10⁻⁴]
pH = 4
<u>Case d:</u> Finding pH for [H⁺] = 1.50 × 10⁻⁵ mol/L :
As we know pH is given as,
pH = -log [H⁺]
Putting value,
pH = -log [1.50 × 10⁻⁵]
pH = 4.82
The suns hear is uneven on the earth
I hope this helps
The answer is B. Flammability