Answer is: 28 kJ.
Chemical reaction: A₂ + B₂ ⇄ 2AB.
Ea(forward) = 105 kJ/mol.
Ea(reverse) = 77 kJ/mol.
ΔH(reaction) = ?
<span>The enthalpy change of reaction is the change in the energy of the reactants to the products.
</span>ΔH(reaction) = Ea(forward) - Ea(reverse).
ΔH(reaction) = 105 kJ/mol - 77 kJ/mol.
ΔH(reaction) = 28 kJ/mol; this is endothermic reaction (ΔH <span>> 0).</span>
It's difficult to write it down, but I'll attach you a good example of hydroboration of indene. I hope you'll find it helpful.
1 mole CO2 = 44.0096 grams CO2
<span>2.1 mol CO2 x (44.0096 grams CO2/1 mole CO2) = 92.4 grams CO2</span>
Answer:
The concentration of monosodium phosphate is 0.1262M
Explanation:
The buffer of H₂PO₄⁻ / HPO₄²⁻ (Monobasic phosphate and dibasic phosphate has a pKa of 7.2
To determine the pH you must use Henderson-Hasselbalch equation:
pH = pKa + log [A⁻] / [HA]
<em>Where [A⁻] is molarity of the conjugate base of the weak acid, [HA].</em>
For H₂PO₄⁻ / HPO₄⁻ buffer:
pH = 7.2 + log [HPO₄⁻² ] / [H₂PO₄⁻]
As molarity of the dibasic phosphate is 0.2M and you want a pH of 7.4:
7.4 = 7.2 + log [0.2] / [H₂PO₄⁻]
0.2 = log [0.2] / [H₂PO₄⁻]
1.58489 = [0.2] / [H₂PO₄⁻]
[H₂PO₄⁻] = 0.1262M
<h3>The concentration of monosodium phosphate is 0.1262M</h3>
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