Consider the halogenation of ethene is as follows:
CH₂=CH₂(g) + X₂(g) → H₂CX-CH₂X(g)
We can expect that this reaction occurring by breaking of a C=C bond and forming of two C-X bonds.
When bond break it is endothermic and when bond is formed it is exothermic.
So we can calculate the overall enthalpy change as a sum of the required bonds in the products:
Part a)
C=C break = +611 kJ
2 C-F formed = (2 * - 552) = -1104 kJ
Δ H = + 611 - 1104 = - 493 kJ
2C-Cl formed = (2 * -339) = - 678 kJ
ΔH = + 611 - 678 = -67 kJ
2 C-Br formed = (2 * -280) = -560 kJ
ΔH = + 611 - 560 = + 51 kJ
2 C-I Formed = (2 * -209) = -418 kJ
ΔH = + 611 - 418 = + 193 kJ
Part b)
As we can see that the highest exothermic bond formed is C-F bond so from bond energies we can found that addition of fluoride is the most exothermic reaction
They speed up when they get hotter and they slow down when they get colder. I think
Answer is: the percent purity of the sodium bicarbonate is 56.83 %.
1. Chemical reaction: 2NaHCO₃ + H₂SO₄ → 2CO₂ + 2H₂O + Na₂SO₄.
2. m(NaHCO₃) = 3.50 g
n(NaHCO₃) = m(NaHCO₃) ÷ M(NaHCO₃).
n(NaHCO₃) = 3.50 g ÷ 84 g/mol.
n(NaHCO₃) = 0.042 mol.
3. From chemical reaction: n(NaHCO₃) : n(CO₂) = 1 : 1.
n(CO₂) = 0.042 mol.
m(CO₂) = 0.042 mol · 44 g/mol.
m(CO₂) = 1.83 g.
4. the percent purity = 1.04 g/1.83 g ·100%.
the percent purity = 56.8 %.
METALS ARE MAGNETIC(and maybe metalloids)
