Answer: c
that is the answer
Answer:
283.725 kJ ⋅ mol − 1
Explanation:
C(s) + 2Br2(g) ⇒ CBr4(g) , Δ H ∘ = 29.4 kJ ⋅ mol − 1
Br2(g) ⇒ Br(g) , Δ H ∘ = 111.9 kJ ⋅ mol − 1
C(s) ⇒ C(g) , Δ H ∘ = 716.7 kJ ⋅ mol − 1
4*eqn(2) + eqn(3) ⇒ 2Br2(g) + C(s) ⇒ 4 Br(g) + C(g) , Δ H ∘ = 1164.3 kJ ⋅ mol − 1
eqn(1) - eqn(4) ⇒ 4 Br(g) + C(g) ⇒ CBr4(g) , Δ H ∘ = -1134.9 kJ ⋅ mol − 1
so,
average bond enthalpy is 
= 283.725 kJ ⋅ mol − 1
Radium and polonium is the answer to this question. I hope I helped out!
However <em>trans</em>-2-Butene does not give a characteristic peak in 1620-1680 cm⁻¹ region but still the presence of carbon double bond carbon can be detected by detecting following peaks in IR Spectrum.
1) 3010-3100 cm⁻¹:
As in trans-2-Butene a hydrogen atoms ate attached to sp² hybridized carbon, therefore the stretching of =C-H (C-H) bond will give a peak of medium intensity in the range of 3010-3100 cm⁻¹.
2) 675-1000 cm⁻¹:
Another peak which is given by the bending of =C-H (C-H) bond with strong intensity will appear in the range of 675-1000 cm⁻¹.
1.Decomposition i think
2.boiling
3.It is a solid at room temperature and pressure.
4.<span>The base donates a hydrogen ion.
5.That causes the oxidation of another element
6.</span>MnO2
7.When a substance is reduced, electrons are lost.
8.True I think
9.False
10.True
Hope these are correct
