Answer:
A. N₂(g) + 3H₂(g) -----> 2NH₃ exothermic
B. S(g) + O₂(g) --------> SO₂(g) exothermic
C. 2H₂O(g) --------> 2H₂(g) + O₂(g) endothermic
D. 2F(g) ---------> F₂(g) exothermic
Explanation:
The question says predict not calculate. So you have to use your chemistry knowledge, experience and intuition.
A. N₂(g) + 3H₂(g) -----> 2NH₃ is exothermic because the Haber process gives out energy
B. S(g) + O₂(g) --------> SO₂(g) is exothermic because it is a combustion. The majority, if not all, combustion give out energy.
C. 2H₂O(g) --------> 2H₂(g) + O₂(g) is endothermic because it is the reverse reaction of the combustion of hydrogen. If the reverse reaction is exothermic then the forward reaction is endothermic
D. 2F(g) ---------> F₂(g) is exothermic because the backward reaction is endothermic. Atomisation is always an endothermic reaction so the forward reaction is exothermic
Answer:
SORRY IM LATE! at least i saw it.
Explanation:
Kingdom is Eucaryota
domain is eucarya
The products. Without a balanced chemical equation, there is no reaction that can take place. The products is the final reation and the reactants help the final reaction take place.
Answer:
1.089%
Explanation:
From;
ν =1/2πc(k/meff)^1/2
Where;
ν = wave number
meff = reduced mass or effective mass
k = force constant
c= speed of light
Let
ν =1/2πc (k/meff)^1/2 vibrational wave number for 23Na35 Cl
ν' =1/2πc(k'/m'eff)^1/2 vibrational wave number for 23Na37 Cl
The between the two is obtained from;
ν' - ν /ν = (k'/m'eff)^1/2 - (k/meff)^1/2 / (k/meff)^1/2
Therefore;
ν' - ν /ν = [meff/m'eff]^1/2 - 1
Substituting values, we have;
ν' - ν /ν = [(22.9898 * 34.9688/22.9898 + 34.9688) * (22.9898 + 36.9651/22.9898 * 36.9651)]^1/2 -1
ν' - ν /ν = -0.01089
percentage difference in the fundamental vibrational wavenumbers of 23Na35Cl and 23Na37Cl;
ν' - ν /ν * 100
|(-0.01089)| × 100 = 1.089%
