Answer:
They curve in opposite directions because of one reason(s) latitude and longitude. Because these are different places, they will experience different weather conditions based on how far they are from sea level.
Explanation:
Answer:
Hey!
0.8730 / 9.650 = 8.42445
Explanation:
8.42445 TO 2 s.f... (best option for rounding)
= 8.4
ANSWER = 8.4 (2 s.f)
HOPE THIS HELPS!!
It is a covalent bond. Whenever a compound uses such suffixes like mono, di, tri, tetra, and so on, it is a covalent compound- thus having covalent bonds.
N2(g) + 3H2(g) =2NH3(g)
Number of moles of reactants > product...
Therfore if pressure is increased.
Equilibrium is disturbed according to LCP....
Equilibrium shift to the right (product)
......
If you want to understant the fundamental concept:
Take for example 2A + 3B = 4C
Reactant: 5 moles (5 volumes)
Product: 4 moles (4 volumes)
When pressure of a gas is increased, volume decreases!
(Vice-versa)
If pressure is increased, volume decreases. Hence number of collisions INCREASES(constrain). Equilibrium shifts in such a direction so as to decrease the number of collision accordinf to LCP...
This happens when number of paeticles decreases as equilibrium shift forward because the forward reaction is accompanied by a decrease in number of particles (5 to 4)
Answer:
ΔH = -135.05 kJ
Explanation:
(1) 2 ClF(g) + O₂(g) → Cl₂O(g) + OF₂(g) ΔHo = 167.5 kJ
(2) 2 F₂(g) + O₂(g) → 2 OF₂(g) ΔHo = −43.5 kJ
(3) 2 ClF₃(l) + 2 O₂(g) → Cl₂O(g) + 3 OF₂(g) ΔHo = 394.1 kJ
We can use Hess' Law to calculate the ΔH of reaction:
ClF(g) + 1/2O₂(g) → 1/2Cl₂O(g) + 1/2OF₂(g) ΔHo = 167.5 x 1/2 = 83.75 kJ
F₂(g) + 1/2O₂(g) → OF₂(g) ΔHo = −43.5 x 1/2 = -21.75 kJ
1/2Cl₂O(g) + 3/2 OF₂(g) → ClF₃(l) + O₂(g) ΔHo = 394.1 x -1 x 1/2 = -197.05kJ
______________________________________________
(4) ClF (g) + F₂ (g) → ClF₃ (l) ΔH = 83.75 kJ-21.75 kJ-197.05kJ
ΔH = -135.05 kJ/mol
