<span>N2(g) + 2 O2(g) → 2 NO2(g)
</span>
(1) N2(g) + O2(g) → 2 NO(g) Δh = +180.7 kJ
(2) 2 NO(g) + O2(g) → 2 NO2(g) Δh = −113.1 kJ
(3) 2 N2O(g) → 2 N2(g) + O2(g) Δh = −163.2 kJ
You want to rearrange the given reactions to match the reaction with the unknown enthalpy. Looking at what we're given, reaction (2) already has 2NO2 on the product side just like our unknown, so let's start there - it can stay as is.
However, it has 2NO on the reactant side - that's not in our unknown reaction, so we need to get rid of it somehow. There's 2NO in reaction (1), and it's on the product side, so it will cancel the 2NO on the reactants side of reaction (2). This means reaction (1) can stay as it is as well.
If you add these two reactions together, you'll get:
N2(g) + O2(g) + 2 NO(g) + O2(g)→ 2 NO(g) + 2 NO2(g)
N2(g) + 2O2(g) + 2 NO(g) → 2 NO(g) + 2 NO2(g)
N2(g) + 2O2(g) → 2 NO2(g)
The 2 NO cancels out and the individual O2's add together to get 2O2. This gives us the unknown reaction so we actually don't need to use reaction (3) at all to answer this question. In fact, it makes things more difficult because it contains N2O, which cannot be cancelled in any way with the reactions we're given.
To find the enthalpy of the unknown reaction, add the enthalpies of the reactions that work. Since we didn't have to modify either of the the first 2 reactions, you can add their original enthalpies as given:
+180.7 kJ + (−113.1 kJ) = +67.6 kJ
265g solution x (84.5g Fe(NO3)2 / 100g solution) x (1 mole Fe(NO3)2 / __g Fe(NO3)2) x (3 moles Fe(s) / 3 moles Fe(NO3)2) x (__g Fe / mole Fe) = __g Fe
<span>plug in molar mass of Fe(NO3)2 and Fe and calculate</span>
I believe that what happens here is that when
those cells carrying those intercellular signals do no pass through the
connections, actually the signal is passed through or are transferred to adjacent
cells for signal transmission, hence the answer is:
<span>B</span>
I think the answer is systems tend to undergo changes towards lower energy and higher entropy.
Usually the point of chemical reactions is to go from a state of higher energy to lower energy (it is possible for reactions to go from lower to higher energy though as in the case of endothermic reactions).
As for entropy, usually it requires energy to lower the entropy of a system since entropy can be thought of as the relative disorder of the system. If you go from a state of higher disorder to lower disorder work usually needs to be done.
This trend can be shown in the equation ΔG=ΔH-TΔs:
ΔG stands for Gibbs free energy
ΔH stands for change in enthalpy
ΔS stands for change in entropy
T stands for temperature
when ever ΔG is negative the reaction is spontaneous which means that it is favored. As you can see ΔG is negative at any temperature when ΔH is negative and ΔS is positive showing that going to a lower energy state and raising the entropy is usually favored in nature.
I hope this helps. Let me know if anything is unclear or if you need further explanation.