Sorry, haven't came across that as of yet.
Answer:
133 g
Explanation:
Step 1: Write the balanced equation
2 Al(s) + 3 Br₂(l) ⇒ 2 AlBr₃(s)
Step 2: Calculate the moles corresponding to 15.0 g of Al
The molar mass of aluminum is 26.98 g/mol. The moles corresponding to 15.0 g of Al are:
Step 3: Calculate the moles of Br₂ that react with 0.556 moles of Al
The molar ratio of Al to Br₂ is 2:3. The moles of bromine that react with 0.556 moles of aluminum are:
Step 4: Calculate the mass corresponding to 0.834 moles of Br₂
The molar mass of bromine is 159.81 g/mol. The mass corresponding to 0.834 moles of Br₂ is:
Sounds good, but would do little to explain why lithium, with 3 electrons, is more reactive than Helium with 2, or why Caesium is more reactive than Sodium, although it clearly has far more electrons with which to shield its nucleus.
Hydrogen is unusual in having a fairly exposed nucleus, but chemistry is not very much about the nucleus, it is about the way the electrons themselves interact. As Lightarrow suggests, it does help if you know the quantum behaviour of electrons in an atom (which I do not claim to know), but it basically boils down to electrons preferring some configurations over others.
At the simplest, the comparison between hydrogen and helium – it is not really to do with the nucleus, it is more to do with electrons liking to be in pairs. Electrons have (like most common particles) two possible spin states, and they are more stable when an electron in one spin state is paired with an electron in the opposite spin state. When two hydrogen atoms meet, the electrons each one of them hold can be shared between them, forming a more stable pair of electrons, and thus binding the two atoms together.
All of the group 1 atoms (hydrogen, lithium, sodium, potassium, caesium; all share the characteristic that they have an odd number of electrons, and that one of those electrons is relatively unstable. The reason that the heavier atoms are more reactive is quite contrary to the argument that Lightarrow put forward – it is not because of a stronger electrical reaction with the nucleus, but because of the larger number of electrons in the bigger atoms, they are actually more weakly attached to their own nucleus, and so more readily interact with the electrons of other atoms.
Another, even more stable configuration for the electrons around an atom requires 8 electrons. This gives the noble gases (apart from Helium) their stability, but it also gives atoms like chlorine and fluorine their reactivity. Atoms like those of chlorine and fluorine are only one electron short of having a group 8 electrons available to them, and so will readily snatch an electron from another atom (particularly if it is an atom that has a single loose electron, such as sodium or caesium) in order to make up that group of 8 electrons.
The above explanation is very crude, and really does need a proper understanding of the quantum states of electrons to give a better quantitative answer (it is probably the kind of answer that might have been acceptable in the 1920s or 1930s – the Bohr orbital model of the atom, but has now been superseded by better explanations of what goes on amongst the electrons of an atom).
Gay-Lussac's law: 
=
We need to convert temperature from Celsius to Kelvin
-55 C = 218 K
30 C = 303 K
V2 = V1*T2 / T1 = 0.550 *303 / 218 = 0.764 l
Answer:
Following are the solution to the given question:
Explanation:
In the aqueous solution of 
ion is used to represent a ![[Hg(H_2O)_4]^{2+}](https://tex.z-dn.net/?f=%5BHg%28H_2O%29_4%5D%5E%7B2%2B%7D)
.
It is used to form a complex with iodide ions as ![[Hgl_4]^{2-}](https://tex.z-dn.net/?f=%5BHgl_4%5D%5E%7B2-%7D)
.
Conversion reaction:
![[Hg(H_2O)_4]^{2+} +4I^{-} \to [HgI_4]^{2-}+4H_20](https://tex.z-dn.net/?f=%5BHg%28H_2O%29_4%5D%5E%7B2%2B%7D%20%2B4I%5E%7B-%7D%20%5Cto%20%5BHgI_4%5D%5E%7B2-%7D%2B4H_20)
Following are the formation constants:
![k_f=\frac{[HgI_4]^{2-}}{[Hg(H_2O)_4]^{2+} [I^{-}]^4}](https://tex.z-dn.net/?f=k_f%3D%5Cfrac%7B%5BHgI_4%5D%5E%7B2-%7D%7D%7B%5BHg%28H_2O%29_4%5D%5E%7B2%2B%7D%20%5BI%5E%7B-%7D%5D%5E4%7D)
Following are the first step to the conversion:
![[Hg(H_2O)_4]^{2+} +I^{-} \to [HgI(H_2O)_3]^{+}+H_20](https://tex.z-dn.net/?f=%5BHg%28H_2O%29_4%5D%5E%7B2%2B%7D%20%2BI%5E%7B-%7D%20%5Cto%20%5BHgI%28H_2O%29_3%5D%5E%7B%2B%7D%2BH_20)
