(i) Assume that CuSO₄ + H₂O was the reaction that happened in that dish. Copper sulfate is anhydrous, so if water is added, it becomes copper sulfate pentahydrate.
CuSO₄(s) + H₂O(l) = CuSO₄ · 5H₂O(s)
Copper sulfate is soluble in water, so copper and sulfate both become aqueous. Copper sulfate has become crystallized, and the water has become attached to it.
(ii) As seen in my equation of (i), all that is left in the dish is a solidified, blue crystallized version which has a octahedral molecular geometry.
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Answer:
Electrons will flow from left to right through the wire.
Pb^2+ ions will be reduccd to Pb metal.
The concentration of Sn2+ ions in the left compartment will increase.
Explanation:
Looking at the relative electrode potentials of the two metals
Sn= -0.14
Pb=-0.13
Tin is expected to function as the anode (left hand half cell) and lead as the anode (right hand half cell) tin oxidizes to sn^2+ hence its concentration increases on the left compartment while lead is reduced to ordinary lead metal on the right hand half cell . since oxidation occurs on the left hand side, electrons flow from left to right.
The volume that will react with methane is 72.24dm3
Scientific methods are used to study living things or investigate questions and solve problems.
The balanced equation is:
![2Na_3N-\ \textgreater \ 6Na + N_2](https://tex.z-dn.net/?f=2Na_3N-%5C%20%5Ctextgreater%20%5C%206Na%20%2B%20N_2)
Then proceed with the following equations.
![100g Na_3N*(\frac{1molNa_3N}{82.98gNa_3N})*(\frac {6mol Na}{2molNa_3N})*(\frac {22.99gNa}{1molNa})=83.12gNa](https://tex.z-dn.net/?f=100g%20Na_3N%2A%28%5Cfrac%7B1molNa_3N%7D%7B82.98gNa_3N%7D%29%2A%28%5Cfrac%20%7B6mol%20Na%7D%7B2molNa_3N%7D%29%2A%28%5Cfrac%20%7B22.99gNa%7D%7B1molNa%7D%29%3D83.12gNa)
The answer is
![83.12gNa](https://tex.z-dn.net/?f=83.12gNa)
.