Oxygen gas produced : 0.7 g
<h3>Further explanation</h3>
Given
10.0 grams HgO
9.3 grams Hg
Required
Oxygen gas produced
Solution
Reaction⇒Decomposition
2HgO(s)⇒2Hg(l)+O₂(g)
Conservation of mass applies to a closed system, where the masses before and after the reaction are the same
mass of reactants = mass of products
mass HgO = mass Hg + mass O₂
10 g = 9.3 g + mass O₂
mass O₂ = 0.7 g
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Answer:
it is easier for them to have an octet of electrons(8e)
Each ion will: obtain the noble gas structure, each atom has high ionization energy
Explanation:
.
Answer:
The traditional electrolyte for aluminium electrolysis is based on molten cryolite (Na3AlF6), acting as solvent for the raw material, alumina (Al2O3).Metals are found in ores combined with other elements. Electrolysis can be used to extract a more reactive metal from the ore.
Aluminum can and is used as both anodes and cathodes in electrochemical cells, but there are some peculiarities to using it as an anode in aqueous solutions. As you note, aluminum forms a passivating oxide layer quite readily, even by exposure to atmosphere. In an aqueous solution, if the potential is high enough, OH− and O2− are generated at the anode, which can then react with the aluminum to produce aluminum oxide. Al^3+ can also be generated directly. The electric field will draw the anions through the growing aluminum oxide layer towards the aluminum surface and the Al^3+ towards the solution, making the oxide layer grow both away from the electrode surface and into the surface of the electrode. In this way, coatings thicker than the normal passivation in air can be produced. However, aluminum oxide is a good electrical insulator, thus if a dense non-porous layer is grown, it will become impossible to pass current through it and growth will stop, leaving a relatively thin oxide layer (this is how the dielectric layers in electrolytic capacitors are made). This is the normal behaviour in aqueous solutions at near-neutral pH (5–7).
However, if a thick aluminum oxide layer is desired (e.g. to produce coatings on aluminum parts for dying or durability), maintaining porosity is necessary to avoid completely blocking access to the surface. One technique that is commonly used is using a low pH solution, which tends to redissolve some of the oxide and neutralize some of the formed OH−, leaving pores in the oxide layer through which the ions can travel and continue to react. These pores also give a good structure to retain dyes or lubricants, but generally need to be sealed after to protect against corrosion.
Answer:
(1) order = 2
(2) R = K [A]²
Explanation:
Given the reaction:
A--------->Product
The rate constant relation for the reaction is given as:
R(i) = K [A]............(*)
Where R(I) is rate constant at different concentration of A.
Taking the rate constant as R1, R2 and R3 for the different concentrations respectively. Then the following equations results
0.011 = K [0.15] ⁿ.........(1)
0.044 = K [0.30]ⁿ .......(2)
0.177 = K [0.60]ⁿ .........(3)
Dividing (2) by (1) and (3) by (1)
Gives:
0.044/0.011 = [0.3/0.15]ⁿ
4 = 2ⁿ; 2² = 2ⁿ; n = 2
Similarly
0.177/0.011 = [0.60/0.15]ⁿ
16.09 = 4ⁿ
16.09 = 16 (approximately)
4² = 4ⁿ ; n = 2
Hence the order of the reaction is 2.
The rate law is R = K [A]²
