Mn+2 is cation and CO3 is anion
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Let us assume that there is a 100g sample of Opal. The masses of each element will be:
29.2g Si
33.3g O
37.5g H2O
Now we divide each constituent's mass by its Mr to get the moles present
Si: (29.2 / 28) = 1.04
O: (33.3 / 16) = 2.08
H2O: (37.5 / 18) = 2.08
Now we divide by the smallest number and obtain:
Si: 1
O: 2
H2O: 2
Thus, the empirical formula of Opal is:
SiO2 . 2H2O
Answer: 
Explanation:
According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products must be same as the mass of reactants.
For the conservation of mass, the number of atoms of each element must be same in reactants and products. Thus we need to balance the reaction by writing appropriate stochiometric coefficients.
All the hydrocarbons burn in oxygen to form carbon dioxide and water.Thus the complete balanced equation is:
Answer:The 1st and 2nd reactions are the example of oxidation -reduction.
Explanation:
Oxidation is basically when a species loses electrons and reduction is basically when the species gains electrons.
A reaction is known as an oxidation -reduction reaction only if oxidation and reduction simultaneously occur in the reaction. It basically means if a species is getting oxidized in the reaction then the other species present in the system must be reduced in the reaction.
Oxidation-reduction reactions are also known as redox reactions.
In the 1st reaction the oxidation state of Na in reactant is 0 and in products is +1 hence Na is oxidized and the oxidation state of chlorine is 0 in reactants and in products is -1 so chlorine is reduced. Hence Na is oxidized and Cl is reduced so the reaction is a example of oxidation-reduction.
2Na(s)+Cl₂(g)→2NaCl(s)
In the second reaction the oxidation state of Na in reactant is 0 and in products is +1 hence Na is oxidized and the oxidation state of Cu is +1 in reactant and 0 in products so Cu is reduced. Hence Na is oxidized and Cu is reduced so the reaction is an example of oxidation-reduction.
Na(s)+CuCl(aq)→NaCl(aq)+Cu(s)
In the third reaction the oxidation state of Na changes from +1 to +1 and that of Cu also changes from +1 to +1. So there is no change in oxidation state of the species present in reactants and products. Hence this reaction is not an example of oxidation and reduction.
