A certain compound on burning in air forms three oxides. One of the oxides turned lime water milky, the other turned anhydrous CuSO4​ blue and third formed a solution of pH=9. Compound is formed of :
A. $S,N,and\;H$
B. $S,Nand\;C$
C. $S,C\;and\;H$
D. $S,H\;and\;Na$

Hint: At the point when substances consume in air they join with oxygen in a burning response to frame oxide compounds. On the off chance that there is a restricted measure of air/oxygen, the consuming must be continued temporarily.

Complete step by step answer:
Compound is a form of $S,H\;and\;Na$ as it responds with lime water and makes it milky.
Water responds with anhydrous $CuS{O_4}$ and turns it blue. The response is
$CuS{O_4} + {H_2}O \to CuS{O_4}.5{H_2}O$
Also, sodium oxide is a straightforward simple strongly basic oxide. It is basic because it contains the oxide particle${O^{2 - }}$ , which is a strong base with a high tendency to combine with hydrogen particles.

So, the correct answer is Option D.

ADDITIONAL INFORMATION:
What are bases?
The ionic compounds that produce negative hydroxide particles when broken down in water are called bases.
What are the types of bases?
Solid base – It is an aggravate that has a capacity to eliminate a proton from an extremely feeble corrosive.
Powerless base – There is inadequate separation when in water. The watery arrangement contains both the frail base just as its form corrosive.
Super base – These bases are better at deprotonation when contrasted with a solid base. These have extremely powerless form acids.
Impartial base – It shapes a bond with a nonpartisan corrosive that offers an electron pair.
Strong base – It is dynamic in strong structure.

Note: Oxides are synthetic mixes with at least one oxygen molecule joined with another component. Oxides are double mixes of oxygen with another element.
The rate or speed at which the reactants consolidate is high, partially due to the idea of the compound response itself, the consuming must be continued temporarily.