The balanced equation for the neutralisation reaction is as follows
2NaOH + H₂SO₄ ---> Na₂SO₄ + 2H₂O
stoichiometry of NaOH to H₂SO₄ is 2:1
the number of moles of NaOH reacted - 0.126 mol/L x 0.0173 L = 0.00218 mol
if 2 mol of NaOH reacts with 1 mol of H₂SO₄
then 0.00218 mol of NaOH reacts with - 0.00218 / 2 = 0.00109 mol of H₂SO₄
molarity is the number of moles of solute in 1 L solution
therefore if 25 mL contains - 0.00109 mol
then 1000 mL contains - 0.00109 mol / 25 mL x 1000 mL = 0.0436 mol/L
therefore molarity of H₂SO₄ is 0.0436 M
The answer is D. A compound
Answer : The energy of the photon emitted is, -12.1 eV
Explanation :
First we have to calculate the 
orbit of hydrogen atom.
Formula used :
where,
= energy of 
orbit
n = number of orbit
Z = atomic number of hydrogen atom = 1
Energy of n = 1 in an hydrogen atom:
Energy of n = 2 in an hydrogen atom:
Energy change transition from n = 1 to n = 3 occurs.
Let energy change be E.
The negative sign indicates that energy of the photon emitted.
Thus, the energy of the photon emitted is, -12.1 eV
Chemical science deals with mater like compound mater
