Pick the correct statement about the pure water. Group of answer choices Pure water contains no ions. Pure water contains equal amounts of hydroxide [OH-] and hydronium [H3O ] ions. Pure water contains larger amounts of hydroxide [OH-] than hydronium [H3O ] ions. Pure water is an electrolyte. Pure water contains smaller amounts of hydroxide [OH-] than hydronium [H3O ] ions.

Pure water contains no ions.  TRUE

Pure water contains equal amounts of hydroxide [OH-] and hydronium [H3O+ ] ions.  TRUE

Explanation:

This is the equilibrium for pure water:

2H₂O  ⇄  H₃O⁺  +  OH⁻          Kw

We see that pure water has no Ions. Pure water can not conduct electricity.

Generally ionized water comes from the water tap.

Another feature of pure water is pH.

Definetely pH of pure water is : 7

As pH = 7, [H₃O⁺] = 1×10⁻⁷

Then, [OH⁻] = 1×10⁻⁷

This is reazonable because Kw is 1×10⁻¹⁴ and Kw = [H₃O⁺] . [OH⁻]

In conclussion:

Pure water contains no ions.  TRUE

Pure water contains equal amounts of hydroxide [OH-] and hydronium [H3O+ ] ions.  TRUE

Pure water contains larger amounts of hydroxide [OH-] than hydronium [H3O+] ions.  FALSE

Pure water is an electrolyte. FALSE

Pure water contains smaller amounts of hydroxide [OH-] than hydronium [H3O ] ions. FALSE

answer is: empirical formula is tl₂o. if we use 100 grams of compound, tha:

m(tl) = 96.2%  · 100 g ÷ 100%.

m(tl) = 96.2 g.

n(tl) = m(tl)  ÷ m(tl). n(tl) = 96.2 g  ÷ 204.4 g/mol.

n(tl) = 0.47 mol. m(o) = 3.77 g. n(o) = m(o)  ÷ m(o). n(o) = 3.77 g  ÷ 16 g/mol. n(ag) = 0.235 mol. n(tl) . n(o) = 0.47 mol : 0.235 mol /0.235. n(tl) . n(o) = 2 : 1.

m(tl₂o) = 204.4 g/mol  · 2 + 16 g/mol = 424.8 g/mol.

m(compound) = 456. 8 g/mol - 424.8 g/mol = 32 g/mol ( 2 oxygens); tl₂o₃.