0.33 g of hydrosulfuric acid, H25, is added to 50.0 mL of water. What is the concentration (molarity) of [H301 of this solution? H25 --> 2H +52
2H+ + 2 H20 --> 2 H30* + 2 OH
[Do not include units in your answer]

The concentration (molarity) of [H₃O⁺]=0.776 M

We balance the reaction first so that the number of atoms and the charge of the product and reactant are the same

Reaction(balanced) :

H₂S⇒2H⁺+S²⁻ (eq 1)

2H⁺+6H₂O⇒4H₃O⁺+2OH⁻ (eq 2)

mass of H₂S=0.33 g

mol H₂S (MW=34 g/mol) :

From equation 1, mol H⁺ = 2 x mol H₂S , so mol H⁺ =

From equation 2, mol H₃O⁺ = 2 x mol H⁺(4 : 2) , so mol H₃O⁺ :

Or simply mol H₃O⁺ = 4 x mol H₂S = 4 x 0.0097 = 0.0388

the molarity of [H₃O⁺] =

i can ! i just need to know what the paragraph is! : p

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm