If the ph of a 1.00-in. rainfall over 1300 miles2 is 3.30, how many kilograms of sulfuric acid, h2so4, are present, assuming that it is the only acid contributing to the ph? for sulfuric acid, ka1 is very large and ka2 is 0.012.

Hey there!:

Volume of rainfall = Area X height

Height = 1 in = 0.254 dm

1mile2 = 2.59 X 10⁸ dm²

so 1300 miles² = 3367 *10⁸ dm²

Volume = 0.254 * 3367 * 10⁸ = 855.218 * 10⁸ dm³

Also,

H2SO4  ionizes as  :

H2SO4 --> H⁺ + HSO4⁻

Ka1 is very high  so it completely dissociates , due to first dissociation [H⁺] = Concentration of H2SO4

HSO4⁻  --> H⁺  + SO₄⁻²

We also consider it as high dissociation :

so each mole of H2SO4 will give 2 moles of H⁺ (proton)

pH = 3.30  = -log[H⁺]

So [H+] = 0.000501 M = 0.000501 moles / L or moles / dm³

Volume = 855.218 * 10⁸dm³

so moles of H+ present in this volume:

= 0.000501 * 855.218 *10⁸ = 4.28* 10⁷moles

So moles of H2SO4 = 4.28 * 10⁷ / 2 = 2.14 * 10⁷ moles

Molecular weight of H2SO4 = 98 g / mol

Mass of H2SO4:

Moles X molecular weight = 98 * 2.14 X 10⁷ grams = 209.72 * 10⁷ grams in kg:  

= 2.09 X 10⁶Kg

Hope that helps!

? you expect us to answer this? ? this is your schoolwork, i don't think we can with this, i think it would considering being

alkaline metals are lithium, sodium, rubidium, potassium, cesium and francium.

alkali metals have electronegativity from 0.7 to 1, lowest in periodic table of elements, which means that alkali metals (i group in periodic table) has positive oxidation number in compounds.

electronegativity is a chemical property that describes the tendency of an atom to attract a shared pair of electrons towards itself.

alkaline metals (far left in main group) have lowest ionizations energy and easy remove valence electrons,

for example, potassium is the alkali metal and has a single valence electron in the outer electron shell.

electron configuration of potassium is: ₁₉k 1s²2s²2p⁶3s²3p⁶4s¹.

they can not exist alone in nature, because they are very reactive.