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3 years ago

How many grams of H2SO4 are needed to prepare 500. mL of a .250M solution?

Chemistry

1 answer:

zavuch27 [327]3 years ago

7 0

Answer:

We need 12.26 grams H2SO4

Explanation:

Step 1: Data given

Volume of a H2SO4 solution = 500 mL = 0.500 L

Concentration of the H2SO4 solution = 0.250 M

Molar mass of H2SO4 = 98.08 g/mol

Step 2: Calculate moles H2SO4

Moles H2SO4 = concentration * volume

Moles H2SO4 = 0.250 M * 0.500 L

Moles H2SO4 = 0.125 moles

Step 3: Calculate mass of H2SO4

Mass of H2SO4 = moles * molar mass

Mass of H2SO4 = 0.125 moles * 98.08 g/mol

Mass of H2SO4 = 12.26 grams

We need 12.26 grams H2SO4

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Answer:

$PO_4^{3-}(aq)+3Ag^+(aq)\rightarrow Ag_3PO_4(s)$

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In this case, since the net ionic equation of a chemical reaction shows up the ionic species that result from the simplification of the spectator ions, which are those at both reactants and products sides, we take into account that aqueous species ionize into ions whereas liquid, solid and gas species remain unionized. In such a way, for the reaction of cesium phosphate and silver nitrate we can write the complete molecular equation:

$Cs_3PO_4(aq)+3AgNO_3(aq)\rightarrow Ag_3PO_4(s)+3CsNO_3(aq)$

Whereas the three aqueous salts are ionized in order to write the following complete ionic equation:

$3Cs^+(aq)+PO_4^{3-}(aq)+3Ag^+(aq)+3NO_3^-(aq)\rightarrow Ag_3PO_4(s)+3Cs^+(aq)+3NO_3^-(aq)$

In such a way, since the cesium and nitrate ions are the spectator ions because of the aforementioned, the net ionic equation turns out:

$PO_4^{3-}(aq)+3Ag^+(aq)\rightarrow Ag_3PO_4(s)$

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