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Ivanshal [37]
3 years ago
11

What mass of Ba3(PO4)2 is contained in 1575 mL of a 0.35M solution of Ba3(PO4)2

Chemistry
1 answer:
LekaFEV [45]3 years ago
7 0

Answer:

= 331.81 g

Explanation:

Molarity is calculated by the formula;

Molarity = Moles/volume in liters

Therefore;

Moles = Molarity ×Volume in liters

          = 0.35 M × 1.575 L

          = 0.55125 Moles

But; Molar mass of Ba3(PO4)2 is 601.93 g/mol

Thus;

Mass = 0.55125 moles × 601.93 g/mol

         <u>=331.81 g</u>

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Determine the volume in mL of 0.37 M HClO4(aq) needed to reach the half-equivalence (stoichiometric) point in the titration of 2
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Answer:

14.3mL you require to reach the half-equivalence point

Explanation:

A strong acid as HClO₄ reacts with a weak base as CH₃CH₂NH₂, thus:

CH₃CH₂NH₂ + HClO₄ → CH₃CH₂NH₃⁺ + ClO₄⁻

As the reaction is 1:1, to reach the equivalence point you require to add the moles of HClO₄ equal to moles CH₃CH₂NH₂ you add originally. Also, half-equivalence point requires to add half-moles of CH₃CH₂NH₂ you add originally.

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As concentration of HClO₄ is 0.37M, volume you require to add 0.005304moles is:

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<h3> 14.3mL you require to reach the half-equivalence point</h3>

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