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dimulka [17.4K]
3 years ago
8

0.415 g of an unknown triprotic acid are used to make a 100.00 mL solution. Then 25.00 mL of this solution is transferred to an

Erlenmeyer flask and is titrated with 0.1029 M NaOH. The initial burette reading is 1.81 mL; when the titration endpoint is reached, the final burette reading is 39.70 mL. How many moles of triprotic acid are neutralized during the titration? Provide your answer in decimal form (e.g. 0.123) to the correct number of significant figures.
Chemistry
1 answer:
Arturiano [62]3 years ago
8 0

Answer:

Explanation:

Initial burette reading = 1.81 mL

final burette reading = 39.7 mL

volume of NaOH used = 39.7 - 1.81 = 37.89 mL .  

37.89 mL of .1029 M NaOH is used to neutralise triprotic acid

No of moles contained by 37.89 mL of .1029 M NaOH

= .03789 x .1029 moles

= 3.89 x 10⁻³ moles

Since acid is triprotic ,  its equivalent weight = molecular weight / 3

No of moles of triprotic acid = 3.89 x 10⁻³ / 3

= 1.30   x 10⁻³ moles .

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Answer:

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b) decreases

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Explanation:

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II) the temperature of the liquid

Hence, when any of these increases, the vapour pressure increases likewise.

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Compare the solubility of silver chromate in each of the following aqueous solutions: Clear All 0.10 M AgCH3COO 0.10 M Na2CrO4 0
slavikrds [6]

Solution :

Comparing the solubility of silver chromate for the solutions :

$0.10 \ M \ AgCH_3COO$    -----     Less soluble than in pure water.

$0.10 \ M \ Na_2CrO_4$   ----- Less soluble than in pure water.

$0.10 \ M \ NH_4NO_3$   -----   Similar solubility as in the pure water

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The silver chromate dissociates to form :

$AgCrO_4 (s) \rightleftharpoons 2Ag^+ (aq) +CrO_4^{2-}(aq)$

When 0.1 M of $AgCH_3COO^-$ is added, the equilibrium shifts towards the reverse direction due to the common ion effect of Ag^+, so the solubility of Ag_2CrO_4 decreases.

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2 years ago
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nydimaria [60]

Answer:

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Serhud [2]

Answer:

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Explanation:

Hello there!

In this case, in agreement to the thermodynamic definition of the Gibbs free energy, in terms of enthalpy of entropy:

\Delta G= \Delta H-T\Delta S

It is possible to calculate the required G by plugging in the given entropy and enthalpy as shown below:

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Best regards!

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