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arsen [322]
2 years ago
6

Suppose you perform a calorimeter experiment to determine the molar heat of neutralization of an unknown acid, H A HA, with sodi

um hydroxide, N a O H NaOH. You mix 37.2 mL of 0.50 M H A HA with 56.8 mL of 0.75 M N a O H NaOH and calculate the heat of reaction as -1.6 kJ. What is the molar heat of neutralization (in kJ/mol) for the unknown acid
Chemistry
1 answer:
Black_prince [1.1K]2 years ago
5 0

Answer:

-86.02 kJ/ mole

Explanation:

The moles of the acid used = Molarity × Volume (L) =

= 0.50 (0.0372 L)

= 0.0186 moles

The heat released = -1.6 kJ

∴ 0.0186 moles neutralization of HA heat is: -1.6 kJ

The molar heat of neutralization due to one mole of the unknown acid = -1.6/0.0186

= -86.02 kJ/ mole

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Leokris [45]
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C: 284.5 x .76 = 216.22
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The empirical formula would be found by dividing out all factors of those subscript numbers. In our case, all of them can be divided by 2. The empirical formula would be C9H18O




7 0
2 years ago
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<span>(19.55 mol Au) / ( 1 ) x (196.97 g Au) / ( 1 mol Au) =
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MrMuchimi

Answer:

El volumen del cuerpo es el mismo al comienzo de la experiencia.

Explicación:

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