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oee [108]
3 years ago
8

A. Sustancia A – Determine el calor específico de una muestra de 1.55 g de acero inoxidable la misma al aumentar la temperatura

a 178°C y que absorbió 141 J.
b. Sustancia B – Determine el calor específico para una muestra de 3.4 g de aceite de oliva a 23°C se le añadió 435 J de calor y aumento su temperatura a 85°C.

II. Utilizando el calor específico de una sustancia (tabla que se encuentra en la presentación) determine la cantidad de calor liberado cuando una muestra de 560 g de agua disminuye la temperatura de 67°C a 22°C.
Chemistry
1 answer:
Ivanshal [37]3 years ago
5 0
Idk tbh nejahxhsjsjns
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Suppose a 0.025M aqueous solution of sulfuric acid (H2SO4) is prepared. Calculate the equilibrium molarity of SO4−2. You'll find
FromTheMoon [43]

<u>Answer:</u> The concentration of SO_4^{2-} at equilibrium is 0.00608 M

<u>Explanation:</u>

As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.

In the second dissociation, the ions will remain in equilibrium.

We are given:

Concentration of sulfuric acid = 0.025 M

Equation for the first dissociation of sulfuric acid:

       H_2SO_4(aq.)\rightarrow H^+(aq.)+HSO_4^-(aq.)

            0.025          0.025       0.025

Equation for the second dissociation of sulfuric acid:

                    HSO_4^-(aq.)\rightarrow H^+(aq.)+SO_4^{2-}(aq.)

<u>Initial:</u>            0.025            0.025      

<u>At eqllm:</u>      0.025-x          0.025+x        x

The expression of second equilibrium constant equation follows:

Ka_2=\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}

We know that:

Ka_2\text{ for }H_2SO_4=0.01

Putting values in above equation, we get:

0.01=\frac{(0.025+x)\times x}{(0.025-x)}\\\\x=-0.0411,0.00608

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of sulfate ion = x = 0.00608 M

Hence, the concentration of SO_4^{2-} at equilibrium is 0.00608 M

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Compounds X and Y are both C6H13Cl compounds formed in the radical chlorination of 3-methylpentane. Both X and Y undergo base-pr
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Answer:

Y is a 3-chloro-3-methylpentane.

The structure is shown in the figure attached.

Explanation:

The radical chlorination of 3-methylpentane can lead to a tertiary substituted carbon (Y) and to a secondary one (X).

The E2 elimination mechanism, as shown in the figure, will happen with a simulyaneous attack from the base and elimination of the chlorine. This means that primary and secondary substracts undergo the E2 mechanism faster than tertiary substracts.

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What relationship does Avogadro's law describe?
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Answer:

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