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3 years ago

Un gas ocupa 24,6 litros a 27°C y 10 moles, que presión posee. R=0.082

Chemistry

1 answer:

Anastasy [175]3 years ago

7 0

Answer:

$P=10 atm$

Explanation:

Hola,

En este caso, usando la ecuación del gas ideal, podemos calcular facilmente la presión a partir de:

$PV=nRT$

Por lo que despejando para la presión obtenemos:

$P=\frac{nRT}{V}$

Así, calculamos:

$P=\frac{10mol*0.082\frac{atm*L}{mol*K}*(27+273)K}{24.6L} \\\\P=10 atm$

¡Saludos!

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Dehydration of 2-methyl-2-pentanol forms one major and one minor organic product. Draw the structures of the two organic product

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Answer:

The major product is 2-methyl-2-pentene [ CH₃-CH₂-CH=C(CH₃)₂ ] and a minor product 2-methyl-1-pentene [ CH₃-CH₂-CH₂-C(CH₃)=CH₂ ].

Explanation:

Dehydration reaction is a reaction in which a molecule loses a water molecule in the presence of a dehydrating agent like sulfuric acid (H₂SO₄).

Dehydration reaction of 2-methyl-2-pentanol gives a major product 2-methyl-2-pentene and a minor product 2-methyl-1-pentene.

CH₃-CH₂-CH₂-C(CH₃)₂-OH (2-methyl-2-pentanol)→ CH₃-CH₂-CH=C(CH₃)₂ (2-methyl-2-pentene, major) + CH₃-CH₂-CH₂-C(CH₃)=CH₂ (2-methyl-1-pentene, minor)

Since more substituted alkene is more stable than the less substituted alkene. So, the trisubstituted alkene, 2-methyl-2-pentene is more stable than the disubstituted alkene, 2-methyl-1-pentene.

Therefore, the trisubstituted alkene, 2-methyl-2-pentene is the major product and the disubstituted alkene, 2-methyl-1-pentene is the minor product.

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3 years ago

Write the balanced chemical equation between H2SO4 and KOH in aqueous solution. This is called a neutralization reaction and wil

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Answer:

0.166M

Explanation:

In a neutralization, the acid, H₂SO₄, reacts with a base, KOH, to produce a salt, K₂SO₄ and water. The reaction is:

H₂SO₄ + 2KOH → K₂SO₄ + 2H₂O

To solve this problem, we need to determine moles of H2SO4 and moles of KOH that reacts to find the moles of sulfuric acid that remains after the reaction:

Moles H2SO4:

0.650L * (0.430mol /L) = 0.2795moles H2SO4

Moles KOH:

0.600L * (0.240mol / L) = 0.144 moles KOH

Moles of sulfuric acid that reacts with 0.144 moles of KOH are:

0.144 moles KOH * (1mol H2SO4 / 2 mol KOH) = 0.072 moles of H2SO4 react.

And remain:

0.2795moles H2SO4 - 0.072moles H2SO4 = 0.2075 moles of H2SO4 reamains.

In 0.650L + 0.600L = 1.25L:

Molar concentration of sulfuric acid:

0.2075 moles of H2SO4 / 1.25L =

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