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Komok [63]
3 years ago
11

A cylinder is labeled \"PENTANE.\" When the gas inside the cylinder is monochlorinated, five isomers of formula C5H11Cl result.

Was the gas pure n-pentane, pure isopentane, pure neopentane, or a mixture of two or all three of these?
Chemistry
1 answer:
melisa1 [442]3 years ago
8 0

Answer:

a mixture of two these

Explanation:

The number of isomeric monochlorides depends on the structure and number of equivalent hydrogen atoms in each isomer of pentane.

n-pentane has three different kinds of equivalent hydrogen atoms leading to three isomeric monochlorides formed.

Isopentane has four different types of equivalent hydrogen atoms hence four isomeric monochlorides are formed.

Lastly, neopentane has only one type of equivalent hydrogen atoms that yields one mono chlorination product.

Hence the cylinder must contain a mixture of isopentane and neopentane which yields four and one isomeric monochlorides giving a total of five identifiable monochloride products as stated in the question.  

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Write the overall, balanced molecular equation and indicate which element is oxidized and which is reduced for the following rea
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Explanation:

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To identify if an element has been reduced or oxidized, the oxidation number is observed in both the reactant and product phase.

An increase in oxidation number denotes that the element has been oxidized.

A decrease in oxidation number denotes that the element has been reduced.

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Answer:

8740 joules are required to convert 20 grams of ice to liquid water.

Explanation:

The amount of heat required (Q), measured in joules, to convert ice at -50.0 ºC to liquid water at 0.0 ºC is the sum of sensible heat associated with ice and latent heat of fussion. That is:

Q = m\cdot [c\cdot (T_{f}-T_{o})+L_{f}] (1)

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c - Specific heat of ice, measured in joules per gram-degree Celsius.

T_{o}, T_{f} - Temperature, measured in degrees Celsius.

L_{f} - Latent heat of fussion, measured in joules per gram.

If we know that m = 20\,g, c = 2.06\,\frac{J}{g\cdot ^{\circ}C}, T_{f} = 0\,^{\circ}C, T_{o} = -50\,^{\circ}C and L_{f} = 334\,\frac{J}{g }, then the amount of heat is:

Q = (20\,g)\cdot \left\{\left(2.06\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot [0\,^{\circ}C-(-50\,^{\circ}C)]+334\,\frac{J}{g} \right\}

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8740 joules are required to convert 20 grams of ice to liquid water.

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