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2 years ago

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Arrange the following substances in order of increasing solubility of water. C6H14, C6H13Br, C6H13OH, C6H12(OH)2.

1 answer:

Juli2301 [7.4K]2 years ago

5 0

Answer:

C6H14 < C6H13Br < C6H13OH < C6H12(OH)2

Explanation:

Hello,

In this case, since the solubility in water is related with the presence of polar bonds in the given molecules we can see that C6H12(OH)2 has the presence two O-H bonds which promote the highest solubility via hydrogen bonds as well as the C6H13OH but in a lower degree as only on O-H bond is present. Next since the bond C-Br in is slightly close to the polar bond C6H13Br rather than the C-C bonds only had by C6H14 we can infer that C6H13Br is more soluble in water than C6H14, therefore the required order is:

C6H14 < C6H13Br < C6H13OH < C6H12(OH)2

Whereas C6H12(OH)2 is the most soluble and C6H14 the least soluble in water.

Best regards.

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When 1.025 g of naphthalene (c10h8) burns in a bomb calorimeter, the temperature rises from 24.25°c to 32.33°c. find δerxn for t

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Answer : The $\Delta E$ for the combustion of naphthalene is $5161.25KJ/mole$

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Mass of naphthalene = 1.025 g

Initial temperature = $24.25^oC$

Final temperature = $32.33^oC$

Specific heat capacity of calorimeter = $5.11KJ/^oC$

Molar mass of naphthalene = 128 g/mole

First, we have to calculate the heat absorbed, $q$

Formula used :

$q=c\times \Delta T=c\times (T_{final}-T_{initial})$

Now put all the given values in this formula, we get

$q=(5.11KJ/^oC)\times (32.33^oCT-24.25^oC)=41.29KJ$

Now we have to calculate the moles of naphthalene.

Moles of $C_{10}H_{8}$ = $\frac{\text{ Mass of }C_{10}H_{8}}{\text{ Molar mass of }C_{10}H_{8}}=\frac{1.025g}{128g/mole}=0.0080moles$

Now we have to calculate the $\Delta E$ for combustion of naphthalene.

$\Delta E=\frac{q}{n}$

where,

q = heat absorbed

n = number of moles

Now put all the values in this formula, we get

$\Delta E=\frac{41.29KJ}{0.0080moles}=5161.25KJ/mole$

Therefore, the $\Delta E$ for the combustion of naphthalene is $5161.25KJ/mole$

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2 years ago