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Lelu [443]
1 year ago
5

31. calculate the amount of energy released as heat by the freezing of 13.3 g of a liquid substance, given that the substance ha

s a molar mass of 82.9 g/mol and a molar enthalpy of fusion of 4.60 kj/mol.
Chemistry
1 answer:
spin [16.1K]1 year ago
5 0

From the calculations, the heat of fusion of the substance is 0.73 kJ

<h3>What is is the heat of freezing?</h3>

The heat of freezing is the energy released when the substance is converted from liquid to solid.

Now we know that the molar mass of the substance is 82.9 g/mol  hence the number of moles of the substance is;  13.3 g /82.9 g/mol  = 0.16 moles

Now the heat of fusion shall be;

H = 4.60 kj/mol * 0.16 moles

H = 0.73 kJ

Learn more about freezing:brainly.com/question/3121416

#SPJ4

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What is the pH of a 0.050 M triethylamine, (C2H5)3N, solution? Kb for triethylamine is 5.3 ´ 10-4. Question options: 1) 11.69 2)
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<span>the pH of a 0.050 M triethylamine, is 11.70
</span>
For triehtylamine, (C_{2}H_{5})_{3}N, the reaction will be
(C_{2}H_{5})_{3}N + H_{2}O ---\ \textgreater \ ( C_{2}H_{5})_{3}NH^{+} + OH^{-}

 and we know, pH = -log[H+] and pOH = -log[OH-]

Also, pOH + pH = 14

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And kb =  \frac{( [( C_{2}H_{5})_{3}NH^{+} ]*  OH^{-} )}{[( C_{2}H_{5})_{3}N]}

 thus, [OH-] =(5.3 ^ 10-4) ^2 / 0.050

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Thus, pOH = 2.30 
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