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Nostrana [21]
3 years ago
11

Ethanol boils at 78.4 °C with \DeltaΔHvap = 38.6 kJ/mol. A 0.200-mol sample of ethanol is heated from some colder temperature up

to 78.4 °C, which requires 1.05 kJ of heat, and then vaporized. What will be the total amount of heat required (for both the heating and the vaporizing)?
Chemistry
1 answer:
NNADVOKAT [17]3 years ago
4 0

Answer:

Q_T=8.77kJ

Explanation:

Hello,

In this case, the total heat is computed via:

Q_T=n*\Delta H_{vap}+1.05 kJ\\

It means, by adding all the involved heat, which simply result:

Q_T=0.200mol*38.6 kJ/mol+1.05 kJ\\Q_T=8.77kJ

Thus, such heat must be provided to the ethanol, that is why it remains positive.

Best regards.

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Given :

Number of molecules of C_4H_2F_7I.

To Find :

How many moles are in given number of molecules.

Solution :

We know, in 1 moles of any element/compound contains 6.022\times 10^{23} at atoms/molecules.

So, number of moles in 5.4\times 10^{24} molecules are :

n = \dfrac{5.4\times 10^{24}}{6.022\times 10^{23}}\\\\n = 8.97 \ moles

Therefore, number of moles are 8.97 .

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A solution is made by combining 500 mL of 0.10 M HF (Ka=7.2 x 10^-4) with 300 mL of 0.15 M NaF. What is the pH of the resulting
n200080 [17]

Answer:

b) 3.10

Explanation:

HF ⇄ H + + F

Using Henderson-Hasselbalch Equation:

pH = pKa + log [A-]/[HA].

Where;

pKa = Dissociation constant = -log Ka

Hence, pKa of HF = -log 7.2 x 10^-4 = 3.14266

[A-] = concentration of conjugate base after dissociation = moles of base/total volume

          = 0.15 x 0.3/0.8

               = 0.05625 M

[HA] = concentration of the acid = moles of acid/total volume

             = 0.10 x 0.5/0.8

                    = 0.0625 M

Note: <em>Total volume = 500 + 300 = 800 mL = 0.8 dm3</em>

pH = 3.14266 + log [0.05625/0.0625]

      = 3.14267 + (-0.04575749056)

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<em>From all the available options below:</em>

<em>a) 2.97 </em>

<em>b) 3.10 </em>

<em>c) 3.19 </em>

<em>d) 3.22 </em>

<em>e) 3.32</em>

The correct option is b.

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