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OLga [1]
3 years ago
11

When 0.5141 g of biphenyl (C12H10) undergoes combustion in a bomb calorimeter, the temperature rises from 25.823 °C to 29.419 °C

. Find ΔrU and ΔrH for the combustion of biphenyl in kJ mol−1 at 298 K. The heat capacity of the bomb calorimeter, determined in a separate experiment, is 5.861 kJ °C−1.
Chemistry
1 answer:
natka813 [3]3 years ago
3 0

Answer:

\Delta_{r}U of the reaction is -6313 kJ/mol

\Delta_{r}H of the reaction is -6312 kJ/mol

Explanation:

Temperature\,\,change= \Delta U = 29.419-25.823 =3.506^{o}C

q_{cal}= C \times \Delta T

=5.861 \times 3.596 = 21.076\,kJ

q_{rxn}= -q_{cal}= -21.076\,kJ

\Delta_{r}U= -21.076 \times \frac{154}{0.5141}= -6313\, kJ/mol

Therefore, \Delta_{r}U of the reaction is -6313 kJ/mol.

The chemical reaction in bomb calorimeter  is as follows.

C_{12}H_{10}(s)+\frac{27}{2}O_{2}(g)\rightarrow 12CO_{2}(g)+5H_{2}O(g)

Number\,of\,moles\Delta n=(12+5)-\frac{27}{2}=3.5

\Delta_{r}H=\Delta E+ \Delta n. RT

=-6313+3.5\times 8.314\times 10^{-3} \times 3.596=-6312\,kJ/mol

Therefore, \Delta_{r}H of the reaction is -6312 kJ/mol.

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Elodia [21]

Answer:

  • <em>Hydration number:</em> 4

Explanation:

<u>1) Mass of water in the hydrated compound</u>

Mass of water = Mass of the hydrated sample - mass of the dehydrated compound

Mass of water = 30.7 g - 22.9 g = 7.8 g

<u>2) Number of moles of water</u>

  • Number of moles = mass in grams / molar mass

  • molar mass of H₂O = 2×1.008 g/mol + 15.999 g*mol = 18.015 g/mol

  • Number of moles of H₂O = 7.9 g / 18.015 g/mol = 0.439 mol

<u>3) Number of moles of Strontium nitrate dehydrated, Sr (NO₃)₂</u>

  • The mass of strontium nitrate dehydrated is the constant mass obtained after heating = 22.9 g

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  • Number of moles of Sr (NO₃)₂ = 22.9 g / 211.63 g/mol =  0.108 mol

<u>4) Ratio</u>

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Which means that the hydration number is 4.

4 0
3 years ago
Read 2 more answers
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Answer:

0.6743 M

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  • moles = Molarity * volume
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<em>One NaOH moles reacts with one acetic acid mole</em>, so <u>the vinegar sample contains 16.86 mmoles of acetic acid as well</u>.

Finally we <u>calculate the concentration (molarity) of acetic acid</u>:

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