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Oxana [17]
2 years ago
5

High-purity benzoic acid (C6H5COOH; ΔHrxn for combustion = −3227 kJ/mol) is used as a standard for calibrating bomb calorimeters

. A 1.221-g sample burns in a calorimeter (heat capacity = 1365 J/°C) that contains exactly 1.430 kg of water. What temperature change is observed?
Chemistry
1 answer:
melamori03 [73]2 years ago
3 0

Answer : The temperature change observed is 4.35^oC

Explanation :

First we have to calculate the heat released by the combustion of benzoic acid.

q=n\times \Delta H

where,

q = heat released by combustion of benzoic acid = ?

n = moles of benzoic acid = \frac{\text{Mass of benzoic acid}}{\text{Molar mass of benzoic acid}}=\frac{1.221g}{122.122g/mole}=0.0099mole

\Delta H = enthalpy of combustion = 3227 kJ/mole

Now put all the given values in the above formula, get:

q=(0.0099mole)\times (3227kJ/mole)=31.9473kJ=31947.3J

Now we have to calculate the heat absorbed.

Heat released by combustion = Heat absorbed by the calorimeter + Heat absorbed by the water

q=[q_1+q_2]

q=[c_1\times \Delta T+m_2\times c_2\times \Delta T]

where,

q = heat released by the combustion = 31947.3 J

q_1 = heat absorbed by the calorimeter

q_2 = heat absorbed by the water

c_1 = specific heat of calorimeter = 1365J/^oC

c_2 = specific heat of water = 4.18J/g^oC

m_2 = mass of water = 1.430 kg  = 1430 g

\Delta T = change in temperature = ?

Now put all the given values in the above formula, we get:

31947.3J=[(1365J/^oC\times \Delta T)+(1430g\times 4.18J/g^oC\times \Delta T)]

\Delta T=4.35^oC

Therefore, the temperature change observed is 4.35^oC

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Which of the following isoelectronic series is correctly ranked from largest ionic radius to smallest ionic radius? 1. N 3−, O 2
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<u>Option 1</u>:  N⁻³ > O⁻² > F⁻ > Na⁺ > Mg⁺²    

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<u>Ionic radius is the radius of an atom´s ion in ionic crystal structure</u>.<u> </u><u><em>In an ion that lose an electron, to form a cation, the radius of the ion gets smaller</em></u><em>, </em>because the repulsion between electrons decrease because fewer electrons are present. Conversely, <u><em>adding on electron to a neutral atom, to form an anion, causes electron - electron repulsions to increase, so the size of the radius of the ion gets bigger.</em></u>                  

<u><em>Isoelectronic species are ions or elements that have the same number of electrons in their electronic shells but have different overall charges, because of their different atomic numbers</em></u>.                        

<u><em>In a isolelectronic series (same number of electrons),</em></u> <u><em>the increase of the positive charge (given by the number of protons in the nucleus), will cause a decrease in radius </em></u>beacuse the greater electrostatic attraction between the electrons and the nucleus. Consequently, the ion with the greatest nuclear charge will have the smallest ionic radius and the ion with the smallest nulear charge will have the largest ionic radius.  

<u>We will use this principle to solve our problem</u>.  

In our case, the given ions are:  

  • N⁻³ :    Z = 7,  e⁻ = 10
  • O⁻²:     Z= 8,   e⁻ =10
  • F⁻:       Z = 9,  e⁻ = 10
  • Na⁺:    Z= 11,   e⁻ = 10
  • Mg⁺²:  Z=12,   e⁻ =10

where Z= number of protons, and e⁻ = number of electrons.

<em><u>Hence the decreasing order of ionic radius is:</u></em>

N⁻³ > O⁻² > F⁻ > Na⁺ > Mg⁺²  

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