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const2013 [10]
3 years ago
6

Carbon tetrachloride, CCl4, was once used as a dry cleaning solvent, but is no longer used because it is carcinogenic. At 55.8°C

, the vapor pressure of CCl4 is 52.7 kPa, and its enthalpy of vaporization is 29.82 kJ/mol. Use this information to estimate the normal boiling point (in °C) for CCl4.
Chemistry
1 answer:
joja [24]3 years ago
4 0
Use the Clausius-Clapeyron equation... 

<span>Let T1 be the normal boiling point, which will occur at standard pressure (P1), which is 101.3 kPa (aka 760 torr or 1.00 atm). You know the vapour pressure (P2) at a different temperature (T2). And you are given the enthalpy of vaporization. Therefore, we can use the Clausius-Clapeyron equation.

</span>ln(P_1/P_2) = \frac{-\delta H_{vap}}{R}  \times [\frac{1}{T_1} - \frac{1}{T_2}]<span>

</span><span>ln(101.3 kPa / 52.7 kPa) = (-29.82 kJ/mol / 8.314x10^{-3} kJ/molK) (1/T - 1/329 K) 
</span>
------ some algebra goes here ----- 

<span>T = 349.99K ...... or ...... 76.8C </span>

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Explanation:

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Therefore, when a 100 g of given pure metal in solid state is heated at its exact melting point which is 215^{o}C then some of the solid will change into liquid state but the temperature will remains the same.  

4 0
3 years ago
Given the following equation, how many grams of PbCO3 will dissolve when exactly 1.0 L of 1.00 M H+ is added to 6.00 g of PbCO3?
9966 [12]
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1.0 L x (1.00 mole / 1 L ) = 1 mole H+

From the given balanced equation, we can use the stoichiometric ratio to solve for the moles of PbCO3:
1 mole H+ x (1 mole PbCO3 / 2 moles H+) = 0.5 moles PbCO3

Converting the moles of PbCO3 to grams using the molecular weight of PbCO3
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4 0
3 years ago
Read 2 more answers
Consider the synthesis of ammonia 3 H2+ N2 à 2 NH3 If a student were to react 38.5 g of nitrogen gas, how many moles of ammonia
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Given data:

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Moles of ammonia produced = ?

Solution:

Chemical  equation:

N₂ + 3H₂    →     2NH₃

Number of moles of nitrogen:

Number of moles = mass/ molar mass

Number of moles = 38.5 g/ 28 g/mol

Number of moles = 1.375 mol

Now we will compare the moles of ammonia and nitrogen from balance chemical equation.

           N₂            :            NH₃

            1              :             2

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Thus 2.75 moles of ammonia  are produced from 38.5 g of nitrogen.

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