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neonofarm [45]
2 years ago
9

A mothball, composed of naphthalene (c10h8), has a mass of 1.64 g . part a how many naphthalene molecules does it contain?

Chemistry
1 answer:
OLga [1]2 years ago
3 0
The molar mass of Naphthalene is 128g/mol
Therefore; a mass of 1.64 g of Naphthalene contains'
   = 1.64g/128 g
    = 0.0128 moles
But, from the Avogadro's law 1 mole of a substance contains 6.022 × 10^23 particles
Therefore 1 mole of Naphthalene contains 6.022×10^23 molecules
Hence; 0.0128 moles × 6.022 ×10^23 molecules
          = 7.716 × 10^21 molecules
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The vapor pressure of ethanol is 30°C at 98.5 mmHg and the heat of vaporization is 39.3 kJ/mol. Determine the normal boiling poi
Gelneren [198K]

Answer : The normal boiling point of ethanol will be, 348.67K or 75.67^oC

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of ethanol at 30^oC = 98.5 mmHg

P_2 = vapor pressure of ethanol at normal boiling point = 1 atm = 760 mmHg

T_1 = temperature of ethanol = 30^oC=273+30=303K

T_2 = normal boiling point of ethanol = ?

\Delta H_{vap} = heat of vaporization = 39.3 kJ/mole = 39300 J/mole

R = universal constant = 8.314 J/K.mole

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

\ln (\frac{760mmHg}{98.5mmHg})=\frac{39300J/mole}{8.314J/K.mole}\times (\frac{1}{303K}-\frac{1}{T_2})

T_2=348.67K=348.67-273=75.67^oC

Hence, the normal boiling point of ethanol will be, 348.67K or 75.67^oC

3 0
3 years ago
Calculate the density of the items below. (D = M/V)<br> 1) M = 15g V = 5.0ml *
grin007 [14]

Answer:

<h3>The answer is 3.0 g/mL</h3>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume} \\

From the question

mass = 15 g

volume = 5 mL

We have

density =  \frac{15}{5}  \\

We have the final answer as

<h3>3.0 g/mL</h3>

Hope this helps you

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