A 3.50 g sample of an unknown compound containing only C , H , and O combusts in an oxygen‑rich environment. When the products have cooled to 20.0 °C at 1 bar, there are 4.41 L of CO2 and 3.26 mL of H2O . The density of water at 20.0 °C is 0.998 g/mL.

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Answer 1
Answer:

Explanation:

First, calculate the moles of CO_(2) using ideal gas equation as follows.

                PV = nRT

or,          n = (PV)/(RT)

                = (1 atm * 4.41 ml)/(0.0821 Latm/mol K * 293 K)      (as 1 bar = 1 atm (approx))

                = 0.183 mol

As,   Density = (mass)/(volume)

Hence, mass of water will be as follows.

                Density = (mass)/(volume)

             0.998 g/ml = (mass)/(3.26 ml)    

                 mass = 3.25 g

Similarly, calculate the moles of water as follows.

        No. of moles = \frac{mass}{\text{molar mass}}

                              =  (3.25 g)/(18.02 g/mol)            

                              = 0.180 mol

Moles of hydrogen = 0.180 * 2 = 0.36 mol

Now, mass of carbon will be as follows.

       No. of moles = \frac{mass}{\text{molar mass}}

          0.183 mol =  (mass)/(12 g/mol)            

                              = 2.19 g

Therefore, mass of oxygen will be as follows.

              Mass of O = mass of sample - (mass of C + mass of H)

                                = 3.50 g - (2.19 g + 0.36 g)

                                = 0.95 g

Therefore, moles of oxygen will be as follows.

          No. of moles = \frac{mass}{\text{molar mass}}

                               =  (0.95 g)/(16 g/mol)            

                              = 0.059 mol

Now, diving number of moles of each element of the compound by smallest no. of moles as follows.

                         C              H           O

No. of moles:  0.183        0.36       0.059

On dividing:      3.1           6.1            1

Therefore, empirical formula of the given compound is C_(3)H_(6)O.

Thus, we can conclude that empirical formula of the given compound is C_(3)H_(6)O.            


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Boyle's law deals with pressure and volume.

What is Boyle's law?

Robert Boyle studied the relationship between the pressure p and the volume V of a confined gas held at a constant temperature.

Boyle observed that the product of the pressure and volume are observed to be nearly constant.

The product of pressure and volume is exactly a constant for an ideal gas.

                                                    PV = constant

As per Boyle’s law, any change in the volume occupied by a gas (at constant quantity and temperature) will result in a change in the pressure exerted by it. In other words, the product of the initial pressure and the initial volume of a gas is equal to the product of its final pressure and final volume (at constant temperature and number of moles).

Therefore, Boyle's law deals with pressure and volume.

Learn more about Boyle's law, here:

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Boyle's law states that P1V1= P2V2.
Pressure= P 
Volume= V

Pressure involves torr, atm, kPa,etc..
Volume involves mls/ milliter, l / liter, etc...

Explain typical physical properties of metals, including electrical conductivity and malleability ?

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