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-Dominant- [34]
3 years ago
11

(C6H6) can be biodegraded by microorganism. if 30 mg of benzene is present, what amount of oxygen required for biodegradation, n

o new cells are formed?
Chemistry
1 answer:
quester [9]3 years ago
4 0

Answer:

36.92 mg of oxygen required for bio-degradation.

Explanation:

5C_6H_6+15O_2\rightarrow 12CO_2+15H_2O

Mass of benzene = 30 mg = 0.03 g (1000 mg = 1 g )

Moles benzene =\frac{0.03 g}{78 g/mol}=0.0003846 mol

According to reaction 5 moles of benzene reacts with 15 moles of oxygen gas.

Then 0.0003846 mol of benzene will react with:

\frac{15}{5}\times 0.0003846 mol=0.0011538 mol of oxygen gas

Mass of 0.0011538 moles of oxygen gas:

0.0011538 mol × 32 g/mol = 0.03692 g = 36.92 mg

36.92 mg of oxygen required for bio-degradation.

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Consider two processes: sublimation of I2(s) and melting of I2(s) (Note: the latter process can occur at the same temperature bu
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Explanation:

Entropy -

In a system, the randomness is measured by the term entropy .

Randomness basically refers as a form of energy that can not be used for any work.

The change in entropy is given by amount heat per change in temperature.

  • When solid is converted to gas entropy increases,

As the molecules in solid state are tightly packed and has more force of attraction between the molecules, but as it is converted to gas, the force of attraction between the molecule decreases and hence entropy increases.

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The particles of the substance , if are tightly held by strong force of attraction will decrease the entropy ,

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If the particles are loosely held , the entropy will increase , i.e. , positive entropy .

Similar with solid converting to liquid , the entropy will increases , i.e. , positive entropy .

Hence ,

The correct sign of entropy for both the process is positive .

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About 75% of living matter is made up of which two essential chemicals?
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A 3.50 g sample of an unknown compound containing only C , H , and O combusts in an oxygen‑rich environment. When the products h
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Explanation:

First, calculate the moles of CO_{2} using ideal gas equation as follows.

                PV = nRT

or,          n = \frac{PV}{RT}

                = \frac{1 atm \times 4.41 ml}{0.0821 Latm/mol K \times 293 K}      (as 1 bar = 1 atm (approx))

                = 0.183 mol

As,   Density = \frac{mass}{volume}

Hence, mass of water will be as follows.

                Density = \frac{mass}{volume}

             0.998 g/ml = \frac{mass}{3.26 ml}    

                 mass = 3.25 g

Similarly, calculate the moles of water as follows.

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

                              =  \frac{3.25 g}{18.02 g/mol}            

                              = 0.180 mol

Moles of hydrogen = 0.180 \times 2 = 0.36 mol

Now, mass of carbon will be as follows.

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

          0.183 mol =  \frac{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}}

                               =  \frac{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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