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Lady bird [3.3K]
3 years ago
6

Find the number of kilograms in 15.40 moles of C5H10O5

Chemistry
1 answer:
OLga [1]3 years ago
5 0
C5H10O5 would weight a total of 2.312 kilograms.
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The temperature is -8 F

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Purple Group Header Warm-Up Why can I smell a chocolate bar even though it is a solid? Read the description of the office myster
likoan [24]

Answer:

The smell of a chocolate is from the presence of volatile compounds present in the chocolate bar which at room temperature readily changes phase from solid to liquid to vapor or gas

Explanation:

There are nearly 600 identified compounds present in a chocolate bar and out of these, there are volatile components which gives the chocolate bar its distinctive aroma.

These volatile chocolate contents readily change phase from solid to vapor, with very short duration liquid phase.

For example, 3 methylbutanal, vanillin, and several organic compounds which are known to be readily volatile.

4 0
4 years ago
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
statuscvo [17]

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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Answer:

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Explanation:

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