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Olenka [21]
3 years ago
9

A sample of ammonia has a mass of 82.9 g. how many molecules are in this sample?

Chemistry
2 answers:
alina1380 [7]3 years ago
8 0
The chemical formula for ammonia is NH3. So first, you need to find the molar mass of ammonia (how many grams in one mole).
N=14g
H3=3g
So one mole of NH3 is 17 grams, you can divide 82.9 grams by 17 grams to find the number of molecules. The answer should be 4.876 moles (molecules) of ammonia. Hope this helps!
Aleksandr [31]3 years ago
5 0

Answer: 29.37\times 10^{23}molecules

Explanation: To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\textMolar mass}}    

Given mass of ammonia NH_3 given = 82.9 g

Molar mass of ammonia NH_3 = 17 g/mol

Putting values in above equation, we get:

\text{Moles of sodium}=\frac{82.9g}{17g/mol}=4.87mol

According to Avogadro's law,

1 mole of any substance contains avogadro's number  6.023\times 10^{23} of particles.

Thus 4.87 moles of ammonia contains=\frac{6.023\times 10^{23}}{1}\times 4.87=29.37\times 10^{23}molecules of ammonia.

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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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