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nevsk [136]
3 years ago
15

neon has an average atomic mass of 20.2 g/mol, whereas argon has an average atomic mass of 40.0 g/mol. How would the number of a

toms in a 1.0 mol sample of neon compare to the number of atoms in a 4.0 mol sample of argon
Chemistry
2 answers:
topjm [15]3 years ago
8 0
Argon sample would have 20.2 divided by 40.0
Greeley [361]3 years ago
6 0

Answer : The number of atoms in a 4.0 mole sample of argon has greater number of atoms or four times as compared to the number of atoms in a 1.0 mole sample of neon.

Explanation :

As we know that the 1 mole of substance contains 6.022\times 10^{23} number of atoms.

As per question,

The 1 mole sample of neon contains 6.022\times 10^{23} number of atoms.

And,

As, 1 mole sample of argon contains 6.022\times 10^{23} number of atoms

So, 4.0 mole sample of argon contains 4.0\times 6.022\times 10^{23}=24.088\times 10^{23} number of atoms

From this we conclude that, the number of atoms in a 4.0 mole sample of argon has greater number of atoms or four times as compared to the number of atoms in a 1.0 mole sample of neon.

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Nicotine, a component of tobacco, is composed of c, h, and n. a 4.200-mg sample of nicotine was combusted, producing 11.394 mg o
Rudiy27

Answer:

            Empirical Formula  =  C₅H₇N₁

Solution:

Data Given:

                      Mass of Nicotine  =  4.20 mg  =  0.0042 g

                      Mass of CO₂  =  11.394 mg  =  0.011394 g

                      Mass of H₂O  =  3.266 mg  =  0.003266 g

Step 1: Calculate %age of Elements as;

                      %C  =  (mass of CO₂ ÷ Mass of sample) × (12 ÷ 44) × 100

                      %C  =  (0.011394 ÷ 0.0042) × (12 ÷ 44) × 100

                      %C  =  (2.7128) × (12 ÷ 44) × 100

                      %C  =  2.7128 × 0.2727 × 100

                      %C  =  73.979 %


                      %H  =  (mass of H₂O ÷ Mass of sample) × (2.02 ÷ 18.02) × 100

                      %H  =  (0.003266 ÷ 0.0042) × (2.02 ÷ 18.02) × 100

                      %H  =  (0.7776) × (2.02 ÷ 18.02) × 100

                      %H  =  0.7776 × 0.1120 × 100

                      %H  =  8.709 %


                      %N  =  100% - (%C + %H)

                      %N  =  100% - (73.979 % + 8.709%)

                      %N  =  100% - 82.688%

                      %N  =  17.312 %

Step 2: Calculate Moles of each Element;

                      Moles of C  =  %C ÷ At.Mass of C

                      Moles of C  = 73.979 ÷ 12.01

                     Moles of C  =  6.1597 mol


                      Moles of H  =  %H ÷ At.Mass of H

                      Moles of H  = 8.709 ÷ 1.01

                      Moles of H  =  8.6227 mol


                      Moles of N  =  %N ÷ At.Mass of O

                      Moles of N  = 17.312 ÷ 14.01

                      Moles of N  =  1.2356 mol

Step 3: Find out mole ratio and simplify it;

                C                                        H                                     N

            6.1597                               8.6227                             1.2356

     6.1597/1.2356                  8.6227/1.2356                 1.2356/1.2356

               4.985                             6.978                                   1

             ≈ 5                                      ≈ 7                                     1

Result:

        Empirical Formula  =  C₅H₇N₁

6 0
3 years ago
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