Question

1 cc $N_2O$ at NTP contains

• A. $\frac {1.8}{224} \times 10^{22} atoms$
• B. $\frac {6.02}{22400} \times 10^{23} molecules$
• C. $\frac {1.32}{224} \times 10^{23} electrons$
• D. All of the above

Answer 1

The Correct Option is D

To determine what 1 cc (cubic centimeter) of N_2O at NTP contains, we need to analyze the possible constituents in terms of atoms, molecules, and electrons.

Step 1: Understanding the context

Nitrous oxide, N_2O, is a molecule composed of 2 nitrogen atoms and 1 oxygen atom. At NTP (Normal Temperature and Pressure), 1 mole of any gas occupies 22,400 cubic centimeters (cc).

Step 2: Calculation of molecules

1 cc of any ideal gas at NTP contains:

\frac{1}{22400} \text{ moles}\times 6.022 \times 10^{23} \text{ molecules/mole} = \frac{6.02 \times 10^{23}}{22400} \text{ molecules}

This calculation aligns with option \frac{6.02}{22400} \times 10^{23} \text{ molecules}.

Step 3: Calculation of atoms

Each molecule of N_2O contains 3 atoms (2 nitrogen + 1 oxygen). Therefore, the number of atoms in 1 cc is:

3 \times \frac{6.02 \times 10^{23}}{22400}

Which is simplified to:

\frac{1.8}{224} \times 10^{22} \text{ atoms}

This calculation confirms option \frac {1.8}{224} \times 10^{22} \text{ atoms}.

Step 4: Calculation of electrons

Each nitrogen atom has 7 electrons, and the oxygen atom has 8 electrons, making:

2 \times 7 + 8 = 22 \text{ electrons per molecule}

The number of electrons in 1 cc is:

22 \times \frac{6.02 \times 10^{23}}{22400}\

Which results in:

\frac{1.32}{224} \times 10^{23} \text{ electrons}

Thus validating option \frac {1.32}{224} \times 10^{23} \text{ electrons}.

Conclusion:

All calculations agree with the options provided. Therefore, the correct answer is "All of the above" because each option accurately represents a constituent of N_2O at NTP for 1 cc in different forms: atoms, molecules, and electrons.