I cc `N_(2)O` at NTP contains

Options

A. `(1.8)/(224)xx10^(22)` atoms
B. `(6.02)/(22400)xx10^(23)` molecules
C. `(1.32)/(224)xx10^(23)` electrons
D. All of the above

Explanation

To solve the question "1 cc of \( N_2O \) at NTP contains", we will break it down step by step.

### Step 1: Understanding the Volume of Gas at NTP
At Normal Temperature and Pressure (NTP), 1 mole of any gas occupies 22400 cm³. Therefore, we can find the number of molecules in 1 cm³ of \( N_2O \).

**Calculation:**
\[ \text{Number of molecules in 1 cm}^3 = \frac{6.02 \times 10^{23} \text{ molecules}}{22400 \text{ cm}^3} \]

### Step 2: Calculate the Number of Molecules in 1 cc of \( N_2O \)
Using the above formula, we can calculate the number of molecules in 1 cc of \( N_2O \).

**Calculation:**
\[ \text{Number of molecules in 1 cm}^3 = \frac{6.02 \times 10^{23}}{22400} \approx 2.68 \times 10^{19} \text{ molecules} \]

### Step 3: Determine the Number of Atoms in 1 Molecule of \( N_2O \)
Each molecule of \( N_2O \) contains:
- 2 Nitrogen atoms
- 1 Oxygen atom

Thus, the total number of atoms in one molecule of \( N_2O \) is:
\[ 2 + 1 = 3 \text{ atoms} \]

### Step 4: Calculate the Total Number of Atoms in 1 cc of \( N_2O \)
To find the total number of atoms in 1 cc of \( N_2O \), we multiply the number of molecules by the number of atoms per molecule.

**Calculation:**
\[ \text{Total number of atoms} = 3 \times \left(\frac{6.02 \times 10^{23}}{22400}\right) = 3 \times 2.68 \times 10^{19} \approx 8.04 \times 10^{19} \text{ atoms} \]

### Step 5: Calculate the Number of Electrons in 1 cc of \( N_2O \)
Next, we need to find the total number of electrons in \( N_2O \). Each Nitrogen atom has 7 electrons, and each Oxygen atom has 8 electrons.

**Calculation:**
\[ \text{Total number of electrons in 1 molecule of } N_2O = (2 \times 7) + 8 = 14 + 8 = 22 \text{ electrons} \]

Now, we can calculate the total number of electrons in 1 cc of \( N_2O \):
\[ \text{Total number of electrons} = 22 \times \left(\frac{6.02 \times 10^{23}}{22400}\right) \approx 22 \times 2.68 \times 10^{19} \approx 5.89 \times 10^{20} \text{ electrons} \]

### Conclusion
Based on the calculations:
1. The number of molecules in 1 cc of \( N_2O \) is approximately \( 2.68 \times 10^{19} \).
2. The number of atoms in 1 cc of \( N_2O \) is approximately \( 8.04 \times 10^{19} \).
3. The number of electrons in 1 cc of \( N_2O \) is approximately \( 5.89 \times 10^{20} \).

Thus, the answer to the question is that 1 cc of \( N_2O \) at NTP contains:
- \( 1.8 \times 10^{22} \) atoms (Option 1)
- \( \frac{6.02}{22400} \times 10^{23} \) molecules (Option 2)
- \( \frac{1.32}{224} \times 10^{23} \) electrons (Option 3)

### Final Answer
The correct answer is **Option 4: All of the above**.