Practice examples for mole conversions, including calculations for atoms, moles, and grams of various elements.

The image you've uploaded contains a set of mole conversion practice problems with solutions. These are common chemistry problems that involve converting between moles, atoms, and grams using Avogadro's number and molar masses.

Let me go through each problem step-by-step, explain the concept, and verify the solutions.

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🔷 1. Determine how many atoms are present in 2.5 moles of Silicon?

Answer: $15.05 \times 10^{23}$ atoms

#### ✔ Step-by-step:
- Type of question: Moles → Atoms
- Equation:
$$
\text{Number of atoms} = \text{moles} \times (6.022 \times 10^{23} \text{ atoms/mol})
$$
- Calculation:
$$
2.5 \, \text{mol} \times \frac{6.022 \times 10^{23} \, \text{atoms}}{1 \, \text{mol}} = 15.055 \times 10^{23} \approx 15.05 \times 10^{23} \, \text{atoms}
$$

✔ Correct!

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🔷 2. How many moles of Silicon is $3.01 \times 10^{23}$ atoms?

Answer: 0.5 moles

#### ✔ Step-by-step:
- Type of question: Atoms → Moles
- Equation:
$$
\text{Moles} = \frac{\text{number of atoms}}{6.022 \times 10^{23}}
$$
- Calculation:
$$
\frac{3.01 \times 10^{23} \, \text{atoms}}{6.022 \times 10^{23} \, \text{atoms/mol}} = 0.5 \, \text{moles}
$$

✔ Correct!

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🔷 3. What is the mass of 4 moles of Iron?

Answer: 224 grams

#### ✔ Step-by-step:
- Type of question: Moles → Grams
- Use molar mass of iron (Fe): 55.85 g/mol (approximately 55.9 g/mol)
- Equation:
$$
\text{Mass} = \text{moles} \times \text{molar mass}
$$
- Calculation:
$$
4 \, \text{mol} \times 55.9 \, \text{g/mol} = 223.6 \, \text{g} \approx 224 \, \text{g}
$$

✔ Correct! (Rounded to nearest gram)

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🔷 4. How many atoms are in 48.6 grams of Magnesium?

Answer: $12.04 \times 10^{23}$ atoms

#### ✔ Step-by-step:
- Type of question: Grams → Atoms
- Steps:
1. Convert grams → moles using molar mass of Mg = 24.3 g/mol
2. Convert moles → atoms using Avogadro’s number

- Equation:
$$
\frac{48.6 \, \text{g}}{1} \times \frac{1 \, \text{mol}}{24.3 \, \text{g}} \times \frac{6.022 \times 10^{23} \, \text{atoms}}{1 \, \text{mol}}
$$

- Calculation:
$$
\frac{48.6}{24.3} = 2 \, \text{moles}
$$
$$
2 \times 6.022 \times 10^{23} = 1.2044 \times 10^{24} = 12.04 \times 10^{23} \, \text{atoms}
$$

✔ Correct!

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🔷 5. How many grams are in $18.06 \times 10^{23}$ atoms of Copper?

Answer: 190.5 grams

#### ✔ Step-by-step:
- Type of question: Atoms → Grams
- Steps:
1. Atoms → moles (divide by Avogadro’s number)
2. Moles → grams (multiply by molar mass of Cu = 63.5 g/mol)

- Equation:
$$
\frac{18.06 \times 10^{23} \, \text{atoms}}{1} \times \frac{1 \, \text{mol}}{6.022 \times 10^{23} \, \text{atoms}} \times \frac{63.5 \, \text{g}}{1 \, \text{mol}}
$$

- Calculation:
$$
\frac{18.06 \times 10^{23}}{6.022 \times 10^{23}} = 3 \, \text{moles}
$$
$$
3 \times 63.5 = 190.5 \, \text{grams}
$$

✔ Correct!

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🔷 6. How many moles of Nickel is 176 grams?

Answer: 3 moles

#### ✔ Step-by-step:
- Type of question: Grams → Moles
- Molar mass of Ni: 58.7 g/mol
- Equation:
$$
\text{Moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{176 \, \text{g}}{58.7 \, \text{g/mol}}
$$

- Calculation:
$$
\frac{176}{58.7} \approx 3.00 \, \text{moles}
$$

✔ Correct!

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## 🧠 Summary of Concepts Used:

| Conversion Type | Formula |
|------------------|--------|
| Moles → Atoms | $\text{moles} \times 6.022 \times 10^{23}$ |
| Atoms → Moles | $\frac{\text{atoms}}{6.022 \times 10^{23}}$ |
| Moles → Grams | $\text{moles} \times \text{molar mass}$ |
| Grams → Moles | $\frac{\text{grams}}{\text{molar mass}}$ |
| Grams → Atoms | $\frac{\text{grams}}{\text{molar mass}} \times 6.022 \times 10^{23}$ |
| Atoms → Grams | $\frac{\text{atoms}}{6.022 \times 10^{23}} \times \text{molar mass}$ |

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## ✔ Final Verdict:
All six problems are solved correctly in your image. The steps follow proper unit cancellation, use correct molar masses, and apply Avogadro’s number appropriately.

If you need help creating similar problems or understanding any specific part further (like why we use 6.022 × 10²³), feel free to ask!