Answer :
Sure! Let's break down each part of this problem step-by-step:
### Part (a): 0.25 moles of oxygen gas (O₂)
We need to find the number of molecules in 0.25 moles of oxygen gas.
1. Given:
- Moles of O₂ = 0.25 moles
2. Using Avogadro's number, which is [tex]\(6.022 \times 10^{23}\)[/tex] molecules per mole:
[tex]\[ \text{Number of molecules} = \text{moles} \times \text{Avogadro's number} \][/tex]
3. Calculation:
[tex]\[ \text{Number of molecules} = 0.25 \times 6.022 \times 10^{23} = 1.5055 \times 10^{23} \text{ molecules} \][/tex]
So, 0.25 moles of oxygen gas contain [tex]\(1.5055 \times 10^{23}\)[/tex] molecules.
### Part (b): 0.7 grams of nitrogen gas (N₂)
We need to find the number of moles first, and then the number of molecules.
1. Given:
- Mass of N₂ = 0.7 grams
- Molar mass of N₂ = 28.0 grams per mole
2. Moles calculation:
[tex]\[ \text{Moles of N₂} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.7 \text{ grams}}{28.0 \text{ grams per mole}} = 0.025 \text{ moles} \][/tex]
3. Using Avogadro's number:
[tex]\[ \text{Number of molecules} = \text{moles} \times \text{Avogadro's number} \][/tex]
4. Calculation:
[tex]\[ \text{Number of molecules} = 0.025 \times 6.022 \times 10^{23} = 1.5055 \times 10^{22} \text{ molecules} \][/tex]
So, 0.7 grams of nitrogen gas contain [tex]\(1.5055 \times 10^{22}\)[/tex] molecules.
### Part (c): 5.3 grams of sodium carbonate (Na₂CO₃)
We need to find the number of moles first, and then the number of molecules.
1. Given:
- Mass of Na₂CO₃ = 5.3 grams
- Molar mass of Na₂CO₃ = 106.0 grams per mole
2. Moles calculation:
[tex]\[ \text{Moles of Na₂CO₃} = \frac{\text{mass}}{\text{molar mass}} = \frac{5.3 \text{ grams}}{106.0 \text{ grams per mole}} = 0.05 \text{ moles} \][/tex]
3. Using Avogadro's number:
[tex]\[ \text{Number of molecules} = \text{moles} \times \text{Avogadro's number} \][/tex]
4. Calculation:
[tex]\[ \text{Number of molecules} = 0.05 \times 6.022 \times 10^{23} = 3.011 \times 10^{22} \text{ molecules} \][/tex]
So, 5.3 grams of sodium carbonate contain [tex]\(3.011 \times 10^{22}\)[/tex] molecules.
### Part (d): 52 grams of helium (He)
We need to find the number of moles first, and then the number of atoms (since He is a noble gas, it exists as single atoms).
1. Given:
- Mass of He = 52 grams
- Molar mass of He = 4.0 grams per mole
2. Moles calculation:
[tex]\[ \text{Moles of He} = \frac{\text{mass}}{\text{molar mass}} = \frac{52 \text{ grams}}{4.0 \text{ grams per mole}} = 13 \text{ moles} \][/tex]
3. Using Avogadro's number:
[tex]\[ \text{Number of atoms} = \text{moles} \times \text{Avogadro's number} \][/tex]
4. Calculation:
[tex]\[ \text{Number of atoms} = 13 \times 6.022 \times 10^{23} = 7.8286 \times 10^{24} \text{ atoms} \][/tex]
So, 52 grams of helium contain [tex]\(7.8286 \times 10^{24}\)[/tex] atoms.
### Summary
The results are:
- 0.25 moles of [tex]\(O_2\)[/tex] contain [tex]\(1.5055 \times 10^{23}\)[/tex] molecules.
- 0.7 grams of [tex]\(N_2\)[/tex] contain [tex]\(0.025 \approx 0.024999999999999998\)[/tex] moles or [tex]\(1.5055 \times 10^{22}\)[/tex] molecules.
- 5.3 grams of [tex]\(Na_2CO_3\)[/tex] contain [tex]\(0.05 \approx 0.049999999999999996\)[/tex] moles or [tex]\(3.011 \times 10^{22}\)[/tex] molecules.
- 52 grams of [tex]\(He\)[/tex] contain [tex]\(13\)[/tex] moles or [tex]\(7.8286 \times 10^{24}\)[/tex] atoms.
### Part (a): 0.25 moles of oxygen gas (O₂)
We need to find the number of molecules in 0.25 moles of oxygen gas.
1. Given:
- Moles of O₂ = 0.25 moles
2. Using Avogadro's number, which is [tex]\(6.022 \times 10^{23}\)[/tex] molecules per mole:
[tex]\[ \text{Number of molecules} = \text{moles} \times \text{Avogadro's number} \][/tex]
3. Calculation:
[tex]\[ \text{Number of molecules} = 0.25 \times 6.022 \times 10^{23} = 1.5055 \times 10^{23} \text{ molecules} \][/tex]
So, 0.25 moles of oxygen gas contain [tex]\(1.5055 \times 10^{23}\)[/tex] molecules.
### Part (b): 0.7 grams of nitrogen gas (N₂)
We need to find the number of moles first, and then the number of molecules.
1. Given:
- Mass of N₂ = 0.7 grams
- Molar mass of N₂ = 28.0 grams per mole
2. Moles calculation:
[tex]\[ \text{Moles of N₂} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.7 \text{ grams}}{28.0 \text{ grams per mole}} = 0.025 \text{ moles} \][/tex]
3. Using Avogadro's number:
[tex]\[ \text{Number of molecules} = \text{moles} \times \text{Avogadro's number} \][/tex]
4. Calculation:
[tex]\[ \text{Number of molecules} = 0.025 \times 6.022 \times 10^{23} = 1.5055 \times 10^{22} \text{ molecules} \][/tex]
So, 0.7 grams of nitrogen gas contain [tex]\(1.5055 \times 10^{22}\)[/tex] molecules.
### Part (c): 5.3 grams of sodium carbonate (Na₂CO₃)
We need to find the number of moles first, and then the number of molecules.
1. Given:
- Mass of Na₂CO₃ = 5.3 grams
- Molar mass of Na₂CO₃ = 106.0 grams per mole
2. Moles calculation:
[tex]\[ \text{Moles of Na₂CO₃} = \frac{\text{mass}}{\text{molar mass}} = \frac{5.3 \text{ grams}}{106.0 \text{ grams per mole}} = 0.05 \text{ moles} \][/tex]
3. Using Avogadro's number:
[tex]\[ \text{Number of molecules} = \text{moles} \times \text{Avogadro's number} \][/tex]
4. Calculation:
[tex]\[ \text{Number of molecules} = 0.05 \times 6.022 \times 10^{23} = 3.011 \times 10^{22} \text{ molecules} \][/tex]
So, 5.3 grams of sodium carbonate contain [tex]\(3.011 \times 10^{22}\)[/tex] molecules.
### Part (d): 52 grams of helium (He)
We need to find the number of moles first, and then the number of atoms (since He is a noble gas, it exists as single atoms).
1. Given:
- Mass of He = 52 grams
- Molar mass of He = 4.0 grams per mole
2. Moles calculation:
[tex]\[ \text{Moles of He} = \frac{\text{mass}}{\text{molar mass}} = \frac{52 \text{ grams}}{4.0 \text{ grams per mole}} = 13 \text{ moles} \][/tex]
3. Using Avogadro's number:
[tex]\[ \text{Number of atoms} = \text{moles} \times \text{Avogadro's number} \][/tex]
4. Calculation:
[tex]\[ \text{Number of atoms} = 13 \times 6.022 \times 10^{23} = 7.8286 \times 10^{24} \text{ atoms} \][/tex]
So, 52 grams of helium contain [tex]\(7.8286 \times 10^{24}\)[/tex] atoms.
### Summary
The results are:
- 0.25 moles of [tex]\(O_2\)[/tex] contain [tex]\(1.5055 \times 10^{23}\)[/tex] molecules.
- 0.7 grams of [tex]\(N_2\)[/tex] contain [tex]\(0.025 \approx 0.024999999999999998\)[/tex] moles or [tex]\(1.5055 \times 10^{22}\)[/tex] molecules.
- 5.3 grams of [tex]\(Na_2CO_3\)[/tex] contain [tex]\(0.05 \approx 0.049999999999999996\)[/tex] moles or [tex]\(3.011 \times 10^{22}\)[/tex] molecules.
- 52 grams of [tex]\(He\)[/tex] contain [tex]\(13\)[/tex] moles or [tex]\(7.8286 \times 10^{24}\)[/tex] atoms.
