Answer :
Certainly! Here is a detailed, step-by-step solution for part c:
### Question c: How many atoms of oxygen are present in 300g of CaCO3?
1. Molar Mass Calculation:
- We know that the molar mass of CaCO3 (calcium carbonate) is \( 100.0869 \) grams per mole. This value indicates how much one mole of CaCO3 weighs.
2. Mass of Sample:
- We are given that the mass of our sample is \( 300 \) grams.
3. Calculating Moles of CaCO3:
- To find out how many moles of CaCO3 are in 300 grams, we use the formula:
[tex]\[ \text{Moles of CaCO3} = \frac{\text{Mass of Sample}}{\text{Molar Mass of CaCO3}} \][/tex]
Substituting in the given values:
[tex]\[ \text{Moles of CaCO3} = \frac{300 \text{ grams}}{100.0869 \text{ grams/mole}} = 2.9973952635160046 \text{ moles} \][/tex]
4. Oxygen Atoms in CaCO3:
- Each molecule of CaCO3 contains 3 oxygen atoms. Therefore, each mole of CaCO3 contains 3 moles of oxygen atoms.
5. Calculating Oxygen Atoms:
- To find the total number of oxygen atoms in the given sample, we multiply the number of moles of CaCO3 by 3 (since each mole of CaCO3 has 3 moles of oxygen atoms) and then by Avogadro's number \( 6.022 \times 10^{23} \) atoms/mole:
[tex]\[ \text{Total Oxygen Atoms} = \text{Moles of CaCO3} \times 3 \times \text{Avogadro's Number} \][/tex]
Substituting in the values:
[tex]\[ \text{Total Oxygen Atoms} = 2.9973952635160046 \times 3 \times 6.022 \times 10^{23} = 5.415094283068014 \times 10^{24} \text{ atoms} \][/tex]
Therefore, there are approximately \( 5.415094283068014 \times 10^{24} \) oxygen atoms in 300 grams of CaCO3.
### Question d: A gas at constant temperature having an initial volume of 600 mL
The problem description is incomplete for part d, as it doesn't specify what is required to be found about the gas. However, based on common gas law problems, it might be asking about changes in volume, pressure, or other related properties under different conditions.
If additional information such as initial and final pressures, or change in volume or amount of gas, is provided, I can help further detailed calculations based on Boyle's Law or other gas laws.
Please provide the complete information or clarify the specific question for part d.
### Question c: How many atoms of oxygen are present in 300g of CaCO3?
1. Molar Mass Calculation:
- We know that the molar mass of CaCO3 (calcium carbonate) is \( 100.0869 \) grams per mole. This value indicates how much one mole of CaCO3 weighs.
2. Mass of Sample:
- We are given that the mass of our sample is \( 300 \) grams.
3. Calculating Moles of CaCO3:
- To find out how many moles of CaCO3 are in 300 grams, we use the formula:
[tex]\[ \text{Moles of CaCO3} = \frac{\text{Mass of Sample}}{\text{Molar Mass of CaCO3}} \][/tex]
Substituting in the given values:
[tex]\[ \text{Moles of CaCO3} = \frac{300 \text{ grams}}{100.0869 \text{ grams/mole}} = 2.9973952635160046 \text{ moles} \][/tex]
4. Oxygen Atoms in CaCO3:
- Each molecule of CaCO3 contains 3 oxygen atoms. Therefore, each mole of CaCO3 contains 3 moles of oxygen atoms.
5. Calculating Oxygen Atoms:
- To find the total number of oxygen atoms in the given sample, we multiply the number of moles of CaCO3 by 3 (since each mole of CaCO3 has 3 moles of oxygen atoms) and then by Avogadro's number \( 6.022 \times 10^{23} \) atoms/mole:
[tex]\[ \text{Total Oxygen Atoms} = \text{Moles of CaCO3} \times 3 \times \text{Avogadro's Number} \][/tex]
Substituting in the values:
[tex]\[ \text{Total Oxygen Atoms} = 2.9973952635160046 \times 3 \times 6.022 \times 10^{23} = 5.415094283068014 \times 10^{24} \text{ atoms} \][/tex]
Therefore, there are approximately \( 5.415094283068014 \times 10^{24} \) oxygen atoms in 300 grams of CaCO3.
### Question d: A gas at constant temperature having an initial volume of 600 mL
The problem description is incomplete for part d, as it doesn't specify what is required to be found about the gas. However, based on common gas law problems, it might be asking about changes in volume, pressure, or other related properties under different conditions.
If additional information such as initial and final pressures, or change in volume or amount of gas, is provided, I can help further detailed calculations based on Boyle's Law or other gas laws.
Please provide the complete information or clarify the specific question for part d.
