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Assuming fully decomposed the volume of CO2 released at Stp on heating 9.85g of BaCO3(at. mass, Ba=137)will be a) 0.84L b) 2.24L c)4.06L d) 1.12L correct answer is option d. Can you explain this answer?
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Assuming fully decomposed the volume of CO2 released at Stp on heating...
Explanation:
To calculate the volume of CO2 released, we need to use the stoichiometry of the reaction between BaCO3 and CO2. The balanced equation for the reaction is as follows:
BaCO3(s) → BaO(s) + CO2(g)
From the equation, we can see that 1 mole of BaCO3 produces 1 mole of CO2. We can calculate the number of moles of BaCO3 using its molar mass:
Molar mass of BaCO3 = (137 g/mol) + (12 g/mol) + (3 * 16 g/mol) = 197 g/mol
Now, we can calculate the number of moles of BaCO3:
Number of moles of BaCO3 = Mass of BaCO3 / Molar mass of BaCO3
= 9.85 g / 197 g/mol
= 0.05 mol
Since 1 mole of BaCO3 produces 1 mole of CO2, the number of moles of CO2 released is also 0.05 mol.
To calculate the volume of CO2 at STP (standard temperature and pressure), we can use the ideal gas law:
PV = nRT
Where:
P = Pressure (STP = 1 atm)
V = Volume
n = Number of moles
R = Ideal gas constant (0.0821 L.atm/mol.K)
T = Temperature (STP = 273 K)
Plugging in the values:
(1 atm) * V = (0.05 mol) * (0.0821 L.atm/mol.K) * (273 K)
V = (0.05 mol * 0.0821 L.atm/mol.K * 273 K) / (1 atm)
= 1.123 L
Therefore, the volume of CO2 released at STP when heating 9.85 g of BaCO3 is approximately 1.12 L, which corresponds to option (d).
To calculate the volume of CO2 released, we need to use the stoichiometry of the reaction between BaCO3 and CO2. The balanced equation for the reaction is as follows:
BaCO3(s) → BaO(s) + CO2(g)
From the equation, we can see that 1 mole of BaCO3 produces 1 mole of CO2. We can calculate the number of moles of BaCO3 using its molar mass:
Molar mass of BaCO3 = (137 g/mol) + (12 g/mol) + (3 * 16 g/mol) = 197 g/mol
Now, we can calculate the number of moles of BaCO3:
Number of moles of BaCO3 = Mass of BaCO3 / Molar mass of BaCO3
= 9.85 g / 197 g/mol
= 0.05 mol
Since 1 mole of BaCO3 produces 1 mole of CO2, the number of moles of CO2 released is also 0.05 mol.
To calculate the volume of CO2 at STP (standard temperature and pressure), we can use the ideal gas law:
PV = nRT
Where:
P = Pressure (STP = 1 atm)
V = Volume
n = Number of moles
R = Ideal gas constant (0.0821 L.atm/mol.K)
T = Temperature (STP = 273 K)
Plugging in the values:
(1 atm) * V = (0.05 mol) * (0.0821 L.atm/mol.K) * (273 K)
V = (0.05 mol * 0.0821 L.atm/mol.K * 273 K) / (1 atm)
= 1.123 L
Therefore, the volume of CO2 released at STP when heating 9.85 g of BaCO3 is approximately 1.12 L, which corresponds to option (d).
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Assuming fully decomposed the volume of CO2 released at Stp on heating 9.85g of BaCO3(at. mass, Ba=137)will be a) 0.84L b) 2.24L c)4.06L d) 1.12L correct answer is option d. Can you explain this answer?
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Assuming fully decomposed the volume of CO2 released at Stp on heating 9.85g of BaCO3(at. mass, Ba=137)will be a) 0.84L b) 2.24L c)4.06L d) 1.12L correct answer is option d. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Assuming fully decomposed the volume of CO2 released at Stp on heating 9.85g of BaCO3(at. mass, Ba=137)will be a) 0.84L b) 2.24L c)4.06L d) 1.12L correct answer is option d. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Assuming fully decomposed the volume of CO2 released at Stp on heating 9.85g of BaCO3(at. mass, Ba=137)will be a) 0.84L b) 2.24L c)4.06L d) 1.12L correct answer is option d. Can you explain this answer?.
Assuming fully decomposed the volume of CO2 released at Stp on heating 9.85g of BaCO3(at. mass, Ba=137)will be a) 0.84L b) 2.24L c)4.06L d) 1.12L correct answer is option d. Can you explain this answer? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about Assuming fully decomposed the volume of CO2 released at Stp on heating 9.85g of BaCO3(at. mass, Ba=137)will be a) 0.84L b) 2.24L c)4.06L d) 1.12L correct answer is option d. Can you explain this answer? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Assuming fully decomposed the volume of CO2 released at Stp on heating 9.85g of BaCO3(at. mass, Ba=137)will be a) 0.84L b) 2.24L c)4.06L d) 1.12L correct answer is option d. Can you explain this answer?.
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