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How many molecules of carbon dioxide are formed when one milligram of hundred percentage pure calcium carbonate is treated with excess hydrochloric acid?
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Calculation of Carbon Dioxide Molecules formed:
Calcium carbonate (CaCO3) reacts with hydrochloric acid (HCl) to form calcium chloride (CaCl2), water (H2O), and carbon dioxide (CO2). The balanced chemical equation for this reaction is:
CaCO3 + 2HCl -> CaCl2 + H2O + CO2

Step 1: Calculate the molar mass of CaCO3
The molar mass of CaCO3 = 40.08 (Ca) + 12.01 (C) + (3 * 16.00) (3 * O) = 100.09 g/mol

Step 2: Calculate the number of moles of CaCO3 in 1 mg
Number of moles = Mass / Molar mass
Number of moles = 0.001 g / 100.09 g/mol = 0.00000999 mol

Step 3: Calculate the number of moles of CO2 formed
From the balanced chemical equation, 1 mol of CaCO3 produces 1 mol of CO2
Therefore, the number of moles of CO2 formed = 0.00000999 mol

Step 4: Calculate the number of molecules of CO2 formed
1 mol of any substance contains Avogadro's number of molecules (6.022 x 10^23)
Number of molecules of CO2 formed = Number of moles * Avogadro's number
Number of molecules of CO2 formed = 0.00000999 mol * 6.022 x 10^23 = 6.02 x 10^19 molecules
Therefore, when 1 milligram of 100% pure calcium carbonate is treated with excess hydrochloric acid, approximately 6.02 x 10^19 molecules of carbon dioxide are formed.
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How many molecules of carbon dioxide are formed when one milligram of hundred percentage pure calcium carbonate is treated with excess hydrochloric acid?
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