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Based on CaCo3 + 2HCl --> H2O + CO2 + CaCl. How many ml of 1.00 M NaOH should be used in the titration. 25ml of HCL and 750mg of Tums are combined to simulate an upset stomach.
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What is the concn of the HCl?
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Based on CaCo3 + 2HCl --> H2O + CO2 + CaCl. How many ml of 1.00 M NaOH should be used in the titration. 25ml of HCL and 750mg of
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You don't have all of the needed information (e.g., 25 mL of what M HCl?). It would help if you
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If 20g of CaCo3is treated with 20gm of HCl how many grams of CO2 can be generated CaCO3+2HCl gives CaCl+H2O+CO2
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Follow your previous CaCO3 problem. Convert 20 g HCl and 20 g CaCO3 to mols first. Remember this is
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A mixture of CaCO3 and MgCO3 weighing 7.85 g was reacted with excess HCL. The reactions are CaCO3+2HCL=CaCl2+H2O+CO2 and
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Two equations in two unknowns; solve simultaneously. Let X = mass CaCO3 Let Y = mass MgCO3
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what is the method to prepartion CO from these equation in the lab
CaCo3 + 2HCl → CaCl2+H2O+ CO2 (g) Or CaCo3 ----(Heat 500 to
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Be patient. See my next response at the initial post.
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how to prepare CO from this equation in the lab CaCo3 + 2HCl → CaCl2+H2O+ CO2 (g)
Or CaCo3 ----(Heat 500 to 600 C0→ Cao +Co2
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I don't see a question here.
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A mixture contained CaCO3 and MgCo3. A sample of this mixture weighing 7.85g was reacted with excess HCl. The reactions are--
CaC
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NEED HELP will NOT get you help. Name it Chemistry or whatever it is. Sra
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A mixture contained CaCO3 and MgCo3. A sample of this mixture weighing 7.85g was reacted with excess HCl. The reactions are--
CaC
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Assume x grams of CaCO3, so 7.85-x grams of MgCO3. Now, you can figure the moles of each (in terms
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what volume of carbondioxide is produced when 10g of Hcl reacts completely with caco3
caco3+2HCl=cacl2+H2o+Co2 moles
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.27 mols HCl agree so .135 mols CO2--- agree at standard temp and pressure a mol is 22.4 liters .135
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which equation represents an oxidation-reduction reaction?
1.HCl+KOH-->KCl+H2O 2.4HCl+MnO2-->MnCl2+2H2O+Cl2
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Well, I could try to give you a serious answer, but where's the fun in that? So, let's crack a joke
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How many milliliters of 0.237 M HCl are needed to react with 52.2g of CaCO3?
2HCl(aq) + CaCO3(s) --> CaCl2(aq) + CO2(g) + H2O(l)
Top answer:
mol wt of CaCO3 = 40.08 + 12.01 + 48 = 100.09 so, you have 52.2/100.09 = .522 moles the equation
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