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Answer:
92.04%
Explanation:
Given:
Mass of CO₂ obtained = 53.0 grams
Mass of calcium carbonate heated = 1.31 grams
Now,
the molar mass of the calcium carbonate = 100.08 grams
The number of moles heated in the problem = Mass / Molar mass
= (1.31 grams) / (100.08 grams/moles)
= 0.013088 moles
now,
1 mol of calcium carbonate yields 1 mol of CO₂
thus,
0.013088 moles of calcium carbonate will yield = 0.013088 mol of CO₂
now,
Theoretical mass of 0.013088 moles of CO₂ will be
= Number of moles × Molar mass of CO₂
= 0.013088 × 44 = 0.5758 grams
Thus, the percent yield for this reaction = 
or
the percent yield for this reaction = 
or
the percent yield for this reaction = 92.04%
Mixtures come in many forms and phases. Most of them can be separated, and the kind of separation method depends on the kind of mixture it is. Below are some common separation methods:
C) Sliver Carbonate AgCO3
Answer:
M HCl sln = 12.0785 M
Explanation:
- molarity (M) [=] mol/L
- %mm = ((mass compound)/(mass sln))*100
∴ mass sln = 100.0 g
∴ δ sln = 1.19 g/mL
∴ % m/m = 37 %
⇒ 37 % =((mass HCl/mass sln))*100
⇒ 0.37 = mass HCl / 100.0 g
⇒ 37 g = mass HCl
∴ molar mass HCl = 36.46 g/mol
⇒ mol HCl = (37 g)*(mol/36.46 g) = 1.015 mol
⇒ volume sln = (100 g sln)*(mL/1.19 g) = 84.034 mL = 0.084034 L
⇒ M HClsln = 1.015 mol/0.084034 L
⇒ M HCl sln = 12.0785 M