Answer:
See explanation
Explanation:
2HCl(aq) + CaCO3(aq) ------->CaCl2(aq) + CO2(g) + H2O(l)
Number of moles of acid present = 50/1000 * 0.15 = 0.0075 moles
Number of moles of calcium carbonate = 0.054g/100 g/mol = 0.00054 moles
2 moles of HCl reacts with 1 mole of calcium carbonate
x moles of HCl reacts with 0.00054 moles of calcium carbonate
x = 2 * 0.00054/1
x = 0.00108 moles of HCl
Amount of acid left = 0.0075 moles - 0.0075 moles = 0.00642 moles
Reaction of HCl and NaOH
HCl(aq) + NaOH(aq) ------> NaCl(aq) + H2O(l)
Since the reaction is in the mole ratio of 1:1
0.00642 moles of HCl is neutralized by 0.00642 moles of NaOH
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Answer:
The six member ring and the position of the -OH group on the carbon (#4) identifies glucose from the -OH on C # 4 in a down projection in the Haworth structure). Fructose is recognized by having a five member ring and having six carbons, a hexose.
