Answer:
74.4 ml
Explanation:
C₆H₈O₇(aq) + 3NaHCO₃(s) => Na₃C₆H₅O₃(aq + 3CO₂(g) + 3H₂O(l)
Given 15g = 15g/84g/mol = 0.1786mole Sodium Bicarbonate
From equation stoichiometry 3moles NaHCO₃ is needed for each mole citric acid or, moles of citric acid needed is 1/3 of moles sodium bicarbonate used.
Therefore, for complete reaction of 0.1786 mole NaHCO₃ one would need 1/3 of 0.1786 mole citric acid or 0.0595 mole H-citrate.
The question is now what volume of 0.8M H-citrate solution would contain 0.0595mole of the H-citrate? This can be determined from the equation defining molarity. That is => Molarity = moles solute / Liters of solution
=> Volume (Liters) = moles citric acid / Molarity of citric acid solution
=> Volume needed in liters = 0.0.0595 mole/0.80M = 0.0744 Liters or 74.4 ml
Nitrogen accounts for 78% of the atmosphere, oxygen 21% and argon 0.9%. Gases like carbon dioxide, nitrous oxides, methane, and ozone are trace gases that account for about a tenth of one percent of the atmosphere.
One of the atoms involved must be a metal and the other a nonmetal
Correct answer is:
,,Atoms can react together, be separated from each other, or be rearranged via chemical reactions. "
:-) ;-)
