Triprotic acid is a class of Arrhenius acids that are capable of donating three protons per molecule when dissociating in aqueous solutions. So the chemical reaction as described in the question, at the third equivalence point, can be show as: H3R + 3NaOH ⇒ Na3R + 3H2O, where R is the counter ion of the triprotic acid. Therefore, the ratio between the reacted acid and base at the third equivalence point is 1:3.
The moles of NaOH is 0.106M*0.0352L = 0.003731 mole. So the moles of H3R is 0.003731mole/3=0.001244mole.
The molar mass of the acid can be calculated: 0.307g/0.001244mole=247 g/mol.
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Explanation:
Answer:
First, find out how many moles of N2I6 you have. Then convert that to grams.
molar mass N2I6 = 789 g
moles N2I6 = 8.2x1022 molecules N2I6 x 1 mole/6.02x1023 molecules = 1.36x10-1 moles = 0.136 moles
grams N2I6 = 0.136 moles x 789 g/mole = 107 g = 110 g (to 2 significant figures)
Well you are at a 86% so t<span>his is considered a "B" grade on an average grade scale. If you take 50 points we would need to know the Total point you could have got in that class to be able to see how much a percent was the assignment work and then take that percentage of the assignment and subtract it from the 86% to see where on the scale you would fall in after the assignment points were taken off. Hope this helps :)
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B. This is because the Hydrogen and Oxygen need balanced out.
Current-
C-1 | C-1
H-4 | H-2
O-2 | O-3
Adding a coefficient of 2 before oxygen in the reactants and H2O in the products would balance this equation
<span>CH4 + 2O2 → CO2 + 2H2O</span>
C-1 | C-1
H-4 | H-4
O-4 | O-4
