Given:
Diprotic weak acid H2A:
Ka1 = 3.2 x 10^-6
Ka2 = 6.1 x 10^-9.
Concentration = 0.0650 m
Balanced chemical equation:
H2A ===> 2H+ + A2-
0.0650 0 0
-x 2x x
------------------------------
0.065 - x 2x x
ka1 = 3.2 x 10^-6 = [2x]^2 * [x] / (0.065 - x)
solve for x and determine the concentration at equilibrium.
<span>the atomic mass of nitrogen is 14. There is 1 nitrogen atom in the molecule so the percentage of N is :
14/35 x100% = 40%</span>
Answer:
Explanation:
C₂H₂ + 2H₂ = C₂H₆
1 mole 2 mole 1 mole
Feed of reactant is 1.6 mole H₂ / mole C₂H₂
or 1.6 mole of H₂ for 1 mole of C₂H₂
required ratio as per chemical reaction written above
2 mole of H₂ for 1 mole of C₂H₂
So H₂ is in short supply . Hence it is limiting reagent .
1.6 mole of H₂ will react with half of 1.6 mole or .8 mole of C₂H₂ to form .8 mole of C₂H₆
a )Calculate the stoichiometric reactant ratio = mole H₂ reacted/mole C₂H₂ reacted
= 1.6 / .8 = 2 .
b )
yield ratio = mole C₂H₆ formed / mole H₂ reacted ) = 0.8 / 1.6 = 1/2 = 0.5 .
C can be the only correct answer - 6.023 x 10^23 is the amount of molecules in a mol of an element. 4.5 x 6.023 x 10^23 can not equal anything but C.
4.5 x 6.023 x 10^23 = 2.71035 x 10^24
