2 KOH(aq) + H2C2O4(s) → K2C2O4(aq) + 2 H2O(l) When a sample of oxalic acid (H2C2O4), which is a diprotic acid (both H-atoms are acidic), is titrated with 0.250 M potassium hydroxide solution, 24.66 mL of the potassium hydroxide solution are required to neutralize the acid. I don’t know is this gonna help but ok ._.
Answer:
Determine the location of the lost significant place value by placing a bar over the digit.
Explanation:
Number of moles of oxygen = x
number of moles of nitrogen = y
x = 2y
initial pressure, p1 = 0.8 atm
final pressure, p2 = 1.10 atm
At constant volume and temperature p1 / n1 = p2 / n2
=> p1 / p2 = n1 / n2
n1 = x + y = 2y + y = 3y
n2 = 0.2 + 3y
=> p1 / p2 = 3y / (0.2 + 3y)
=> 0.8 / 1.10 = 3y / (0.2 + 3y)
=> 0.8 (0.2 + 3y) = 1.10 (3y)
0.16 + 2.4y = 3.3y
=> 3.3y - 2.4y = 0.16
=> 0.9y = 0.16
=> y = 0.16 / 0.9
=. x = 2*0.16/0.9 = 0.356
Answer: 0.356 moles O2
D. is the answer
hope i could help
