The compound is Molecular
Hey there!:
V1 = 3.05 L
V2 = 3.00 L
P1 = 724 mmHg
P2 = to be calculated
T1 = 298 K
T2 = 273 K
Therefore:
P1*V1 / T1 = P2*V2 / T2
P2 = ( P1*V1 / T1 ) * T2 / V2
P2 = 724 * 3.05 * 273 / 298 * 3.00
P2 = 602838.6 / 894
P2 = 674.31 mmHg
1 atm ----------- 760 mmHg
atm ------------- 674.31 mHg
= 674.31 * 1 / 760
= 0.887 atm
Hope this helps!
Answer:
pH = 4.09
Explanation:
molarity of oxalic acid in the solution
= 0.1 x 25 / (25 + 35)
= 0.0417 M
molarity of NaOH in the solution
= 0.1 x 35 / (25 +35)
= 0.0583 M
H2C2O4 + NaOH -------------------> NaHC2O4 + H2O
0.0417 0.0583 0 0
0 0.0166 0.0417
now second acid -base titration
NaHC2O4 + NaOH -------------------> Na2C2O4 + H2O
0.0417 0.0166 0 0
0.0251 0 0.0166 ---
now
pH = pKa2 + log [Na2C2O4 / NaHC2O4]
pH = 4.27 + log (0.0166 / 0.0251)
pH = 4.09
3.5 M has 3.5 moles per litre
so we have one litre, so we need 3.5 moles
moles = mass/molarmass
3.5 * 23 = 80.5
