Answer:
At the second equivalent point 200 mL of NaOH is required.
Explanation:
at the first equivalent point:
H2A + OH- = HA- + H2O
initial mmoles y*100 y*100 - -
final mmoles 0 0 y*100 y*100
at the second equivalent point:
HA- + OH- = A2- + H2O
initial mmoles y*100 y*100 - -
final mmoles - - y*100 y*100
at the second equivalent point we have that y*100 mmoles of NaOH or 100 mL of NaOH ir required, thus:
at the second equivalent point 200 mL of NaOH is required.
Answer:
The answer is D..i.e phosphate. it consists of 1p and 4oxygen atoms
1) 2C₆H₆ + 15O₂ = 6H₂O + 12CO₂
2) n(C₆H₆)/2=n(CO₂)/12
n(CO₂)=6n(C₆H₆)
n(CO₂)=6*12.8 mol = 76.8 mol
Answer:
105 grams PbI₂
Explanation:
Pb(NO₃)₂ + 2KI => 2KNO₃ + PbI₂(s)
moles Pb(NO₃)₂ = 0.265L(1.2M) = 0.318 mole
moles KI = 0.293(1.55M) = 0.454 mole => Limiting Reactant
moles PbI₂ from mole KI in excess Pb(NO₃)₂ = 1/2(0.454 mole) = 0.227 mol PbI₂
grams PbI₂ = 0.227 mol PbI₂ x 461 g/mole = 104.68 g ≈ 105 g PbI₂(s)
