Since we are told that 1L of air contains 0.21L of oxygen, you can use the conversion (0.21L O₂)/(1L air). That means that you can just multiply 6.0L by 0.21L to get 1.26L of O₂.
that means that the lungs can hold about 1.26L of oxygen.
I hope this helps. Let me know if anything is unclear.
Answer:
Mole fraction Ar = 0.31
Explanation:
Remember that the sum of the mole fractions in a mixture of gases = 1
Mole fraction = Moles from a gas / Total moles
Mole fraction N₂+ Mole fraction O₂+ Mole fraction SO₂+ Mole fraction CO₂ + Mole fraction Ar = 1
N₂ = 0.21
O₂= 0.16
CO₂ = 0.23
SO₂ = 0.09
Mole fraction Ar = 1 - 0.21 - 0.16 - 0.23 - 0.09
Mole fraction Ar = 0.31
PH=-log[H⁺]
pH=-log(1.87×10⁻¹³)
pH=12.72
I hope this helps. Let me know if anything is unclear.
The pH of the solution : 12
<h3>Further explanation</h3>
Reaction
HCOOH + NaOH ⇒ HCOONa + H₂O
mol HCOOH =
mol NaOH =
Mol NaOH>mol HCOOH ⇒ at the end of the reaction there will be a strong base remains from mol NaOH, so that the pH is determined from [OH⁻]
ICE method :
HCOOH + NaOH ⇒ HCOONa + H₂O
4 5
4 4 4 4
0 1 1 1
Concentration of [OH⁻] from NaOH :
pOH=-log[OH⁻]
pOH=-log 10⁻²=2
pH+pOH=14
pH=14-2=12
