Answer:
[N₂] = 0.032 M
[O₂] = 0.0086 M
Explanation:
Ideal Gas Law → P . V = n . R . T
We assume that the mixture of air occupies a volume of 1 L
78% N₂ → Mole fraction of N₂ = 0.78
21% O₂ → Mole fraction of O₂ = 0.21
1% another gases → Mole fraction of another gases = 0.01
In a mixture, the total pressure of the system refers to total moles of the mixture
1 atm . 1L = n . 0.082L.atm/mol.K . 298K
n = 1 L.atm / 0.082L.atm/mol.K . 298K → 0.0409 moles
We apply the mole fraction to determine the moles
N₂ moles / Total moles = 0.78 → 0.78 . 0.0409 mol = 0.032 moles N₂
O₂ moles / Total moles = 0.21 → 0.21 . 0.0409 mol = 0.0086 moles O₂
Henderson–Hasselbalch equation is given as,
pH = pKa + log [A⁻] / [HA]
-------- (1)
Solution:
Convert Ka into pKa,
pKa = -log Ka
pKa = -log 1.37 × 10⁻⁴
pKa = 3.863
Putting value of pKa and pH in eq.1,
4.29 = 3.863 + log [lactate] / [lactic acid]
Or,
log [lactate] / [lactic acid] = 4.29 - 3.863
log [lactate] / [lactic acid] = 0.427
Taking Anti log,
[lactate] / [lactic acid]
= 2.673
Result:
2.673 M
lactate salt when mixed with 1 M Lactic acid produces a buffer of pH = 4.29.
Answer: Heating the hydrated forms of cobalt chloride reverses the reactions above, returning cobalt chloride to the blue, water-free, or anhydrous, state. Water is "liberated" in these reactions, known as dehydration reactions.
Explanation:
Answer:
Explanation:
1) Find number of each of the type of atom that is present in the compound, using the chemical formula .
2) Then multiply number of atoms of each element that is present in the compound with the atomic weight of each of the element
3) Add everything together and add the units (grams/mole ) after the number
Let finds that of water
Chemical formula of water is (H20 )
hydogens atoms= 2
oxygen atom= 1
Atomic weight for Hydrogen= 1
Atomic weight for Oxygen= 16
Total number of atoms of Hydrogen from the formula (H2O)= 2
Total number of atoms of Oxygen from the formula (H2O)= 1
the molar mass=
Hydrogen: ( 2 x 1)= 2
Oxygen: ( 1 x 16)= 16
Add together= (16+2)
= 18
Then add the unit, we have(18 g/mol.)
Answer:
36.51%.
Explanation:
First find the percentage of iron in pure Fe2O3 using the atomic masses of the elements:
= (2 * 55.845) * 100 / (2*55.845+ 3*15.999)
= 69.94 %.
So the percentage of iron in the mixture
= 52.2 * 0.6994
= 36.51 (answer).