Answer: I found the answer it is.
Answer:
12.78
Explanation:
pOH= -log(0.0600) = 1.22184875
pH= 14-1.22184875 = 12.78
No. of moles in 134g of Mg, n=134/24 =5.58 moles
No. of atoms = n* Avogadro number
= 5.58 X 6.022 X 10^23
=33.60 X 10^23 atoms
You would have to use the ideal gas law for this:
PV=nRT
Pressure, Volume, n=moles, R gas constant, Temperature in Kelvin
P=nRT/V
(1.8mol)(62.36)(309K)/43.0L = 805mm Hg
Answer:
The mass of oxygen needed to be pumped in the aquarium is 10 g
Explanation:
Here the value of the degree of solubility of oxygen in water is sought
The solubility of oxygen in fresh water is approximately 1.22 × 10⁻³ mol·dm⁻³ which is equivalent to approximately 40 mg/L
The volume of water the aquarium holds = 250 L
The mass of oxygen required = The solubility of oxygen in water × (The volume of water in the aquarium)
Therefore, the mass of oxygen needed to be pumped in the aquarium = 40 mg/L × 250 L = 10,000 mg
1,000 mg = 1 g
∴ 10,000 mg = 10 g
The mass of oxygen needed to be pumped in the aquarium = 10,000 mg = 10 g.
