According to Henry's law, solubility of solution is directly proportional to partial pressure thus,

Solubility at pressure 3.08 atm is 72.5/100, solubility at pressure 8 atm should be calculated.
Putting the values in equation:

On rearranging,

Therefore, solubility will be 1.88 mg of
gas in 1 g of water or, 188 mg of tex]N_{2}[/tex] gas in 100 g of water.
The third question requires you to solve for the weight of sodium (Na) and weight of Chloride (Cl) from the calculated moles of each element Na, and Cl.
So, you need to multiply the calculated moles of Na with its molar mass (23 g/ mol) to get the answer for Na. And multiply the calculated moles of Cl with its molar mass (35.45 g/mol) to get the answer for Cl.
(7.3 x 10^29 atoms) / (24 atoms/molecule) / (6.022 x 10^23 molecules/mol) =
5.1 x 10^4 mol C6H12O6
We have that the the density of FeS is mathematically given as
From the question we are told
Iron(II) sulfide has a primitive <em>cubic</em> unit cell with <em>sulfide</em> ions at the <em>lattice points.</em>
The ionic radii of iron(II) ions and sulfide ions are 88 pm and 184 pm, respectively.
What is the density of FeS (in g/cm3)?
<h3>
Density</h3>
Generally the equation for the Velocity is mathematically given as

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Answer:
0.33 mol/kg NH₃
Explanation:
Data:
b(NH₃) = 0.33 mol/kg
b(Na₂SO₄) = 0.10 mol/ kg
Calculations:
The formula for the boiling point elevation ΔTb is

i is the van’t Hoff factor — the number of moles of particles you get from a solute.
(a) For NH₃,
The ammonia is a weak electrolyte, so it exists almost entirely as molecules in solution.
1 mol NH₃ ⟶ 1 mol particles
i ≈ 1, and ib = 1 × 0.33 = 0.33 mol particles per kilogram of water
(b) For Na₂SO₄,
Na₂SO₄(aq) ⟶ 2Na⁺(aq) + 2SO₄²⁻(aq)
1 mol Na₂SO₄ ⟶ 3 mol particles
i = 1 and ib = 3 × 0.10 = 0.30 mol particles per kilogram of water
The NH₃ has more moles of particles, so it has the higher boiling point.