Answer:
2.02 g
Explanation:
Molality (m) is the <em>number of moles of solute dissolved in 1 kg (1000 g) of solvent- </em>that is,
<em>
</em>
Our first step is to calculate the moles of NH₃ in 475 grams of methanol, using <u>molality</u> as a conversion factor.
0.250 mol/kg × 0.475 kg = 0.119 mol
This solution contains <u>0.119 moles of NH₃</u>. Now we must convert the moles to grams, using the <u>molar mass of NH₃</u> as a conversion factor
.
17.031 g/mol × 0.119 mol = 2.02 g
Therefore, 2.02 grams of NH₃ must be dissolved in 475 grams of methanol to make a 0.250 m solution.
At pH 3.70, H2SO4 can be regarded as being fully dissociated. The dissociation reaction is
H₂SO₄(l) --> 2H⁺(aq) + SO₄²⁻(aq)
Since pH = -log₁₀[H⁺]
[H⁺] = 
= 
[H⁺] = 2 x 10⁻⁴ M
Since there are 2 moles of H⁺ for every mole of H₂SO₄
[H₂SO₄] = 0.5 * [H⁺]
=0.5 * 2 x 10⁻⁴ M
[H₂SO₄] = 1 x 10⁻⁴ M
Convert the amount of rain and the area into SI units
Rain = 1.00 in * 0.0254 m/in
Rain = 0.0254 m
Area = 1600 miles² * 2589988.11 m²/miles² = 4.661 x 10⁹ m²
Calculate the volume of water
Volume = Rain * Area
=0.0254 m * 4.661 x 10⁹
Volume = 1.184 x 10⁸ m³
Calculate the number of kilo moles of H₂SO₄
kilo moles H₂SO₄ = [H₂SO₄] * Volume
=1.0 x 10⁻⁴ M * 1.184 x 10⁸ m³
kilo moles H₂SO₄ = 11,839 k mole
Calculate the mass of H₂SO₄
mass H₂SO₄ = kilo moles H₂SO₄ * MW H₂SO₄
= 11839 k mol * 98.07848 kg/kmol
mass H₂SO₄ = 1.16 x 10⁶ kg H₂SO₄
The mass of electrons , i just did the same test and got it right
Upon a constant pressure (P), volume (V) of a gas will vary in direct proportion to changes in temperature (T). So V1/T1 = V2/T2
V2 = V1T2/T1 = (500)(300)/150
V2 = 150000/150 = 1000 mL
