Answer:
The molarity of the sulfuric acid is 0.018 M
Explanation:
The molarity of a solution is the number of moles of the solute (sulfuric acid in this case) in a 1-liter solution.
Every 100 g of the solution, we have 95 g sulfuric acid because its concentration is 95% w/w.
With the density, we can calculate how many liters are 100 g of solution:
density = mass / volume
1.85 g / ml = 100 g / volume
volume = 100 g / 1.85 g/ml
volume = 54.1 ml or 0.0541 l
Now, we know that we have 95 g sulfuric acid in 0.0541 l solution. In 1 l, we have then:
1 l * 95g / 0.0541 l = 1.756 g sulfuric acid.
But we want to know how many moles sulfuric acid we have per liter. Then, using the molar mass, we can calculate how many moles there are in 1.756 g sulfuric acid:
1.756 g * 1 mol / 98.08 g = 0.018 mol
The molarity is 0.018 M
Answer:
8.55 × 10²⁴ Ions
Explanation:
Ammonium Chloride is an ionic compound which contains a monatomic anion (Cl⁻ ; Chloride) and a polyatomic cation (NH₄⁺ ; Ammonium).
Hence, when added in water Ammonium Chloride ionizes as;
NH₄Cl → NH₄⁺ + Cl⁻
Hence, we can say that it produces two ions when dissolved in water.
Also,
We know that 1 mole of any substance contains exactly 6.022 × 10²³ particles which is also called as Avogadro's Number. So in order to calculate the number of ions contained by 7.1 moles of NH₄Cl, we will use following relation to first calculate the number of molecules as;
Moles = Number of Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹
Solving for Number of Molecules,
Number of Molecules = Moles × 6.022 × 10²³ Molecules.mol⁻¹
Putting values,
Number of Molecules = 7.1 mol × 6.022 × 10²³
Number of Molecules = 4.27 × 10²⁴ Molecules
So,
As,
1 Molecule of NH₄Cl contained = 2 Ions
So,
4.27 × 10²⁴ Molecules of NH₄Cl will contain = X ions
Solving for X,
X = 2 Ions × 4.27 × 10²⁴ Molecules / 1 Molecule
X = 8.55 × 10²⁴ Ions
I also think it’s B but not quite sure
Answer:
a) Ksp = 7.9x10⁻¹⁰
b) Solubility is 6.31x10⁻⁶M
Explanation:
a) InF₃ in water produce:
InF₃ ⇄ In⁺³ + 3F⁻
And Ksp is defined as:
Ksp = [In⁺³] [F⁻]³
4.0x10⁻²g / 100mL of InF₃ are:
4.0x10⁻²g / 100mL ₓ (1mol / 172g) ₓ (100mL / 0.1L) = <em>2.3x10⁻³M InF₃. </em>Thus:
[In⁺³] = 2.3x10⁻³M InF₃ × (1 mol In⁺³ / mol InF₃) = 2.3x10⁻³M In⁺³
[F⁻] = 2.3x10⁻³M InF₃ × (3 mol F⁻ / mol InF₃) = 7.0x10⁻³M F⁻
Replacing these values in Ksp formula:
Ksp = [2.3x10⁻³M In⁺³] × [7.0x10⁻³M F⁻]³ = <em>7.9x10⁻¹⁰</em>
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b) 0.05 moles of F⁻ produce solubility of InF₃ decrease to:
7.9x10⁻¹⁰ = [x] [0.05 + 3x]³
Where x are moles of In⁺³ produced from solid InF₃ and 3x are moles of F⁻ produced from the same source. That means x is solubility in mol / L
Solving from x:
x = -0.018 → False solution, there is no negative concentrations.
x = 6.31x10⁻⁶M → Right answer.
Thus, <em>solubility is 6.31x10⁻⁶M</em>
Answer:
6 x 10⁵ kg Hg
Explanation:
The mass of mercury in the entire lake is found by multiplying the concentration of the mercury by the volume of the lake.
The volume of the lake is calculated in cubic feet:
V = (SA)x(depth) = (100mi²)(5280ft/mi)² x (20ft) = 5.57568 x 10¹⁰ ft³
Cubic feet are then converted to mL (1cm³=1mL)
(5.57568 x 10¹⁰ ft³) x (12in/ft)³ x (2.54cm/in)³ = 1.578856752 x 10¹⁵ mL
The mass of mercury is then found:
m = CV = (0.4μg/mL)(1g/10⁶μg)(1kg/1000g) x (1.578856752 x 10¹⁵ mL) = 6 x 10⁵ kg Hg
