Answer:
2.8
Explanation:
First, we will calculate the molarity of the acetylsalicylic acid solution.
M = mass of solute (g) / molar mass of solute × volume of solution (L)
M = 0.327 g / 180.158 g/mol × 0.237 L
M = 7.66 × 10⁻³ M
For a weak acid such as acetylsalicylic acid, we can find the concentration of H⁺ using the following expression.
[H⁺] = √(Ca × Ka)
where,
Ca: concentration of the acid
Ka: acid dissociation constant
[H⁺] = √(7.66 × 10⁻³ × 3.3 × 10⁻⁴)
[H⁺] = 1.6 × 10⁻³ M
The pH is:
pH = -log [H⁺]
pH = -log 1.6 × 10⁻³ = 2.8
4 mol / 205g H2O = 4/.205 = 19.5 mol/kg boiling point = 100 + 19.5 • 0.51 = 109 ºC
Biotic - animals anything tht is living abiotic anything tht is non-living
Answer:
5,844 grams of NaCl
Explanation:
Knowing the molecular weight 58,44 g/mole and saying 1 molar solution is 58,44 of NaCl in 1 liter of solution. 100 mL means 10% of the whole solution then we are going to have 10% of NaCl
58,44 x 0,1 = 5,844 grams of NaCl
Answer:
0.08 mol L-1
Explanation:
Sulfuric acid Formula: H2SO4
Ammonia Formula: NH3
Ammonium sulfate Formula: (NH₄)₂SO₄
H2SO4 + 2NH3 = 2NH4+ + SO4 2-
H2SO4 + 2NH3 = (NH₄)₂SO₄
H2SO4 = (1/2)x (32.8 x 10^-3 L x 0.116 mol L-1)/25 x 10^-3 L
= 0.08 mol L-1