Find the molar mass of CaCO3 then subtract the molar mass what it originally weighed and the loss of mass. Hopefully this works!
The pH of a neutral aqueous solution at 37°C is 6.8.
<h3>What is Kw? </h3>
Kw is defined as the dissociation, which is also known as self-ionization, constant of water. this is an equilibrium constant, and its expression is:
Kw = [OH⁻] . [H₃O⁺]
Neutral pH determines that the concentrations of OH⁻ and H₃O⁺ are equal.
<h3>Calculation</h3>
Let us suppose concentration of OH and H₃O⁺ is x, to calculate it:
Kw =[OH⁻] . [H₃O⁺] = x²
x² = 2.4 × 10⁻¹⁴ M²
x = 1.5919 × 10⁻⁷ M
Hence, the concentration of OH and H₃O⁺ (x) = [H₃O⁺] = [OH⁻] = 1.5919×10⁻⁷ M
pH = -log[H₃O⁺] = -log( 1.5919×10⁻⁷ M)
pH = 6.8
Thus, we find that the pH of a neutral aqueous solution at 37 °c (which is the normal human body temperature) is 6.8.
learn more about pH:
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For an aqueous solution of MgBr2, a freezing point depression occurs due to the rules of colligative properties. Since MgBr2 is an ionic compound, it acts a strong electrolyte; thus, dissociating completely in an aqueous solution. For the equation:
ΔTf<span> = (K</span>f)(<span>m)(i)
</span>where:
ΔTf = change in freezing point = (Ti - Tf)
Ti = freezing point of pure water = 0 celsius
Tf = freezing point of water with solute = ?
Kf = freezing point depression constant = 1.86 celsius-kg/mole (for water)
m = molality of solution (mol solute/kg solvent) = ?
i = ions in solution = 3
Computing for molality:
Molar mass of MgBr2 = 184.113 g/mol
m = 10.5g MgBr2 / 184.113/ 0.2 kg water = 0.285 mol/kg
For the problem,
ΔTf = (Kf)(m)(i) = 1.86(0.285)(3) = 1.59 = Ti - Tf = 0 - Tf
Tf = -1.59 celsius
If you are asking which is the most abundant, Uranium-238 is
