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Margaret [11]
2 years ago
13

"Calculate the number of millimoles in 500 mg of each of the following amino acids: alanine (MW = 89), leucine (131), tryptophan

(204), cysteine (121), and glutamic acid (147)."
Chemistry
1 answer:
Cloud [144]2 years ago
5 0

Answer:

- Alanine =  5.61 mmoles

- Leucine = 3.81 mmoles

- Tryptophan = 2.45 mmoles

- Cysteine = 4.13 mmoles

- Glutamic acid = 3.40 mmoles

Explanation:

Mass / Molar mass = Moles

Milimoles = Mol . 1000

500 mg / 1000 = 0.5 g

- Alanine = 0.5 g / 89 g/m → 5.61x10⁻³ moles  . 1000 = 5.61mmoles

- Leucine = 0.5 g / 131 g/m → 3.81 x10⁻³ moles  . 1000 = 3.81 mmoles

- Tryptophan = 0.5 g / 204 g/m →  2.45x10⁻³ moles . 1000 = 2.45 mmoles

- Cysteine = 0.5 g / 121 g/m → 4.13x10⁻³ moles . 1000 = 4.13 mmoles

- Glutamic acid = 0.5 g 147 g/m → 3.40x10⁻³ moles . 1000 =  3.4 mmoles

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Julli [10]

The solubility product, ksp of PbBr₂ is 2.102 × 10⁻⁶

<h3>What is solubility?</h3>

The solubility of a solute is defined as the maximum amount of that solute that can be dissolved in a known quantity of solvent at a given temperature.

<h3>What is a solubility product? </h3>

Some salts are sparingly soluble in a solvent. For them, we calculate the solubility product.

It is an equilibrium constant that defines the relationship between a solid and its respective ions in an aqueous solution in equilibrium.

The greater the solubility product, the greater the solubility and vice versa.

Here, the solubility of PbBr₂ = 2.96 g/l

Molar solubility of  PbBr₂ = \frac{2.96}{molar mass of PbBr_2} = 2.96/367 = 8.07 × 10⁻³

At equilibrium,

PbBr_2\rightarrow Pb^{2+} + 2Br^-

1 mole of PbBr2 dissociates into 1 mole of Pb²⁺ ions and 2 moles of Br⁻

Let the molar concentration of Pb²⁺ be x, then the molar concentration of Br⁻ is 2x

Ksp = x.(2x)²

       = 4x³

Substitute, x = 8.07 × 10⁻³

Ksp = 4 (8.07 × 10⁻³)³

      = 2.102 × 10⁻⁶

Thus, The ksp of PbBr₂ is 2.102 × 10⁻⁶

Learn more about solubility product:

brainly.com/question/1419865

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B. K+ has 18 electrons

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Explanation:

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