It evaporates as the CO2 transitions from solid to gas form.
The answer is false. The correct answer would be efficiency.
Answer:
The answer to your question is 6.2 %
Explanation:
Data
mass of glucose (solute) = 27 g
mass of solvent = 410 g
Formula
Percent by mass = mass of solute/mass of solution x 100
-Substitution
Percent by mass = 27/(410 + 27) x 100
- Simplification
Percent by mass = 27/437 x 100
-Result
Percent by mass = 6.18 ≈ 6.2 %
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