B) experiments that Darwin's original hypotheses
I believe is the right answer :)
Answer:
After 3 half lives 10 g of radio active material left.
Explanation:
Given data:
Total amount of radio active material = 80 g
Amount left after 3 half lives = ?
Solution:
At time zero = 80 g
At first half life = 80 g/2 = 40 g
At 2nd half life = 40 g/2 = 20 g
At 3rd half life = 20 g/2 = 10 g
Thus, after 3 half lives 10 g of radio active material left.
Answer:
Mass of solute in Kg = 0.07 Kg
Explanation:
Given
Molecular weight of the chemical = 28 grams per mole
Solution consists of 2.5 moles of solution
Mass of solute in the solution = 2.5 *28 = 70 grams
Mass of solute in Kg = 70 grams/1000 = 0.07 Kg
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