Answer:
See image below
Explanation:
The image is labeled according to the sequence N'-Trp-Ser-Asg-Gly-Cys-His-COOH' which means that in the main chain of the peptide, the amino group of the Tryptophan and the carboxylic group of the Histidine are free and thus its charge depends on the pH; other groups that rely on the pH are the side groups of the Cysteine and the Histidine.
Overall, ionizable groups in this peptide are:
- Amino Group of the Tryptophan (pKa = 9.39)
- SH group of the Cysteine (pKa = 8.18)
- Secondary amine of the Histidine (pKa = 6.00)
- Carboxylic Group of the Histidine (pKa = 1.82)
Then, the amino group of Trp and SH group of Cys are protonated since the peptide is at a pH below the pKa. The secondary amine of the Histidine is deprotonated because the pH is greater than the pKa, as well with its carboxylic acid group.
Answer: The FitnessGram PACER Test is a multistage aerobic capacity test that progressively gets more difficult as it continues. The test is used to measure a student's aerobic capacity as part of the FitnessGram assessment. Students run back and forth as many times as they can, each lap signaled by a beep sound.
Explanation:
Answer:
0.0905 M
Explanation:
Let's consider the neutralization reaction between H2SO4 and KOH.
H₂SO₄ + 2 KOH → K₂SO₄ + 2 H₂O
22.87 mL of 0.158 M KOH react. The reacting moles of KOH are:
0.02287 L × 0.158 mol/L = 3.61 × 10⁻³ mol
The molar ratio of H₂SO₄ to KOH is 1:2. The reacting moles of H₂SO₄ are 1/2 × 3.61 × 10⁻³ mol = 1.81 × 10⁻³ mol
1.81 × 10⁻³ moles of H₂SO₄ are in 20.0 mL. The molarity of H₂SO₄ is:
M = 1.81 × 10⁻³ mol / 0.0200 L = 0.0905 M
Answer:
Ver las respuestas abajo.
Explanation:
Este problema se puede resolver conociendo la relacion entre horas y minutos, sabemos que:
1 hora [h] → 60 minutos [min]
De esta manera:
2 [min] = 2/60 = 0.033 [h]
15 [min] = 15/60 = 0.25 [h]
30 [min] = 30/60 = 0.5 [h]
10 [min] = 10/60 = 0.166 [h]
6 [min] = 6/60 = 0.1 [h]
20 [min] = 20/60 = 0.33 [h]
5 [min] = 5/60 = 0.0833 [h]
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