Answer:
Tyrosine is a polar and aromatic compound. its side chain acidity and basicity is neutral
if a peptide contain only a string of tyrosine residue especially l tyrosine the solubility increases more
Explanation:
even tyrosine number remains constant, tyrosine containing peptide will be more soluble. This peptide is soluble in 1 M HCl (100 mg/ml), with heating. The solubility in water (25 °C) is 0.45 mg/ml in the pH range 3.2 - 7.5.
2.0 mg/ml; at pH 9.5, the solubility is 1.4 mg/ml; and at pH 10, the solubility is 3.8 mg/ml.
Answer:
0.0894 M
Explanation:
So you first want to change the grams into moles (since molarity is moles solute per Liter solution).
The molar mass of NaNO3 is 84.995 g/mol, so we divide 0.38g of NaNO3 by 84.995 g/mol to get approximately 0.004471 mol NaNO3.
Now, all you have to do is to divide by the total Liters of water (your solution).
Since the volume is expressed in mL, you just need to divide by 1000 (1 L = 1000mL) to get 0.050 L solution.
Therefore, to get molarity, you divide 0.004471 mol by 0.050 L to get your answer of 0.0894 M.
At 55g of Glucose/ 1000ml
1g of Glucose =(1000/55)ml
and 13g of Glucose will be 13*1000/55= 236.4ml
The patient should be given 236.4ml of the solution.