The answer is option C.
That is it is a heterogeneous mixture.
Heterogeneous mixture have the following properties:
1. Different components could be observed in the substance.
2. Different samples of the substance appeared to have different proportions of the components.
3.The components could be easily separated using filters and sorting.
Answer:
a) Unsaturated
b) Supersaturated
c) Unsaturated
Explanation:
A saturated solution contains the <u>maximum amount of a solute that will dissolve in a given solvent at a specific temperature</u>.
An unsaturated solution contains <u>less solute than it has the capacity to dissolve. </u>
A supersaturated solution, <u>contains more solute than is present in a saturated solution</u>. Supersaturated solutions are not very stable. In time, some of the solute will come out of a supersaturated solution as crystals.
According to these definitions and considering that the solubility of KCl in 100 mL of H₂O at <u>20 °C is 34 g</u>, and at <u>50 °C is 43 g</u> we can label the solutions:
a) 30 g in 100 mL of H₂O at 20 °C ⇒ unsaturated
b) 65 g in 100 mL of H₂O at 50 °C ⇒ supersaturated
c) 42 g in 100 mL of H₂O at 50 °C and slowly cooling to 20 °C to give a clear solution <u>with no precipitate</u> ⇒ unsaturated (if it were saturated it would have had precipitate)
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.
The mass of Copper deposited at the cathode : 0.296 g
<h3>Further explanation</h3>
Given
time = t = 10 min=600 s
current = i = 1.5 A
F = 96500 C
charge Cu=+2
Required
The mass of Copper
Solution
Faraday's Law
e = Ar/valence(valence Cu=2, Ar=63.5 g/mol)
Input the value :
