Practice Problem 1: The following reagents are combined in a test tube: o 2.00 mL of 0.20 M potassium iodide o 1.00 mL of 1% starch o 0.50 mL of 0.20 M ammonium persulfate o 0.50 mL of 0.012 M sodium thiosulfate o 2.00 mL of 0.20 M potassium nitrate o 2.00 mL of 0.20 M ammonium sulfate What is the concentration of iodide in the solution in the test tube

0.05 M is the concentration of iodide in the solution in the test tube.

Explanation:

Molarity of KI = 0.20 M

Volume of KI solution = 2.00 mL = 0.002 L ( 1 mL = 0.001 L)

Moles of KI = n

Volume of solutions formed by  mixing all the solutions in a test tube = V

V = 2.00 mL+1.00 mL+0.50 mL+0.50 mL+2.00 mL+2.00 mL = 8.00 ml

V = 8.00 mL = 0.008 L

Concentration of KI in the test tube = [KI]

0.05 M is the concentration of iodide in the solution in the test tube.

nanotechnology is a powerful tool in biomaterials research as it creates numerous possibilities for the engineering of functional molecules. nano-bio refers to projected ability to incorporate biomimicry into various materials at nanoscale.  

answer: option (b) is the correct answer.

explanation:

a covalent bond is formed by sharing of electrons. whereas polarity arises due to difference in electronegativity of two atoms.

therefore, more is the electronegativity of an atom which combines with the titanium, more readily it will induce a partial positive and negative charge.

thus, out of the given options fluorine is most electronegative and has an incomplete octet. therefore, it will readily combine with titanium to share the electrons and complete its octet.