They exist in the outer orbitals
Answer:
The total mass of D-Glucose dissolved in a 2μL aliquot is 1 E-4 g
Explanation:
providing a solution to 5% weight-volume as found in commerce:
⇒ % 5 = (5g d-glucose/ 100 mL sln)×100
⇒ 0.05 = g C6H12O6/mL sln
⇒ g C6H12O6 = (2 μL sln)×(0.001 mL/μL)×(0.05 g C6H12O6/mL sln)
⇒ g C6H12O6 = 1 E-4 g C6H12O6
Possible answer could be: P/T = k
