Robert A. Millikan and Harvey Fletcher performed the oil drop experiment.
Answer:
Save the *minerals* used to extract metals.
Answer:
C) sp2 and sp2
Explanation:
The hybridization depens on the ammount and type of bonds the atom analized has in the molecule.
For example:
- A C atom bonded to 4 H atoms has a sp3 hybridization.
- A C atom bonded to 2 H atoms and to 1 C with a double bond (like in ethene) has a sp2 hybridization
- A C bonded to 1 H and 1 C with a triple bond (like in ethyne) has a sp hybridization.
Analyzing the type and amount of unions of the nitrogen and the carbonyl you will be able to determine the hybridization.
In the imine, the N atom has a double bond to a C and a simple bond two other C, plus the lone pair of electrons (counts as a bond) so it will have a sp2 hybridization.
In the carbonyl, the C has two simple bonds to other C and a double bond to an oxygen atom. It will also have a sp2 hybridization
Answer is: <span>the molarity of this glucose solution is 0.278 M.
m</span>(C₆H₁₂O₆<span>) = 5.10 g.
n</span>(C₆H₁₂O₆) = m(C₆H₁₂O₆) ÷ M(C₆H₁₂O₆<span>) .
</span>n(C₆H₁₂O₆) = 5.10 g ÷ 180.156 g/mol.
n(C₆H₁₂O₆<span>) = 0.028 mol.
</span>V(solution) = 100.5 mL ÷ 1000 mL/L.
V(solution) = 0.1005 L.
c(C₆H₁₂O₆) = n(C₆H₁₂O₆) ÷ V(solution).
c(C₆H₁₂O₆) = 0.028 mol ÷ 0.1005 L.
c(C₆H₁₂O₆<span>) = 0.278 mol/L.</span>
