This is false. An alcohol does indeed have a polar C-O single bond, but what we should really be focusing on is the extraordinarily polar O-H single bond. When oxygen, fluorine, or nitrogen is bound to a hydrogen atom, there is a small (but not negligible) charge separation, where the eletronegative N, O, or F has a partial negative charge, and the H has a partial positive charge. Water has two O-H single bonds in it (structure is H-O-H). The partially negative charge on the O of the water molecule (specifically around the lone pair) can become attracted either a neighboring water molecule's partially positive H atom, or an alcohol's partially positive H atom. This is weak (and partially covalent) attraction is called a hydrogen bond. This is stronger than a typical dipole-dipole attraction (as would be seen between neighboring C-O single bonds), and much stronger than dispersion forces (between any two atoms). When the solvent (water) and the solute (the alcohol) both exhibit similar intermolecular forces (hydrogen bonding being the most important in this case), they can mix completely in all proportions (i.e. they are miscible) in water.
Answer:
0.24 M
Explanation:
Molarity = Moles solute / Liters solution
Step 1: Identify variables
400 mL = Liters solution
0.60 moles = Moles solute
Step 2: Identify conversions
1 L = 1000 mL
Step 3: Convert mL to L
400mL(1 L/1000mL) = 0.4 L
Step 4: Find molarity
M = (0.4 L)(0.60 mol) = 0.24 M
Answer:
The nitrogens are both sp3 hybridized. Their bonds are formed by sp overlaps. The carbon and oxygen are sp2 hybridized. The double bond with oxygen is produced by a sp2 overlap to form the sigma component and a probital overlap to form the pi component. The bonds with hydrogen are formed by sp2 overlaps.
Explanation:
Answer:
8.0 mol O₂
Explanation:
Let's consider the complete combustion reaction of C₉H₁₂.
C₉H₁₂ + 12 O₂ → 9 CO₂ + 6 H₂O
The molar ratio of C₉H₁₂ to O₂ is 1:12. The moles of O₂ required to react with 0.67 moles of C₉H₁₂ are:
0.67 mol C₉H₁₂ × (12 mol O₂/1 mol C₉H₁₂) = 8.0 mol O₂
8.0 moles of O₂ are required to completely react with 0.67 moles of C₉H₁₂.
<span>FIXATION.In the nitrogen cycle, nitrogen is converted into many chemical forms through biological and physical processes like nitrogen fixation, ammonification, nitrification and denitrification.During ammonification, ammonium ions are released after decomposition of plants.During nitrogen fixation, nitrogen gas is changed to nitrates by nitrogen fixing bacteria in the soil.These are used by vegetation.</span>
