1. Calcium hydrogen phosphate
2. Iron(II) Hydrogen Sulfate
3. Calcium hydroxide
4. Aluminum chloride hydroxide
Answer:
V O2 = 1.623 L
Explanation:
- 1 mol ≡ 6.022 E23 molecules
∴ molecules O2 = 4.00 E22 molecules
⇒ moles O2 = (4.00 E22 molecules O2)×(mol O2/6.022 E23 molecules)
⇒ moles O2 = 0.0664 moles
at STP:
∴ T = 25°C ≅ 298 K
∴ P = 1 atm
assuming ideal gas:
∴ V = RTn/P
⇒ V O2 = ((0.082 atm.L/K.mol)(298 K)(0.0664 mol))/( 1 atm)
⇒ V O2 = 1.623 L
The ball travels 10m/s
Speed = distance/time
Answer:
Here's what I get
Explanation:
(a) Intermediates
The three structures below represent one contributor to the resonance-stabilized intermediate, in which the lone pair electrons on the heteroatom are participating (the + charge on the heteroatoms do not show up very well).
(b) Relative Stabilities
The relative stabilities decrease in the order shown.
N is more basic than O, so NH₂ is the best electron donating group (EDG) and will best stabilize the positive charge in the ring. However, the lone pair electrons on the N in acetanilide are also involved in resonance with the carbonyl group, so they are not as available for stabilization of the ring.
(c) Relative reactivities
The relative reactivities would be
C₆H₅-NH₂ > C₆H₅-OCH₃ > C₆H₅-NHCOCH₃