Answer:
The NMR spectrum that corresponds best to p-bromoaniline is the one that is attached in the image below.
Explanation:
For the p-bromoaniline 3 types of hydrogen are observed. The first signal that appears at 3.7 ppm would be from the hydrogens of the NH2 group, the hydrogens in ortho position with respect to the NH2 group give a double at approximately 6.54 ppm, and finally the characteristic 7.21 ppm signal is observed for the hydrogens in meta position with with respect to the NH2 group.
Answer:
that are formed from hybridized orbitals
Explanation:
The chemical concept of resonance is when a change in the position of the electrons occurs, without changing the position of the atoms.
The structure obtained in the resonance will not be any of the previous ones, but a hybrid of resonance between those structures.
Answer:
Sr(s) + C(s) + 3/2 O₂(g) → SrCO₃(s)
Explanation:
The standard enthalpy of formation (ΔH°f) is the energy involved in the formation of 1 mole of a substance from its elements in their most stable states. The chemical equation for the formation of SrCO₃(s) is the following.
Sr(s) + C(s) + 3/2 O₂(g) → SrCO₃(s)
Using the combined gas law, where PV/T = constant, we first solve for PV/T for the initial conditions: (4.50 atm)(36.0 mL)/(10.0 + 273.15 K) = 0.57213.
Remember to use absolute temperature.
For the final conditions: (3.50 atm)(85.0 mL)/T = 297.5/T
Since these must equal, 0.57213 = 297.5/T
T = 519.98 K
Subtracting 273.15 gives 246.83 degC.
