Answer:
i dont see it but i hope it looks like a dna molecule or something would be wrong
Explanation:
there is no pic
However <em>trans</em>-2-Butene does not give a characteristic peak in 1620-1680 cm⁻¹ region but still the presence of carbon double bond carbon can be detected by detecting following peaks in IR Spectrum.
1) 3010-3100 cm⁻¹:
As in trans-2-Butene a hydrogen atoms ate attached to sp² hybridized carbon, therefore the stretching of =C-H (C-H) bond will give a peak of medium intensity in the range of 3010-3100 cm⁻¹.
2) 675-1000 cm⁻¹:
Another peak which is given by the bending of =C-H (C-H) bond with strong intensity will appear in the range of 675-1000 cm⁻¹.
Answer:

Explanation:
Hello there!
In this case, for these problems about collecting a gas over water, we must keep in mind that once the gas has been collected, the total pressure of the system is given by the atmospheric pressure, in this case 1.01 atm. Next, since we also have water in the mixture, we can write the following equation:

Thus, by solving for the pressure of nitrogen and using consistent units, we obtain:

The Answer is D: Reactor products charge huge batteries.
Answer : The electron configurations consistent with this fact is, (b) [Kr] 4d¹⁰
Explanation :
Electronic configuration : It is defined as the representation of electrons around the nucleus of an atom.
Number of electrons in an atom are determined by the electronic configuration.
Paramagnetic compounds : They have unpaired electrons.
Diamagnetic compounds : They have no unpaired electrons that means all are paired.
The given electron configurations of Palladium are:
(a) [Kr] 5s²4d⁸
In this, there are 2 electrons in 's' orbital and 8 electrons in 'd' orbital. From the partial orbital diagrams we conclude that 's' orbital are paired but 'd' orbital are not paired. So, this configuration shows paramagnetic.
(b) [Kr] 4d¹⁰
In this, there are 10 electrons in 'd' orbital. From the partial orbital diagrams we conclude that electrons in 'd' orbital are paired. So, this configuration shows diamagnetic.
(c) [Kr] 5s¹4d⁹
In this, there are 1 electron in 's' orbital and 9 electrons in 'd' orbital. From the partial orbital diagrams we conclude that 's' orbital and 'd' orbital are not paired. So, this configuration shows paramagnetic.