Answer:
Explanation:
A. Attached is a page which contain the structural formula of the three compounds of C3H80.
Condensed structural formula of C3H8O:
Propan-1-ol: CH3CH2CH2OH
Propan-2-ol: CH3CH(OH)CH3
Methoxy methane: CH3OCH2CH3
B. Attached are is a page which contain the structural formula of the three compounds of C3H60.
Condensed structural formula of C3H6O:
Propanal: CH3CH2CHO
Propanone: CH3COCH3
Cyclopropanol: (C3H5)OH
2-propen-1-ol: CH2CHCH2OH
1-propenol: CH3CHCHOH
Climate change or it could just be Climate,
Hopefully that helps ❤
a)
A: Copper
B: CuO
C: 
D: $\mathrm{CuCO_3}$
E: $\mathrm{CO_2}$
F: $\mathrm{Cu(NO_3)_2}$
b)
$\mathrm{CuO+ H_2SO_4}\rightarrow \mathrm{CuSO_4 + H_2O}$
c)
$\mathrm{CuCO_3+ 2HNO_3}\rightarrow \mathrm{Cu(NO_3)_2+ CO_2+ H_2O}$
Answer:
a) 
b) entropy of the sistem equal to a), entropy of the universe grater than a).
Explanation:
a) The change of entropy for a reversible process:


The energy balance:
![\delta U=[tex]\delta Q- \delta W](https://tex.z-dn.net/?f=%5Cdelta%20U%3D%5Btex%5D%5Cdelta%20Q-%20%5Cdelta%20W)
If the process is isothermical the U doesn't change:
![0=[tex]\delta Q- \delta W](https://tex.z-dn.net/?f=0%3D%5Btex%5D%5Cdelta%20Q-%20%5Cdelta%20W)


The work:

If it is an ideal gas:


Solving:

Replacing:


Given that it's a compression: V2<V1 and ln(V2/V1)<0. So:

b) The entropy change of the sistem will be equal to the calculated in a), but the change of entropy of the universe will be 0 in a) (reversible process) and in b) has to be positive given that it is an irreversible process.
The wavelength is the distance between one crest/trough to another crest/trough. On the image, it's basically the length between each peak of the wave. You can see that the distance between peaks in wave A are much shorter than the distance between the peaks in wave B.
Thus, wave B has the longer wavelength.