Data:
Q = 402.7 J → releases → Q = - 402.7 J
m = 16.25 g
T initial = 54 ºC
adopting: c = 4.184J/g/°C
ΔT (T final - T initial) = ?
Solving:
Q = m*c*ΔT
-402.7 = 16.25*4.184*ΔT
-402.7 = 67.99*ΔT
If: ΔT (T final - T initial) = ?
Answer:
Methacrylaldehyde
Explanation:
The first step is the calculation of the <u>IHD</u> (index hydrogen deficiency):
This value indicates that we have <u>2 double bonds</u>. Now, if we check the IR info we can conclude that we have an <u>oxo group</u> (C=O) due to the signal in <u>1705 cm^-1 </u>. So, the options that we can have are <u>aldehyde or ketone</u>.
If we analyze the NMR info we have a signal in 194.7 <u>with only 1 hydrogen</u>. This indicates that necessary we have an <u>aldehyde due to the hydrogen</u>. Also, for the signal in 14 we will have a , for the signal at 134.2 we will have a and for the signal at 146.0 we will have a quaternary carbon (no hydrogens present).
So, we will have a , , C (without hydrogens), an aldehyde group and a double bond.
When we put all this together we will obtain the <u>Methacrylaldehyde</u> (see figure).
Answer: chemical substance
Explanation:
Answer:
Explanation:
Water vapor, carbon dioxide, methane, and other trace gases in Earth's atmosphere absorb the longer wavelengths of outgoing infrared radiation from Earth's surface. These gases then emit the infrared radiation in all directions, both outward toward space and downward toward Earth