I have attached an image of the IR spectrum required to answer this question.
Looking at the IR, we can look for any clear major stretches that stand out. Immediately, looking at the spectrum, we see an intense stretch at around 1700 cm⁻¹. A stretch at this frequency is due to the C=O stretch of a carbonyl. Therefore, we know our answer must contain a carbonyl, so it could still be a ketone, aldehyde, carboxylic, ester, acid chloride or amide. However, if we look in the 3000 range of the spectrum, we see some unique pair of peaks at 2900 and 2700. These two peaks are characteristic of the sp² C-H stretch of the aldehyde.
Therefore, we can already conclude that this spectrum is due to an aldehyde based on the carbonyl stretch and the accompanying sp² C-H stretch.
The original sample was a compound because it was composed of two different elements and was not purely one element
<span>Among the benefits; enlarging so that you can see it visually, making something complex and hard to understand to simplify it. One draw back is that not entirely accurate because it is basic. Additionally; models help us to visualize smaller structures that are too small to see properly. However; they do not take all variables into account and thus they may be inaccurate. </span>
Answer:
The correct option is;
a) As a juvenile
Explanation:
An high school student who is charged with drunk driving will be tried as a juvenile in the juvenile court and are likely to receive more penalties than normally prescribed for a defendant charged with driving under the influence.
The penalties that could be faced includes
1) Having to spend time in a juvenile detention center
2) Be put on probation
3) Recommendation for drug and alcohol counseling
4) Driving license suspension for a duration from 90 days to up to 2 years
5) Having driving under the influence intervention program.
Answer:
161.57 K
Explanation:
V1/T1 = v2/T2
(5.24 L)/(411 K) = (2.06 L)/T2
T2 = 161.57 K