1. 5 ethyl, 2 methyl octane
2. 1 ethyl, 2 methyl cyclopentane
3. 3,3,5,5- tetrafluoro heptane
4. 3,4-dimethyl hexene
5. 3,4-dimethyl cyclobutene
6. 3,5 diisopropyl cyclohexene
7. 3,3,4 trimethyl pentyne
8. 2,6 dibromo phenol
keep in mind that between 4-7, there could be #1 in front of the main name. for example with #4: 3,4-dimethyl-1- hexene. this honestly depends on the professor how he/she likes it. It is not necessary because if the number is not specified, it is assumed is #1
Answer: 2.8
Explanation: just took the quiz
Δmc
2
For one reaction:
Mass Defect =Δm
=2(m
H
)−m
He
−m
n
=2(2.015)−3.017−1.009
=0.004 amu
1 amu=931.5 MeV/c
2
Hence,
E=0.004×931.5 MeV=3.724 MeV
E=3.726×1.6×10
−13
J=5.96×10
−13
J
For 1 kg of Deuterium available,
moles=
2g
1000g
=500
N=500N
A
=3.01×10
26
Energy released =
2
N
×5.95×10
−13
J
=8.95×10
13
It would be D.the food chain
Answer:
Explanation:
Take a random sample of nuts from the jar. Let's take two handfuls, after shaking the jar and mixing the nuts thoroughly. Separate the nuts into almonds and cashews. Count each pile, then do the following calculation (these numbers are random, for example only).
<u> Count</u> <u>Percentage %</u>
Almonds 38 (38)/(87)x100
Cashews <u> 49</u> 49/87x100
87 87/87 = 100%
Ratio of Almonds to Cashews: <u>38/49</u>
