Attached is the MOT of NO molecule.
1. Bond order is calculated is
In present case, number of e- in bonding orbital = 6
number of e- in anti-bonding orbital = 1
∴ Bond-order =
= 2.5
2. From the attached figure, it can be seen that NO has one unpaired electron in π* orbital.
Physical: The chemist could try to bend it to find out how malleable it is. He could also try to pull it into wires to find out how ductile it is.
Chemical: The chemist could put the metal into contact with other substances to get an idea of how reactive it is, and he could try to burn it and find out how flammable it is.
Answer:
The answer to your question is: letter D. 1.33 L
Explanation:
Data
V1 = 50 ml
C1 = 19.3
To solve this problem use the formula C₁V₁ = C₂V₂
C2 = C1V1 / V2
C = concentration
V = volume
a) 1.15 L
C2 = (19.3)(50) / 1150
C2 = 0.84 M
b) No right answer
c) V2= 0.80 L
C2 = (19.3)(50) / 800
C2 = 1.2 M
d) V2 = 1.33 L
C2 = (19.3)(50) / 1330
C2 = 0.72 M
e) V2 = 350 ml
C2 = (19.3)(50) / 350
C2 = 2.75 M
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Answer:
I believe that it is the 2nd option.
Explanation:
My reasonings are because C4H10O has 7 isomers. In which 4 are alcohol and the other 3 are ether.
The first option is ethers, specifically ethoxyethane.
The third option is ethers, specifically 1-methoxypropane.
The fourth option is an alcohol, specifically 1- butanol.
Therefore, leads us to the 2nd option that it is NOT an isomer of C4H10O
