Answer:
1st one is right..it helps filter waste inside the cell.
The answer is D. Okay l hope this helps
Answer:
Explanation:1. NaNH2 (1-Butene)
CH3CH2CH2CH2Cl --------------> CH3CH2CH=CH2 + HCl (elimination reaction)
2. Br2, CCl4
CH3CH2CH=CH2 ---------------> CH3CH2CH(Br)CH2Br (Simple addition Reaction)
3. NaNH2 (1-Butyne)
CH3CH2CH(Br)CH2Br ----------------> CH3CH2C≡CH + 2HBr
Sodamide (NaNH2) is a very strong base and generally results in Terminal Alkynes when treated with Vicinal Dihalides.
Alcoholic KOH on the other hand results in the formation of Alkynes with triple bonds in the middle of the molecule.
I would say 2 because co2 goes out and o goes in
Density is equal to mass divided by volume so the densest object will be the object that has the largest mass in the smallest area.
In this case object A is the densest with a density of 10g/cm^3.
I hope this helps. Let me know if anything is unclear.
