The correct option is (b)
NaNH2 is an effective base. It can be a good nucleophile in the few situations where its strong basicity does not have negative side effects. It is employed in elimination reactions as well as the deprotonation of weak acids.Alkynes, alcohols, and a variety of other functional groups with acidic protons, such as esters and ketones, will all be deprotonated by NaNH2, a powerful base.Alkynes are deprotonated with NaNH2 to produce what are known as "acetylide" ions. These ions are powerful nucleophiles that can react with alkyl halides to create carbon-carbon bonds and add to carbonyls in an addition reaction.Acid/base and nucleophilic substitution are the two types of reactions.Using the right base, terminal alkynes can be deprotonated to produce a carbanion.A good C is the acetylide carbanion.The acetylide carbanion can undergo nucleophilic substitution reactions because it is a potent C nucleophile. (often SN2) with 1 or 2 alkyl halides with electrophilic C to create an internal alkyne (Cl, Br, or I).Elimination is more likely to occur with 3-alkyl halides.It is possible to swap either one or both of the terminal H atoms in ethylene (acetylene) to create monosubstituted (R-C-C-H) and symmetrical (R = R') or unsymmetrical (R not equal to R') disubstituted alkynes (R-C-C-R').
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Answer:
1218.585
Explanation:
Looking at the subscripts we know there are 2 atoms of Fe, 3 atoms of C, and 6 of O.
Take the molar mass of each atom (from the periodic table) and multiply by the # of atoms
Fe: 55.845×2= 111.69
C: 12.011×3= 36.033
O:15.999×6=95.994
Add the values together: 243.717 g/mol
That is 1 mole of the molecule. Multiply by 5 for the final answer.
243.717×5=1218.585
Answer:
Zinc nitrate gives white ppt. which dissolves in excess ammonium hydroxide and produce a colorless solution whereas lead nitrate gives a chalky white ppt. of lead hydroxide which doesnot dissolve.
Explanation:
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Explanation:
The given data is as follows.
= 250 mL, 
= 750 mL
= 
= 35 + 273 K = 308 K
= 35 + 273 K = 308 K
= 0.55 atm, 
= 1.5 atm
P = ? , V = 10.0 L
Since, temperature is constant.
So, 
= PV
Now, putting the given values into the above formula as follows.
= PV
= 
P = 0.126 atm
As, 1 atm = 760 torr. So, 
= 95.76 torr.
Thus, we can conclude that the final pressure, in torr, of the mixture is 95.76 torr.
