Answer:
One nucleophilic center
Explanation;
Nucleophile:
Nucleophile is a substance which is nucleus loving in nature (<em>Nucleo</em>; Nucleus , <em>phile</em>; Loving). It is known as a specie which donates a lone pair of electrons to electrophile (electron loving) in a chemical reaction.
Thus, Nucleophile is the region of higher electron density in a molecule and attacks on the lower electron density region of another molecule. Also, the nucleophile can also contain a negative charge.
Number of Nucleophilic centers in Methanol:
The chemical structure of Methanol is attached below and it can be observed that the oxygen atom is containing two lone pair of electrons. Hence, the oxygen atom can act as a nucleophilic center. Therefore, there is only one nucleophilic center in methanol.
H₃C-OH + H₃C-Br → H₃C-O-CH₃ + HBr
In above reaction methanol is acting as a nucleophile and is attacking on electrophilic center (Carbon) of methyl bromide yielding dimethyl ether.
Answer:
Graphs should be titled.
Although the bunnies feed foxes, if there are TOO many foxes, they don’t go well; so that looked cyclic to me. It is a good argument.
STUDY VS GRADES is INCREASINGLY
OBVIOUS! I already know what hummingbirds like. Rain can help wash the smoke from forest fires from the air; do you pick one, OK?
Don’t be nervous - study more. I’ve had a great many students, and they not only survived my classes - most THRIVED!
Explanation:
Answer:
Three possible blood type alleles are Iᴬ, Iᴮ and i
Explanation:
Iᴬ, Iᴮ and i are three possible blood type alleles.
Iᴬ and Iᴮ are known as co-dominant, and The i allele is recessive.
Thus, Three possible blood type alleles are Iᴬ, Iᴮ and i
<u>-TheUnknownScientist</u>
Answer:
Total worth of gold in the ocean = $5,840,000,000,000,000
Explanation:
As stated in the question above, 4.0 x 10^-10 g of gold was present in 2.1mL of ocean water.
Therefore, In 1 L of ocean water there will be,
(4.0 x 10^-10)/0.0021
= 1.9045 x 10^-7 g of gold per Liter of ocean water.
So in 1.5 x 10^-21 L of ocean water, there will be
(1.9045 x 10^-7) * (1.5 x 10^-21)
= 2.857 x 10^14 g of gold in the ocean.
1 gram of gold costs $20.44, that is 20.44 dollars/gram. The total cost of the gold present in the ocean is
20.44 * (2.857 x 10^14)
= $5,840,000,000,000,000
