Molecule
The smallest unit in a compound is its molecule.
<span>HNO2 =====> H+ + NO2-
</span>I<span>nitial concentration</span> = 0.311
<span>C = -x,x,x </span>
<span>E = 0.311-x,x,x
</span>KNO2 ====>K+ + NO2-
<span>Initial concentration = 0.189 </span>
<span>C= -0.189,0.189,0.189 </span>
E = 0,0.189,0.189
Answer:
Four substitution products are obtained. The carbocation that forms can react with either nucleophile (H2O or CH3OH) from either the top or bottom side of the molecule
Explanation:
An SN1 reaction usually involves the formation of a carbocation in the slow rate determining step. This carbocation is now attacked by a nucleophile in a subsequent fast step to give the desired product.
However, the product is obtained as a racemic mixture because the nucleophile may attack from the top or bottom of the carbocation hence both attacks are equally probable.
The attacking nucleophile in this case may be water or CH3OH
Responder:
27
Explicación:
Dado que:
Número de protones en el átomo X = 29
Carga en el átomo X = +2
Si no hay cargo neto;
número de protones = número de electrones
Sin embargo, dado que el átomo X tiene una carga de +2 (dando 2 electrones).
Por lo tanto,
Número de electrones = número de protones - número de carga en el átomo)
Número de electrones = (29 - 2) = 27
Yes it could, but you'd have to set up the process very carefully.
I see two major challenges right away:
1). Displacement of water would not be a wise method, since rock salt
is soluble (dissolves) in water. So as soon as you start lowering it into
your graduated cylinder full of water, its volume would immediately start
to decrease. If you lowered it slowly enough, you might even measure
a volume close to zero, and when you pulled the string back out of the
water, there might be nothing left on the end of it.
So you would have to choose some other fluid besides water ... one in
which rock salt doesn't dissolve. I don't know right now what that could
be. You'd have to shop around and find one.
2). Whatever fluid you did choose, it would also have to be less dense
than rock salt. If it's more dense, then the rock salt just floats in it, and
never goes all the way under. If that happens, then you have a tough
time measuring the total volume of the lump.
So the displacement method could perhaps be used, in principle, but
it would not be easy.
