Answer:
[EtOH] = 2.2M and Wt% EtOH = 10.1% (w/w)
Explanation:
1. Molarity = moles solute / Volume solution in Liters
=> moles solute = mass solute / formula weight of solute = 9.8g/46g·mol⁻¹ = 0.213mol EtOH
=> volume of solution (assuming density of final solution is 1.0g/ml) ...
volume solution = 9.81gEtOH + 87.5gH₂O = 97.31g solution x 1g/ml = 97.31ml = 0.09731 Liter solution
Concentration (Molarity) = moles/Liters = 0.213mol/0.09731L = 2.2M in EtOH
2. Weight Percent EtOH in solution (assuming density of final solution is 1.0g/ml)
From part 1 => [EtOH] = 2.2M in EtOH = 2.2moles EtOH/1.0L soln
= {(2.2mol)(46g/mol)]/1000g soln] x 100% = 10.1% (w/w) in EtOH.
Because they are different they all show different traits
A low electronegativity
Explanation:
Potassium is a metal that is expected to have a very low electronegativity value.
Electronegativity is the relative tendency by which an atom attracts valence electrons in a chemical bond.
Potassium is an element in the first group on the periodic table.
The common trend is that electronegativity increases from left to right and decreases down a group.
- Potassium as metal will prefer to lose electrons rather than attracting because that will make it achieve the octet configuration that will ensure its stability.
- This is why it will have low electronegativity.
Learn more:
Electronegativity brainly.com/question/11932624
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Answer:
a) 2-bromopyrrole
Explanation:
Our options for this questions are:
a) 2-bromopyrrole
b) 2,3-dibromopyrrole
c) N-bromopyrrole
d) 3-bromopyrrole
To understand how the reaction works we have to start with the <u>resonance structures</u>. (Figure 1), on these structures, we will obtain a n<u>egative charge on carbon 2</u> in the pyrrole ring, therefore on this carbon we can generate an attack to an electrophile.
The second step is to check how the mechanism take place. An <u>electrophile is generated</u> by the
and
. This electrophile can be <u>attacked</u> by the negative charge on carbon 2 producing the 2-bromopyrrole. (See figure 2).
I hope it helps!