Hello!
To solve this problem we're going to use the
Charles' Law. This Law describes the relationship between
Volume and Temperature in an ideal gas. Applying this law we have the following equation:
So, the final temperature is
54,23 °CHave a nice day!
Answer:
Here's what I get
Explanation:
(a) Intermediates
The three structures below represent one contributor to the resonance-stabilized intermediate, in which the lone pair electrons on the heteroatom are participating (the + charge on the heteroatoms do not show up very well).
(b) Relative Stabilities
The relative stabilities decrease in the order shown.
N is more basic than O, so NH₂ is the best electron donating group (EDG) and will best stabilize the positive charge in the ring. However, the lone pair electrons on the N in acetanilide are also involved in resonance with the carbonyl group, so they are not as available for stabilization of the ring.
(c) Relative reactivities
The relative reactivities would be
C₆H₅-NH₂ > C₆H₅-OCH₃ > C₆H₅-NHCOCH₃
The many electron diamonds around the central carbon are 1689
The empirical formula of the following compounds 0.903 g of phosphorus combined with 6.99 g of bromine.
<h3>What is empirical formula?</h3>
The simplest whole number ratio of atoms in a compound is the empirical formula of a chemical compound in chemistry. Sulfur monoxide's empirical formula, SO, and disulfur dioxide's empirical formula, S2O2, are two straightforward examples of this idea. As a result, both the sulfur and oxygen compounds sulfur monoxide and disulfur dioxide have the same empirical formula.
<h3>
How to find the empirical formula?</h3>
Convert the given masses of phosphorus and bromine into moles by multiplying the reciprocal of their molar masses. The molar masses of phosphorus and bromine are 30.97 and 79.90 g/mol, respectively.
Moles phosphorus = 0.903 g phosphorus 
= 0.0293 mol
Moles bromine 6.99 g bromine
=0.0875 mol
The preliminary formula for compound is P0.0293Bro.0875. Divide all the subscripts by the subscript with the smallest value which is 0.0293. The empirical formula is P1.00Br2.99 ≈ P₁Br3 or PBr3
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