Answer:
density equals weigt divided by volume
Explanation:

2.43*1000.
There are 1000 grams in a kilogram
Answer: Tin (Sn)
Explanation: The electron configuration for tin (Sn) is shown in the picture. It's last electrons are:
5s^2 4d^10 5p^2
The valence electrons are in the 5th electron shell and include 2 each in the 5s and 5p orbitals.
<h3><u>Full Question:</u></h3>
The following compound has been found effective in treating pain and inflammation (J. Med. Chem. 2007, 4222). Which sequence correctly ranks each carbonyl group in order of increasing reactivity toward nucleophilic addition?
A) 1 < 2 < 3
B) 2 < 3 < 1
C) 3 < 1 < 2
D) 1 < 3 < 2
<h3><u>Answer: </u></h3>
The rate of nucleophilic attack of carbonyl compounds is 2<3 <1.
Option B
<h3><u>Explanation. </u></h3>
Nucleophilic attack is explained as the attack of an electron rich radical to a carbonyl compound like aldehyde or a ketone. A nucleophile has a high electron density, so it searches for a electropositive atom where it can donate a portion of its electron density and become stable.
A carbonyl compound is a
hybridized carbon atom with a double bonded oxygen atom in it. The oxygen atom pulls a huge portion of electron density from carbon being very electropositive.
In a ketone, there are two factors that make it less likely to undergo a nucleophilic attack than aldehyde. Firstly, the steric hindrance of two carbon groups being attached with the carbonyl carbon makes it harder for the nucleophile to approach. Secondly, the electron push by the carbon groups attached makes the carbonyl carbon a bit less electropositive than the aldehyde one. So aldehydes are more reactive towards a nucleophilic addition reaction.
Answer:
0
Explanation:
Given parameters:
Half-life = 8.08days
Unknown:
What fraction is left unchanged after 16.16days = ?
Solution:
The half - life of a substance is the time taken for the half of a radioactive material to decay to half.
Day 0 Day 8.08 Day 16.16
100% 50% 0% Parent
0% 50% 100% Daughter
After 16.16 days, non of the original sample will remain unchanged.