Topscore Questions...Help!

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

jetjesse

Member
7+ Year Member
15+ Year Member
20+ Year Member
Joined
Aug 22, 2003
Messages
28
Reaction score
0
Points
0
Age
56
Location
Nashville, TN
I have a couple of questions I got wrong on the chem. section of topscore....
Why are these right?

1)Which of the following atoms could have an expanded valence shell and form compounds w/5 or more valence electrons?
S
Si
N
Br
The answer: S, N, and Br

2)What is the value of the rate constant for a reaction that has a 1/2 life of 1.5 hours?
The answer: A. 4.6x10-1h-1

3)Which of the following statements correctly identifies the difference between 1st order and 2nd order reactions when considering 1/2 lives?
The answer: t 1/2 is independent of concentration for all 1st order reactions.

4) Which of the following expressions correctly describes the relationship between solubility, s, and Ksp for MgF2? Ksp=?
The answer: 4s^3

Why are these right? Help!
 
Originally posted by jetjesse
I have a couple of questions I got wrong on the chem. section of topscore....
Why are these right?

2)What is the value of the rate constant for a reaction that has a 1/2 life of 1.5 hours?
The answer: A. 4.6x10-1h-1


4) Which of the following expressions correctly describes the relationship between solubility, s, and Ksp for MgF2? Ksp=?
The answer: 4s^3

Why are these right? Help!


for #2, use the formula: t1/2=ln2/k or t1/2=.693/k, where k is the first order rate constant.

for #4, write out the dissociation reaction, and you'll get:
MgF2(s)--> Mg2+(aq) +2F-(aq), and the expression for a saturated solution at equilibrium is Ksp=[Mg][F2]^2. let s=the concentration of Mg, then 2s= the concentration of F2 in the saturated solution at equilibrium (since [F2] is two times [Mg]. therefore, Ksp=(s)(2s)^2, and that equals 4s^3.
- just remember that every slightly soluble salt of general formula MX2 will have Ksp=4s^3, where s is the molar solubility.

hope that helps.
 
for #3 the equation for the half life of a first order rxn is t1/2=ln2/K <-- as you can see this equation is independent of [A]. A second order rxn has a equation corresponding to: t1/2=1/k[A] which as you can see depends on the conc of the reactant.
 
Id like to know number 1 as well. I got that wrong today. I always thought that the answer was SI,S, and Br. I don't ever remember N being able to expand beyond the octet.


croco, that makes sense, but doesnt that go against the definition of what a 1st-order reaction is? Isnt it suppose to depend of conc. of 1 reactant? I don't understand how talking about 1/2 life is different.
 
Thank you very much. t1/2=.693k is obviously very useful.

What other formulas would you remember concerning exponential decay?
 
For # 1 Si cannot have more than 5 valence electrons on its valence shell because it will give it an unfavorable formal charge. Imagine Si had 6 valence electrons giving it a formal charge of 4-6-0=-2 which is unfavorable. Nitrogen is able have more than 5 because imagine it has 6 valence electrons in its valence shell. That would give it a formal charge of 5-6-0=-1. This could be favorable in a molecule because nitrogen is fairly electronegative. Never forget that electrons in the D-orbital is always unfavorable because that states that these electrons are in a higher energy state and hence are very reactive and unstable. The reason you typically see many bonds when it comes to Si is because it is typically seen with oxygen which as also needs its formal charge nuetralized.
 
I forgot to reply to the second part. No a first order reaction "half life" is not dependent on any concentration. If you mean it depends on the equation ln[A]=ln[A]i-kt, then you are wrong. This equation is not a half life equation. This equation shows how the the initial and final concentrations of a reactant change with time.
 
Top Bottom