The University of OklahomaCHEM 1415

1)The absorbance of a 2.1x10^-3 M solution was determine to be 0.325 at 350 nm. If the path length was 10 mm, what is the molar extinction coefficient (in M^-1cm^-1).

Group of answer choices

6.8x10-3 M-1cm-1

15.5 M-1cm-1

1.5x103 M-1cm-1

154.8 M-1cm-1

2)Consider the following reaction which is initially at equilibrium in a reaction container:

M(s) + H2O(g) ? MO(g) + H2(g)

What would happen as a result of increasing the container volume?

Group of answer choices

No Change

MO (g) and H2 (g) concentrations would increase

MO (g) concentration would decrease

Equilibrium constant K would decrease

3)Lithium carbonate has a molar solubility of 2.23x10^-3 M. What is Ksp for Lithium carbonate?

Group of answer choices

5.54x10-9

4.97x10-6

4.44x10-8

2.77x10-9

4)The rate constant for a reaction was determined to be 0.32 M-1s-1. What is the half-life of this reaction if the initial concentration is 0.5 M?

Group of answer choices

0.78 s

2.17 s

3.12 s

6.25 s

5)What is the oxidation state for S in Al2(SO4)3?

Group of answer choices

+2

-2

-6

+6

Answer:

1. 

The formula for absorbance is A = eCl where e is molar extinction coefficient C is concentration and l is the path length  using this and putting the values given in the question we get e as 1.548 x 10^2 = 154.8

2. 
When there is an increase in volume, the equilibrium will shift to favor the direction that produces more moles of gas. hence MO and H2 concentrations will increase

3. the formula of salt is Li2CO3 so formula for Ksp will be 4x^3 where x is the molar solubility of the given salt. the formula is so because when the compound will dissociate it will give 2 moles of Li and one mole of carbonate. the  molar solubility of Li will be double the  molar solubility of the salt and that of carbonate will be equal to that of the salt. here it is given that x is 2.23x10^-3 M so Ksp will be 4 x ( 2.23x 10^-3)^3 = 4.44 x 10^-8 

4. based on the units of rate constant we can see that  reaction order is second, the formula of half life for second order reaction is  t1/2 = 1/( k[A]_o    ) where [A]o is the initial concentration using this we get half-life as 6.25s

5. since the overall compound is neutral, let us assume the oxidation state of S is x now we know oxidation state of O and Al are -2 and +3 respectively. 

now we know 2*3 +3* x + 3*4*(-2) = 0 we have multiplied each elements oxidation state with the number of times each element appears in the compound for example the number of atoms of Al in the compound is 2 that of S is three and of O is 12  from here we get value of x as +6