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jessecatty jessecatty
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10 years ago
I need help with the following.  I have the answers just not sure
1) Which of the following substances crystallizes as a molecular solid? a. Kl b. SiO2 c. Sn d. CH3OH e. AI2(SO4)3
The answer I chose is CH3OH, but another class mate told me SiO2, is he right?  I think I am the others I know are not correct.
2) For the reaction
H2(g) + I2(g)==2Hl(g), Kc=50.2 at 445oC if [H2]=[I2]=[Hl]=1.75 x 10-3 M at 445o, which of these statements are true?
a. The system is at equilibrium, thus no concentration changes will occur
b. The concentrations of H2 and Hl will fall as the system moves toward equilibrium
c. The concentrations of H2 and I2 will increase as the system approaches  equilibrium
d. The concentrations of Hl and I2 will increase as the system approaches equilibrium
e. The concentration of Hl will increase as the system approaches equilibrium.
I chose d for the answer but again someone else comes up with a different answer, please explain.
3) For the following equilibrium:
H2O(g)+CO(g)-->H2(g)+CO2(g)  H=-9.8kcal
Which of the following will increase the amount of CO2(g)?
a. decrease the pressure
b. decrease the volume of the container
c. decrease the temperature
d. decrease the concentration of CO2(g)
e. Increase the concentration of H2(g)
I put increase the concentration of H2 but someone told me its decrease the temp?  I am confused
4) Find the pH of a 0.135 M aqueous solution of nitrous acid (HNO2), for which Ka=4.5x10-4
I ended up with 2.1 for this answer but it took me forever.  Is it correct?  Help please.
5)What is the pH of a 0.50 M solution of NaOH3? The Ka for HNO2 is 4.5x10-4
For this one I ended up with 8.5 for the pH.  A classmate told me it was 7pH.  I need  help ASAP please

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AlexxAlexx
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10 years ago
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10 years ago
4)
We have HNO2 <=> H+ + NO2
and Ka=4.5x10-4.
So, we have: Ka=[H+]*[NO2-]/[HNO2]
The acid is pretty weak, so we can assume that [NO2]<<[HNO2], and [HNO2]=0.135 M.
In addition, [H+]=[NO2-]=x.

We have: Ka=[H+]*[NO2-]/[HNO2] <=> 4.5*10-4=x2/0.135 <=> x2=0,6075*10-4 and x=0,78*10-2 .....
pH=-log[H+]=-logx=-log(0,78*10-2)=2,1 (with a slight help of a calculator).. So yes, it's correct.

5) It's the first time I see NaOH3. Are you sure you don't mean NaOH? If you are sure, then this molecule might not be a base/acid, and so the pH might be 7 (as your classmate told you). I'm not sure though. How did you get the result anyways?
jessecatty Author
wrote...
10 years ago
I checked my work again and I ended up with 8.5pH when it was NaOH2 not NaOH3.  That answer was on one of my quizzes and it was correct.  How much of a difference would the 2 and the 3 make?  I am still learning Chem so it is not my strong subject.
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10 years ago
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How much of a difference would the 2 and the 3 make?
One different number can can make a huge difference. However, I still don't know about this NaOH2. Maybe you meant Na(OH)2? Are there any more data that the exercise provides (of that you know) like the Kb of NaOH2, or the name of this compound?
Can you briefly explain how you solved this?
(I ask out of curiosity, since you already told that the correct answer was 8,5.)
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