Please Help-Chem Questions.

[Jimbo1]

G'Day fellow FSN'ers. I am really poor at chem  :buck2:  and would like your kind assistance in these questions if possible:

1) A bottle contains NaOCl labelled at 153ppm. 750mL of this bottle will contain:
A-9.3 x 10*20 Na+ ions
B-1.2 x 10*21 molecules of NaOCl
C-1.5 mol of OCl- ions
D-204 mg of NaOCl

2)Which of the following contains the greatest number of sulfur atoms:
A- 68.9g of CuSO4.5H20
B-29.4L of SO2 at SLC
C-2.0L of 0.40M H2SO4 solution
D-3.0 x 10*24 molecules of H2S

3) Glucose undergoes a combustion reaction to produce Co2 and H20.
a) Write a balanced equation to represent this reaction
b)What volume of C02 at 37*C and 101.3kPa would be produced when 10.0g of the glucose reacts.
c) What is the name given to this reaction

4) 125 mL of 0.250M copeer(II) nitrate solution is added to 90.0mL of 0.560M sodium hydroxide solution resulting in the formation of a precipitate of Calcium Hydroxide.
a) Write a balanced chemical equation for this reaction
b) Calculate amount in mol of both copper(II) nitrate and sodium hydroxide present at the start of the reaction and hence determine which reactant is in excess.
c) Calculate the mass of precipitate formed.

Please help.
Thank You 

[Toothpaste]

3) Glucose undergoes a combustion reaction to produce Co2 and H20.
a) Write a balanced equation to represent this reaction
b)What volume of C02 at 37*C and 101.3kPa would be produced when 10.0g of the glucose reacts.
c) What is the name given to this reaction

Shot-gun the biology-like one.

a) C6H12O6 (aq) + 6O2 (g) --> 6CO2 (g) + 6H2O (l)

b) n(C6H12O6) = 10g / 180g
                    = .0555556 mol
=> n(CO2) = 6 * n(C6H12O6)
             = 0.333 mol

37 degrees C =  310 K

PV = nRT

V(CO2) = (0.333mol * 8.31 * 310K)/ 101.3kPa
= 8.48 L

c) Cellular (aerobic) respiration

[Collin Li]

LOL, "respiration" will do for VCE Chemistry. Nice work.

[midas_touch]

4) 125 mL of 0.250M copeer(II) nitrate solution is added to 90.0mL of 0.560M sodium hydroxide solution resulting in the formation of a precipitate of Calcium Hydroxide.
a) Write a balanced chemical equation for this reaction
b) Calculate amount in mol of both copper(II) nitrate and sodium hydroxide present at the start of the reaction and hence determine which reactant is in excess.
c) Calculate the mass of precipitate formed.

a) Cu(NO3)2  + 2NaOH  ->  2NaNO3  +  Cu(OH)2
b) Use n = cv which is a formula you must be familiar with come the unit 3 exam.

       Cu(NO3)2:
         n = 0.125*0.25
            = 0.03125 mol
      NaOH:
         n = 0.09*0.56
           = 0.0504 mol

However, you need to consider mole ratios of the reactants:
n(NaOH) reacting/n(Cu(NO3)2) reacting = 2

But, n(NaOH) present/n(Cu(NO3)2) present = 0.0504/0.03125 = 1.6128

Hence NaOH will run out first, making it the limiting reactant, and the Cu(NO3)2 in excess

c) Now we know which reactant is the limiting reactant, we can calculate the amount of the precipitate formed (Cu(OH)2):

n(NaOH): 0.0504 mol

Once again consider mole ratios in the equation
n(NaOH) reacted/n(Ca(OH)2) formed = 2

Hence n(Ca(OH)2) formed = 0.0504/2 = 0.0252 mol

Use m = nM (Another forumla you must learn to use) to figure out the mass of Ca(OH)2 formed

m = 0.0252*74.1
= 1.87g (correct to 3 sig figs)

[cara.mel]

What's wrong with the first two?

Disclaimer: I am doing these on 5 hours of sleep so sorry if I skip lines in working or whatever

1. 153ppm = 153mg/L = 0.153g/L => in 3/4 of a litre there is .153*.75=0.115g of NaOCl
n(NaOCl) = m/Mr = 0.115/74.5 = 1.54*10^-3mol
Then unfortunately it's just a matter of going through each option to see if they work
a) n(Na+) = n(NaOCl) = 1.54*10^-3mol
number of ions = n(Na+)*6.023*10^23 (avogadro's constant) = 9.3*10^20 ions => a is right

Confirming the other 3 are wrong:
b) number of NaOCl molecules = number of Na+ ions => wrong
c) n(OCl-) = n(NaOCl) => wrong
d) worked out there was 115mg of it earlier when working out the mole.

First step you do in any question you are unsure of is try and find the mole of something, it will usually take you where you want to go



2) Although the question says number of atoms, by looking at the question it would take less steps overall to get them all into moles than # of atoms (and seeing as they are directly proportional, we can do that). So we want to find the one with the higest amount (in mol) of S
a) n(CuSO4.5H2O) = m/Mr = 68.9/249.5 = 0.276mol
n(S) = n(CuSO4) = 0.276mol
b) At SLC, 1 mol = 24.5L
=> n(SO2) = 29.4/24.5 = 1.2mol
n(S) = n(SO2) = 1.2mol
c) n(H2SO4) = cV = 2*.4=.8mol
n(S) = n(H2SO4) = 0.8mol
d) n(H2S) = 3*10^24/6.023*10^23 = 5mol
n(S) = n(H2S) = 5mol

Therefore d is the biggest => answer 

[J.Z-M]

For some reason I got :
1.B
(Probably wrong though)

[J.Z-M]

Oh wait up, i think silly me did a mistake

[midas_touch]

What's wrong with the first two?

Disclaimer: I am doing these on 5 hours of sleep so sorry if I skip lines in working or whatever

1. 153ppm = 153mg/L = 0.153g/L => in 3/4 of a litre there is .153*.75=0.115g of NaOCl
n(NaOCl) = m/Mr = 0.115/74.5 = 1.54*10^-3mol
Then unfortunately it's just a matter of going through each option to see if they work
a) n(Na+) = n(NaOCl) = 1.54*10^-3mol
number of ions = n(Na+)*6.023*10^23 (avogadro's constant) = 9.3*10^20 ions => a is right

Confirming the other 3 are wrong:
b) number of NaOCl molecules = number of Na+ ions => wrong
c) n(OCl-) = n(NaOCl) => wrong
d) worked out there was 115mg of it earlier when working out the mole.

First step you do in any question you are unsure of is try and find the mole of something, it will usually take you where you want to go



2) Although the question says number of atoms, by looking at the question it would take less steps overall to get them all into moles than # of atoms (and seeing as they are directly proportional, we can do that). So we want to find the one with the higest amount (in mol) of S
a) n(CuSO4.5H2O) = m/Mr = 68.9/249.5 = 0.276mol
n(S) = n(CuSO4) = 0.276mol
b) At SLC, 1 mol = 24.5L
=> n(SO2) = 29.4/24.5 = 1.2mol
n(S) = n(SO2) = 1.2mol
c) n(H2SO4) = cV = 2*.4=.8mol
n(S) = n(H2SO4) = 0.8mol
d) n(H2S) = 3*10^24/6.023*10^23 = 5mol
n(S) = n(H2S) = 5mol

Therefore d is the biggest => answer 

I forgot how to do ppm , I was trying to do it on a mole basis.

[Ninox]

quoting your sig J.Z-M, to get an ENTER of 95, as of 2007, you would need an average study score of 41.52 scaled.
Good luck! 

and impeccable work guys!

[bec]

I've just got a quick question too:

In each of the following equations, identify the acids and bases and name the conjugate acid-base pairs
a) HSO₄^- + H₂O --> H₃O⁺ + SO₄^2-
b) NH₄⁺ + S^2- --> NH₃+HS^-
c) CH₃COO^- + H₃O⁺ --> H₂O +CH₃COOH

I've done it but despite correctly identifying the acids/bases, some of my conjugate acid-base pairs were the wrong way around. I said:
a) HSO₄^-/SO₄^2- (the book agrees with me here) and H₃O⁺ /H₂O (book has it other way around)
b) The pair I got wrong was HS^- / S^2-
c) I said CH₃COOH[/b] / CH₃COO^-

I've always been taught to list them acid/base (ie, not base/acid) so did I get these wrong because I'm identifying acids and bases wrong, or does order just not matter?

Edit: just realised the book lists them in the order they appear in the original equation...which is correct, that or ensuring they're in the order acid/base?

[Collin Li]

I don't think you need to be worried about it too much. You should probably write it in the order it appears in a reaction when possible.

[bec]

thanks coblin
can anyone explain this to me: "The decomposition of a compound into its elements must be a redox reaction."
I think i get it but i need confirmation....is it because in a compound, for the oxidation numbers to total to zero there must be at least one element that has a positive oxidation number and >1 element with a negative oxidation number - then, when the compound decomposes into its elements, each product will have the oxidation number of zero.... meaning that both oxidation and reduction have occurred?
like if CO₂ --> C + O₂ (if that can even happen, i wouldn't know)
even writing this has sort of solidified it in my mind actually (despite the rambling way i've written it...)
but a yay or nay would be good anyway so i know i'm on the right track!
thanks!

[Collin Li]

Yep, you're pretty much right. In any compound, the oxidation states of two or more atoms are non-zero. The oxidation states of elements are zero, so the decomposition of a compound results in a redox reaction.

[bec]

another question:
in a compound containing elements more electronegative than oxgen, the more electronegative element "gets the negative oxidation number"  (i quote from the notes my teacher gave us)
can someone rephrase that?
I'm assuming that means that in F₂O, F has oxidation no. -1, O is +2? and in ClO, Ox. no of Cl = -1, O=+1?

is that right?

[Collin Li]

Yeah. The same applies for electropositive atoms (the opposite of electronegative) like hydrogen. This rule takes priority over the rule about electronegativity, so in , H is +1 first, then therefore O is -1.