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hoochiscrazy

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aussiechick007 said:
^ thats not very accurate, lol
It is the accepted standardised method though,
  • measured using a turbidity tube (same method as above)
  • Measured in NTU (nephelometric turbidity units)

Edit: Azreil wat was the answer for the neutralisation reaction question. we seemed to be getting some conflicting answers lol
 
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:::a::: said:
Please help; explain how to get the answer for this:

Calculate the molar heat of combustion of ethanol

delta H = -m C delta t

m = 300 g (water)
C = 4.18 x10^3 (J/kg)
t = 20.5 degrees

mass of ethanol = 1.15 g

particularly convertering the delta H value from -25707kJ to kj/mol
H = -300 x 4.18 x 10^3 x 20.5
= -25707kJ per 1.15 g

then, molar heat = H x MW/1.15
= 25707 x 45.06/1.15
= 1007267.3 kj/mol

(i think this is right...)

Analyse the relationship between the position of elements in the periodic table and the acid-base behaviour of their oxides
 

jkwii

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u shouldnt be incorrect unless you're calculator fails
 

Undermyskin

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n OH- = 0.2 * 30 * 10^-3 = 6 * 10^-3 mol
n H+ = 0.1 * 40 * 10^-3 * 2 = 8 * 10^-3 mol

--> n H+ in excess = 2 * 10^-3 mol

--> [H+] = 2* 10^-3/70 * 10^-3 = 1/35

--> pH = 1.54

Yep, sorry that I always forget to sum the last volume.
----

Delta H = 300 * 4.18 * 20.5 = 25707 J = 25.707 kJ

n C2H5OH = 1.15/ 46.068 = 0.0249... mol = a mol

--> Molar heat = 25.707/ a = 1029.800066 kJ/mole = 1030 kJ/mol
 
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Azreil

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Amount OH- = 0.2 * 0.03 = 0.006
Amount H+ = 2 * 0.1 * 0.04 = 0.008
0.002 H+ in excess in 70mL
/70 * 1000 = 0.028571428571428571428571428571429
-log[0.028571428571428571428571428571429] = 1.544...
= 1.54 (2sf)

I made this question up and everyone's getting different things ><

EDIT: I'm almost certain 1.54 is correct.
 
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:::a:::

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aussiechick007 said:
H = -300 x 4.18 x 10^3 x 20.5
= -25707kJ per 1.15 g

then, molar heat = H x MW/1.15
= 25707 x 45.06/1.15
= 1007267.3 kj/mol

(i think this is right...)

Analyse the relationship between the position of elements in the periodic table and the acid-base behaviour of their oxides
worked it out, should be

H = -25.707 KILOJOULES (per 1.15g of ethanol)

molar heat is simply calculated by:

n = m/M
= 1.15 / 46.02

then molar heat = H / n
=~ 1029 kj / mol
 

hoochiscrazy

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Azreil said:
Amount OH- = 0.2 * 0.03 = 0.006
Amount H+ = 2 * 0.1 * 0.04 = 0.008
0.002 H+ in excess in 30mL
/30 * 1000 = 0.0666666
-log[0.0666666] = 1.176...
= 1.18 (2sf)

I made this question up and everyone's getting different things ><
Isnt it excess in 70ml?
 

Azreil

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hoochiscrazy said:
Isnt it excess in 70ml?
YOU SAW NOTHING.

And by that I mean I realised and fixed my mistake but not fast enough for you to miss it >>
 
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dont know what went wrong with mine.... oh i put 10^3 into the original H calculation... i don't think should've done that
 

axlenatore

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nandayo said:
Think you've calculated the pOH. Pretty sure you take that from 14 and get the final pH which is basic at 12.46, because H2SO4 is the limiting reagent and there are excess OH ions.
No because sulfuric acid is diprotic and therefore the concentration of h+ is double so the answer is - log (0.1 x 0.01)

Edit didnt work it out probably but still
 
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Azreil

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Three people have got 1.54

I'm nearly positive that's right.

MOVING ON.

Outline the industrial and natural sources of NOx and SO2
 

:::a:::

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How many Core First-hand investigations are there?

what are they? (just need to ensure i haven't entirely left something out)





ta!:santa:
 

Undermyskin

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NOx:
- Electrical devices
- Combustion of fuels
- Bacteria

SO2:
- Combustion of fuels
- Volcanoes
- Extraction of metals
 
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Azreil said:
Outline the industrial and natural sources of NOx and SO2
nitrogen oxides produced by:
- internal combustion engine reactions at high temperatures
- lightning causing the gases in the atmosphere to react and produce nitrogen oxide

sulfur oxides produced by:
- bacterial action
- volcanic eruptions
- burning fossil fuels
- smelting of sulfide ores in the production of copper
- bushfire smoke

Assess the impact of atomic absorption spectroscopy (AAS) on the scientific understanding of the effects of trace elements.
 

Undermyskin

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Across a period, except for the 1st one, oxides are less basic, more acidic with the middle one amphoteric.

Down a group, more basic, hence less acidic.
 

:::a:::

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thread has died.

RIP

one last thing, anion / cation tests - how the fuck do you remember them (all). honestly, is there a way of cheating this in the exam?

halp!!
 

danz90

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I'm off people...


GOOD LUCK TO CHEMISTRY CLASS OF '08!

I know you guys will kill it.... all the best for tomorrow!

Can't wait for discussion time after the exam ;)
 

j-3-s-5

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:::a::: said:
thread has died.

RIP

one last thing, anion / cation tests - how the fuck do you remember them (all). honestly, is there a way of cheating this in the exam?

halp!!
Wondering the same thing lol. Just trying to memorise them now but having no luck at all.
 

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