Login

Search the forums now!

Welcome, Guest. Please login or register.

November 08, 2025, 04:06:38 am
Pages: [1]   Go Down

Author Topic: help please  (Read 1250 times)  Share 

0 Members and 1 Guest are viewing this topic.

lacoste

  • Victorian
  • Forum Leader
  • ****
  • Posts: 973
  • Respect: +2
help please
« on: July 20, 2009, 05:59:17 pm »
0
Calculate the energy density of octane, C8H18, in MJ kg–1 if the heat of combustion of octane is –5470 kJ mol–1.?



how do i do that, could you also please explain your steps too


:)
Logged

TrueTears

  • TT
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 16363
  • Respect: +667
Re: help please
« Reply #1 on: July 20, 2009, 06:06:04 pm »
0
First work out for 1 mol of octane how much does it weight in grams, let this be say x g. Convert this to kilograms say y kg.

then we know that for y kg of octane the heat of combustion is 5470 kJ

y kg : 5470 kJ

kg :

there fore 1 kg will release amount of energy, make sure you put your answers in MJ
Logged
PhD @ MIT (Economics).

Interested in asset pricing, econometrics, and social choice theory.

lacoste

  • Victorian
  • Forum Leader
  • ****
  • Posts: 973
  • Respect: +2
Re: help please
« Reply #2 on: July 20, 2009, 06:08:50 pm »
0
The ans is  (5470/110) MULTIPLY 1000, the thing is, that i dont know where the 110 came from?

ANS is 47.98 MJ kg–1
Logged

TrueTears

  • TT
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 16363
  • Respect: +667
Re: help please
« Reply #3 on: July 20, 2009, 06:12:52 pm »
0
Yeap I got 47.98 MJ

Just use the steps I showed above.
Logged
PhD @ MIT (Economics).

Interested in asset pricing, econometrics, and social choice theory.

lacoste

  • Victorian
  • Forum Leader
  • ****
  • Posts: 973
  • Respect: +2
Re: help please
« Reply #4 on: July 20, 2009, 06:18:26 pm »
0
how many grams do i use when its not stated in the qst?

Logged

TrueTears

  • TT
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 16363
  • Respect: +667
Re: help please
« Reply #5 on: July 20, 2009, 06:18:43 pm »
0
Okay

= 114

in 1 mol = 114 g = 0.114 kg

therefore 0.114 kg releases 5470 000 J of energy

so 1 kg releases = 47982456 J of energy

therefore in the correct units it should be 47.98 MJ

yes, no maybe?
Logged
PhD @ MIT (Economics).

Interested in asset pricing, econometrics, and social choice theory.

lacoste

  • Victorian
  • Forum Leader
  • ****
  • Posts: 973
  • Respect: +2
Re: help please
« Reply #6 on: July 20, 2009, 06:22:25 pm »
0
why does the soln use 110? not 114
Logged

TrueTears

  • TT
  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 16363
  • Respect: +667
Re: help please
« Reply #7 on: July 20, 2009, 06:23:29 pm »
0
Quote from: lacoste on July 20, 2009, 06:22:25 pm
why does the soln use 110? not 114
Ring em up and ask em
Logged
PhD @ MIT (Economics).

Interested in asset pricing, econometrics, and social choice theory.

lacoste

  • Victorian
  • Forum Leader
  • ****
  • Posts: 973
  • Respect: +2
Re: help please
« Reply #8 on: July 20, 2009, 06:25:03 pm »
0
Quote from: TrueTears on July 20, 2009, 06:23:29 pm
Quote from: lacoste on July 20, 2009, 06:22:25 pm
why does the soln use 110? not 114
Ring em up and ask em


OKAY!

kkk typo, thanks TT!!!!!!!!!!!!!!!!
Logged

Over9000

  • Victorian
  • Part of the furniture
  • *****
  • Posts: 1468
  • Loves the banter
  • Respect: +20
Re: help please
« Reply #9 on: July 20, 2009, 06:25:14 pm »
0
Quote from: TrueTears on July 20, 2009, 06:23:29 pm
Quote from: lacoste on July 20, 2009, 06:22:25 pm
why does the soln use 110? not 114
Ring em up and ask em

Do they have ph. numbers on the book?
Logged
Gundam 00 is SOOOOOOOOHHHHHHHHH GOOOOOOOOOOODDDDDDDDDDDD I cleaned my room

VCE 200n(where n is an element of y): Banter 3/4, Swagger 3/4, Fresh 3/4, Fly 3/4

lacoste

  • Victorian
  • Forum Leader
  • ****
  • Posts: 973
  • Respect: +2
Re: help please
« Reply #10 on: July 20, 2009, 07:38:11 pm »
0
A calorimeter containing 100 mL of water is calibrated by passing a 3.00 A current through the instrument for 36.0 s at a potential difference of 3.50 V. The temperature rises by 0.82°C.
Potassium hydroxide weighing 0.654 g is added to the calorimeter and dissolved rapidly by stirring. The temperature rises from 20.82°C to 22.23°C.
   1. Calculate the calibration factor for the calorimeter and water.
   2 . Determine *triangle*H for the equation:
KOH(s)---->   KOH(aq)

No need to show working, just want to know the final answer to the 2.?


Many thanks.
Logged
Pages: [1]   Go Up