tiny q (1 Viewer)

[angelxtearz]

this is partly my fault for not entirely studying for the prelim but
for calculations of gas volumes isnt it just (conditions x no of moles)?

and i don't really understand the effect of pressure and all the other stimuli in correlation to the use or effectiveness of a catalyst.


thankyoz

 

[mitsui]

i dont get the Q u r on about. ><''

but isnt effect of pressure meaning diff conditions?? so it is still (condition x no. moles)

...no idea for the catalyst

 

[Dreamerish*~]

this is partly my fault for not entirely studying for the prelim but
for calculations of gas volumes isnt it just (conditions x no of moles)?

and i don't really understand the effect of pressure and all the other stimuli in correlation to the use or effectiveness of a catalyst.


thankyoz

At 0°C and the pressure of 1 atmosphere, one mole of any gas will occupy a volume of 22.41 L.

At 25°C and the pressure of 1 atmosphere, one mole of any gas will occupy a volume of 24.79 L.

So yes. You just shove the number of moles into the formula. Moles x 22.41 or 24.79 = volume.

The second question depends on what equation you're talking about. For fermentation, for example, excess temperature will kill yeast - the catalyst. For the Haber process, presence of CO and sulfur will poison the catalyst.

 

[pLuvia]

Yeh what are you trying to ask?

Here's an easy way to do the calculations

 

[Dreamerish*~]

those were 2 qz.

one on
-volume
-catalyst- how it is affected by different stimuli eg temp, pressure

ne1?

At 0°C and the pressure of 1 atmosphere, one mole of any gas will occupy a volume of 22.41 L.

At 25°C and the pressure of 1 atmosphere, one mole of any gas will occupy a volume of 24.79 L.

So yes. You just shove the number of moles into the formula. Moles x 22.41 or 24.79 = volume.

The second question depends on what equation you're talking about. For fermentation, for example, excess temperature will kill yeast - the catalyst. For the Haber process, presence of CO and sulfur will poison the catalyst.

Is this what you're talking about?

 

[mitsui]

another "huh" from me. LOL

 

[angelxtearz]

do i HAF ta put a title.....

with the moles, i was thinking that for a volume of a gas it wud b no of moles x conditions. (gay lussac principle) but the answers for da 2004 papr . q 9 disagreed
well this is wht i gut.
9.C
no. of moles for 03- <O
0.72g/48g= 0.02 mol<O</O
no. of moles for NO-<O</O
0.66g/30g=0.02<O</O
volume of NO2(g) produced = 0.02 x 22.71<O</O
=0.45 L
and...i wuz wrong...

 

[Dreamerish*~]

with the moles, i was thinking that for a volume of a gas it wud b no of moles x conditions. (gay lussac principle) but the answers for da 2004 papr . q 9 disagreed
well this is wht i gut.
9.Cfficeffice" /><o>></o>>
no. of moles for 03- <o>></o>>
0.72g/48g= 0.02 mol<o>></o>>
no. of moles for NO-<o>></o>>
0.66g/30g=0.02<o>></o>>
volume of NO2(g) produced = 0.02 x 22.71<o>></o>>
=0.45 L
and...i wuz wrong...

Wow...

Can you post up the question and the correct answer?

 

[angelxtearz]

so weird

thts so weird, y r were there all these random emoticons, i swear i didnt do it, i copied it from mi word doc...............................ScArY!

 

[angelxtearz]

questions

9​

​

Ozone reacts with nitric oxide according to the equation
NO(g) + O3(g) → NO2(g) + O2(g)
0.66 g NO(g) was mixed with 0.72 g O3(g).
What is the maximum volume of NO2(g) produced at 0°C and 100 kPa?
(A) 0.34 L
(B) 0.37 L
(C) 0.45 L
(D) 0.50 L

derz da qz​

 

[Riviet]

0.72g/48g= 0.02
0.66g/30g=0.02

DO NOT ROUND YOUR ANSWERS IN CHEMISTRY UNLESS ASKED TO DO SO IN THE QUESTION!!!

By calculator, 0.72/48=0.015, which is 7.5% less than your rounded answer! This is most likely why you are ending up with the wrong answer.

Also 0.66/30=0.022, use this exact decimal in your calculation.

Try it again, you should get it right now.

 

[Dreamerish*~]

9​

Ozone reacts with nitric oxide according to the equation
NO(g) + O3(g) → NO2(g) + O2(g)
0.66 g NO(g) was mixed with 0.72 g O3(g).
What is the maximum volume of NO2(g) produced at 0°C and 100 kPa?
(A) 0.34 L
(B) 0.37 L
(C) 0.45 L
(D) 0.50 L

derz da qz​

NO_2(g) + O_3(g) → NO_2(g) + O_2(g)

0.66 g of NO is reacted with 0.72 g of O₃.

n = m/M

The molar mass of NO is 14 + 16 = 30 g, therefore we have 0.66/30 = 0.022 moles. (Note that I didn't round off)

The molar mass of O₃ is 16 x 3 = 48 g, therefore we have 0.72/48 = 0.015 moles. (Note that I didn't round off here, either)

The molar ratio of NO to O₃ is 1:1. We have 0.022 moles of NO but only 0.015 moles of O₃, therefore NO is in excess - only 0.015 moles of it reacts.

The molar ratio of NO to NO₂ is 1:1. Therefore 0.015 moles of NO₂ is produced.

At the temperature of 0°C and 100 kPa, one mole of any gas occupies 22.41 L.

0.015 x 22.41 = 0.33615

Now, round off to two decimal places: 0.34.

. : The answer is (A).

 

[angelxtearz]

ha..tht is so weird. oh well at least i was using the right method well thankez for showing me..lol