AS Chemistry- helping each other out!
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Original post by locket0511
I have been researching to try and find the correct way to convert a number from mol/dm^3 to g/dm^3 but I am really struggling. Which is the correct way to do it accurately?
Could you provide an exam so I can see how you do it.
thanks
Could you provide an exam so I can see how you do it.
thanks
multiply the moldm-3 by the mr(relative molar mass) of whatever it is to get it to gdm-3
When cracking takes place, a large number of products are formed.Suggest why a large number of products are formed.
can't do chem yay
(edited 8 years ago)
Original post by ThatGuyRik
multiply the moldm-3 by the mr(relative molar mass) of whatever it is to get it to gdm-3
so is this correct?
A sample of vinegar contains 0.1 mol/dm3 ethanoic acid. What is its concentration in g/dm3? ( Mr, of ethanoic acid is 60)
concentration in g/dm3 = concentration in g/dm3 × Mr
concentration = 0.1 × 60 = 6 g/dm3
Original post by Kadak
When cracking takes place, a large number of products are formed.Suggest why a large number of products are formed.
Large carbon chain breaks into many smaller carbon chains?
Assuming from what i remember last year.
(edited 8 years ago)
Original post by locket0511
I have been researching to try and find the correct way to convert a number from mol/dm^3 to g/dm^3 but I am really struggling. Which is the correct way to do it accurately?
Could you provide an exam so I can see how you do it.
thanks
Could you provide an exam so I can see how you do it.
thanks
From the amount I know (at As level), you simply multiply the concentration (mol dm-3) by the Mr of the substance (in g mol-1) to get the concentration in g dm-3. There may well be more advanced/accurate ways of doing it.
Original post by _NMcC_
From the amount I know (at As level), you simply multiply the concentration (mol dm-3) by the Mr of the substance (in g mol-1) to get the concentration in g dm-3. There may well be more advanced/accurate ways of doing it.
Thanks a lot!!!
so is this correct?
A sample of vinegar contains 0.1 mol/dm3 ethanoic acid. What is its concentration in g/dm3? ( Mr, of ethanoic acid is 60)
concentration in g/dm3 = concentration in g/dm3 × Mr
concentration = 0.1 × 60 = 6 g/dm3
Original post by locket0511
Thanks a lot!!!
so is this correct?
A sample of vinegar contains 0.1 mol/dm3 ethanoic acid. What is its concentration in g/dm3? ( Mr, of ethanoic acid is 60)
concentration in g/dm3 = concentration in g/dm3 × Mr
concentration = 0.1 × 60 = 6 g/dm3
so is this correct?
A sample of vinegar contains 0.1 mol/dm3 ethanoic acid. What is its concentration in g/dm3? ( Mr, of ethanoic acid is 60)
concentration in g/dm3 = concentration in g/dm3 × Mr
concentration = 0.1 × 60 = 6 g/dm3
Yes that looks fine. I'm getting the same answer. It's acceptable to have the units either as g/dm3 or g dm-3
(edited 8 years ago)
Original post by _NMcC_
Yes that looks fine. I'm getting the same answer.
Thank you very much for your help!!!
Original post by locket0511
Thank you very much for your help!!!
No worries.
hils?
(edited 8 years ago)
Hi
I do not understand ionic half equations, was wondering if you guys could help me understand them.
(i)So instance it asks to 'Write the ionic equation to show the oxidation of Calcium, Ca, to Calcium ions, Ca^2+'.
Ca ---> Ca2+ + 2e^-
(correct or not?)
(ii) And 'Write the ionion half-equation to show the reduction of water to hydrogen, H2, and hydroxide ions, OH^-'.
(could some one walk me through this, baby steps
)
(iii) Combine the two ionic half-equations.
(I think I know how to do this lol, I THINK)
Also other questions:
Complete and balance the following equations:
(ii) Na2O + H2O --> ?
(iii) Na2O + HCl ---? ?
(idk because I've only learnt about group 2 and 7.. not group 1 :s )
Thanks
I do not understand ionic half equations, was wondering if you guys could help me understand them.
(i)So instance it asks to 'Write the ionic equation to show the oxidation of Calcium, Ca, to Calcium ions, Ca^2+'.
Ca ---> Ca2+ + 2e^-
(correct or not?)
(ii) And 'Write the ionion half-equation to show the reduction of water to hydrogen, H2, and hydroxide ions, OH^-'.
(could some one walk me through this, baby steps
) (iii) Combine the two ionic half-equations.
(I think I know how to do this lol, I THINK)
Also other questions:
Complete and balance the following equations:
(ii) Na2O + H2O --> ?
(iii) Na2O + HCl ---? ?
(idk because I've only learnt about group 2 and 7.. not group 1 :s )
Thanks
(edited 8 years ago)
More questions 
Sodium:
1st ionisation energy: 496
2nd ionisation energy: 4563
Magnesium:
1st .. : 738
2nd.. : ?
Predict a value for the second Ionisation energy of Magnesium. Explain your choice.
So I was thinking:
Sodium's 1st is lower than Magnesium due to the atomic size and nuclear attraction. (But wouldn't it's first be high because it does lose a whole electron shell?) It's 2nd is significantly higher in IE (Is it because of the loss of an electron shell?)
Magnesium's 1st IE is greater due to it being higher in electron density (having a larger charge and smaller radius), which causes it to have a strong nuclear attraction. It's 2nd IE would result in it losing an electron shell. So wouldn't the jump/difference between 1st and 2nd be significantly great? (or is the jump at the 3rd IE?)
I'm not sure if the 2nd IE of Magnesium is suppose to be greater or less than the 2nd IE of Sodium.
Thanks
Sodium:
1st ionisation energy: 496
2nd ionisation energy: 4563
Magnesium:
1st .. : 738
2nd.. : ?
Predict a value for the second Ionisation energy of Magnesium. Explain your choice.
So I was thinking:
Sodium's 1st is lower than Magnesium due to the atomic size and nuclear attraction. (But wouldn't it's first be high because it does lose a whole electron shell?) It's 2nd is significantly higher in IE (Is it because of the loss of an electron shell?)
Magnesium's 1st IE is greater due to it being higher in electron density (having a larger charge and smaller radius), which causes it to have a strong nuclear attraction. It's 2nd IE would result in it losing an electron shell. So wouldn't the jump/difference between 1st and 2nd be significantly great? (or is the jump at the 3rd IE?)
I'm not sure if the 2nd IE of Magnesium is suppose to be greater or less than the 2nd IE of Sodium.
Thanks
Lololol more questions (these are questions I missed out from numerous past paper booklets, I swear )
Cars are fitted with catalytic converters in order to reduce the pollution caused by the combustion of petrol. Potential pollutant gases include CO, NO and unburnt hydrocarbons. The first two compounds are removed by passing the hot gases over a platinum catalyst.
CO(g) + NO(g) --> CO2(g) + 1/2N2(g) Delta H = -373kJmol-1
(iii) Comment on the relative value of the Activation energy of this reaction compared with the much fast reaction of NO with O2.
Lol, after typing that out I THINK I get it? Or.. maybe not.
Would you just say that:
The activation energy would be lower in the reaction of NO + O2 because less energy is required for the reaction to take place?
Thanks again
(these are questions I missed out from numerous past paper booklets, I swear )Cars are fitted with catalytic converters in order to reduce the pollution caused by the combustion of petrol. Potential pollutant gases include CO, NO and unburnt hydrocarbons. The first two compounds are removed by passing the hot gases over a platinum catalyst.
CO(g) + NO(g) --> CO2(g) + 1/2N2(g) Delta H = -373kJmol-1
(iii) Comment on the relative value of the Activation energy of this reaction compared with the much fast reaction of NO with O2.
Lol, after typing that out I THINK I get it? Or.. maybe not.
Would you just say that:
The activation energy would be lower in the reaction of NO + O2 because less energy is required for the reaction to take place?
Thanks again
(edited 8 years ago)
Original post by Dinaa
More questions
Sodium:
1st ionisation energy: 496
2nd ionisation energy: 4563
Magnesium:
1st .. : 738
2nd.. : ?
Predict a value for the second Ionisation energy of Magnesium. Explain your choice.
So I was thinking:
Sodium's 1st is lower than Magnesium due to the atomic size and nuclear attraction. (But wouldn't it's first be high because it does lose a whole electron shell?) It's 2nd is significantly higher in IE (Is it because of the loss of an electron shell?)
Magnesium's 1st IE is greater due to it being higher in electron density (having a larger charge and smaller radius), which causes it to have a strong nuclear attraction. It's 2nd IE would result in it losing an electron shell. So wouldn't the jump/difference between 1st and 2nd be significantly great? (or is the jump at the 3rd IE?)
I'm not sure if the 2nd IE of Magnesium is suppose to be greater or less than the 2nd IE of Sodium.
Thanks
Sodium:
1st ionisation energy: 496
2nd ionisation energy: 4563
Magnesium:
1st .. : 738
2nd.. : ?
Predict a value for the second Ionisation energy of Magnesium. Explain your choice.
So I was thinking:
Sodium's 1st is lower than Magnesium due to the atomic size and nuclear attraction. (But wouldn't it's first be high because it does lose a whole electron shell?) It's 2nd is significantly higher in IE (Is it because of the loss of an electron shell?)
Magnesium's 1st IE is greater due to it being higher in electron density (having a larger charge and smaller radius), which causes it to have a strong nuclear attraction. It's 2nd IE would result in it losing an electron shell. So wouldn't the jump/difference between 1st and 2nd be significantly great? (or is the jump at the 3rd IE?)
I'm not sure if the 2nd IE of Magnesium is suppose to be greater or less than the 2nd IE of Sodium.
Thanks
It's probably best not to think of it in terms of electron 'shells'. When you ionise a substance, the electron removed will be the outer most electron, which, on average, will be the electron with the most energy, in the case of sodium the one in the 3s orbital. After removing that electron, you are then having to remove an electron from the 2p orbitals, which on average will be closer so greater attraction and there is also less shielding between nucleus and the electron so there is a big jump in ie. With mg there are 2 electrons in the 3s so you can remove 2 before the large jump so hence 2nd ie of mg will be lower
Posted from TSR Mobile
Need some help with why dilute nitric acid is used when identifying metal halides.
Is it to remove any carbonate/hydroxide ions so they do not interfere.
I read something on chemrevise that I don't fully understand.
It said that : "The role of nitric acid is to react with any carbonates present to prevent formation of the precipitate Ag2CO3. This would mask the desired observations".
Then it has an equation: 2HNO3 + Na2CO3 ------> 2NaHNO3 + H2O + CO2.
Anyone care to explain the equation and the issue with silver carbonate, by that I mean how does that mask the observations?
Thanks
Is it to remove any carbonate/hydroxide ions so they do not interfere.
I read something on chemrevise that I don't fully understand.
It said that : "The role of nitric acid is to react with any carbonates present to prevent formation of the precipitate Ag2CO3. This would mask the desired observations".
Then it has an equation: 2HNO3 + Na2CO3 ------> 2NaHNO3 + H2O + CO2.
Anyone care to explain the equation and the issue with silver carbonate, by that I mean how does that mask the observations?
Thanks
Original post by Super199
Need some help with why dilute nitric acid is used when identifying metal halides.
Is it to remove any carbonate/hydroxide ions so they do not interfere.
I read something on chemrevise that I don't fully understand.
It said that : "The role of nitric acid is to react with any carbonates present to prevent formation of the precipitate Ag2CO3. This would mask the desired observations".
Then it has an equation: 2HNO3 + Na2CO3 ------> 2NaHNO3 + H2O + CO2.
Anyone care to explain the equation and the issue with silver carbonate, by that I mean how does that mask the observations?
Thanks
Is it to remove any carbonate/hydroxide ions so they do not interfere.
I read something on chemrevise that I don't fully understand.
It said that : "The role of nitric acid is to react with any carbonates present to prevent formation of the precipitate Ag2CO3. This would mask the desired observations".
Then it has an equation: 2HNO3 + Na2CO3 ------> 2NaHNO3 + H2O + CO2.
Anyone care to explain the equation and the issue with silver carbonate, by that I mean how does that mask the observations?
Thanks
Silver carbonate would give you a yellowy - grey precipitate as it is poorly soluble in water. This is a problem because it then becomes pretty much impossible to tell if your precipitate is because of agco3 or confirms the presence of a halide ion because they would look basically identical.
Posted from TSR Mobile
Guys which molecules have permanent dipole-permanent dipole bonds? Is it only the polar ones?
Original post by Dinaa
Hi
I do not understand ionic half equations, was wondering if you guys could help me understand them.
(i)So instance it asks to 'Write the ionic equation to show the oxidation of Calcium, Ca, to Calcium ions, Ca^2+'.
Ca ---> Ca2+ + 2e^-
(correct or not?)
I do not understand ionic half equations, was wondering if you guys could help me understand them.
(i)So instance it asks to 'Write the ionic equation to show the oxidation of Calcium, Ca, to Calcium ions, Ca^2+'.
Ca ---> Ca2+ + 2e^-
(correct or not?)
Yep, charges are balanced
(ii) And 'Write the ionion half-equation to show the reduction of water to hydrogen, H2, and hydroxide ions, OH^-'.
(could some one walk me through this, baby steps
) Reduction of Water so splitting water into Hydrogen and Hydroxide ions
H2O --> H2 + OH-
(this is the equation at first - unbalanced)
I can't really explain it but you're trying to split water into hydrogen and hydroxide anions, so therefore you need to balance it now
If we multiply the water moles by 2 so 2H2O instead of H2O, you realise that they are 2 oxygen atoms and 4 hydrogen atoms.
Let's balance the remainder now. So let's leave the H2 for now but focus on the OH- though. Multiply by 2 so therefore
2H2O --> H2 + 2OH-
However the charges aren’t balanced L
Add 2 electrons to the reactant to balance the charges on both sides. Therefore the equation becomes
2H2O +e- --> H2 + 2OH-
4 hydrogens on both sides, 2 oxygens on both sides. Charges are also balanced.
(iii) Combine the two ionic half-equations.
(I think I know how to do this lol, I THINK)
2H2O + Ca +2e- = H2 + 2OH- + Ca2+ 2e-
That’s the combined equation at first. Immediately you can cancel out the electrons so the equation becomes
2H2O + Ca +2e- = H2 + 2OH- + Ca2+ 2e-
2H2O + Ca = H2 + 2OH- + Ca2+
I’m pretty sure that the hydroxide anions will attract the calcium cations so the final equation becomes
2H2O + Ca = H2 + Ca(OH)2
Also other questions:
Complete and balance the following equations:
(ii) Na2O + H2O --> ?
Na2O + H2O à 2NaOH + ½H2
Pretty much the same thing as previous. You should recall from Group 2 that dissolving a metal oxide in water produces a hydroxide and hydrogen. Balancing here is slightly different though.
(iii) Na2O + HCl ---? ?
(idk because I've only learnt about group 2 and 7.. not group 1 :s )
Thanks
Na2O + 2HCl = 2NaCl + H2O
Again, pretty much the same thing as Group 2 metals J
Original post by Dinaa
Lololol more questions (these are questions I missed out from numerous past paper booklets, I swear )
Cars are fitted with catalytic converters in order to reduce the pollution caused by the combustion of petrol. Potential pollutant gases include CO, NO and unburnt hydrocarbons. The first two compounds are removed by passing the hot gases over a platinum catalyst.
CO(g) + NO(g) --> CO2(g) + 1/2N2(g) Delta H = -373kJmol-1
(iii) Comment on the relative value of the Activation energy of this reaction compared with the much fast reaction of NO with O2.
Lol, after typing that out I THINK I get it? Or.. maybe not.
Would you just say that:
The activation energy would be lower in the reaction of NO + O2 because less energy is required for the reaction to take place?
Thanks again
(these are questions I missed out from numerous past paper booklets, I swear )Cars are fitted with catalytic converters in order to reduce the pollution caused by the combustion of petrol. Potential pollutant gases include CO, NO and unburnt hydrocarbons. The first two compounds are removed by passing the hot gases over a platinum catalyst.
CO(g) + NO(g) --> CO2(g) + 1/2N2(g) Delta H = -373kJmol-1
(iii) Comment on the relative value of the Activation energy of this reaction compared with the much fast reaction of NO with O2.
Lol, after typing that out I THINK I get it? Or.. maybe not.
Would you just say that:
The activation energy would be lower in the reaction of NO + O2 because less energy is required for the reaction to take place?
Thanks again
I wouldn’t really say that about this case because you’re not given the values.
I’m not sure, but hopefully if I’m wrong, someone will correct me
The reaction (NO + O2) has a lower activation energy as it undergoes an alternative pathway which requires less energy.
It may be so since the CO bond is stronger than the bond in diatomic oxygen so less energy (hence a lower activation energy) is required.
(edited 8 years ago)
Original post by EMPA2015
I have done AS BIOLOGY I CAN HELP GIVE YOU ANSWERS< DO YOU HAVE AS CHEMISTRY ?
This has been said billions of times across TSR and a few other people have said on this forum already, you're not allowed to provide answers for EMPA or ISAs AT ALL!
That would be classed as cheating which wouldn't be fair
You're not allowed to share answers at all or shared locked/unavailable papers.
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