AS Chemistry- helping each other out!
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Original post by samb1234
Fragments are indeed formed by the bombarding of electrons which takes place during ionisation. If the electron hits the molecule with enough energy it can potentially break the bonds forming a fragment. The fingerprint region is the region of an ir spec which is due to the stretching of the various bonds. The fingerprint region is formed of a large number of small absorptions and as far as I'm aware all you need to know is that this pattern is unique
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Got it! Thanks
I have been given a question to complete at home in preparation of an exam and I am struggling with some parts.
A student investigated the reaction between calcium carbonate CaCO3 and nitric acid, HNO3 using the following apparatus.(Reaction mixture in a conical flask going to a gas syringe) They used a balance to measure out 2.00g of calcium carbonate lumps which was placed in the conical flask. They also added 50.0cm^3 of 0.250 moldm^-3 nitric acid to the conical flask before replacing the bung and using the gas syringe to measure the volume of gas produced.
1)Write an equation for the reaction between calcium carbonate and nitric acid.
I got CaCO3 + 2HNO3 -> CO2 + H20 + Ca(CO3)2
2) Calculate the number of moles of calcium carbonate added to the flask.
I got 2/(40.1+12+48) = 0.02mol
3) Calculate the number of moles of nitric acid that was added to the flask.
I got (50*0.250)/1000 = 0.0125mol
4) Which Reagent was in excess?
I got CaCO3 1:2 Mol ratio
0.02:0.0125
0.00625:0.0125
So CaCO3 in excess (WHY?)
5) Calculate the excess mass of CaCO3 that was added to the conical flask?
How do i do this?
6) How many moles of CO2 would you expect to be produced?
Help Please?
7) What volume of Carbon dioxide would be produced at RTP(room temp)
Help Please.
8) In each of the following cases state how the rate and volume of carbon dioxide would be affected by;
increasing the mass of Calcium Carbonate.
Using 50Cm3 of 0.5 moldm nitric acid rather than 0.25 moldm of nitric acid.
Using the same mass of powdered caco3 rather than lumps.
Any help I ould be greatful for sorry for such a large question but i only had one night to prepare.
A student investigated the reaction between calcium carbonate CaCO3 and nitric acid, HNO3 using the following apparatus.(Reaction mixture in a conical flask going to a gas syringe) They used a balance to measure out 2.00g of calcium carbonate lumps which was placed in the conical flask. They also added 50.0cm^3 of 0.250 moldm^-3 nitric acid to the conical flask before replacing the bung and using the gas syringe to measure the volume of gas produced.
1)Write an equation for the reaction between calcium carbonate and nitric acid.
I got CaCO3 + 2HNO3 -> CO2 + H20 + Ca(CO3)2
2) Calculate the number of moles of calcium carbonate added to the flask.
I got 2/(40.1+12+48) = 0.02mol
3) Calculate the number of moles of nitric acid that was added to the flask.
I got (50*0.250)/1000 = 0.0125mol
4) Which Reagent was in excess?
I got CaCO3 1:2 Mol ratio
0.02:0.0125
0.00625:0.0125
So CaCO3 in excess (WHY?)
5) Calculate the excess mass of CaCO3 that was added to the conical flask?
How do i do this?
6) How many moles of CO2 would you expect to be produced?
Help Please?
7) What volume of Carbon dioxide would be produced at RTP(room temp)
Help Please.
8) In each of the following cases state how the rate and volume of carbon dioxide would be affected by;
increasing the mass of Calcium Carbonate.
Using 50Cm3 of 0.5 moldm nitric acid rather than 0.25 moldm of nitric acid.
Using the same mass of powdered caco3 rather than lumps.
Any help I ould be greatful for sorry for such a large question but i only had one night to prepare.
Original post by EwaHum
I have been given a question to complete at home in preparation of an exam and I am struggling with some parts.
A student investigated the reaction between calcium carbonate CaCO3 and nitric acid, HNO3 using the following apparatus.(Reaction mixture in a conical flask going to a gas syringe) They used a balance to measure out 2.00g of calcium carbonate lumps which was placed in the conical flask. They also added 50.0cm^3 of 0.250 moldm^-3 nitric acid to the conical flask before replacing the bung and using the gas syringe to measure the volume of gas produced.
1)Write an equation for the reaction between calcium carbonate and nitric acid.
I got CaCO3 + 2HNO3 -> CO2 + H20 + Ca(CO3)2
2) Calculate the number of moles of calcium carbonate added to the flask.
I got 2/(40.1+12+48) = 0.02mol
3) Calculate the number of moles of nitric acid that was added to the flask.
I got (50*0.250)/1000 = 0.0125mol
4) Which Reagent was in excess?
I got CaCO3 1:2 Mol ratio
0.02:0.0125
0.00625:0.0125
So CaCO3 in excess (WHY?)
5) Calculate the excess mass of CaCO3 that was added to the conical flask?
How do i do this?
6) How many moles of CO2 would you expect to be produced?
Help Please?
7) What volume of Carbon dioxide would be produced at RTP(room temp)
Help Please.
8) In each of the following cases state how the rate and volume of carbon dioxide would be affected by;
increasing the mass of Calcium Carbonate.
Using 50Cm3 of 0.5 moldm nitric acid rather than 0.25 moldm of nitric acid.
Using the same mass of powdered caco3 rather than lumps.
Any help I ould be greatful for sorry for such a large question but i only had one night to prepare.
A student investigated the reaction between calcium carbonate CaCO3 and nitric acid, HNO3 using the following apparatus.(Reaction mixture in a conical flask going to a gas syringe) They used a balance to measure out 2.00g of calcium carbonate lumps which was placed in the conical flask. They also added 50.0cm^3 of 0.250 moldm^-3 nitric acid to the conical flask before replacing the bung and using the gas syringe to measure the volume of gas produced.
1)Write an equation for the reaction between calcium carbonate and nitric acid.
I got CaCO3 + 2HNO3 -> CO2 + H20 + Ca(CO3)2
2) Calculate the number of moles of calcium carbonate added to the flask.
I got 2/(40.1+12+48) = 0.02mol
3) Calculate the number of moles of nitric acid that was added to the flask.
I got (50*0.250)/1000 = 0.0125mol
4) Which Reagent was in excess?
I got CaCO3 1:2 Mol ratio
0.02:0.0125
0.00625:0.0125
So CaCO3 in excess (WHY?)
5) Calculate the excess mass of CaCO3 that was added to the conical flask?
How do i do this?
6) How many moles of CO2 would you expect to be produced?
Help Please?
7) What volume of Carbon dioxide would be produced at RTP(room temp)
Help Please.
8) In each of the following cases state how the rate and volume of carbon dioxide would be affected by;
increasing the mass of Calcium Carbonate.
Using 50Cm3 of 0.5 moldm nitric acid rather than 0.25 moldm of nitric acid.
Using the same mass of powdered caco3 rather than lumps.
Any help I ould be greatful for sorry for such a large question but i only had one night to prepare.
1. Pretty sure it's typo but it's Ca(NO3)2 made
4. To ensure all acid has reacted
5. Calculate the mole that has reacted from the equation (which you have done). Subtract from 0.02 mol you calculated and multiply by the Mr of CaCO3.
6. Ratio from the equation
7. A mole of gas in standard condition occupies 24dm3
8a no effect as the reaction is limited by moles of the acid
b. Higher rate as higher concentration and more gas produced as the excess CaCO3 will now then react
C. Higher rate as more surface area etc etc...
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(edited 9 years ago)
Original post by C0balt
1. Pretty sure it's typo but it's Ca(NO3)2 made
4. To ensure all acid has reacted
5. Calculate the mole that has reacted from the equation (which you have done). Subtract from 0.02 mol you calculated and multiply by the Mr of CaCO3.
6. Ratio from the equation
7. A mole of gas in standard condition occupies 24dm3
8a no effect as the reaction is limited by moles of the acid
b. Higher rate as higher concentration and more gas produced as the excess CaCO3 will now then react
C. Higher rate as more surface area etc etc...
Posted from TSR Mobile
4. To ensure all acid has reacted
5. Calculate the mole that has reacted from the equation (which you have done). Subtract from 0.02 mol you calculated and multiply by the Mr of CaCO3.
6. Ratio from the equation
7. A mole of gas in standard condition occupies 24dm3
8a no effect as the reaction is limited by moles of the acid
b. Higher rate as higher concentration and more gas produced as the excess CaCO3 will now then react
C. Higher rate as more surface area etc etc...
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Just awnsered in another thread
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1) It should be Ca(NO3)2
4) It was important that the calcium carbonate was in excess so the nitric acid fully reacted.
5) You multiply the number of moles from the nitric acid by the molar mass of the CaCO3 to work out the exact mass required, rather than adding in excess..... 0.0125 / 2 (1:2 ratio) x 100 = 0.625g As 2g was used, 2-0.625=1.375g in excess
6)Refer back to original equation. 2HNO3:1Co2, 2:1 ratio, meaning for every 2 moles of HNO3 used only 1 mole of CO2 is produced. 0.0125 moles of HNO3 were used, therefore 0.0125/2 = 0.00625 moles of CO2 produced.
7) Moles x 24000 = volume in cm3, therefore 0.00625 x24000 = 150cm3
8) Doubling the volume of HNO3 means that 50*0.5 /1000 = 0.025 moles of HNO3 used and 0.025/2 = 0.0125 moles of CO2 produced. 0.0125 x 24000 = 300cm3 of CO2 produced (doubles)
Powdered CaCO3 would have no effect on the volume of CO2 produced, the reaction would just happen faster
Hope this helps!!!
4) It was important that the calcium carbonate was in excess so the nitric acid fully reacted.
5) You multiply the number of moles from the nitric acid by the molar mass of the CaCO3 to work out the exact mass required, rather than adding in excess..... 0.0125 / 2 (1:2 ratio) x 100 = 0.625g As 2g was used, 2-0.625=1.375g in excess
6)Refer back to original equation. 2HNO3:1Co2, 2:1 ratio, meaning for every 2 moles of HNO3 used only 1 mole of CO2 is produced. 0.0125 moles of HNO3 were used, therefore 0.0125/2 = 0.00625 moles of CO2 produced.
7) Moles x 24000 = volume in cm3, therefore 0.00625 x24000 = 150cm3
8) Doubling the volume of HNO3 means that 50*0.5 /1000 = 0.025 moles of HNO3 used and 0.025/2 = 0.0125 moles of CO2 produced. 0.0125 x 24000 = 300cm3 of CO2 produced (doubles)
Powdered CaCO3 would have no effect on the volume of CO2 produced, the reaction would just happen faster
Hope this helps!!!
Original post by C0balt
1. Pretty sure it's typo but it's Ca(NO3)2 made
4. To ensure all acid has reacted
5. Calculate the mole that has reacted from the equation (which you have done). Subtract from 0.02 mol you calculated and multiply by the Mr of CaCO3.
6. Ratio from the equation
7. A mole of gas in standard condition occupies 24dm3
8a no effect as the reaction is limited by moles of the acid
b. Higher rate as higher concentration and more gas produced as the excess CaCO3 will now then react
C. Higher rate as more surface area etc etc...
Posted from TSR Mobile
4. To ensure all acid has reacted
5. Calculate the mole that has reacted from the equation (which you have done). Subtract from 0.02 mol you calculated and multiply by the Mr of CaCO3.
6. Ratio from the equation
7. A mole of gas in standard condition occupies 24dm3
8a no effect as the reaction is limited by moles of the acid
b. Higher rate as higher concentration and more gas produced as the excess CaCO3 will now then react
C. Higher rate as more surface area etc etc...
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So the moles of the CO2 would be 0.00625?
How would i use the 24dm in question 7?
Thanks for the help with the other questions.
Original post by EwaHum
So the moles of the CO2 would be 0.00625?
How would i use the 24dm in question 7?
Thanks for the help with the other questions.
How would i use the 24dm in question 7?
Thanks for the help with the other questions.
Yes.
24dm3/mol so just multiply by the moles of CO2
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Original post by mathscot
Well thanks for letting me know....?
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Original post by C0balt
you both answered at the same time sorry for taking both of your time up but thanks a lot to both all of you.
Original post by EwaHum
you both answered at the same time sorry for taking both of your time up but thanks a lot to both all of you.
Np
I check this thread more frequently than chemistry forums as I have it on my subscription
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Happy yo help
Id love to know how everybody is revising? I'm finding it slightly overwhelming especially unit 2, although I put that down to my awful teacher. She taught us the last module in 30 minutes!!!!!
Original post by Georgiam247
Id love to know how everybody is revising? I'm finding it slightly overwhelming especially unit 2, although I put that down to my awful teacher. She taught us the last module in 30 minutes!!!!!
Lol, the last module is very small though.
All you really need to know is that co2 isnt all bad, you need to know how the ozone layer works and about catalytic converters and ccs. That's all
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(edited 9 years ago)
Hey has any one done the evaluative 15 mark chemistry practical yet? I'm confused on what we need to prepare for. My friends said that they went over calculations on moles, molar mass etc and just went through the basics. Does anyone have an idea? Thanks.
when you say sth is being oxidised or reduced, do you say the element of the compound or the compound?
e.g. the S in SO2 is reduced / SO2 is reduced?
thank you!
e.g. the S in SO2 is reduced / SO2 is reduced?
thank you!
Can someone explain the free radical substitution of the following reaction:
CHF3 + Cl2 ----->CClF3 +HCl
Don't understand the propagation stage.
CHF3 + Cl2 ----->CClF3 +HCl
Don't understand the propagation stage.
Original post by Super199
Can someone explain the free radical substitution of the following reaction:
CHF3 + Cl2 ----->CClF3 +HCl
Don't understand the propagation stage.
CHF3 + Cl2 ----->CClF3 +HCl
Don't understand the propagation stage.
Do you know what the propagation stages are? If so post them so I can check I've done them right then I will explain
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Original post by samb1234
Do you know what the propagation stages are? If so post them so I can check I've done them right then I will explain
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Na I don't sorry.
Original post by Super199
Na I don't sorry.
My guess is CHF3 +Clradical goes to HCl + CF3 radical.
CF3 radical +cl2 goes to CClF3 +cl radical
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Original post by Super199
Can someone explain the free radical substitution of the following reaction:
CHF3 + Cl2 ----->CClF3 +HCl
Don't understand the propagation stage.
CHF3 + Cl2 ----->CClF3 +HCl
Don't understand the propagation stage.
Initiation: Cl-Cl -> 2Cl.
propagation I: CHF3 + Cl. -> HCl + CF3.
propagation II: CF3. + Cl2 -> CF3Cl + Cl.
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