Identifying element X in a hydrated salt
I've been going through this question and seem unable to get the right answer:
X(NO3)2 . 4 H20 is a hydrated compound in which X is a group 2 element.
4.722g of the substance was heated until no futher change in mass, and weighed again. the new mass is 3.282g
Identify element X
If anyone could show me a method for working this out, it'd be great.
X(NO3)2 . 4 H20 is a hydrated compound in which X is a group 2 element.
4.722g of the substance was heated until no futher change in mass, and weighed again. the new mass is 3.282g
Identify element X
If anyone could show me a method for working this out, it'd be great.
Original post by MusicNerd
I've been going through this question and seem unable to get the right answer:
X(NO3)2 . 4 H20 is a hydrated compound in which X is a group 2 element.
4.722g of the substance was heated until no futher change in mass, and weighed again. the new mass is 3.282g
Identify element X
If anyone could show me a method for working this out, it'd be great.
X(NO3)2 . 4 H20 is a hydrated compound in which X is a group 2 element.
4.722g of the substance was heated until no futher change in mass, and weighed again. the new mass is 3.282g
Identify element X
If anyone could show me a method for working this out, it'd be great.
Using that data you can work out what proportion molecular mass is the 4 H2O. As you know the molecular mass of water you then know the mass of the X(NO3)2. As you know the mass of NO3, then you know the mass of X....
TA DA!
Original post by MusicNerd
I've been going through this question and seem unable to get the right answer:
X(NO3)2 . 4 H20 is a hydrated compound in which X is a group 2 element.
4.722g of the substance was heated until no futher change in mass, and weighed again. the new mass is 3.282g
Identify element X
If anyone could show me a method for working this out, it'd be great.
X(NO3)2 . 4 H20 is a hydrated compound in which X is a group 2 element.
4.722g of the substance was heated until no futher change in mass, and weighed again. the new mass is 3.282g
Identify element X
If anyone could show me a method for working this out, it'd be great.
mol=mass/mr
4.722g-3.282g=1.44g (mass of water that is evaporated)
mol of this water = 1.44/18.0 = 0.08 mol
The formula tells you that water is present in this ratio : 4 mol water: 1 mol X(NO3)2 so....
MOD EDIT: full answer.
(edited 10 years ago)
Original post by Mr Big Al
mol=mass/mr
4.722g-3.282g=1.44g (mass of water that is evaporated)
mol of this water = 1.44/18.0 = 0.08 mol
The formula tells you that water is present in this ratio : 4 mol water: 1 mol X(NO3)2 so....
4.722g-3.282g=1.44g (mass of water that is evaporated)
mol of this water = 1.44/18.0 = 0.08 mol
The formula tells you that water is present in this ratio : 4 mol water: 1 mol X(NO3)2 so....
nooo, don't answer it for her. Not the point of this forum.
MOD EDIT: full answer.
(edited 10 years ago)
Original post by JMaydom
nooo, don't answer it for her. Not the point of this forum.
" If anyone could show me a method for working this out, it'd be great."
It is best to be shown an example when you are stuck, so from now on any similar types of questions and exam questions can now be answered.
I was simply answering the question.
Original post by Mr Big Al
" If anyone could show me a method for working this out, it'd be great."
It is best to be shown an example when you are stuck, so from now on any similar types of questions and exam questions can now be answered.
I was simply answering the question.
It is best to be shown an example when you are stuck, so from now on any similar types of questions and exam questions can now be answered.
I was simply answering the question.
I explained a method. The difference is that is still required the person to actually do the work. With your way you can just ctl-c ctl-v the answer across. No guarantee the OP has actually understood it when they hand in the answer.
+ it promotes lazy 'can't be bothered to do my homework myself' attitude
Original post by JMaydom
I explained a method. The difference is that is still required the person to actually do the work. With your way you can just ctl-c ctl-v the answer across. No guarantee the OP has actually understood it when they hand in the answer.
+ it promotes lazy 'can't be bothered to do my homework myself' attitude
+ it promotes lazy 'can't be bothered to do my homework myself' attitude
Actually, The question I used was an example from notes i'd taken at college, so I already knew the answer. However, using the method given in the example, i couldn't get calcium, so I came on here to see if the method i was given had an error in it (it did) and if anyone had a better method.
I agree that just giving an answer doesn't help learning, explaining it does. In any case, this is for my revision, not homework and I am using the corrected method for answering different questions, that incidentally, my teachers won't see.
Thanks for trying to prevent me being lazy, but I posted this to aid my understanding, not to cheat.
(oh, love your profile pic by the way! )
(edited 10 years ago)
Original post by Mr Big Al
mol=mass/mr
4.722g-3.282g=1.44g (mass of water that is evaporated)
mol of this water = 1.44/18.0 = 0.08 mol
The formula tells you that water is present in this ratio : 4 mol water: 1 mol X(NO3)2 so....
0.08 mol water : 0.02 mol X(NO3)2
Mr of X(NO3)2 = mass/mol = 3.282g/0.02 = 164.1
Mr of (NO3)2 = 124.0
164.1-124.0= 40.1
X= Calcium
Hope that's right
4.722g-3.282g=1.44g (mass of water that is evaporated)
mol of this water = 1.44/18.0 = 0.08 mol
The formula tells you that water is present in this ratio : 4 mol water: 1 mol X(NO3)2 so....
0.08 mol water : 0.02 mol X(NO3)2
Mr of X(NO3)2 = mass/mol = 3.282g/0.02 = 164.1
Mr of (NO3)2 = 124.0
164.1-124.0= 40.1
X= Calcium
Hope that's rightAh, get it now! thanks!
Original post by JMaydom
nooo, don't answer it for her. Not the point of this forum.
MOD EDIT: full answer.
MOD EDIT: full answer.
With that attitude no one would get anywhere! Everyone needs a little help now and then! Ever thought she has already tried doing it herself and that she's not being lazy!
Posted from TSR Mobile
Original post by lauraMarieBrooks
With that attitude no one would get anywhere! Everyone needs a little help now and then! Ever thought she has already tried doing it herself and that she's not being lazy!
Posted from TSR Mobile
Posted from TSR Mobile
I think quiet the opposite. What is more useful?...... learning the method with which to approach problems like this or being handed an answer?
Now i admit, the OP has explained their situation and I this case a full answer is indeed useful. However you see so many people obviously just looking for people to do the work for them, usually by simply copying the question straight across.
Also, what's with that Mod Edit thing?
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