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secondary 3 | E Maths
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ITS SCIENCE CHEM YALL IGNORE THE SUBJ.
Date Posted: 4 years ago
Views: 631
Wait
For your part c working, you need to consider one of these (both will lead to the same conclusion):
Do I have enough Na2SO4 to react fully with all the available Pb(NO3)2?
Do I have enough Pb(NO3)2 to react fully with all the available Na2SO4?
This is based on the stoichiometric ratio 1 : 1.
Think of baking.
Suppose a classroom goes like this.
1 table + 4 chairs = 1 seating area
(and the ratio remains fixed).
If I have 4 tables and 10 chairs to be placed in the fixed ratio,
- do I have enough tables?
- do I have enough chairs
- how many groups of seating area can I have at most?
The idea for this question is very similar.
Do I have enough Na2SO4 to react fully with all the available Pb(NO3)2?
Do I have enough Pb(NO3)2 to react fully with all the available Na2SO4?
This is based on the stoichiometric ratio 1 : 1.
Think of baking.
Suppose a classroom goes like this.
1 table + 4 chairs = 1 seating area
(and the ratio remains fixed).
If I have 4 tables and 10 chairs to be placed in the fixed ratio,
- do I have enough tables?
- do I have enough chairs
- how many groups of seating area can I have at most?
The idea for this question is very similar.
See 2 Answers
done
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As a start, I send you this for the state symbols. Wait while I write my workings.
Date Posted:
4 years ago
amazing explanation and example, thank you so much!
done
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Let me know if my workings and my explanation sound Greek to you.
Date Posted:
4 years ago
um i dont really understand ionic equations bc my teacher havent really taught me how to
Ok, let me explain what ionic equations are all about. Give me half an hour to write them on this chat box.
Ionic equations are equations involving a productive change in the reaction system. By this, I mean a change in state, because something useful will be formed only when there is a change.
Imagine you are an athlete training up for a championship. You are training with your coach.
Would you want your performance level to stay the same at the end of the months-long training period? Or would you want your performance to improve drastically to help you earn a better result?
For most athletes, they will know the obvious.
Now, in this ionic equation, we would like to see a change happening in the system. This is seen by a change in physical states (solid, liquid, aqueous, gaseous).
Before you even attempt to write down an ionic equation, you will need to know the solubilities of your compounds, which I have attached in the first post.
Hang on for my next part.
Imagine you are an athlete training up for a championship. You are training with your coach.
Would you want your performance level to stay the same at the end of the months-long training period? Or would you want your performance to improve drastically to help you earn a better result?
For most athletes, they will know the obvious.
Now, in this ionic equation, we would like to see a change happening in the system. This is seen by a change in physical states (solid, liquid, aqueous, gaseous).
Before you even attempt to write down an ionic equation, you will need to know the solubilities of your compounds, which I have attached in the first post.
Hang on for my next part.
An ionic equation is possible when there is a change in state to/from an aqueous state, because ions are formed in the reaction or removed from the reaction.
Now, here the two reactants, sodium sulfate and lead nitrate are soluble in water. They don't react with each other as solids, so they have to be in a dissolved form, usually in water. We call this aqueous. Similarly, one of the compounds formed. sodium nitrate, is soluble in water, and is thus in an aqueous state since the water of reaction is still present.
(If you react hydrochloric acid with sodium hydroxide to form sodium chloride, the sodium chloride won't pop out as a white solid correct?)
So, here's the thing. When dissolved in water, these soluble compounds will ALWAYS split into their respective ions, because the water of dissolution overcomes the ionic lattice structure of the ionic compound. In contrast, insoluble compounds cannot be split into their respective ions as the water of dissolution cannot break the ionic lattice structure.
As such,
- sodium sulfate exists as sodium ions (Na +) and sulfate (SO4 2-) ions in water due to its solubility
- lead nitrate exists as lead (Pb 2+) ions and nitrate (NO3 -) ions in water due to its solubilty
- sodium nitrate exists as sodium (Na +) and nitrate (NO3 -) ions in water due to its solubility
- lead sulfate exists as a plain solid itself without breaking down into ions, due to its lack of solubility
Wait for my next post.
Now, here the two reactants, sodium sulfate and lead nitrate are soluble in water. They don't react with each other as solids, so they have to be in a dissolved form, usually in water. We call this aqueous. Similarly, one of the compounds formed. sodium nitrate, is soluble in water, and is thus in an aqueous state since the water of reaction is still present.
(If you react hydrochloric acid with sodium hydroxide to form sodium chloride, the sodium chloride won't pop out as a white solid correct?)
So, here's the thing. When dissolved in water, these soluble compounds will ALWAYS split into their respective ions, because the water of dissolution overcomes the ionic lattice structure of the ionic compound. In contrast, insoluble compounds cannot be split into their respective ions as the water of dissolution cannot break the ionic lattice structure.
As such,
- sodium sulfate exists as sodium ions (Na +) and sulfate (SO4 2-) ions in water due to its solubility
- lead nitrate exists as lead (Pb 2+) ions and nitrate (NO3 -) ions in water due to its solubilty
- sodium nitrate exists as sodium (Na +) and nitrate (NO3 -) ions in water due to its solubility
- lead sulfate exists as a plain solid itself without breaking down into ions, due to its lack of solubility
Wait for my next post.
So now, let's consider the cases "before" and "after" the reaction.
-----------------------------------------------
BEFORE
The species present before the reaction are
- Na + (aq),
- SO4 2- (aq)
- Pb 2+ (aq)
- NO3 - (aq)
-----------------------------------------------
AFTER
The species present after the reaction are
- PbSO4 (s) (a full solid, no more ions)
- Na + (aq) (since the resultant compound with NO3- is still aq)
- NO3 - (aq) (since the resultant compound with Na+ is still aq)
-----------------------------------------------
It's kind of like dancing. Suppose I am in a dance room with 3 other people (as an example).
The four of us started out without partners (aq). The objective at the end is to link up with someone else for the dance.
I start out as Na +. At the end of the dance session, I am still aq, meaning I have not found a suitable partner for the dance. Mission failed.
A second person starts out as NO3 -. At the end of the dance session, the person is still aq as well, so the person also fails the mission.
We are both out of the ionic equation.
The other two people, Pb 2+ and SO4 2-, manage to find their dance partners (i.e. each other), so it's a success (they no longer stay as aq and become s).
These "successes" give rise to their presence in the ionic equation.
-----------------------------------------------
So in summary,
- Pb 2 + and SO4 2- link up together and are therefore in the ionic equation
- Na + and NO3 - remain as Na + and NO3 - even at the end, so they are NOT in the ionic equation
-----------------------------------------------
Therefore, the ionic equation is
Pb 2+ (aq) + SO4 2- (aq) --> PbSO4 (s)
-----------------------------------------------
Let me know if you need more explanation or a second example.
-----------------------------------------------
BEFORE
The species present before the reaction are
- Na + (aq),
- SO4 2- (aq)
- Pb 2+ (aq)
- NO3 - (aq)
-----------------------------------------------
AFTER
The species present after the reaction are
- PbSO4 (s) (a full solid, no more ions)
- Na + (aq) (since the resultant compound with NO3- is still aq)
- NO3 - (aq) (since the resultant compound with Na+ is still aq)
-----------------------------------------------
It's kind of like dancing. Suppose I am in a dance room with 3 other people (as an example).
The four of us started out without partners (aq). The objective at the end is to link up with someone else for the dance.
I start out as Na +. At the end of the dance session, I am still aq, meaning I have not found a suitable partner for the dance. Mission failed.
A second person starts out as NO3 -. At the end of the dance session, the person is still aq as well, so the person also fails the mission.
We are both out of the ionic equation.
The other two people, Pb 2+ and SO4 2-, manage to find their dance partners (i.e. each other), so it's a success (they no longer stay as aq and become s).
These "successes" give rise to their presence in the ionic equation.
-----------------------------------------------
So in summary,
- Pb 2 + and SO4 2- link up together and are therefore in the ionic equation
- Na + and NO3 - remain as Na + and NO3 - even at the end, so they are NOT in the ionic equation
-----------------------------------------------
Therefore, the ionic equation is
Pb 2+ (aq) + SO4 2- (aq) --> PbSO4 (s)
-----------------------------------------------
Let me know if you need more explanation or a second example.
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