help (1 Viewer)

[eternallyboreduser]

Does this q only require knowledge from mod 2? Because i am very stuck on it.

 

[Test-king12]

im good off that

 

[gammahydroxybutyrate]

Does this q only require knowledge from mod 2? Because i am very stuck on it.View attachment 42434

i don't know the new chem syllabus but this is just a matter of balancing reactions/working with molar masses no?

 

[eternallyboreduser]

im good off that

wdym

 

[Test-king12]

wdym

Say again!?!?

 

[IhateEnglish123]

wdym

he means his good off that

 

[eternallyboreduser]

he means his good off that

why

 

[IhateEnglish123]

why

his clearly good off that mate can you stop pressing into his personal matters?

 

[eternallyboreduser]

his clearly good off that mate can you stop pressing into his personal matters?

chill dawg it aint that deep i was joking

 

[Luukas.2]

Does this q only require knowledge from mod 2? Because i am very stuck on it.View attachment 42434

Excess HCl liberates CO₂, so C and O are two of the four elements in the mineral... and they are present as carbonate ions, CO₃^2-, or as hydrogencarbonate ions, HCO₃^- (or possibly both)... but the latter options are only possible if hydrogen could be one of the four elements in the mineral.

The uranyl ion is UO₂²⁺ and thus uranyl zinc acetate is UO₂Zn(CH₃COO)₄. The only place that sodium could have come from in the precipitate is the mineral, so that element must also be present.

From this, we can hypothesise that the mineral is Na_xH_y(CO₃)_z where x + y = 2z (to balance the charge).

Now, use the reaction

Na_xH_y(CO₃)_z + 2z HCl ------> z CO₂ + z H₂O + x Na⁺ + yH⁺ + 2z Cl^-​

with 10.000 g of Na_xH_y(CO₃)_z and 4.917 g of CO₂ to find z.

Then, use the reaction

Na_xH_y(CO₃)_z + Xs UO₂Zn(CH₃COO)₄ ------> x (UO₂)₃ZnNa(CH₃COO)₉.6H₂O + other products​

with 0.1000g of the mineral and 2.148 g of the precipitate to find x.

Then, use y = 2z - x.

The answers should be x = 5, y = 3, z = 4.

Hence, the mineral is Na₅H₃(CO₃)₄, or more likely Na₅(HCO₃)₃CO₃, or a double salt Na₂CO₃.3NaHCO₃.

 

[eternallyboreduser]

Excess HCl liberates CO₂, so C and O are two of the four elements in the mineral... and they are present as carbonate ions, CO₃^2-, or as hydrogencarbonate ions, HCO₃^- (or possibly both)... but the latter options are only possible if hydrogen could be one of the four elements in the mineral.

The uranyl ion is UO₂²⁺ and thus uranyl zinc acetate is UO₂Zn(CH₃COO)₄. The only place that sodium could have come from in the precipitate is the mineral, so that element must also be present.

From this, we can hypothesise that the mineral is Na_xH_y(CO₃)_z where x + y = 2z (to balance the charge).

Now, use the reaction

Na_xH_y(CO₃)_z + 2z HCl ------> z CO₂ + z H₂O + x Na⁺ + yH⁺ + 2z Cl^-​

with 10.000 g of Na_xH_y(CO₃)_z and 4.917 g of CO₂ to find z.

Then, use the reaction

Na_xH_y(CO₃)_z + Xs UO₂Zn(CH₃COO)₄ ------> x (UO₂)₃ZnNa(CH₃COO)₉.6H₂O + other products​

with 0.1000g of the mineral and 2.148 g of the precipitate to find x.

Then, use y = 2z - x.

The answers should be x = 5, y = 3, z = 4.

Hence, the mineral is Na₅H₃(CO₃)₄, or more likely Na₅(HCO₃)₃CO₃, or a double salt Na₂CO₃.3NaHCO₃.

R concepts from other modules other than 2 needed because i have no clue what some of these stuff r… never ever heard of a uranyl ion beforr

 

[IhateEnglish123]

chill dawg it aint that deep i was joking

how am i suppose to chill?!? you clearly have no respect for others personal space, get it together mate and pull your finger out!

 

[Test-king12]

chill dawg it aint that deep i was joking

 

[Luukas.2]

R concepts from other modules other than 2 needed because i have no clue what some of these stuff r… never ever heard of a uranyl ion beforr

Very few HSC students will have heard of the uranyl ion, nor the precipitation described. It's a challenging question because it tests if you can deduce such things from the information given. The product given has nine acetate anions with a total 9- charge, so the cations must add to 9+. Since the sodium is always a 1+ cation and zinc is always a 2+ cation, and there is only one of each in the formula of the product, the remaining three uranyl units must amount to a total of a 6+ charge, and so the ion must be UO₂²⁺. Then, you can treat it as any other cation, like ammonium, for example.

There are parts of module 3 that will help you to know why some of this is as it is, and more familiarity with using precipitation and acid / carbonate reactions will develop over time, but there isn't anything that is strictly beyond modules 1 and 2 that is necessary.

One way to make a challenging question is to offer unfamiliar chemistry and to ask you to interpret / predict based on your present understanding. There are plenty of other cations and anions that could be used, some of which you won't formally encounter properly until module 8, but so long as you can sort out the charge and composition, you can go from there. For example, hydrazinium hydrochloride (N₂H₅Cl) will cause a precipitate if added to silver nitrate solution. So long as you can deduce what the hydrazinium cation must be, you can then write:

N₂H₅Cl (aq) + AgNO₃ (aq) --------> AgCl (s) + N₂H₅⁺ (aq) + NO₃^- (aq)​

or, alternatively, N₂H₅NO₃ (aq).

In this question:

• every CO₂ (g) molecule produced came from a carbonate or hydrogencarbbonate ion... and in either case, the ratio of C in CO₂ to C in the mineral is 1:1, so the data given must tell us about z.
• All the sodium in the precipitation product comes from the mineral, so that information tells us about x (possibly as the molar mass of the mineral).
• We know the relationship between x, y, and z.
• Thus, we have three pieces of information and three unknowns, so the problem should be solvable as a set of simultaneous equations at worst.

Technically, we are making some assumptions (such as that there are no other ions (like oxide or hydroxide) that might react with the acid, and that the precipitation does precipitate all of the sodium. We are also dealing with an unusual product as a salt with a sodium cation that is not soluble, which comes as a surprise. Now, it would help to have met double salts - ammonium iron(II) sulfate, for example, is (NH₄)₂Fe(SO₄)₂, while magnetite, Fe₃O₄, makes more sense if recognised as reflecting a composition made of both iron(II) oxide and iron(III) oxide... that is, Fe₃O₄ = FeO.Fe₂O₃ - but it isn't strictly necessary in order to deduce the formula of wegscheiderite as Na₅H₃(CO₃)₄... though getting from there to Na₅(HCO₃)₃CO₃ or to Na₂CO₃.3NaHCO₃ is an extra challenge.

 

[eternallyboreduser]

Excess HCl liberates CO₂, so C and O are two of the four elements in the mineral... and they are present as carbonate ions, CO₃^2-, or as hydrogencarbonate ions, HCO₃^- (or possibly both)... but the latter options are only possible if hydrogen could be one of the four elements in the mineral.

The uranyl ion is UO₂²⁺ and thus uranyl zinc acetate is UO₂Zn(CH₃COO)₄. The only place that sodium could have come from in the precipitate is the mineral, so that element must also be present.

From this, we can hypothesise that the mineral is Na_xH_y(CO₃)_z where x + y = 2z (to balance the charge).

Now, use the reaction

Na_xH_y(CO₃)_z + 2z HCl ------> z CO₂ + z H₂O + x Na⁺ + yH⁺ + 2z Cl^-​

with 10.000 g of Na_xH_y(CO₃)_z and 4.917 g of CO₂ to find z.

Then, use the reaction

Na_xH_y(CO₃)_z + Xs UO₂Zn(CH₃COO)₄ ------> x (UO₂)₃ZnNa(CH₃COO)₉.6H₂O + other products​

with 0.1000g of the mineral and 2.148 g of the precipitate to find x.

Then, use y = 2z - x.

The answers should be x = 5, y = 3, z = 4.

Hence, the mineral is Na₅H₃(CO₃)₄, or more likely Na₅(HCO₃)₃CO₃, or a double salt Na₂CO₃.3NaHCO₃.

How do yk this?

 

[Luukas.2]

How do yk this?

Standard reaction in general chemistry:

acid (aq) + metal carbonate (s or aq) ----> salt (aq if soluble) + carbon dioxide (g) + water (l)​

For example, neutral species equation:

2HCl (aq) + Na₂CO₃ (s) -----> 2NaCl (aq) + CO₂ (g) + H₂O (l}​

As a net ionic equation:

2H⁺ (aq) + Na₂CO₃ (s) -----> 2Na⁺ (aq) + CO₂ (g) + H₂O (l}​

Second example, neutral species equation:

2HNO₃ (aq) + CuCO₃ (s) -----> Cu(NO₃)₂ (aq) + CO₂ (g) + H₂O (l}​

As a net ionic equation:

2H⁺ (aq) + CuCO₃ (s) -----> Cu²⁺ (aq) + CO₂ (g) + H₂O (l}​

Third example, neutral species equation:

H₂SO₄ (aq) + Li₂CO₃ (aq) -----> Li₂SO₄ (aq) + CO₂ (g) + H₂O (l}​

As a net ionic equation:

2H⁺ (aq) + CO₃^2- (aq) -----> CO₂ (g) + H₂O (l}​

Similar examples can be written for the general reaction

acid (aq) + metal hydrogencarbonate (s or aq) ----> salt (aq if soluble) + carbon dioxide (g) + water (l)​

In short, the evolution of carbon dioxide gas from a mineral is evidence for the presence of carbonate and/or hydrogencarbonate anions. The problem with the question as written is that there could be other anions present that react with acid but without the evolution of carbon dioxide - like oxide or hydroxide - but the question can't be answered if they are present (due to insufficient information) and, in any case, they aren't present in this mineral.

 

[eternallyboreduser]

Standard reaction in general chemistry:

acid (aq) + metal carbonate (s or aq) ----> salt (aq if soluble) + carbon dioxide (g) + water (l)​

For example, neutral species equation:

2HCl (aq) + Na₂CO₃ (s) -----> 2NaCl (aq) + CO₂ (g) + H₂O (l}​

As a net ionic equation:

2H⁺ (aq) + Na₂CO₃ (s) -----> 2Na⁺ (aq) + CO₂ (g) + H₂O (l}​

Second example, neutral species equation:

2HNO₃ (aq) + CuCO₃ (s) -----> Cu(NO₃)₂ (aq) + CO₂ (g) + H₂O (l}​

As a net ionic equation:

2H⁺ (aq) + CuCO₃ (s) -----> Cu²⁺ (aq) + CO₂ (g) + H₂O (l}​

Third example, neutral species equation:

H₂SO₄ (aq) + Li₂CO₃ (aq) -----> Li₂SO₄ (aq) + CO₂ (g) + H₂O (l}​

As a net ionic equation:

2H⁺ (aq) + CO₃^2- (aq) -----> CO₂ (g) + H₂O (l}​

Similar examples can be written for the general reaction

acid (aq) + metal hydrogencarbonate (s or aq) ----> salt (aq if soluble) + carbon dioxide (g) + water (l)​

In short, the evolution of carbon dioxide gas from a mineral is evidence for the presence of carbonate and/or hydrogencarbonate anions. The problem with the question as written is that there could be other anions present that react with acid but without the evolution of carbon dioxide - like oxide or hydroxide - but the question can't be answered if they are present (due to insufficient information) and, in any case, they aren't present in this mineral.

Am i supposed to know this in mod2 chem because i have not yet come across anything like that. I have no clue what a net ionic or neutral species eqn is

 

[Luukas.2]

Am i supposed to know this in mod2 chem because i have not yet come across anything like that. I have no clue what a net ionic or neutral species eqn is

These concepts are spread between modules 2 and 3, technically. Net ionic equations are covered in the module 2 worksheets for KISS, for example. The reactions above are in Module 3 topic tests from Hegarty, rather than module 2.

I wouldn't expect a question this hard until perhaps a Yr 11 annual exam, so after both modules are done.

 

[Test-king12]

How do yk this?