Author Topic: HSC Chemistry Question Thread (Read 1268833 times) Tweet Share

jakesilove · « **Reply #855 on:** October 05, 2016, 01:16:25 pm »

Quote from: anotherworld2b on October 05, 2016, 01:12:55 pm

Thank you for your help
Could i get some help with disocisation? Im confused on how H+ ions and OH- ions are produced when water is added to acid and base respectively.

Absolutely! For example, let's look at something like Sulfuric acid, and see what happens. Note: I don't know how to do the equilibrium symbol, so I'm using a right arrow instead.

Let's add water to Sulfuric acid
$H_2O_l+H_2SO_{4aq} \rightarrow H_3O^+_{aq}+HSO_{4aq}$

Clearly, the Acid has DONATED a proton here, and the water has accepted it. The water has therefore acted as a base, by the B-L definition. You sort of just need to know if a substance is an acid or a base (based on your studies this far), and if you don't just make up an equation that makes sense. If there are lots of Hydrogen atoms, it's usually easier to lose a hydrogen than to gain one. If there is an OH in the chemical structure, it is generally a base. It's also a little bit intuitive; just to past questions, and if you're really not sure, guess and write out an equation that makes sense!

anotherworld2b · « **Reply #856 on:** October 05, 2016, 01:38:08 pm »

Oh okay then for a base + watee reaction the base will recieve OH- leaving behind a H?

Quote from: jakesilove on October 05, 2016, 01:16:25 pm

Absolutely! For example, let's look at something like Sulfuric acid, and see what happens. Note: I don't know how to do the equilibrium symbol, so I'm using a right arrow instead.

Let's add water to Sulfuric acid
$H_2O_l+H_2SO_{4aq} \rightarrow H_3O^+_{aq}+HSO_{4aq}$

Clearly, the Acid has DONATED a proton here, and the water has accepted it. The water has therefore acted as a base, by the B-L definition. You sort of just need to know if a substance is an acid or a base (based on your studies this far), and if you don't just make up an equation that makes sense. If there are lots of Hydrogen atoms, it's usually easier to lose a hydrogen than to gain one. If there is an OH in the chemical structure, it is generally a base. It's also a little bit intuitive; just to past questions, and if you're really not sure, guess and write out an equation that makes sense!

anotherworld2b · « **Reply #857 on:** October 05, 2016, 01:48:41 pm »

Could i get help with these two questions as well?
I am really sturggling with pH calculations

RuiAce · « **Reply #858 on:** October 05, 2016, 02:31:26 pm »

Quote from: anotherworld2b on October 05, 2016, 01:48:41 pm

Could i get help with these two questions as well?
I am really sturggling with pH calculations

$\text{The acid is not named in Q8, but the pH is given.}\\ \text{By the formula }[H^+]=10^{-pH}\text{ we have}\\ [H^+]=10^{-5.5}mol \, L^{-1}$
$\text{We don't work in concentrations; we works in moles.}\\ \text{We use the formula }n=CV\text{ between conc. and moles}\\ n=10^{-5.5} \times 15000=0.04743\dots$
$\textbf{Write out the chemical equation otherwise you cannot do anything.}\\ H^+ + NaOH \to Na^+ + H_2O$
__________________________________
$\text{The question becomes harder. Now we must consider the moles of NaOH}\\ \text{They give us the mass }m=10g\text{, so obviously we need }n=\frac{m}{M}\\ n=\frac{10.0}{22.99+16.00+1.008}=0.2500\dots$
$\text{When we mix the NaOH and acid, }\textbf{clearly}\text{ the moles of NaOH is greater.}\\ \text{NaOH is the }\textit{excess reagent}\text{, whereas the acid is the }\textit{limiting reagent.}\\ \text{We need to find the moles of NaOH }\textit{leftover}\\ n_{NaOH^+\text{ final}}=0.2500\dots -0.04743\dots = 0.202578\dots$
$\text{Finally, working backwards}\\ [NaOH]=0.202578/15000 = 1.350...\times 10^{-5}$
$\text{We can then find }pOH\text{ and hence }pH\\ \text{Or find }[H^+]\text{ and then }pH$
I don't have the time to check the values.

RuiAce · « **Reply #859 on:** October 05, 2016, 02:32:39 pm »

Quote from: Bayley_Long on October 05, 2016, 02:18:55 pm

Do you have the answers? (I'm guessing not).

If you do: For the first question I got a pH of 13.4 as my final answer. I don't want to share my working out if its wrong :3
Also what is that second question?

Let the unknown volume be v.

Then the question is just a question involving neutralisation asking you to work both backwards (and maybe forwards) to find a required volume.

anotherworld2b · « **Reply #860 on:** October 05, 2016, 05:18:08 pm »

Can i please get help with these questions

marynguyen18 · « **Reply #861 on:** October 06, 2016, 11:59:59 am »

I'm still a bit unsure about the nuclear chemistry, whats the difference between alpha, beta and gamma decay?

RuiAce · « **Reply #862 on:** October 06, 2016, 12:07:38 pm »

Quote from: marynguyen18 on October 06, 2016, 11:59:59 am

I'm still a bit unsure about the nuclear chemistry, whats the difference between alpha, beta and gamma decay?

The difference is that what's emitted is completely different.

In alpha decay, what's emitted is an entire helium nucleus. This means that 2 protons and neutrons are both emitted together.
Alpha decay occurs when the nucleus is too large. It has to do with the weak nuclear force but all you care is that it's too large and thus the nucleus needs to be made smaller.

In beta decay, what's emitted is an electron from the nucleus. The neutron emits the electron, and turns into a proton. This happens when the n:p ratio is too large so that we can rebalance it out.

In gamma decay, what's emitted is gamma radiation. This is EMR. This simply occurs because the nucleus has too much energy in it and needs to be radiated away.

marynguyen18 · « **Reply #863 on:** October 06, 2016, 01:49:48 pm »

thank you so much RuiAce

ml125 · « **Reply #864 on:** October 06, 2016, 10:58:40 pm »

Quote from: anotherworld2b on October 05, 2016, 05:18:08 pm

Can i please get help with these questions

17. This is a double replacement reaction. To start off, write out the equation. You will then need to find the moles of the precipitate (BaSO₄) using the given mass. From the equation, you can then find the moles of sulfuric acid using the molar ratio, and thus find [H₂SO₄]. You will then need to consider the proticity of the acid and base to determine the value of [H⁺] and [OH^-] to be used in neutralisation. Then, sub in these values and the given volume of acid into C₁V₁=C₂V₂ to solve for the volume of NaOH.

1. With the information you have, you can find the moles of aluminium hydroxide (Al(OH)₃). Looking at the composition of the compound, the moles of OH^- will be triple the moles of Al(OH)₃. From this, you can calculate a value for [OH^-] which will be equivalent to both [H⁺] and [HCl](as HCl is monoprotic). You will then be able to find the moles and thus the mass of HCl.

2. First, you will need to find the moles of H⁺ using the given value for [H⁺] and the volume required. This will be equal to the moles of OH^- required for neutralisation. Observe the formula for calcium hydroxide: Ca(OH)₂. The moles of Ca(OH)₂ will be half the moles of OH^-. Using the attained molar value, you will be able to find the mass of Ca(OH)₂ required for neutralisation.

anotherworld2b · « **Reply #865 on:** October 07, 2016, 10:44:09 am »

Can i get help with these questions please? Also i am kind of confused about how to do equilibrium equations?

:3 · « **Reply #866 on:** October 07, 2016, 11:37:40 am »

Quote from: anotherworld2b on October 07, 2016, 10:44:09 am

Can i get help with these questions please? Also i am kind of confused about how to do equilibrium equations?

1.
A) Basically remember these formulas as they're commonly asked in exams:
- SO2 + H2O --> H2SO3 (sulfurous acid)
- 2H2SO3 + O2 --> 2H2SO4 (sulfuric acid).
B) Like above, remember this:
- H2SO4 + CaCO3 (marble) --> CaSO4 (crumbly mineral) + H2O + CO2.

2.
A) Find the limiting reagent (i.e. find both of the moles for each compound, then the one with least moles/insufficent moles to react with the other compound = used up one).
B) Find excess moles (i.e. you minus the one with the least moles from above from the one with the most moles), and then convert it into grams using the formula: g = n x mm.
C) Use mole to mole ratios (using the compound that is completely used up) to find the moles of the nitrogen dioxide, then times it by the STP number to get the volume in litres.

Hopefully this isn't too confusing.

nibblez16 · « **Reply #867 on:** October 07, 2016, 11:53:53 am »

Hello. Since we have other subjects coming up for chem, how can we stay up to date with it?

RuiAce · « **Reply #868 on:** October 07, 2016, 12:19:02 pm »

Quote from: nibblez16 on October 07, 2016, 11:53:53 am

Hello. Since we have other subjects coming up for chem, how can we stay up to date with it?

What do you mean by other subjects coming up for chem? Did you mean a ton of exams before chemistry?

Gregs · « **Reply #869 on:** October 07, 2016, 01:30:43 pm »

Hi, I was just wondering if anyone had any word documents or something of the like with heaps of the Chemistry-maths equation questions that we can be given. Thanks

p.s. not a chemical equation list with all of the equations we need to know

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