Chemistry HL/SL help

[LMaxwell]

(iv) At a certain temperature and pressure, 1.1 dm3 of N2(g) reacts with 3.3 dm3 of H2(g). Calculate the volume of NH3(g), that will be produced.

(if more info is needed, I can provide the entire question)

Thanks

[Drake Glau]

This is an Ideal Gas Law (PV=nRT) question, does it not give you the P and T? Or maybe something that indicates it is at STP where we can use avagadro's constant of 22.4L/mole of gas?

^That would make this easier, but here goes an attempt without it...

N₂ is the limiting reactant. Since P and T are constant, it's just following the stoichiometry of the reaction using the volumes.

N₂+3H₂=2NH₃

You make 2 times the amount of N₂ so 1.1*2=2.2dm³

[LMaxwell]

Calculate the pH of a 1.00 x 10^-3 mol dm^-3 aqueous solution of ethylamine, C₂H₅NH₂, given that the pK_a value for ethylamine is 10.73 at 298K.

Please help?

[ILM]

Okay, you want to find pH, so lets start with the general rule of pH which is ;

pH =-log (H⁺)

So you want the concentration of the hydrogen ions, and you can do this using the pKa value given to you:

pKa=-log₁₀Ka, so Ka= 1.9 * 10^-11

K_a= (H⁺) (CH₅NH^-)/ (con. of ethylamine) (since it is 1:1 mole the concentration of (H⁺) is equal to the concentration of (CH₅NH^-))

1.9 * 10^-11=x²/1.00 x 10^-3

x²= 1.00 x 10^-3*1.9 * 10^{-11 =} 1.9*10-14

by taking square root, (H⁺)= 1.4* 10^-7

now you have nothing except putting it in the rule

pH= -log (1.4* 10^-7) = 6.86

If you don't understand anything please don't hesitate to ask

[LMaxwell]

Okay, you want to find pH, so lets start with the general rule of pH which is ;

pH =-log (H⁺)

So you want the concentration of the hydrogen ions, and you can do this using the pKa value given to you:

pKa=-log₁₀Ka, so Ka= 1.9 * 10^-11

K_a= (H⁺) (CH₅NH^-)/ (con. of ethylamine) (since it is 1:1 mole the concentration of (H⁺) is equal to the concentration of (CH₅NH^-))

1.9 * 10^-11=x²/1.00 x 10^-3

x²= 1.00 x 10^-3*1.9 * 10^{-11 =} 1.9*10-14

by taking square root, (H⁺)= 1.4* 10^-7

now you have nothing except putting it in the rule

pH= -log (1.4* 10^-7) = 6.86

If you don't understand anything please don't hesitate to ask

Yeah that's what I got at first but it's wrong.

The question is from the course companion and can be found on the following link (scroll down a bit): http://aura.edu.in/read/IB/CHEMISTRY_COURS_COMPANION_(OXFORD)/original/Page-152.html

I see how they did it but I thought there would be an easier way?

Thanks for the help

[ILM]

Okay, you want to find pH, so lets start with the general rule of pH which is ;

pH =-log (H⁺)

So you want the concentration of the hydrogen ions, and you can do this using the pKa value given to you:

pKa=-log₁₀Ka, so Ka= 1.9 * 10^-11

K_a= (H⁺) (CH₅NH^-)/ (con. of ethylamine) (since it is 1:1 mole the concentration of (H⁺) is equal to the concentration of (CH₅NH^-))

1.9 * 10^-11=x²/1.00 x 10^-3

x²= 1.00 x 10^-3*1.9 * 10^{-11 =} 1.9*10-14

by taking square root, (H⁺)= 1.4* 10^-7

now you have nothing except putting it in the rule

pH= -log (1.4* 10^-7) = 6.86

If you don't understand anything please don't hesitate to ask

Yeah that's what I got at first but it's wrong.

The question is from the course companion and can be found on the following link (scroll down a bit): http://aura.edu.in/r...l/Page-152.html

I see how they did it but I thought there would be an easier way?

Thanks for the help

Okay, me and you forgot the most important thing in solving those problems which is the dissociation of ethylamine. Ethylamine is basic in its nature so it will not at all give H ions this is because (which both you and me forgot) has an amine group (like ammonia) meaning that it will increase the OH concentrations (it will take a hydrogen from water) so the way that i will use to solve now is by finding Kb.

pKa+pKb=14 so pKb= 3.27 and Kb= 5.37*10^-4

then the square root of kb* conc. of ethylamine =7.33*10^-4

which means the pOH= 3.135 so pH= 10.865

[DropBoite]

hello all! i cant get this questions answer right, please help!

find the net equation:

2h+ + 2e- --- h2

4oh- --- 2h2o + o2 + 4e-

[ILM]

You should first of all balance the number of electrons, so multiply equation 1 by 2, the equations will look as follows:

4H+4e ---------> 2H2

4OH---------> 2H2O+O2+4e

So, by adding both:

4OH+4H------> 2H2O+O2+2H2 (electrons cancelled)

However, chemically OH+H------> water

so,

4H+4OH---------> 4 H2O

[DropBoite]

You should first of all balance the number of electrons, so multiply equation 1 by 2, the equations will look as follows:

4H+4e ---------> 2H2

4OH---------> 2H2O+O2+4e

So, by adding both:

4OH+4H------> 2H2O+O2+2H2 (electrons cancelled)

However, chemically OH+H------> water

so,

4H+4OH---------> 4 H2O

the answer is 2h20 ---> 2H2 +O2 How did they get this even?

Pearson HL also has some other weird ones, like the electrolysis of NaCl:

2cl- --> Cl2 + 2e-

4OH- -->2h20 + 02 + 4e-

gets the net equation: 2 NaCl + 2h2o --> H2 + Cl2 + 2Na + +2OH-

Edited April 14, 2012 by DropBoite

[ILM]

By the way what i put up is correct, but you didn't told us the context of the question. Your reaction indicates that we are talking abou****er electrlysis, in which each 2 molecules electrolysed would give 2 moles of hydrogen and one mole of oxygen. I will use the last answer i gave you

4OH+4H------> 2H2O+O2+2H2

and since

4H+4OH---------> 4 H2O

then

4 H2O------->2H2O+O2+2H2

then 2 water molecules will cancel the 2 water molecules in the right side to give

2H20---------> O2 + 2H2

[DropBoite]

1) Find the net equation, chlorine gas is liberated when potassium manganate(VII) is added to hydrochloric acid

:

1st: MnO4- + 8H+ + 5e- --> Mn2+ + 4H2O

2nd: 1/2 Cl2 + e- ---> Cl-

Correct answer:

10Cl- + 2MnO4- + 16H + --> 5Cl2 + 2Mn2+ + 8H2O;

2) Iodide ions, I–(aq), react with iodate ions, IO3–(aq), in an acidic solution to form molecular iodine and water.

Write an ionic equation for the reaction of I– with IO3– in an acidic solution:

I just don't know how to get this, please help!

Answer: 5I-(aq) + IO3–(aq) + 6H+(aq) ® 3I2(aq) + 3H2O(l);

3) In voltaic cells why doesn't the concept of preferential discharge even exist? And yes they too are aqueous.

4) Identify a substance that will oxidize bromide ions but not chloride ions. Explain your choice, and write an equation for the redox reaction you have chosen.

Apparently this is how it works:

Agent: O2/Cr2O72–;

E

ο

value for the reaction with Br– is positive

E

ο

value for the reaction with Cl– is

negative( This I want clarified, as it is not negative*)

/Cl2 stronger oxidizing agent than

O2/Cr2O72–;

4Br– + O2 + 4H+ → 2Br2 + 2H2O/Cr2O72– + 14H + 6Br– → 2Cr3+ + 7H2O + 3Br2;

I thank you a lot!

Edited April 15, 2012 by DropBoite

[LMaxwell]

1)

1st: MnO₄^- + 8H⁺ + 5e^- --> Mn²⁺ + 4H₂O

2nd: 1/2 Cl₂ + e^- --> Cl^-

Firstly, times the 2nd equation by two to get whole numbers. This gives: Cl₂ + 2e^- --> 2Cl^-

You'll also want to switch that equation around in order to get the electrons on the opposite side. This gives: 2Cl^- --> Cl₂ + 2e^-

Now, you want to equal the number of electrons on each side in order to cancel them out from the redox reaction.

Times the 1st equation by two: 2MnO₄^- + 16H⁺ + 10e^- --> 2Mn²⁺ + 8H₂O

Times the 2nd equation by five: 10Cl^- --> 5Cl₂ + 10e^-

We can add these equations together, to give: 2MnO₄^- + 16H⁺ + 10e^- + 10Cl^- --> 2Mn²⁺ + 8H₂O + 5Cl₂ + 10e^-

Now you can cancel out the common species on both sides (in this case, only the electrons).

Resulting in: 2MnO₄^- + 16H⁺ + 10Cl^- --> 2Mn²⁺ + 8H₂O + 5Cl₂

[LMaxwell]

2) Iodide ions, I^–(aq), react with iodate ions, IO₃^–(aq), in an acidic solution to form molecular iodine and water.

Note: two abbreviations will be used, LHS (left hand side of the reaction) and RHS (right hand side of the reaction).

I^- + IO₃^- --> I₂ + H₂O (however, this isn't balanced)

We need two more hydrogen ions on LHS to balance the hydrogens, therefore: I^- + IO₃^- + 2H⁺ --> I₂ + H₂O

However, now we need three oxygens on the RHS to balance the oxygens, therefore: I^- + IO₃^- + 2H⁺ --> I₂ + 3H₂O

However, now we need six hydrogen ions overall on the LHS to balance the hydrogens, therefore: I^- + IO₃^- + 6H⁺ --> I₂ + 3H₂O

However, now we have an overall charge of ⁺⁴ on the LHS, therefore we can increase the number of iodine ions to give: 5I^- + IO₃^- + 6H⁺ --> I₂ + 3H₂O

Although, now we need to balance the iodines to also have six on the RHS, therefore: 5I^- + IO₃^- + 6H⁺ --> 3I₂ + 3H₂O

p.s. I'm not sure if I did it the proper IB way, but this way seems clearer to me.

Edited April 15, 2012 by brofessional

[LikeA'13OSS]

What happens to the rate constant (k)and activation energy (Ea) of a reaction when the temperature is increased?

Thanks

[ILM]

Nothing for Ea however k is a temperature dependent, increasing temp. would increase k.

[LikeA'13OSS]

The question is in the file i attached, please help

[ILM]

The change of color is usually means that iodine is started to be produced (initial rate) so time is like an indication for rate, notice the following:

1- When you halved the concentration of I- , with the stability of concentrations of other reactants, the reaction rate halved. This indicates that the rate is directly proportional to the I- concentration thus 1 order.

2- compare experiment 1 and 3, the H2O2 concentration is halved, and the rate decreased by 4 so it is 2nd order.

The right answer is A

Note: I am moving this to chem Hl/ sl help, if you have other questions post it there.

[CkyBlue]

@Manani, you should not be posting questions for us to answer. Show us the work you have done and where you are having problems understanding the question. The previous two questions are simple concepts that should have been grasped by just looking over your chemistry readings.

[Guest dylaanz]

Hello guys. So recently we have just started learning about stoichiometry and now we are doing solution stoich. I missed the lesson on it and now I have no idea how to do the hw the teacher assigned us. Help anyone? D;

So it would be nice if someone can please help me do this sample question with some explanations:

What is the concentration of a KOH(aq) solution if 12.8mL of this solution if required to react with 25.0mL of 0.110mol/L H2SO4(aq)?

Edited April 17, 2012 by dylaanz

[Drake Glau]

Step 1: Make a balanced equation. To save time and effort you could just quickly notice that 2 KOH will react with 1 H2SO4.

Step 2: How many moles of H2SO4 do you have? To do this you will use the equation c=n/v (concentration=moles/volume).

Volume must be in liters also.

That gives us 0.110=x/0.025. Find X and that will be your amount of moles.

We know from step 1 that every mole of acid will react with 2 moles of KOH. This would be 2x.

Now you know that c=n/v and you now have n and you were given v.

c=2x/0.0128

or c=(2(0.110*0.025))/0.0128

To some mod: Move this?

Edited April 17, 2012 by Drake Glau