Question

Help please. #38 (Question is attached)

Answer 1

hmm multiply the 3rd equation by 3/2 and then add them up.

Answer 2

-1888.7, which is the correct answer. Can you please explain how you did that

Answer 3

I know for these types of questions they usually give another equation for you to flip the other ones to make it into that particular one but what confused me here was they asked the change in H of Tm2O3... how do you get that alone?

Answer 4

Hm start off by identifying what you're looking for, the product has to be Tm2O3, so that needs to be on the right side. Overall, the point in these equations is to cancel out compounds that aren't needed, so i looked for discrepancies in the coefficients.

For example, the first equation has 6 HCl on the right, and the third equation has 4 HCl in the left.

So to cancel these out these need to be the same. To find what number you needed to multiply to get the coefficients the same i just used some simple algebra:

4x=6, x=6/4=3/2

Answer 5

Awesome! Thank you once again :)

Answer 6

I actually have another question. I need some insight on how to do it. Can you help me out with that?

Answer 7

Don't need the answer just any easier way to do it and/or what I'd need to know to answer it

Answer 8

insight on Hess's law type of problems?

Answer 9

It's actually given reagents, indicate which reagents would be mixed to give certain compounds described

Answer 10

so the question at the top?

Answer 11

Do you see the attached file I just sent?

Answer 12

question 1 is: Cu(OH)2 (s)

Answer 13

i do. So, basically you wanna identify the elements/polyatomic ions that make up the compounds. Can you do that for the one you just said? (CuSO4)

Answer 14

i mean Cu(OH)2

Answer 15

That is where i got confused. Based on that i thought i'd need CuCO3 and Cr(OH)3 but the answer is different and confused me

Answer 16

the problem with your choice is that CuCO3 is insoluble in water, so when you introduce them into a solution they won't react.

Answer 17

you would want to use something soluble, like CuSO4

Answer 18

and i would need 2 aqueous to start off with, right?

Answer 19

this is where i get confused because nothing else (used with CuSO4) gives an OH

Answer 20

\(Cr(OH)_3 + CuSO_4\) probably makes Cu(OH)_2, since it's the only possible way to add hydroxide to the copper

Answer 21

Funny thing is, the answer says it's CuSO4 and NH3.. O_o

Answer 22

lol that's impossible.. if you mixed those you'd make \([Cu(NH3)_4SO_4]\)

Answer 23

also how come CuSO4 isn't Cu2SO4? Thought sulfate had a 2- charge

Answer 24

it does, copper here is \(Cu^{2+}\)

Answer 25

how come it's not written out though

Answer 26

like they have Na2SO4 and they have the 2 written but for this they have it as CuSO4

Answer 27

because it's implied. sodium always makes the ion \(Na^+\), but most transition metals can have several oxidation states, it's written out explicitly in the same copper(II) sulcate is \(CuSO_4\) while copper(I)sulcate is \(Cu_2SO_4\).

Answer 28

it autocorrected sulfate to sulcate

Answer 29

Ah I see. So when I write out the products (using Cr(OH)3 + CuSO4), that would make CrSO4 + Cu(OH)2 even though it's Cr(OH)3?

Answer 30

it would make \(Cr_2(SO_4)_3\) you have to cancel out the charges

Answer 31

why does it make it (SO4)3?

Answer 32

OH nvm i see that. So it's better to put these equations at complete net ionic and then place the products?

Answer 33

yeah, that would be a good idea. write them as ions first

Answer 34

for the second part, questions 3, can you give me some insight on how to do these, please

Answer 35

actually number 4, sorry!

Answer 36

for the first one (combustion) you have to know what CAN burn, and what you need to make it burn.

for the second, you need to use something that will be able to dissolve calcium hydroxide (by reacting with the OH^-, \(\color{red}{which\;is\;a\;base}\))

Answer 37

How would you know which one to chose? There are 2 with OH^-

Answer 38

choose*

Answer 39

you would use an acid to react with the OH because Ca(OH)2 is not very soluble

so by adding acid you drive the reaction forward

\(Ca(OH)_2 + H_3O^+ \rightleftharpoons Ca^{2+}+2H_2O \)

Answer 40

Where exactly did you get the H3O^+ from since it's not in the given substances?

Answer 41

\(HNO_3\;and\;H_2SO_4\) are both (strong) acids

Answer 42

I'm so confused. We weren't even taught this :(

Answer 43

Sorry, don't meant to take up so much of your time

Answer 44

it's okay. idk, it's about driving equilibriums forward (kind of like what you do in "common ion effect" type problem if you've done those).

Answer 45

i'll stare at it and see if i get it lol thank you very much!

Answer 46

this video might help you. it's not specific to the problem but it's the same concept. http://www.youtube.com/watch?v=pskvC5ROCdc

Answer 47

I will look into that, thank you!

Answer 48

I actually had emailed my professor about number 1, Cu(OH)2 and she wrote back this: "Ammonia is a base so it produces OH- in water solution, thus forming Cu(OH)2(s). At low concentrations ammonia molecules actually form a dark blue complex with copper ions, which was used as a test for those ions in the copper cycle lab. I hope this clarifies."

Answer 49

That makes sense, but \(NH_{3(aq)}\) is not concentrated ammonia, \(NH_{3(l)}\) which is what you would need to do that.

Answer 50

Now i'm even more confused.. hahaa

Answer 51

i thought only those ending with OH are base

Answer 52

the ones with OH are strong bases, NH3 is weak base, meaning that it only abstracts protons from other compounds as dictated by it's pKb.

Answer 53

:S the only way to get less confused about this is to read about it

Answer 54

I was afraid of that! haha I will get on that now. Thank you once again!

Answer 55

yeah, you've tons of reading ahead dude. no probs!