8.99

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Glendy Gonzalez 1A
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8.99

Postby Glendy Gonzalez 1A » Thu Jan 25, 2018 8:54 pm

Can someone please explain me how can you calculate the enthalpy of formation for 2HCl(aq)+Zn(s) ---> H2(g)+ZnCl2(aq). I can't find the enthalpy value for ZnCl2.

Hena Sihota 1L
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Re: 8.99

Postby Hena Sihota 1L » Thu Jan 25, 2018 9:18 pm

The problem states that the reaction produces hydrogen gas and chloride ions, so the chemical equation can be rewritten as 2HCl(aq)+Zn(s)-->H2(g)+Zn2+(aq)+2Cl-(aq). You can now find all the enthalpies of formation needed to find the enthalpy of the reaction in appendix 2A.

David Zhou 1L
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Re: 8.99

Postby David Zhou 1L » Fri Jan 26, 2018 9:47 pm

Also, if you're ever unable to find the enthalpy value or any other kind of experimental value, you can always Google it. There's another earlier homework problem in which you needed to use the specific heat capacity of Cu, and for that I just Ok Google'd it because that's really the fastest way to find it if it's not given to you in the problem.

Sara Varadharajulu
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Re: 8.99

Postby Sara Varadharajulu » Fri Jan 26, 2018 11:07 pm

ZnCl2 is aq, which means you can think of it as zn 2+ ions and 2 cl- ions (in the solution they appear as ions anyway). look up the enthalpies of formation for these two ions (these are in the book).


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