HW E.23

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Ramneet Sandhu 3D
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Joined: Fri Aug 30, 2019 12:16 am

HW E.23

Postby Ramneet Sandhu 3D » Thu Oct 03, 2019 9:45 am

Can someone please help me out with part (a) and (c) from E.23? Do you have to do the problem a different way when ions are involved? Thanks!

Fiona Latifi 1A
Posts: 102
Joined: Sat Sep 14, 2019 12:16 am

Re: HW E.23

Postby Fiona Latifi 1A » Thu Oct 03, 2019 9:51 am

For parts A and C, you basically do it the same way. You just have to determine the number of ions in the compound. In part A, there is only 1 Cu2+ ion. In part C, there are 6 F- ions because it has a subscript of 6. So you would have to first divide by the molar mass for both parts A and C. Then, multiply by the number of ions. I hope this helps!

Fatemah Yacoub 1F
Posts: 114
Joined: Thu Jul 11, 2019 12:16 am

Re: HW E.23

Postby Fatemah Yacoub 1F » Thu Oct 03, 2019 9:55 am

Basically when you are dealing with ions, the very end of your dimensional analysis is multiplying by the number of ions(the one you are trying to solve for) in your formula. So in E. 23 part a. you start with 3.00g of CuBr2 use the mole ratio using the molar mass and then since the formula is CuBr2 there is only 1 mole of Cu per mole of CuBr2. Therefore your dimensional analysis would end with 1 mol of Cu/ 1 mole CuBr2.
For part C.) you would do the same dimensional analysis to convert grams to moles but this time there is 10 moles of H2O per 1 mole of Na2CO3x 10 H20 making the conversion factor 10 moles H2O/ 1 mole Na2CO3 so you would basically multiply by 10.
Hope this helps!

Andrew Liang 1I
Posts: 105
Joined: Fri Aug 30, 2019 12:18 am

Re: HW E.23

Postby Andrew Liang 1I » Thu Oct 03, 2019 10:14 am

Here is my work process for E23

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