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o ok that makes sense sorry kind of slow sometimes but this will help a lot thank. now that im back home I can work on this a bit more soo that means I should be able to get the concentration by doing this 1.84g*0.025L=0.046=46*10^-2g/L or should it look more like this seeing as it's all ways put like this g/L 1.84g/0.025L=73.6g/L and should I convert the grams in to moles seeing as that's just molar mass not moles MM=g??????.....nvm yes yes I should or am I completely wrong lol

Thx you soo much for your help I defiantly neeed to review some of the notes. I was looking over the first question and ya I could not remember why I squared that one. I realise that I should have put more sub script in to maybe make it easier to read and understand the steps that I went thought. Also maybe change the name of the post too something that would help people understand what kind of help I needed. Which would be help with equations and math not just explaining theory or fact. can i ask for the second question if i do need to use the 250 ml sample in any of the equations to get the right Ksp for the AgCH3COO. like can i use it to get any useful information that would help me calculate the Ksp???? Or were you telling me that I should disregard it.

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ok i need some help i been doing one of my last cores for my grade 12 online and i been running in to some problem most of the time i can think my way out of it but some time i would like a little help. now you would say well go ask the teacher and i have the problem is he never gets back to me he allways just marks my stuff and i get full marks idk if im doing stuff right or if the way i figured it out works all the time because he never msg is me back i even emailed him and nothing soo i turn to you guys the science forums for help. soo i was wondering if you could tell me if i was doing a few of theys questions right and if i need to be doing something different first up: 1. A saturated solution of BaSO4 is given to patients needing digestive tract x-rays. a) Write an equation that represents the solubility equilibrium. (1 mark) b) Calculate the [ba 2+] present in the saturated solution. (2 marks) and this is my assure 1) A. BaSO4 <- -> Ba + SO4 B. ksp(BaSO4) = 1.1*10^-10= Ba^2 – SO4= S^2 S=/ 1.1*10^-10 =1.05*10^-8 S=1.05*10^-8 Ba = 1.05*10^-8 (M/L) i had some problems with this and i dont feel that i have done it right but im not sure could you give me some pointers??? second:A saturated solution of AgCH3 COO was evaporated to dryness. The 250.0 mL sample was found to contain 1.84 g AgCH3 COO. Calculate the solubility product constant for AgCH3 COO. now this one i didnt use the 250.0 mL sample in any of the calculations soo i dont think i got it right but here what i did 4. AgCH3COO <- -> (AgCH3) (CO2) MM= 166.9 = MM= 122.9 + MM=44 1.84g/166.9g * 1 mole=0.0110245656=0.0110 (conversion factor) AgCH3=0.0110 CO2=0.0110 (using mole ratio) Ksp(AgCH3COO)=0.0220 third:Will a precipitate form when 90.0 mL of 1. 00 × 10 −2 M Cu(NO3)2 and 10.0 mL of 1. 00 × 10 −2 M NaIO3 are mixed? Explain using appropriate calculations. (3 marks) i think i did this one right but want to make sure. A. Ksp(Cu(IO3)2)= 6.9*10^-8 Cu(IO3)2 <-  Cu2 + IO3 90.0 mL of 1. 00 10 ^2 M CuNO32. 10.0 mL of 1.00*10^2 M NaIO3. 90.0ml+10.0ml= 100(FV) Cu2(FC)=1.00*10^-2 * 90/100 = 0.9*10^-3. IO3(FC) =1.00*10^2 * 10/100 =10 tiralKsp= (0.9*10^-3) *(10)= 0.9*10^-3 trialKsp(0.9*10^-3) < Ksp(6.9*10^-8 ) a precipitate will not form think i have to change thins to will form thats about it. any help would be greatly appreciated.