genepeer Posted April 19, 2011 Report Share Posted April 19, 2011 Or you can use mass percentage. The ore has Mr=160, 112 of which are Iron. (110/160)*80 tonnes = 56.wait, so 1 tonne is 2kg? Reply Link to post Share on other sites More sharing options...
Drake Glau Posted April 19, 2011 Report Share Posted April 19, 2011 (edited) Fe2O3+3CO -> 2Fe+3CO21 Iron(III) Oxide makes 2 FeThat means each mole of Iron(III) Oxide will form 2 Fe so your ratio is 2:1. 2Fe per 1Fe2O3Apparently a tonne is actually 1,000kg...but my math still worked. Probably because I used the same conversion factor to go from tonnes to grams and grams to tonnes... Edited April 19, 2011 by Drake Glau 1 Reply Link to post Share on other sites More sharing options...
Hammerdown Posted April 19, 2011 Report Share Posted April 19, 2011 Also Bromine exists as the isotopes 79Br and 81Br. What is the percentage of 79Br in a sample witha relative atomic mass of 79.9?A. 40 %B. 45 %C. 50 %D. 55 %I think it is 45 but not exactly sure how to do this properly thanks again Drake your awesome Reply Link to post Share on other sites More sharing options...
Hammerdown Posted April 19, 2011 Report Share Posted April 19, 2011 No Drake you seem to have gotten MrFe=55.845 How ? Reply Link to post Share on other sites More sharing options...
Drake Glau Posted April 19, 2011 Report Share Posted April 19, 2011 (edited) No Drake you seem to have gotten MrFe=55.845 How ?Periodic table Also Bromine exists as the isotopes 79Br and 81Br. What is the percentage of 79Br in a sample witha relative atomic mass of 79.9?A. 40 %B. 45 %C. 50 %D. 55 %I think it is 45 but not exactly sure how to do this properly thanks again Drake your awesomeRelative atomic mass is calculated using a weighted average. In this case (with two isotopes concerned at the moment) this will be %1(79)+%2(81)=79.9With this you can see that it's 0.1 less than the median so that tells you your percentage of the smaller massed isotope needs to be larger. That there rules out A B and C immediately Edited April 19, 2011 by Drake Glau 2 Reply Link to post Share on other sites More sharing options...
Hinuku Posted April 19, 2011 Report Share Posted April 19, 2011 Could anyone explain the graphs of zero, first, and second order of reaction? Concentration against time. I have no idea what causes the different shapes. Thanks,Matt Reply Link to post Share on other sites More sharing options...
Sandwich Posted April 19, 2011 Report Share Posted April 19, 2011 http://writer.zoho.com/image.do?imgurl=43f284c1e0ba6646e2d2a98b8a52495191ec6d7cf7d22a0458f6d938d98b4eb12d5d7740d23ecd6ad6c4d78bc2fc7605http://writer.zoho.com/image.do?imgurl=43f284c1e0ba6646e2d2a98b8a52495191ec6d7cf7d22a0458f6d938d98b4eb1244339962b518fb455c870188bf16ebbYou might find those useful. Literally, zero order means that if concentration is doubled, rate doesn't change, first order means that if concentration doubles, the rate doubles and second order means that if the concentration doubles, rate quadruples.The order of something within a rate equation represents how many times it is involved in the reaction before the rate-determining step (going off the top of my head here, but if memory serves that's how the definition goes. Check it in your book to be sure!). The order is expressed as a "power" so if X were zero order it'd be [X]0 (not involved pre-RDS), [X]1 would be first order and [X]2 is second order (so it's involved twice).You then use these to come up with imaginary, hypothetical mechanisms which would fit with your findings. So if you know that X appears to be second order, you know that your hypothetical mechanism will have to involve it twice before the RDS and so on.If somebody could confirm that my memory has got all of that right, it would be fab Been a long time since May 09! 1 Reply Link to post Share on other sites More sharing options...
Hinuku Posted April 19, 2011 Report Share Posted April 19, 2011 One more issue here If I have a rate expression as follows: r= k[H2(g)][NO(g)]2Than how come a suitable mechanism for a reaction would be:2NO(g) ----> N2O2(g)N2O2(g) + H2(g) ----> N2O(g) + H2O(g)N2O(g) + H2(g) ----> N2(g) + H2O(g)What I don't get, is that H2(g) is used twice whereas the reaction expression shows that it should be only used once since it is 1st order.Any idea? Reply Link to post Share on other sites More sharing options...
Drake Glau Posted April 20, 2011 Report Share Posted April 20, 2011 (edited) Been a while since I've done kinetics but I'll take a stab at an explanation.If your second step is the rate determining step (meaning it's the slowest of the 3) then your rate would be based off that which would involve the 2NO but then only one H2Again, been about 7 months or so and I haven't reviewed for exams yet, but it makes sense in my head Just read Sandwich's post also and she said the same thing sort of..."The order of something within a rate equation represents how many times it is involved in the reaction before the rate-determining step"So only 1H2 is used before the rate determining step (it's actually used in the step in this case) Edited April 20, 2011 by Drake Glau 1 Reply Link to post Share on other sites More sharing options...
Hinuku Posted April 20, 2011 Report Share Posted April 20, 2011 Been a while since I've done kinetics but I'll take a stab at an explanation.If your second step is the rate determining step (meaning it's the slowest of the 3) then your rate would be based off that which would involve the 2NO but then only one H2Again, been about 7 months or so and I haven't reviewed for exams yet, but it makes sense in my head Just read Sandwich's post also and she said the same thing sort of..."The order of something within a rate equation represents how many times it is involved in the reaction before the rate-determining step"So only 1H2 is used before the rate determining step (it's actually used in the step in this case)\So what you mean is that what happens after the rate determining step doesn't matter? The rate expression needs to be consistent only until the RDS? Reply Link to post Share on other sites More sharing options...
Drake Glau Posted April 20, 2011 Report Share Posted April 20, 2011 Yep, after the RDS it really doesn't matter. If you have 3 steps (like this example) and the 2nd is slowest, it really doesn't matter how fast that 3rd one goes does it? It's faster than the 2nd (because it's the RDS) so it has to wait on it whether it's SUPER FAST or just a little faster than the RDS, still has to wait. 1 Reply Link to post Share on other sites More sharing options...
Hinuku Posted April 20, 2011 Report Share Posted April 20, 2011 Yep, after the RDS it really doesn't matter. If you have 3 steps (like this example) and the 2nd is slowest, it really doesn't matter how fast that 3rd one goes does it? It's faster than the 2nd (because it's the RDS) so it has to wait on it whether it's SUPER FAST or just a little faster than the RDS, still has to wait.And how did you actually know which was the RDS? I just check which step satisfies the rate equation but maybe there is a different method? Plus, could you explain this question to me?Consider the reaction between solid CaCO3 and aqueous HCl. The reaction will be speeded up by an increase in which of the following conditions?I. Concentration of the HClII. Size of the CaCO3 particlesIII. TemperatureThe answer is 1 and 2, how come size doesn't matter if it is solid? Reply Link to post Share on other sites More sharing options...
Drake Glau Posted April 20, 2011 Report Share Posted April 20, 2011 Yep, after the RDS it really doesn't matter. If you have 3 steps (like this example) and the 2nd is slowest, it really doesn't matter how fast that 3rd one goes does it? It's faster than the 2nd (because it's the RDS) so it has to wait on it whether it's SUPER FAST or just a little faster than the RDS, still has to wait.And how did you actually know which was the RDS? I just check which step satisfies the rate equation but maybe there is a different method? Plus, could you explain this question to me?Consider the reaction between solid CaCO3 and aqueous HCl. The reaction will be speeded up by an increase in which of the following conditions?I. Concentration of the HClII. Size of the CaCO3 particlesIII. TemperatureThe answer is 1 and 2, how come size doesn't matter if it is solid?RDS follows the rate expression, yes.You said the answer is I and II but your question suggests you meant I and III The question specifically asks what will speed up the reaction. If the particles get bigger the total surface area of the reactant actually decreases and since it's a solid a smaller surface area=less particles that can react and according to collision theory less particles=less chance of collision=slower rate, not faster.I and III both increase the rate. I adds particles and III increases their speed, both of which increase the chance for collisions with proper geometric configuration and energy thus increasing your rate. 1 Reply Link to post Share on other sites More sharing options...
Markee Posted April 20, 2011 Report Share Posted April 20, 2011 EASY QUESTION GUYS! If you were to revise, which would you use as revision? CHemistry text book or CHemistry revision guide? Reply Link to post Share on other sites More sharing options...
nametaken Posted April 20, 2011 Report Share Posted April 20, 2011 EASY QUESTION GUYS! If you were to revise, which would you use as revision? CHemistry text book or CHemistry revision guide? I'd say revision guide. It's got concise notes in it as opposed to the textbook. I wish I had a revision guide.... Reply Link to post Share on other sites More sharing options...
Markee Posted April 20, 2011 Report Share Posted April 20, 2011 EASY QUESTION GUYS! If you were to revise, which would you use as revision? CHemistry text book or CHemistry revision guide? I'd say revision guide. It's got concise notes in it as opposed to the textbook. I wish I had a revision guide....Yahoo! Thank you buddy! Imma be flying! lol.Use your notes. Those, I guess, are as close to revision guides as it seems that sir has summed it up for us. 1 Reply Link to post Share on other sites More sharing options...
Markee Posted April 22, 2011 Report Share Posted April 22, 2011 Easy question guys.If you go across the periodic table, is it called GROUP or PERIOD? Reply Link to post Share on other sites More sharing options...
Keel Posted April 22, 2011 Report Share Posted April 22, 2011 Period. Group is going down. 1 Reply Link to post Share on other sites More sharing options...
Rigel Posted April 26, 2011 Report Share Posted April 26, 2011 Hello guys, i need help on these two questions:3. What amount of solute ions, in moles, is present in 50 cm3 of 0.10 mol dm-3 sodium hydroxide solution?A. 2.5 × 10–3B. 5.0 × 10–3C. 1.0 × 10–2D. 5.0 × 10–2I think it is B, but the book says that the answer is C 4. Complete combustion of a hydrocarbon produces 0.44 g of CO2 and 0.18 g of H2O. What is theempirical formula of the hydrocarbon?A. CHB. CH2C. CH3D. CH4OK i don't really understand this question hehe. I chose A, but the actual answer is B.Thank you for your help! Reply Link to post Share on other sites More sharing options...
kiwi.at.heart Posted April 26, 2011 Report Share Posted April 26, 2011 (edited) Hello guys, i need help on these two questions:3. What amount of solute ions, in moles, is present in 50 cm3 of 0.10 mol dm-3 sodium hydroxide solution?A. 2.5 × 10–3B. 5.0 × 10–3C. 1.0 × 10–2D. 5.0 × 10–2I think it is B, but the book says that the answer is C 4. Complete combustion of a hydrocarbon produces 0.44 g of CO2 and 0.18 g of H2O. What is theempirical formula of the hydrocarbon?A. CHB. CH2C. CH3D. CH4OK i don't really understand this question hehe. I chose A, but the actual answer is B.Thank you for your help! 3. If the question was asking how many moles of NaOH there was, the answer would be B, but its asking for ions and for every NaOH molecule there is 2 ions, Na+ and OH-. Therefore you need to times the amount of moles by 2 and you get C as the answer.4. First you get a general unbalanced formula for the combustion reaction which is CHx + O2 => CO2 + H2O. Then we can figure out the mole ratio of CO2 and H2O by the values we are given which you can figure out to be 1:1. From this, we can figure out that the coefficients of CO2 and H20 are the same so we can balance the equation to 2CHx + 3O2 => 2CO2 + 2H2O. Obviously, for the equation to balance right x must equal 2 so the empirical formula is CH2 and B. Edited April 26, 2011 by Kiwiatheart 2 Reply Link to post Share on other sites More sharing options...
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