Mam I really dont understand because you said for 6.1 that NaHCO3 is the limiting Neh? Which doesnt make sense because we have 0.02 moles of ethanoic acid and then I left I was like "maybe I'm wrong" then 6.3 says calculate the mass for excess so mam if you concluded that sodium bicarbonate is limiting then it means ethanoic is excess so we just calculate the mass but now mam I'm trying to understand why did you did for 6.3 its very difficult to comprehend😢
@@tshepangBaptista-ls6iqat 11:18, ma'am calculated that NaHCO3 was in excess. There were 0.119 moles of NaHCO3 and 0.02 moles of CH3COOH. She then started to explain how to work out the mass of the NaHCO3 in excess after the reaction has occured at 16:30. Because there is only 0.02 moles of Ethanoic Acid that can react with the Sodium Bicarbonate (and the mol ratio is 1:1) only 0.02 mols of NaHCO3 was used up in the reaction which left 0.09 moles of it after the reaction finished. At 18:36 she uses that to work out the mass of the NaHCO3 in excess. Hope this helps😁
Within a question (so between steps) try not to round off. If you want to though make sure you keep at least 6 decimal places. Only round off properly at the end ❤️❤️
I love the past paper videos! It helps so much seeing how teachers approach questions😅❤️
Glad you like them!😅
I agree!!!
When I tried this paper I couldn't understand how they got some of the answers😢 Thank you so much for explaining it😊 It makes so much sense now😅
Glad it helped!
@missmartinsmathsscience it helped a lot! Thank you sooooo much❤❤❤
Mam I really dont understand because you said for 6.1 that NaHCO3 is the limiting Neh? Which doesnt make sense because we have 0.02 moles of ethanoic acid and then I left I was like "maybe I'm wrong" then 6.3 says calculate the mass for excess so mam if you concluded that sodium bicarbonate is limiting then it means ethanoic is excess so we just calculate the mass but now mam I'm trying to understand why did you did for 6.3 its very difficult to comprehend😢
@@tshepangBaptista-ls6iqat 11:18, ma'am calculated that NaHCO3 was in excess. There were 0.119 moles of NaHCO3 and 0.02 moles of CH3COOH. She then started to explain how to work out the mass of the NaHCO3 in excess after the reaction has occured at 16:30. Because there is only 0.02 moles of Ethanoic Acid that can react with the Sodium Bicarbonate (and the mol ratio is 1:1) only 0.02 mols of NaHCO3 was used up in the reaction which left 0.09 moles of it after the reaction finished. At 18:36 she uses that to work out the mass of the NaHCO3 in excess. Hope this helps😁
Wow thanks a lot Miss Martins ❤😊
The way you make science very easy is lit 🤗
Mam can you plz help us with physical science practical
Which one?🎉
@@missmartinsmathsscienceboyles law please
mem if we use our liming reagent in a mol ratio with a product, do we use the mol that we need or the mol that we have?
The Mol that we have ❤️
Kevinmathscience wife??
Omdss 😂
😂😂😂 could be true🤔
to what decimal place should i round within steps in exam?
Within a question (so between steps) try not to round off. If you want to though make sure you keep at least 6 decimal places. Only round off properly at the end ❤️❤️