Hello sir, for 1.7 the memo says the answer is C, after also choosing the option A then checking your video for explanation and saw that we chose the same answer, I consulted with my teacher, and this was her comment. In the statement they tell you that they have the same concentration and volume meaning that they will require the same amount of mols of KOH when using the n=cv formula.
remember that the concentration of a solution remains the same unless another substance is added to it. In experiment 1, they show us that there was nothing added to sodium thiosulphate, like in experiment 2 and 3 (volumes of HCl were added but not in experiment 1).This already tells us that the concentration of sodium thiosulphate without anything added is 0.13. You should then use that information to help you calculate the moles in the experiment before the HCl was added: Before: n=cv = (0.13)(0.03) =3.9×10^-3 moles before HCl was added then you want to calculate the concentration by using the above moles this case, adding the volume of HCl and adding the volume of sodium thiosulphate, which gives us 0.05. c=n/v =3.9×10^-3/0.05 =0.078 mol•dm^-3
On 7.2.2) It is said we must calculate the pH of the solution after step 1 has occurred. The pH will be determined by the access HNO3, so we have to calculate the concentration of HNO3 before it is neutralized by Ba(OH)2 So the volume we'll use is 25cm³
trauma 😢
Hello sir, for 1.7 the memo says the answer is C, after also choosing the option A then checking your video for explanation and saw that we chose the same answer, I consulted with my teacher, and this was her comment. In the statement they tell you that they have the same concentration and volume meaning that they will require the same amount of mols of KOH when using the n=cv formula.
Thanks uncle ❤
You're the best teacher 👍👍
question 9.3 it’s R to Q not Q to R
because electrons flow from positive terminal to negative terminal ❤️
but thank you sir 🔥
on the memo sir the answer is ph=o.64
57:22
Sir I'm not sure if i do understand 7.2.2. Ain't we suppose to calculate the amount of HNO3 that was unused
i know we should, but why do we need the excess to work out pH?
I am not able to understand why the conc of 0,13 went with 30cm³ instead of 50cm³ for Q5 no.5,3. Can someone pls explain it to me
remember that the concentration of a solution remains the same unless another substance is added to it. In experiment 1, they show us that there was nothing added to sodium thiosulphate, like in experiment 2 and 3 (volumes of HCl were added but not in experiment 1).This already tells us that the concentration of sodium thiosulphate without anything added is 0.13. You should then use that information to help you calculate the moles in the experiment before the HCl was added:
Before:
n=cv
= (0.13)(0.03)
=3.9×10^-3 moles before HCl was added
then you want to calculate the concentration by using the above moles this case, adding the volume of HCl and adding the volume of sodium thiosulphate, which gives us 0.05.
c=n/v
=3.9×10^-3/0.05
=0.078 mol•dm^-3
On 7.2.2) It is said we must calculate the pH of the solution after step 1 has occurred. The pH will be determined by the access HNO3, so we have to calculate the concentration of HNO3 before it is neutralized by Ba(OH)2
So the volume we'll use is 25cm³
Exactly. After step 1 means at the end of reaction 1 but still the language used should be made more clear.
Mmmmh I wanted this one so badly 1:04:51
50:44
1:04:20
Plz explain question four
It is the reason why I'm upgrading 🤦♂️🥲
Lol,same.🥲🫠
Bruhh 😢
Dawg😂
@Relebohilem288 😭🤣🤣we will laugh about this year next year though 🤣
@@mosheseshoka110 trust 🤣🤦🏿♂️
51:11