ohh! thank you, i forgot , i will go back and revise that unit.

Let us check
81g HF * 1mol /20g HF * 540kJ /2 mol HF
Yes you are right it is 1100 kJ

In this question we need to work out the energy of fuel for 153 g  from the information of 100% C mass  and 100% H mass.
Charcoal – 100% carbon so 34kJ/g*153g
Hydrogen – 100% H so 142kJ/g * 153g
Wood pine – 50% C and 6% H = ( (50*153)/100*34kJ/g + (6*153 )/100*142) kJ
Bituminous coal  – 7% C and5% H = ( (7*153)/100*34kJ/g + (5*153 )/100*142) kJ

Crude oil  – 85% C and 12 % H = ( (85*153)/100*34kJ/g + (12*153 )/100*142) kJ

Natural gas  –70% C and 23 % H = ( (70*153)/100*34kJ/g + (23*153 )/100*142) kJ

Now Check which one is giving maximum energy

Aditi Molarity is a way of expressing how much solute is present in the given solution. Basically is is the way we express concentration of solute in solution.
By definition
Molarity = moles/ volume in L = mmol /volume in mmol
If you remember we did questions based on it in unit 4.

Hi Dr. Uma,

I was wondering on how I should solve this problem?

Hi Dr. Uma,

I was wondering about the question linked to this post. Somehow I keep getting around 1100 kJ, however, the answer says it is 2200 kJ.

Hi, I kind of forgot how to solve for molarity, and I came across this question, but I checked the solutions, and, it didn't make sense. Could you help me with this question? Thank you!

Thanks Aditi
Check I have corrected it.

so the quiz I was talking about in class that wasn't working is , Quiz : 7.3 Atomic Spectra of Hydrogen and Bohr Model it shows error

Similar question we did in the class.
Now it is first order kinetics so we can use
ln[A]_t=-kt+ln[A]₀
You can rearrange this equation as
t = 1/k x  ln([A]₀/[A]_t
I discussed this format also.
[A]₀= 100 (assume because percent is given)
[A]_t= 1 (100-99)
k = 0.693/_t1/2 This we calculated in part i as 8ms
Just plug in all values and work out the answer as ms.