Thanks

if someone could help with this question

a. From the graph determine the instantaneous rate when the concentration of o2 is at 8x 10-3M

b. Between 60 seconds and 300seconds, which instantaneous rate was higher? why?

For b would this be right

60 seconds is steeper that means it has a higher gradient than the 300 secs. Higher gradient means higher rate

For c from the graph determine the instantaneous rate at 60 seconds and at 300 seconds

Would this be right

300- 6/500= 0.012

60-13/40= 0.325

Would a be pretty much the same as the 60 secs

Mod edit: Merged Double posts

To work out the instantaneous reaction rate at a given point on the graph you will want to draw a tangent line (a straight line that just touches this particular given point) and then work out the gradient of this tangent line (the gradient of the line is equal to the rate of the reaction - rise/run == change in product/time). Once the tangent line has been drawn (i have shown a quick example for this at ~320 seconds) you can calculate the gradient by essentially forming a right-angle triangle and obtaining the change in O2 concentration (rise) for a particular time interval (run) and divide this concentration value by the time to determine the change in concentration per second (the reaction rate). I am not entirely sure what you have done in your calculations, but the method i have explained here should give you the correct answers. Its also important to note that the units of the y-axis (the product concentration) are 10^-3M and that the O2 concentration is decreasing over time so the reaction rate should be a negative value.

And when doing question like these, don't worry too much about getting the tangent line 100% perfect - it really is just an estimation and in question like these the answer key will often accept a range of values to account for this error in estimation.

Excuse the scrappiness of the example above - my ms paint skills are lacking

Need a bit of a refresher regarding SF If we were calculating the amount in mole of 3.45g of H_{2}, would we only be able to go to 2SF due to the molar mass of H being 1.0 in the data booklet?

On this page

here you can find VCAA 'advice to teachers' on sig figs.

For your specific problem, since the molar mass of hydrogen listed in periodic table in the data booklet (or whatever resource you are using) is an approximation of the molar of hydrogen (which is more accurately 1.00784), I think its important to take it into account.

In this particular instance depending on if you use the approximation of hydrogen as 1.0 g/mol compared to the more accurate 1.00784 g/mol, you will get different values for the 3rd significant figure. It is for this exact reason why when using a value approximated to 2 sig figs, the answer should be written to two sig figs. This being said, as colline has pointed out, if this is for a SAC which your teacher is marking, you should definitely ask your teacher what she is looking for.